Samacheer Kalvi Class 11 Physics Solutions Chapter 8 Heat and Thermodynamics

Part - I:

I. Multiple choice questions:

Question 1. In hot summer after a bath, the body's:
(a) internal energy decreases
(b) internal energy increases
(c) heat decreases
(d) no change in internal energy and heat
Answer: (a) internal energy decreases
In simple words: After a bath in hot weather, our body tries to cool down. This process uses up some of our stored energy, causing the internal energy to go down.

🎯 Exam Tip: Remember that sweating and evaporation cause a cooling effect, which requires energy from the body, thus decreasing its internal energy.

Question 2. The graph between volume and temperature in Charle's law is:
(a) an ellipse
(b) a circle
(c) a straight line
(d) a parabola
Answer: (c) a straight line
In simple words: Charles's law says that if you keep the pressure the same, the volume of a gas grows bigger as it gets hotter, and this relationship can be drawn as a straight line on a graph.

🎯 Exam Tip: Recall that Charles's Law (V ∝ T) indicates a direct proportionality, which is always represented by a straight line graph when temperature is in Kelvin.

Question 3. When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is:
(a) isothermal
(b) adiabatic
(c) isobaric
(d) isochoric
Answer: (b) adiabatic
In simple words: When a tire bursts, the air rushes out so quickly that there isn't enough time for it to exchange heat with the outside world. This type of rapid process, where no heat enters or leaves, is called adiabatic.

🎯 Exam Tip: The key to identifying an adiabatic process is its rapid nature, which prevents heat exchange with the surroundings, leading to a temperature change.

Question 4. An ideal gas passes from one equilibrium state (\( P_1, V_1, T_1, N \)) to another equilibrium state (\( 2P_1, 3V_1, T_2, N \)). Then:
(a) \( T_1 = T_2 \)
(b) \( T_1 = \frac{T_2}{6} \)
(c) \( T_1 = 6T_2 \)
(d) \( T_1 = 3T_2 \)
Answer: (b) \( T_1 = \frac{T_2}{6} \)
In simple words: We can use the ideal gas law, \( PV = NKT \), to relate the initial and final states of the gas. By plugging in the given values for pressure and volume, we can find the relationship between the initial and final temperatures.

🎯 Exam Tip: Always use the ideal gas law \( PV = nRT \) or \( PV = NKT \) (where \( K \) is Boltzmann constant) for problems involving changes in state for an ideal gas. Ensure all units are consistent.

Question 5. When a uniform rod is heated, which of the following quantity of the rod will increase:
(a) mass
(b) weight
(c) centre of mass
(d) moment of inertia
Answer: (d) moment of inertia
In simple words: When a rod gets hotter, it expands, meaning its size increases slightly. Since its mass stays the same but it spreads out more, it becomes harder to make it spin, which means its moment of inertia goes up.

🎯 Exam Tip: Moment of inertia depends on both mass and how that mass is distributed around the axis of rotation. When a body expands due to heating, its dimensions increase, leading to a larger moment of inertia.

Question 6. When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process:
(a) Q > 0, W > 0
(b) Q < 0, W > 0
(c) Q > 0, W < 0
(d) Q < 0, W < 0
Answer: (a) Q > 0, W > 0
In simple words: When food cooks, heat is added to the steam, so \( Q \) is positive. As the steam pushes the lid, it does work on the surroundings, so \( W \) is also positive.

🎯 Exam Tip: Remember the sign conventions: heat added to the system is positive (\( Q > 0 \)), and work done by the system is positive (\( W > 0 \)).

Question 7. When you exercise in the morning, by considering your body as a thermodynamic system, which of the following is true?
(a) \( \Delta U > 0, W > 0 \)
(b) \( \Delta U < 0, W > 0 \)
(c) \( \Delta U < 0, W < 0 \)
(d) \( \Delta U = 0, W > 0 \)
Answer: (b) \( \Delta U < 0, W > 0 \)
In simple words: When you exercise, your body uses its stored energy, so your internal energy (\( \Delta U \)) goes down. At the same time, your muscles do work, which means your body performs work on its surroundings (\( W \)) so \( W \) is positive.

🎯 Exam Tip: In thermodynamics, work done *by* the system (like your body exercising) is positive, and if energy is being used, internal energy decreases, making \( \Delta U \) negative.

Question 8. A hot cup of coffee is kept on the table. After some time it attains a thermal equilibrium with the surroundings. By considering the air molecules in the room as a thermodynamic system, which of the following is true?
(a) \( \Delta U > 0, Q = 0 \)
(b) \( \Delta U > 0, W < 0 \)
(c) \( \Delta U > 0, Q > 0 \)
(d) \( \Delta U = 0, Q > 0 \)
Answer: (c) \( \Delta U > 0, Q > 0 \)
In simple words: As the hot coffee cools down, it transfers heat to the air around it, so the air gains heat (\( Q > 0 \)). This added heat increases the energy of the air molecules, causing the air's internal energy to also increase (\( \Delta U > 0 \)).

🎯 Exam Tip: When a system receives heat, \( Q \) is positive. This heat typically increases the internal energy of the system, making \( \Delta U \) positive, unless work is done by the system.

Question 9. An ideal gas is taken from (\( P_i, V_i \)) to (\( P_f, V_f \)) in three different ways. Identify the process in which the work done on the gas the most.
(a) Process A
(b) Process B
(c) Process C
(d) Equal work is done in Process A, B & C
Answer: (a) Process A
In simple words: The work done on a gas in a PV diagram is the area under the curve. Looking at the graph, Process A has the largest area underneath its curve, meaning the most work is done on the gas in that process.

🎯 Exam Tip: In a P-V diagram, the work done by the gas is the area under the curve. For work done *on* the gas, you consider the area with the opposite sign, or simply look for the largest magnitude of area when comparing different processes.

Question 10. The V-T diagram of an ideal gas which goes through a reversible cycle A \( \rightarrow \) B \( \rightarrow \) is shown below. (Processes D \( \rightarrow \) A and B \( \rightarrow \) C are adiabatic) The corresponding PV diagram for the process is (all figures are schematic)
(a) P
A
D
C
V
(b) P
A
D
C
V
(c) P
A
D
C
V
(d) P
A
D
C
V
Answer: (b) P
D
C
A
B
V
In simple words: The original V-T diagram shows how volume and temperature change. By understanding what each process (isothermal, adiabatic, etc.) looks like on a P-V graph, we can find the matching P-V diagram. Process B shows the correct changes in pressure and volume that match the given V-T cycle.

🎯 Exam Tip: Understanding how different thermodynamic processes (isothermal, adiabatic, isobaric, isochoric) are represented on both V-T and P-V diagrams is crucial for solving these types of conversion problems.

Question 11. A distant star emits radiation with maximum intensity at 350 nm. The star is:
(a) 8280 K
(b) 5000 K
(c) 7260 K
(d) 9044 K
Answer: (a) 8280 K
In simple words: Wien's displacement law tells us that hotter objects glow with light that has shorter wavelengths. By using this law, we can calculate the star's temperature from the peak wavelength of its light.

🎯 Exam Tip: Wien's displacement law is \( \lambda_m T = b \), where \( \lambda_m \) is the peak wavelength, \( T \) is the temperature, and \( b \) is Wien's displacement constant (\( 2.898 \times 10^{-3} \text{ m K} \)). Ensure units are consistent (meters for wavelength).
Here, \( T = \frac{b}{\lambda_m} = \frac{2.898 \times 10^{-3}}{350 \times 10^{-9}} = 0.008280 \times 10^6 = 8280 \text{ K} \).

Question 12. Identify the state variables given here.
(a) Q, T, W
(b) P,T,U
(c) Q, W
(d) P,T,Q
Answer: (b) P,T,U
In simple words: State variables are properties that describe the condition of a system at a particular moment, regardless of how it got there. Pressure, temperature, and internal energy are examples because their values depend only on the current state of the system, not on its history.

🎯 Exam Tip: State variables (or state functions) depend only on the initial and final states of a system, not on the path taken. Heat (Q) and work (W) are path-dependent, so they are not state variables.

Question 13. In an isochoric process, we have:
(a) W = 0
(b) Q = 0
(c) \( \Delta U = 0 \)
(d) \( \Delta T = 0 \)
Answer: (a) W = 0
In simple words: An isochoric process is when the volume of a system stays the same. If the volume does not change, then no work is done by or on the system.

🎯 Exam Tip: The work done (\( W \)) in a thermodynamic process is given by \( P\Delta V \). In an isochoric process, \( \Delta V = 0 \), hence \( W = 0 \).

Question 14. The efficiency of a heat engine working between the freezing point and boiling point of water is: (NEET2018)
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Answer: (b) 20%
In simple words: The efficiency of an ideal heat engine depends on the temperatures of its hot and cold reservoirs. By converting the freezing and boiling points of water to Kelvin, we can calculate the engine's maximum possible efficiency.

🎯 Exam Tip: To calculate the efficiency (\( \eta \)) of a Carnot engine, use the formula \( \eta = 1 - \frac{T_L}{T_H} \), where \( T_L \) is the cold reservoir temperature and \( T_H \) is the hot reservoir temperature, both in Kelvin. Freezing point is 0°C (273 K) and boiling point is 100°C (373 K).

Question 15. An ideal refrigerator has a freezer at a temperature of -12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is:
(a) 50°C
(b) 45.2°C
(c) 40.2°C
(d) 37.5°C
Answer: (c) 40.2°C
In simple words: A refrigerator's performance is linked to the temperatures it works between. We can use the formula for coefficient of performance to find the temperature of the air where heat is released, once we know the freezer temperature and the performance value.

🎯 Exam Tip: The coefficient of performance (COP) for a refrigerator is given by \( \text{COP} = \frac{T_L}{T_H - T_L} \). Remember to convert all temperatures from Celsius to Kelvin before using the formula. Here, \( T_L = -12 + 273 = 261 \text{ K} \). If \( \text{COP} = 5 \), then \( 5 = \frac{261}{T_H - 261} \). Solving for \( T_H \) gives \( T_H = 313.2 \text{ K} \), which is \( 313.2 - 273 = 40.2^\circ \text{C} \).

I. Short Answer Questions:

Question 1. 'An object contains more heat'- is it a right statement? If not why?
Answer: No, the statement 'An object contains more heat' is not right. When an object is heated, it receives heat from an external source, which increases its internal energy. Heat is a form of energy that is transferred *between* objects due to a temperature difference, it is not something an object "contains". We should say "an object contains more thermal energy" or "has higher internal energy" instead. Heat flows when there is a temperature difference, always from a hotter body to a colder one. So, heat is energy in transit.
In simple words: No, this statement is wrong. Heat is not something an object holds; it's the energy that moves from a hot object to a cold one. Instead, an object has internal energy.

🎯 Exam Tip: Clearly differentiate between 'heat' (energy in transit) and 'internal energy' (energy stored within a system). An object cannot 'contain' heat, but it can gain or lose internal energy through heat transfer.

Question 2. Obtain an ideal gas law from Boyle's and Charles'law.
Answer: We can get the ideal gas law by combining Boyle's Law and Charles's Law.
Boyle's Law says that at a constant temperature, the pressure of a gas is inversely proportional to its volume. This means \( P \propto \frac{1}{V} \).
Charles's Law states that when the pressure is kept constant, the volume of a gas is directly proportional to its temperature. This means \( V \propto T \).
When we combine these two laws, we get \( PV \propto T \), which can be written as \( PV = CT \), where \( C \) is a constant.
For a gas with \( N \) particles, \( C \) can be written as \( kN \), where \( k \) is Boltzmann's constant.
So, the equation becomes \( PV = NkT \). This is the ideal gas law. For \( \mu \) moles of gas, \( N = \mu N_A \) (where \( N_A \) is Avogadro's number), and \( R = N_A k \) is the universal gas constant.
Therefore, the ideal gas law for \( \mu \) moles is \( PV = \mu RT \). This law describes how pressure, volume, and temperature relate for an ideal gas under equilibrium conditions.
In simple words: Boyle's law says pressure and volume are opposite (when one goes up, the other goes down). Charles's law says volume and temperature go up and down together. Put these two ideas together, and you get the ideal gas law, which tells us how pressure, volume, and temperature are all linked for a perfect gas.

🎯 Exam Tip: To derive the ideal gas law, clearly state Boyle's Law (\( P \propto \frac{1}{V} \)) and Charles's Law (\( V \propto T \)) separately, then combine them to get \( PV \propto T \), and finally introduce the constant to form \( PV = NkT \) or \( PV = \mu RT \).

Question 3. Define one mole.
Answer: One mole of any substance is defined as the specific amount of that substance which contains the Avogadro number (\( N_A \)) of particles. These particles can be atoms, molecules, or any other defined elementary entities. For instance, one mole of water contains \( 6.022 \times 10^{23} \) water molecules.
In simple words: A mole is a way to count a very large number of tiny particles, like atoms or molecules. One mole of anything always has the same special number of particles, called Avogadro's number.

🎯 Exam Tip: Remember Avogadro's number, \( 6.022 \times 10^{23} \), as the key quantity defining a mole. Emphasize that it applies to any type of particle.

Question 4. Define specific heat capacity and give its unit.
Answer: Specific heat capacity of a substance is defined as the amount of heat energy needed to raise the temperature of 1 kilogram of that substance by 1 Kelvin (or 1°C). Different materials require different amounts of heat to change their temperature. The SI unit for specific heat capacity is Joules per kilogram per Kelvin (\( \text{J kg}^{-1} \text{ K}^{-1} \)).
In simple words: Specific heat capacity tells us how much heat energy is needed to warm up 1 kilogram of a material by just one degree. Its unit is Joules per kilogram per Kelvin.

🎯 Exam Tip: State the definition clearly and accurately, ensuring you include both the mass (1 kg) and temperature change (1 K or 1°C) in the definition. Also, provide the correct SI unit.

Question 5. Define molar specific heat capacity.
Answer: Molar specific heat capacity is the amount of heat energy required to increase the temperature of one mole of a substance by 1 Kelvin or 1°C. This is similar to specific heat capacity, but instead of using mass, it uses the number of moles. For example, the molar specific heat of water is the energy needed to heat one mole of water by one degree.
In simple words: Molar specific heat capacity is the amount of heat energy needed to warm up one mole of a substance by one degree of temperature.

🎯 Exam Tip: The definition of molar specific heat capacity is identical to specific heat capacity, but with "one mole of substance" instead of "1 kg of substance."

Question 6. What is a thermal expansion?
Answer: Thermal expansion is the natural tendency of matter to change in size, shape, area, and volume when its temperature changes. Most materials get larger when they heat up and shrink when they cool down. This is because the atoms vibrate more vigorously at higher temperatures, pushing each other further apart.
In simple words: Thermal expansion is when things get bigger or smaller as their temperature changes. Usually, they get bigger when they get hotter.

🎯 Exam Tip: Define thermal expansion as the change in dimensions (shape, area, volume) due to temperature change. Mention that this typically involves expansion upon heating and contraction upon cooling.

Question 7. Give the expressions for linear, area and volume thermal expansions.
Answer: The expressions for linear, area, and volume thermal expansions are as follows:
(i) Linear expansion: This describes how the length of an object changes. The coefficient of linear expansion (\( \alpha_L \)) is given by \( \alpha_L = \frac{\Delta L}{L \Delta T} \). Here, \( \frac{\Delta L}{L} \) is the fractional change in length and \( \Delta T \) is the change in temperature.
(ii) Area expansion: This describes how the surface area of an object changes. The coefficient of area expansion (\( \alpha_A \)) is given by \( \alpha_A = \frac{\Delta A}{A \Delta T} \). We also know that for isotropic materials, \( \alpha_A = 2\alpha_L \).
(iii) Volume expansion: This describes how the overall volume of an object changes. The coefficient of volume expansion (\( \alpha_V \)) is given by \( \alpha_V = \frac{\Delta V}{V \Delta T} \). For isotropic materials, \( \alpha_V = 3\alpha_L \). These formulas help us predict how much materials will expand or contract with temperature changes.
In simple words: For linear expansion, \( \alpha_L = \frac{\text{change in length}}{\text{original length} \times \text{change in temperature}} \). For area, \( \alpha_A = \frac{\text{change in area}}{\text{original area} \times \text{change in temperature}} \). For volume, \( \alpha_V = \frac{\text{change in volume}}{\text{original volume} \times \text{change in temperature}} \).

🎯 Exam Tip: Provide the mathematical expressions for each type of expansion and clearly define the terms in each formula. Also, include the relationships between the coefficients (\( \alpha_A = 2\alpha_L \) and \( \alpha_V = 3\alpha_L \)).

Question 8. Define latent heat capacity. Give its unit.
Answer: Latent heat capacity of a substance is defined as the amount of heat energy required to change the state of a unit mass of the material without changing its temperature. For example, when ice melts into water, its temperature stays at 0°C, but it needs energy to change its state. The SI unit for latent heat capacity is Joules per kilogram (\( \text{J kg}^{-1} \)).
In simple words: Latent heat capacity is the amount of heat energy needed to make 1 kilogram of a substance change its state (like melting or boiling) without changing its temperature. Its unit is Joules per kilogram.

🎯 Exam Tip: Emphasize that latent heat capacity is associated with a change of state (phase transition) and occurs without a change in temperature. Clearly state its SI unit.

Question 9. State Stefan-Boltzmann law.
Answer: The Stefan-Boltzmann law states that the total amount of heat energy radiated per second per unit area by a black body is directly proportional to the fourth power of its absolute temperature. This means that even a small increase in temperature can lead to a large increase in the energy radiated. Mathematically, this is expressed as \( P = \sigma A T^4 \), where \( P \) is the total power radiated, \( A \) is the surface area, \( T \) is the absolute temperature, and \( \sigma \) is the Stefan-Boltzmann constant.
In simple words: The Stefan-Boltzmann law says that very hot objects give off a lot more heat as radiation. The amount of heat they radiate goes up quickly (to the power of four) with their temperature.

🎯 Exam Tip: State the law clearly, mentioning "black body," "heat energy radiated per second per unit area," and "fourth power of its absolute temperature." Including the formula \( P = \sigma A T^4 \) is also helpful.

Question 10. What is Wien's law?
Answer: Wien's law states that the wavelength at which a black body emits the maximum intensity of radiation is inversely proportional to its absolute temperature. This means hotter objects emit light at shorter wavelengths, which is why very hot objects appear blue-white, while cooler objects appear red.
In simple words: Wien's law tells us that the hotter something is, the shorter the wavelength of light it shines brightest. So, very hot things glow blue, and cooler things glow red.

🎯 Exam Tip: Focus on the inverse relationship between the peak wavelength of emission and the absolute temperature. The formula \( \lambda_m T = b \) is a concise way to state this law.

Question 11. Define thermal conductivity. Give its unit.
Answer: Thermal conductivity is a measure of how well a material can transfer heat. It is defined as the quantity of heat transferred per unit time through a unit length of a material in a direction normal to a unit surface area, due to a unit temperature difference under steady-state conditions. Good conductors like metals have high thermal conductivity, while insulators like wood have low thermal conductivity. The SI unit for thermal conductivity is Watts per meter per Kelvin (\( \text{W m}^{-1} \text{ K}^{-1} \)).
In simple words: Thermal conductivity tells us how easily heat moves through a material. If heat moves fast, it has high conductivity. Its unit is Watts per meter per Kelvin.

🎯 Exam Tip: Ensure your definition includes all key aspects: quantity of heat transferred, unit time, unit length, unit surface area, and unit temperature difference, under steady-state conditions. Provide the correct SI unit.

Question 12. What is a black body?
Answer: A black body is an idealized physical body that absorbs all electromagnetic radiation (light, heat, etc.) that falls on it, without reflecting or transmitting any. It also emits radiation perfectly when heated. A perfect black body is a theoretical concept, but objects like the sun behave very similarly to a black body.
In simple words: A black body is a perfect absorber of all light and heat, and when it's hot, it gives off the maximum possible light and heat. It doesn't reflect or let any light pass through.

🎯 Exam Tip: The two key characteristics of a black body are that it absorbs all incident radiation and emits the maximum possible radiation for its temperature.

Question 13. What is a thermodynamic system? Give examples.
Answer: A thermodynamic system is a specific finite part of the universe that we choose to study. It contains a large number of particles, such as atoms and molecules, whose state can be described by certain parameters like pressure (P), volume (V), and temperature (T). The part of the universe outside the system is called the surroundings, and the system and surroundings are separated by a boundary. For example, a thermodynamic system can be a liquid (like water in a cup), a solid (like an ice cube), a gas (like air in a balloon), or even radiation.
In simple words: A thermodynamic system is a specific part of the world we are looking at to study its energy. It could be water, air, or a solid. Everything else around it is called the surroundings.

🎯 Exam Tip: Define a thermodynamic system as a specific part of the universe under study. Emphasize that its state is defined by macroscopic parameters (P, V, T) and provide diverse examples.

Question 14. What are the different types of thermodynamic systems?
Answer: There are three main types of thermodynamic systems, classified by how they interact with their surroundings:
(i) Open system: An open system can exchange both matter and energy with its environment. For example, an open cup of hot coffee can lose steam (matter) and heat (energy) to the air.
(ii) Closed system: A closed system can exchange energy but not matter with its environment. For instance, a sealed can of soda can get hot or cold (exchange energy), but no soda can leak out (no matter exchange).
(iii) Isolated system: An isolated system cannot exchange either energy or matter with its environment. A perfectly insulated thermos containing hot coffee is an example, as it tries to keep both heat and the coffee inside completely separated from the outside world. This type of system is theoretical, as perfect insulation is impossible.
In simple words: There are three types of systems: open (exchanges matter and energy), closed (exchanges only energy), and isolated (exchanges nothing).

🎯 Exam Tip: Clearly define each type of system (open, closed, isolated) based on its ability to exchange matter and energy with the surroundings, and provide a simple example for each.

Question 15. What is meant by 'thermal equilibrium'?
Answer: Thermal equilibrium is a state where two or more systems in contact with each other have reached the same temperature. At this point, there is no net flow of heat energy between them, and their temperatures will no longer change over time. It's like two objects eventually becoming equally warm or cold when placed together.
In simple words: Thermal equilibrium means that two things touching each other are at the same temperature, so no more heat moves between them.

🎯 Exam Tip: The key idea of thermal equilibrium is that systems in contact have the same temperature, and there is no net heat transfer between them.

Question 16. What are the quantities that are used to describe the equilibrium states of a Thermodynamic system? Give example.
Answer: The quantities used to describe the equilibrium states of a thermodynamic system are called state variables (or thermodynamic variables). These variables define the system's condition and do not depend on how the system reached that state. Common examples include pressure (P), volume (V), and temperature (T). These three are often referred to as the primary state variables.
In simple words: We use quantities like pressure, volume, and temperature to describe a thermodynamic system when it is balanced and not changing.

🎯 Exam Tip: List pressure, volume, and temperature as the primary state variables. Emphasize that these variables define the "state" of the system at equilibrium.

Question 17. What are intensive and extensive variables? Give examples.
Answer: Thermodynamic variables can be divided into two types:
Extensive variables: These are properties that depend on the size or mass of the system. If you change the amount of material, these properties will change. Examples include volume, total mass, entropy, internal energy, and heat capacity.
Intensive variables: These are properties that do not depend on the size or mass of the system. They remain the same regardless of how much material you have. Examples include temperature, pressure, specific heat capacity, and density. To illustrate, a small cup of hot water and a large pot of hot water can both have the same temperature, but the pot will have a larger volume and total mass.
In simple words: Extensive variables change with the amount of stuff you have (like volume or total mass), while intensive variables do not change with the amount of stuff (like temperature or pressure).

🎯 Exam Tip: Provide a clear definition for both extensive and intensive variables. Give at least two distinct examples for each type to illustrate the difference effectively.

Question 18. What is an equation of state? Give an example.
Answer: An equation of state is a mathematical relationship that connects various state variables of a thermodynamic system, such as pressure, volume, and temperature, in a specific manner. It helps describe the physical properties of a substance under different conditions. A classic example is the ideal gas law, which is given by \( PV = NkT \) (or \( PV = \mu RT \)). Another example is Van der Waals' equation for real gases, which is \( (P + \frac{a}{V^2})(V - b) = RT \). These equations allow us to predict how a gas will behave when its conditions change.
In simple words: An equation of state is a math rule that shows how things like pressure, volume, and temperature are connected for a substance. The ideal gas law (\( PV = NkT \)) is a good example.

🎯 Exam Tip: Define an equation of state as a relation between state variables. Provide the ideal gas law as a primary example and briefly mention Van der Waals' equation for real gases.

Question 19. State Zeroth law of thermodynamics.
Answer: The Zeroth Law of Thermodynamics states that if two systems, A and B, are each in thermal equilibrium with a third system, C, then systems A and B are also in thermal equilibrium with each other. This law helps define the concept of temperature. For example, if a thermometer (System C) reads the same temperature when placed in water (System A) and then in milk (System B), it means the water and milk are at the same temperature as each other.
In simple words: If two things are both the same temperature as a third thing (like a thermometer), then those first two things are also the same temperature as each other.

🎯 Exam Tip: Clearly state the condition involving three systems (A, B, and C) and their thermal equilibrium. Emphasize that this law establishes the transitive property of thermal equilibrium, which allows for the definition of temperature.

Question 20. Define the internal energy of the system.
Answer: The internal energy of a system is the total energy of all its molecules. This includes the kinetic energy (energy of motion) and potential energy (stored energy due to position) of these molecules. Understanding internal energy helps us analyze how heat and work interact with a system.
In simple words: Internal energy is all the energy hidden inside a system, like the movement and stored energy of its tiny parts.

🎯 Exam Tip: When defining internal energy, remember to mention both kinetic and potential energies of the molecules, not just one.

Question 21. Are internal energy and heat energy the same? Explain.
Answer: Internal energy and heat energy are not the same thing. Internal energy is the energy stored *within* a body, which increases when its temperature rises or when its state changes (like melting from solid to liquid). Heat, on the other hand, is the *transfer* of energy from one body to another because of a temperature difference. Heat is energy *in transit*, while internal energy is energy *possessed*. For example, a hot cup of coffee has high internal energy, but "heat" describes the energy leaving it.
In simple words: No, they are different. Internal energy is the energy a body has inside it, while heat is energy moving from a hotter place to a colder place.

🎯 Exam Tip: Distinguish between heat as an energy transfer and internal energy as energy stored within a system to score full marks.

Question 22. Define one calorie.
Answer: One calorie is the amount of heat energy needed to raise the temperature of 1 gram of water by 1 degree Celsius. In modern physics, 1 calorie is equal to 4.186 Joules (J). This unit is commonly used for measuring energy content in food.
In simple words: One calorie is the heat needed to warm up one gram of water by one degree Celsius. It equals 4.186 J.

🎯 Exam Tip: Always remember the numerical equivalent of 1 calorie in Joules (4.186 J) as it is a standard conversion factor.

Question 23. Did joule converted mechanical energy to heat energy? Explain.
Answer: Yes, James Joule showed that mechanical energy can be converted into heat energy. In his experiment, falling masses turned a paddle wheel in water. The friction between the paddle wheel and water made the water's temperature rise. This means the potential energy of the masses changed into the internal energy (heat) of the water. Joule's work was crucial in establishing the law of conservation of energy.
In simple words: Yes, Joule showed mechanical energy can become heat. He used falling weights to spin a paddle in water, and the friction heated the water.

🎯 Exam Tip: When explaining Joule's experiment, emphasize the conversion of gravitational potential energy to the internal energy of water through friction.

Question 24. State the first law of thermodynamics.
Answer: The first law of thermodynamics states that the change in a system's internal energy (\( \Delta U \)) is equal to the heat supplied to the system (\( Q \)) minus the work done *by* the system on its surroundings (\( W \)). This law is essentially a statement of energy conservation.
In simple words: The first law says that the energy inside a system changes by how much heat you add, minus the work the system does.

🎯 Exam Tip: Clearly state the relationship \( \Delta U = Q - W \) and explain each term, remembering the sign convention for work done *by* the system.

Question 25. Can we measure the temperature of the object by touching it?
Answer: No, we cannot accurately measure the temperature of an object just by touching it. Touching an object only tells us its "hotness" or "coolness" relative to our own body temperature, which is a subjective feeling. Temperature is a precise physical quantity that requires a thermometer for accurate measurement. Our sense of touch can easily be fooled by materials that conduct heat differently.
In simple words: No, touching only tells us if something feels hot or cold compared to us, not its exact temperature.

🎯 Exam Tip: Highlight that human touch provides a subjective sensation, not an objective measurement of temperature.

Question 26. Give the sign convention for Q and W.
Answer: The sign conventions for heat (\( Q \)) and work (\( W \)) in thermodynamics are:

🎯 Exam Tip: Always remember that work done *by* the system is positive in physics, while work done *on* the system is negative. This is a common source of error.

Question 28. Give the expression for work done by the gas.
Answer: In general, the work \( W \) done by a gas when its volume increases from an initial volume \( V_i \) to a final volume \( V_f \) is given by the integral of pressure \( P \) with respect to volume \( V \). This integral represents the area under the pressure-volume (\( PV \)) curve.
\[ W = \int_{V_{i}}^{V_{f}} P \, dV \]
In simple words: Work done by a gas is found by adding up all the small pressure multiplied by small changes in volume as the gas expands.

🎯 Exam Tip: The expression \( W = \int_{V_{i}}^{V_{f}} P \, dV \) is fundamental; clearly state the limits of integration and what P and dV represent.

Question 29. What is PV diagram?
Answer: A \( PV \) diagram, also known as a pressure-volume diagram, is a graph that shows the relationship between the pressure (\( P \)) and volume (\( V \)) of a thermodynamic system. This type of diagram is very useful for visualizing and calculating the amount of work done by or on a gas during expansion or compression. Each point on the diagram represents a specific state of the system.
In simple words: A \( PV \) diagram is a chart showing how pressure and volume change in a system. It helps to see how much work is done.

🎯 Exam Tip: Emphasize that a PV diagram is a graphical representation and its area under the curve represents the work done.

Question 30. Explain why the specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume.
Answer: The specific heat capacity at constant pressure (\( C_P \)) is greater than at constant volume (\( C_V \)) because when a gas is heated at constant pressure, it expands. This expansion means the gas does work on its surroundings. So, the heat supplied not only increases the internal energy of the gas but also provides the energy needed for this external work. However, when a gas is heated at constant volume, no work is done because there's no change in volume. All the supplied heat goes directly into increasing the internal energy. Therefore, more heat is required to raise the temperature by the same amount at constant pressure than at constant volume.
In simple words: \( C_P \) is bigger than \( C_V \) because at constant pressure, the gas expands and does work, needing extra heat. At constant volume, no work is done, so all heat just warms the gas.

🎯 Exam Tip: The key difference lies in the work done during expansion. Mentioning that heat supplied at constant pressure goes into both increasing internal energy and doing work is crucial.

Question 32. Give an expression for work done in an isothermal process.
Answer: For an isothermal process, where the temperature (\( T \)) remains constant, the work \( W \) done by a gas when its volume changes from \( V_i \) to \( V_f \) is given by the expression:
\[ W = \mu RT \ln \left(\frac{V_{f}}{V_{i}}\right) \]
Here, \( \mu \) is the number of moles of the gas, \( R \) is the ideal gas constant, and \( \ln \) denotes the natural logarithm. The constant temperature ensures that the internal energy does not change during the process.
In simple words: In an isothermal process (constant temperature), the work done by the gas is calculated using a formula involving the number of moles, gas constant, temperature, and the natural logarithm of the volume change.

🎯 Exam Tip: Remember that in an isothermal process, the internal energy change (\( \Delta U \)) is zero, so all heat input goes into work done by the system.

Question 33. Express the change in internal energy in terms of molar specific heat capacity.
Answer: The change in internal energy (\( dU \)) for a gas can be expressed in terms of its molar specific heat capacity at constant volume (\( C_V \)) and the change in temperature (\( dT \)). When a gas is heated at constant volume, no work is done, so all the heat added goes into increasing the internal energy.
\[ dU = \mu C_V dT \]
Here, \( \mu \) represents the number of moles of the gas. This formula shows how internal energy is directly related to temperature change and the specific heat capacity at constant volume.
In simple words: The change in internal energy is found by multiplying the number of moles, specific heat at constant volume, and the change in temperature.

🎯 Exam Tip: Note that molar specific heat capacity at constant volume (\( C_V \)) is used because it directly reflects changes in internal energy without work being done.

Question 34. Apply first law for (i) an isothermal (ii) adiabatic (iii) isobaric processes.
Answer: The first law of thermodynamics, \( \Delta U = Q - W \), can be applied differently to various thermodynamic processes:
(i) **Isothermal Process:** In an isothermal process, the temperature (\( T \)) is kept constant. This means the change in internal energy (\( \Delta U \)) is zero for an ideal gas.
\( \Delta U = 0 \)
\( \implies Q - W = 0 \)
\( \implies Q = W \)
So, any heat added to the system is entirely converted into work done by the system.
(ii) **Adiabatic Process:** In an adiabatic process, there is no heat exchange between the system and its surroundings, so \( Q = 0 \).
\( \Delta U = Q - W \)
\( \implies \Delta U = 0 - W \)
\( \implies \Delta U = -W \)
This means that if work is done *by* the system, its internal energy decreases. If work is done *on* the system, its internal energy increases.
(iii) **Isobaric Process:** In an isobaric process, the pressure (\( P \)) remains constant. Work is done by or on the system, and heat can be exchanged. The work done is \( W = P \Delta V \).
\( \Delta U = Q - W \)
\( \implies \Delta U = Q - P \Delta V \)
For isobaric expansion (\( Q > 0 \)), the system gains heat and does work.
For isobaric compression (\( Q < 0 \)), work is done on the system, and heat leaves.
In simple words: For isothermal, internal energy stays same so heat equals work. For adiabatic, no heat exchange, so internal energy changes because of work. For isobaric, pressure stays same, and internal energy changes based on heat and work done by volume change.

🎯 Exam Tip: Clearly state the defining condition for each process (constant T, Q=0, constant P) and how it simplifies the first law of thermodynamics.

Question 35. Give the equation of state for an adiabatic process.
Answer: For an adiabatic process, where no heat is exchanged with the surroundings, the equation of state relating pressure (\( P \)) and volume (\( V \)) is given by:
\[ PV^\gamma = \text{Constant} \]
Here, \( \gamma \) (gamma) is the adiabatic exponent, which is the ratio of the specific heat capacity at constant pressure (\( C_P \)) to the specific heat capacity at constant volume (\( C_V \)), i.e., \( \gamma = \frac{C_P}{C_V} \). This equation shows how pressure and volume change during an adiabatic process.
In simple words: For an adiabatic process, the pressure multiplied by the volume raised to the power of gamma always stays the same.

🎯 Exam Tip: Remember to define \( \gamma \) as the adiabatic exponent or ratio of specific heats when stating the adiabatic equation of state.

Question 36. State an equation state for an isochoric process.
Answer: In an isochoric process, the volume (\( V \)) of the system remains constant. For an ideal gas undergoing an isochoric process, the equation of state relates pressure (\( P \)) and temperature (\( T \)):
\[ P = \left(\frac{\mu R}{V}\right) T \]
Since \( \mu \), \( R \), and \( V \) are all constant, this equation simplifies to \( P \propto T \). This means that for a fixed volume, the pressure of an ideal gas is directly proportional to its absolute temperature. This relationship is often referred to as Gay-Lussac's Law.
In simple words: In an isochoric process (constant volume), the pressure of a gas is directly proportional to its temperature.

🎯 Exam Tip: The key takeaway for an isochoric process is \( V = \text{Constant} \), which implies that \( P \propto T \) for an ideal gas.

Question 37. If the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? If not, why?
Answer: No, if the piston of a container is pushed fast inward, the ideal gas equation will **not** be valid in the intermediate stage. This is because the ideal gas equation \( PV = nRT \) (or \( PV = NkT \)) describes a system in thermodynamic equilibrium. When a piston is compressed very quickly, the gas inside does not have enough time to exchange heat with its surroundings or to reach a uniform temperature and pressure throughout. Temperature and pressure gradients develop, meaning the system is not in equilibrium. Such a rapid process is called a non-quasi-static process.
In simple words: No, the ideal gas equation won't work if the piston is pushed too fast. This is because the gas won't be settled or balanced (in equilibrium) during the quick change.

🎯 Exam Tip: Emphasize that the ideal gas law only applies to systems in thermodynamic equilibrium; rapid processes disrupt this equilibrium.

Question 38. Draw the PV diagram for,
(a) isothermal process
(b) Adiabatic process
(c) Isobaric process
(d) Isochoric process
Answer: Here are the pressure-volume (\( PV \)) diagrams for different thermodynamic processes:
(a) **Isothermal Process:** The temperature remains constant, so \( PV = \text{constant} \). The curve is a hyperbola.

(b) **Adiabatic Process:** No heat exchange, \( PV^\gamma = \text{constant} \). The curve is steeper than an isothermal curve.

(c) **Isobaric Process:** The pressure remains constant, so \( P = \text{constant} \). This is shown as a horizontal line on the \( PV \) diagram.

(d) **Isochoric Process:** The volume remains constant, so \( V = \text{constant} \). This is shown as a vertical line on the \( PV \) diagram.

These diagrams help visualize how gases behave under different conditions and calculate the work involved.
In simple words: A \( PV \) diagram for isothermal is a curve; for adiabatic, it's a steeper curve. For isobaric, it's a flat horizontal line, and for isochoric, it's a straight vertical line.

🎯 Exam Tip: Practice drawing these \( PV \) diagrams clearly, paying attention to the shape of the curves (hyperbola for isothermal, steeper for adiabatic) and straight lines (horizontal for isobaric, vertical for isochoric).

Question 39. What is a cyclic process?
Answer: A cyclic process in thermodynamics is a series of changes that brings a system back to its initial state. This means that after undergoing several transformations, all the properties of the system (like pressure, volume, and temperature) return to their original values. Because the system ends up in the same state it started, the net change in its internal energy over a complete cycle is zero. This is a key concept in understanding heat engines.
In simple words: A cyclic process is when a system goes through different steps but always ends up exactly where it started, with no overall change inside.

🎯 Exam Tip: Remember that for a cyclic process, the net change in internal energy (\( \Delta U \)) is zero, simplifying the first law of thermodynamics to \( Q = W \).

Question 40. What is meant by a reversible and irreversible processes?
Answer:
**Reversible Processes:** A reversible process is an ideal thermodynamic process that can be reversed in such a way that both the system and its surroundings return to their original states without any net change in the universe. This means the process occurs infinitesimally slowly, allowing the system to remain in equilibrium at all times. Real-world processes are never perfectly reversible, but this concept helps understand ideal efficiency.
**Irreversible Processes:** An irreversible process is a real-world thermodynamic process that cannot be reversed without causing some permanent change in the surroundings. All natural processes, like heat flow from hot to cold or friction, are irreversible. These processes lead to an increase in the total entropy of the universe. They cannot be accurately plotted on a \( PV \) diagram because the system is not in equilibrium at every stage.
In simple words: A reversible process can be perfectly undone without leaving any trace. An irreversible process cannot be undone completely, always leaving some change behind in the world.

🎯 Exam Tip: The main distinction is that a reversible process can return to its initial state with no change to the surroundings, while an irreversible one always causes a net change in the universe.

Question 41. State Clausius form of the second law of thermodynamics.
Answer: The Clausius form of the second law of thermodynamics states: "Heat energy always flows from a hotter object to a colder object spontaneously." This means that heat will naturally move from a region of higher temperature to a region of lower temperature without any external help. It is impossible for heat to spontaneously flow from a colder body to a hotter body. This directionality is fundamental to how natural processes occur.
In simple words: Heat always moves by itself from warm things to cold things, never the other way around, says Clausius.

🎯 Exam Tip: Remember to use the keyword "spontaneously" to correctly state the Clausius form, emphasizing that no external work is needed for this natural heat flow.

Question 42. State Kelvin-Planck statement of second law of thermodynamics.
Answer: The Kelvin-Planck statement of the second law of thermodynamics declares that: "It is impossible to construct a heat engine that operates in a cycle and converts all the heat absorbed from a single reservoir completely into work." This means that every heat engine must reject some amount of heat to a colder reservoir. There is always some energy that cannot be converted into useful work. This principle establishes the limits of engine efficiency.
In simple words: The Kelvin-Planck statement says you can't build an engine that turns all its heat into work; some heat must always be wasted.

🎯 Exam Tip: Key phrases are "single reservoir" and "converts heat completely into work"; their impossibility is the core of the Kelvin-Planck statement.

Question 43. Define heat engine.
Answer: A heat engine is a device that takes heat energy from a high-temperature source, converts some of that heat into mechanical work, and then rejects the remaining heat to a low-temperature sink, all while operating in a continuous cycle. Examples include internal combustion engines and steam turbines. This process allows us to derive useful work from temperature differences.
In simple words: A heat engine is a machine that uses heat from a hot place to do work, and then sends some leftover heat to a colder place.

🎯 Exam Tip: A good definition includes the key components: heat source, work output, heat sink, and cyclic operation.

Question 44. What are processes involved in a Carnot engine?
Answer: A Carnot engine is an ideal heat engine that operates through a theoretical cycle known as the Carnot cycle. This cycle involves four reversible processes:
1. **Isothermal Expansion:** The working substance absorbs heat from a hot reservoir and expands at a constant high temperature.
2. **Adiabatic Expansion:** The working substance expands further, but without exchanging heat, causing its temperature to drop.
3. **Isothermal Compression:** The working substance rejects heat to a cold reservoir and is compressed at a constant low temperature.
4. **Adiabatic Compression:** The working substance is compressed further, again without exchanging heat, returning it to its initial high-temperature state.
These four steps make the Carnot cycle the most efficient possible cycle for converting heat into work.
In simple words: A Carnot engine uses four steps: hot expansion (isothermal), cool expansion (adiabatic), cool compression (isothermal), and hot compression (adiabatic).

🎯 Exam Tip: List the four processes in order, specifying whether each is isothermal or adiabatic, and whether it involves expansion or compression.

Question 45. Can the given heat energy be completely converted to work in a cyclic process? If not, why?
Answer: No, according to the second law of thermodynamics, heat energy cannot be completely converted into work in a cyclic process. This is because every heat engine operating in a cycle must reject some heat to a colder reservoir. If it were possible to convert all heat into work in a cycle, it would violate the Kelvin-Planck statement of the second law. However, in a non-cyclic process, it is possible to convert heat completely into work, for instance, by simply expanding a gas isothermally.
In simple words: No, heat cannot be fully turned into work in a cycle because some heat must always be released to a colder place, based on the second law of thermodynamics.

🎯 Exam Tip: Clearly state "No" for a cyclic process and briefly mention the second law (Kelvin-Planck statement) as the reason. Note the distinction with non-cyclic processes.

Question 46. State the second law of thermodynamics in terms of entropy.
Answer: In terms of entropy, the second law of thermodynamics states: "For all processes that occur naturally (irreversible processes), the total entropy of the universe always increases. For reversible processes, the total entropy of the universe remains constant." Entropy is a measure of disorder or randomness in a system. This law indicates that natural processes tend towards greater disorder.
In simple words: The second law says that the total disorder (entropy) in the universe always goes up for natural changes, and stays the same for ideal, reversible changes.

🎯 Exam Tip: When stating the second law in terms of entropy, differentiate between irreversible (natural) processes causing an increase in entropy and reversible (ideal) processes resulting in no change.

Question 47. Why does heat flow from a hot object to a cold object?
Answer: Heat flows spontaneously from a hot object to a cold object because this direction of flow increases the total entropy (disorder) of the universe, which is consistent with the second law of thermodynamics. If heat were to flow from a cold object to a hot object spontaneously, the total entropy of the universe would decrease, violating the second law. Therefore, nature favors the increase in disorder, leading to heat flowing from higher to lower temperatures until thermal equilibrium is reached.
In simple words: Heat moves from hot to cold naturally because it increases the overall disorder (entropy) in the world, which is what the universe tends to do.

🎯 Exam Tip: Connect the natural flow of heat (hot to cold) directly to the increase in total entropy of the universe as dictated by the second law of thermodynamics.

Question 48. Define the coefficient of performance.
Answer: The coefficient of performance (\( \beta \)) for a refrigerator or a heat pump is defined as the ratio of the heat extracted from the cold body (or sink) to the external work done by the compressor. It tells us how effective the device is at moving heat for a given amount of work input.
\[ \beta = \frac{Q_L}{W} \]
Here, \( Q_L \) is the heat removed from the cold reservoir, and \( W \) is the work input. A higher coefficient of performance means better efficiency.
In simple words: The coefficient of performance shows how good a refrigerator is. It's the amount of heat it pulls out from inside, divided by the energy it uses to do that work.

🎯 Exam Tip: Remember the formula \( \beta = \frac{Q_L}{W} \) and that a higher value indicates better energy efficiency for cooling or heating.

III. Long Answer Questions:

Question 1. Explain the meaning of heat and work with suitable examples.
Answer:
**Heat:** Heat is the spontaneous transfer of thermal energy from an object at a higher temperature to an object at a lower temperature when they are in contact. This energy transfer happens until both objects reach the same temperature. Heat is energy *in transit*, not a property an object "contains." For example, if you place a hot cup of coffee on a table, heat will transfer from the coffee to the cooler table and the surrounding air until everything reaches the same temperature. It's more accurate to say "the coffee is hot" rather than "the coffee has heat."
**Work:** Work in thermodynamics is the transfer of energy by means other than heat, typically involving a force causing a displacement. When a system does work, its volume often changes against an external pressure, like a piston moving in a cylinder. For instance, when you rub your hands together, you do mechanical work on them, increasing their internal energy and making them warmer. If you then place your warm hands on your chin, heat will transfer from your hands to your chin due to the temperature difference. Both heat and work are forms of energy transfer, but they occur through different mechanisms.
In simple words: Heat is energy moving from hot to cold things by itself. Work is energy moving because of a force causing movement, like pushing something. Both are ways energy is transferred.

🎯 Exam Tip: Clearly differentiate heat and work as forms of energy transfer, not as energy stored. Use examples that highlight this distinction, such as a hot object cooling or rubbing hands together.

Question 2. Discuss the ideal gas laws.
Answer: The ideal gas laws describe how the pressure, volume, and temperature of an ideal gas are related. Here are the key laws:
(i) **Boyle's Law:** When the temperature of a gas is kept constant, the pressure (\( P \)) of the gas is inversely proportional to its volume (\( V \)). This means if you reduce the volume, the pressure goes up.
\[ P \propto \frac{1}{V} \quad (\text{at constant } T \text{ and number of moles}) \]
(ii) **Charles's Law:** When the pressure of a gas is kept constant, its volume (\( V \)) is directly proportional to its absolute temperature (\( T \)). So, if you heat a gas, it expands.
\[ V \propto T \quad (\text{at constant } P \text{ and number of moles}) \]
(iii) **Combining the Laws (Ideal Gas Law):** By combining Boyle's and Charles's laws, we get the ideal gas law. Initially, this can be written as \( PV = C T \), where \( C \) is a constant. This constant \( C \) can be expressed as \( C = k \times N \), where \( k \) is Boltzmann's constant and \( N \) is the total number of particles.
\( \implies PV = NkT \)
If we consider \( \mu \) moles of gas, then \( N = \mu N_A \), where \( N_A \) is Avogadro's number. Since \( N_A k = R \) (the ideal gas constant), the ideal gas law for \( \mu \) moles becomes:
\[ PV = \mu RT \]
This is the equation of state for an ideal gas, connecting its pressure, volume, and temperature under equilibrium conditions.
In simple words: Ideal gas laws tell us how pressure, volume, and temperature of a gas are linked. Boyle's law says pressure goes up if volume shrinks (at same temperature). Charles's law says volume grows if temperature rises (at same pressure). Putting them together gives the main ideal gas equation, \( PV = \mu RT \).

🎯 Exam Tip: Clearly state the relationship for Boyle's Law and Charles's Law separately before combining them into the ideal gas equation. Define all variables and constants like \( k \) and \( R \).

Question 3. Explain in detail the thermal expansion.
Answer: Thermal expansion is the tendency of matter to change its dimensions (shape, area, or volume) in response to a change in temperature. When a substance is heated, its atoms and molecules vibrate more intensely and spread out, causing the material to expand. The extent of expansion varies for different states of matter:
* **Solids:** When a solid is heated, its atoms vibrate with larger amplitudes around their fixed positions. The relative change in size for solids is generally small because the intermolecular forces are strong, holding the atoms tightly in place.
* **Liquids:** Liquids expand more than solids because their intermolecular forces are weaker, allowing molecules to move more freely. This means liquid molecules can spread out more easily when heated.
* **Gases:** Gases expand much more than liquids or solids. This is because the intermolecular forces in gases are almost negligible, allowing the molecules to move with great freedom and spread out significantly when their kinetic energy increases due to heating.
This phenomenon is used in many applications, like bimetallic strips in thermostats or the expansion gaps in bridges.
In simple words: Thermal expansion is when things get bigger because they get hotter. Atoms move more and spread out. Gases expand the most, then liquids, and solids expand the least because their parts are held tightly.

🎯 Exam Tip: Provide a clear definition of thermal expansion and then explain its behavior in solids, liquids, and gases, relating it to intermolecular forces and molecular motion.

Question 4. Describe the anomalous expansion of water. How is it helpful in our lives?
Answer: Water exhibits an unusual behavior called anomalous expansion. Unlike most liquids that continuously contract as they cool, water contracts normally when cooled from room temperature down to 4°C. However, when cooled *below* 4°C, it starts to expand instead of contracting, reaching its maximum density at 4°C. Below 4°C, its volume increases, and its density decreases.
This anomalous expansion is very helpful for life on Earth, especially in cold regions:
* **Aquatic Life Survival:** In winter, as a lake cools, the densest water (at 4°C) sinks to the bottom, while colder water (below 4°C) and ice (0°C) stay at the surface because they are less dense. This ice layer insulates the water below, keeping it at or near 4°C. This allows aquatic plants and animals to survive in the unfrozen water at the bottom. Without this property, lakes would freeze solid from the bottom up, killing most aquatic life.
* **Weathering:** The expansion of water when it freezes in cracks of rocks also contributes to physical weathering, breaking down rocks over time.
In simple words: Water acts strangely; it shrinks when cooling until 4°C, then it starts to grow bigger as it gets colder. This helps fish and other water creatures live in cold places because lakes freeze only on top, keeping the bottom warm.

🎯 Exam Tip: Focus on the specific temperature range (0°C to 4°C) where water expands upon cooling and explain how this property allows aquatic life to survive in frozen bodies of water.

Question 5. Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.
Answer:
**Calorimetry:** Calorimetry is the scientific process of measuring the amount of heat exchanged (released or absorbed) by a thermodynamic system during physical changes or chemical reactions. It is based on the principle of heat conservation: in an isolated system, the heat lost by hotter objects equals the heat gained by colder objects. Calorimeters are devices designed to achieve thermal isolation for accurate measurements. The fundamental idea is that no heat escapes to the surroundings, simplifying the energy balance.

**Derivation for Final Temperature:**
Consider two substances, A and B, initially at different temperatures, \( T_1 \) and \( T_2 \), respectively, where \( T_1 > T_2 \). When they are mixed in a calorimeter, they exchange heat until they reach a common final equilibrium temperature, \( T_f \).

According to the principle of calorimetry (conservation of heat):
Heat lost by substance A = Heat gained by substance B
We know that the heat gained or lost (\( Q \)) by a substance of mass \( m \), specific heat capacity \( s \), and temperature change \( \Delta T \) is given by \( Q = ms \Delta T \).

Let:
\( m_1 \) = mass of substance A
\( s_1 \) = specific heat capacity of substance A
\( T_1 \) = initial temperature of substance A

\( m_2 \) = mass of substance B
\( s_2 \) = specific heat capacity of substance B
\( T_2 \) = initial temperature of substance B

\( T_f \) = final equilibrium temperature of the mixture

Heat lost by substance A: \( Q_{lost} = m_1 s_1 (T_1 - T_f) \)
Heat gained by substance B: \( Q_{gain} = m_2 s_2 (T_f - T_2) \)

According to the principle of calorimetry:
\( Q_{lost} = Q_{gain} \)
\( m_1 s_1 (T_1 - T_f) = m_2 s_2 (T_f - T_2) \)

Now, we expand and rearrange to solve for \( T_f \):
\( m_1 s_1 T_1 - m_1 s_1 T_f = m_2 s_2 T_f - m_2 s_2 T_2 \)
\( m_1 s_1 T_1 + m_2 s_2 T_2 = m_2 s_2 T_f + m_1 s_1 T_f \)
\( m_1 s_1 T_1 + m_2 s_2 T_2 = T_f (m_1 s_1 + m_2 s_2) \)

Therefore, the final temperature \( T_f \) is:
\[ T_f = \frac{m_1 s_1 T_1 + m_2 s_2 T_2}{m_1 s_1 + m_2 s_2} \]
This expression allows us to calculate the equilibrium temperature when two substances at different initial temperatures are mixed.
In simple words: Calorimetry is about measuring how much heat is exchanged. When two things mix, the heat lost by the hotter one equals the heat gained by the colder one. We can use a formula that balances these heat amounts to find their final temperature.

🎯 Exam Tip: Clearly state the principle of calorimetry (heat lost = heat gained). Show all algebraic steps when deriving the final temperature expression, and define each variable used.

III. Long Answer Questions:

Question 1. Explain the meaning of heat and work with suitable examples.
Answer: Heat is the natural flow of energy from a hotter object to a colder one when they are in contact. This process is called heating. For example, a hot cup of coffee loses heat to the cooler air around it until both are at the same temperature. It's not correct to say "a hot cup of coffee has more heat"; instead, we say "the coffee is hot," meaning it has higher internal energy. Work, in physics, happens when a force causes an object to move or displace. It also involves the transfer of energy through mechanical means, like a piston moving in a cylinder of gas. For instance, when you rub your hands together, friction does work, which increases the temperature of your hands. If you then place your warm hands on your chin, heat transfers from your hands to your chin, making your chin feel warmer. This shows how work can be converted into thermal energy, which then transfers as heat.
In simple words: Heat is energy moving from hot to cold things when they touch. Work is when energy is used to move something or cause a change.

🎯 Exam Tip: Clearly define both heat and work, emphasizing that heat is energy in transit. Always include a relatable example for each concept to illustrate your understanding.

Question 2. Discuss the ideal gas laws.
Answer: The ideal gas laws describe how an ideal gas behaves under different conditions of pressure, volume, and temperature.
(i) **Boyle's Law:** When the temperature of a gas is kept constant, the pressure of the gas is inversely proportional to its volume. This means if you push on a gas to make its volume smaller, its pressure will go up. We can write this as \( P \propto \frac { 1 }{ V } \).
(ii) **Charles's Law:** When the pressure of a gas is kept constant, its volume is directly proportional to its absolute temperature. This means if you heat a gas, its volume will get bigger. This is written as \( V \propto T \).
(iii) **Combining the Laws:** When we combine Boyle's and Charles's laws, we get \( PV = CT \), where C is a constant. This equation helps us understand how these properties relate together.
(iv) **Ideal Gas Equation:** The constant C is actually equal to \( kN \), where \( k \) is Boltzmann's constant and \( N \) is the total number of particles (atoms or molecules) in the gas. If the gas has \( \mu \) moles of particles, then \( N = \mu N_A \), where \( N_A \) is Avogadro's number. By substituting these, we get \( PV = \mu R T \), where \( R = N_A k \) is the ideal gas constant. This is the ideal gas law, also called the equation of state for an ideal gas.
In simple words: Ideal gas laws tell us how gases act. Boyle's law says pressure and volume are opposites if temperature is steady. Charles's law says volume grows with temperature if pressure is steady. Together, they form the ideal gas equation, \( PV = \mu R T \), which links pressure, volume, temperature, and the amount of gas.

🎯 Exam Tip: Remember to state the conditions (constant temperature or pressure) for Boyle's and Charles's laws clearly. For the ideal gas law, ensure all terms (P, V, T, \( \mu \), R) are correctly identified.

Question 3. Explain in detail the thermal expansion.
Answer: Thermal expansion is when matter changes its size, area, or volume because its temperature changes. When a solid is heated, its atoms vibrate more intensely around their fixed positions, causing the material to expand slightly. Liquids expand more than solids because their particles have weaker forces between them, allowing them to move more freely. Gases expand the most because their particles have very weak intermolecular forces.
(i) **Linear Expansion:** For solids, if the temperature changes by a small amount \( \Delta T \), the change in length \( \Delta L \) is directly proportional to \( \Delta T \). The formula is \( \frac { \Delta L }{ L } = \alpha_L \Delta T \), where \( \alpha_L \) is the coefficient of linear expansion.
(ii) **Area Expansion:** Similarly, for a small change in temperature \( \Delta T \), the change in area \( \Delta A \) is directly proportional to \( \Delta T \). The formula is \( \frac { \Delta A }{ A } = \alpha_A \Delta T \), where \( \alpha_A \) is the coefficient of area expansion.
(iii) **Volume Expansion:** For a small change in temperature \( \Delta T \), the change in volume \( \Delta V \) is directly proportional to \( \Delta T \). The formula is \( \frac { \Delta V }{ V } = \alpha_V \Delta T \), where \( \alpha_V \) is the coefficient of volume expansion. Each type of expansion happens because the average distance between atoms increases with temperature.
In simple words: Thermal expansion is when things get bigger as they get hotter. Solids expand a little, liquids more, and gases the most. There are three types: linear (length), area (surface), and volume (space it takes up).

🎯 Exam Tip: Remember to define thermal expansion clearly and explain the behavior of solids, liquids, and gases. Ensure you state the formulas and the meaning of each coefficient for linear, area, and volume expansion.

Question 4. Describe the anomalous expansion of water. How is it helpful in our lives?
Answer: Most liquids expand when heated and contract when cooled. However, water shows unusual behavior, called anomalous expansion, between 0°C and 4°C. When water at 0°C is heated, its volume *decreases* until it reaches 4°C. At 4°C, water has its minimum volume and therefore its maximum density. If heated beyond 4°C, it expands like other liquids. Conversely, when water is cooled from room temperature, its volume keeps shrinking until 4°C, then it starts expanding as it cools from 4°C to 0°C.

This anomalous expansion is very helpful for life in cold countries. In winter, as the air temperature drops, the surface water in lakes and ponds cools down. When the surface water reaches 4°C, it becomes denser and sinks to the bottom. Water colder than 4°C (e.g., 3°C, 2°C, 1°C, 0°C) is less dense, so it stays near the surface. When the surface water freezes at 0°C, the ice floats on top, forming an insulating layer. This layer prevents the water below from freezing completely, allowing aquatic animals and plants to survive in the unfrozen deeper parts of the lake. Without this property, lakes would freeze solid from bottom to top, killing all life.
In simple words: Water is special because it shrinks when heated from 0°C to 4°C, then expands normally. It is densest at 4°C. This helps fish and other water creatures survive in winter, as ice floats on top, keeping the deeper water from freezing.

🎯 Exam Tip: When explaining anomalous expansion, clearly state the temperature range (0°C to 4°C) and how density changes. For its importance, describe the insulating effect of ice on water bodies in cold climates.

Question 5. Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.
Answer: Calorimetry is the science of measuring the amount of heat energy absorbed or released by a thermodynamic system during a heating process. When a hotter body touches a colder body, heat flows from the hot body to the cold body until they reach a common temperature. In an insulated system, the heat lost by the hot body equals the heat gained by the cold body. This can be expressed mathematically.
We can say: \( Q_{gain} = -Q_{lost} \) So, \( Q_{gain} + Q_{lost} = 0 \)
Heat changes are measured using a calorimeter. Imagine a hot sample at temperature \( T_1 \) put into a calorimeter with water at room temperature \( T_2 \). Over time, both the sample and water reach a final equilibrium temperature \( T_f \). Because the calorimeter is insulated, the total heat in the system stays the same. The heat gained by the water (and calorimeter) equals the heat lost by the hot sample.
According to the definition of specific heat capacity: \( Q_{gain} = m_2 s_2 (T_f - T_2) \) \( Q_{lost} = m_1 s_1 (T_f - T_1) \)
Since \( Q_{gain} = -Q_{lost} \), we can write: \( m_2 s_2 (T_f - T_2) = -m_1 s_1 (T_f - T_1) \) \( m_2 s_2 T_f - m_2 s_2 T_2 = -m_1 s_1 T_f + m_1 s_1 T_1 \) Now, gather the terms with \( T_f \): \( m_2 s_2 T_f + m_1 s_1 T_f = m_1 s_1 T_1 + m_2 s_2 T_2 \) Factor out \( T_f \): \( T_f (m_1 s_1 + m_2 s_2) = m_1 s_1 T_1 + m_2 s_2 T_2 \) So, the final temperature \( T_f \) is: \( T_f = \frac { m_1 s_1 T_1 + m_2 s_2 T_2 }{ m_1 s_1 + m_2 s_2 } \) Here, \( m_1 \) and \( s_1 \) are the mass and specific heat capacity of the hot sample, and \( m_2 \) and \( s_2 \) are for the water. This formula is vital for calculating final temperatures in mixing problems.
In simple words: Calorimetry measures heat transfer. When hot and cold things mix, the heat lost by one equals the heat gained by the other. The final temperature can be found using a special formula that balances the heat changes.

🎯 Exam Tip: When deriving the final temperature, clearly state the principle of heat exchange (heat lost = heat gained). Ensure all variables are defined and the algebraic steps are correct. Remember to mention that the system must be insulated.

Question 6. Discuss various modes of heat transfer.
Answer: Heat energy can move from one place to another through three main ways: conduction, convection, and radiation. Understanding these modes helps explain how warmth spreads in different situations.
(i) **Conduction:** This is when heat moves directly through a material, usually solids, because of a temperature difference. The particles in the hotter part vibrate more and pass this energy to the slower-moving particles in the cooler part through direct contact. For example, when you hold a metal spoon in hot soup, the heat travels up the spoon to your hand. This is how heat travels through a metal spoon put in hot soup.
(ii) **Convection:** This mode involves the transfer of heat through the actual movement of fluids (liquids or gases). When a fluid is heated, it becomes less dense and rises, while cooler, denser fluid sinks. This creates a continuous flow or current, distributing heat. A common example is boiling water in a pot, where hot water rises and cooler water falls. This natural movement helps distribute heat throughout the fluid.
(iii) **Radiation:** This is the transfer of heat energy through electromagnetic waves, like infrared radiation, without needing any medium (matter) to travel through. This is why we can feel the warmth from the sun even though there's empty space between the sun and Earth. It's also how a bonfire warms you even if you're not touching it. Each mode plays a specific role in how heat flows in different environments.
In simple words: Heat can move in three ways: conduction (touching), convection (moving fluids), and radiation (invisible waves).

🎯 Exam Tip: For each mode, provide a clear definition and a distinct example. Highlight the key difference between them: conduction requires contact, convection involves fluid movement, and radiation uses electromagnetic waves.

Question 7. Explain Newton's law of cooling.
Answer: Newton's law of cooling states that the rate at which an object loses heat is directly proportional to the temperature difference between the object and its surroundings. This means a very hot object will cool down much faster than a warm object in the same environment.
Mathematically, this is expressed as: \( \frac { dQ }{ dt } \propto - (T - T_s) \) The negative sign shows that heat is lost, and the object's temperature decreases over time.
Let \( T \) be the temperature of the object and \( T_s \) be the temperature of the surroundings.
The graph shows that the cooling rate is highest when the temperature difference is largest, and it slows down as the object gets closer to the surrounding temperature.

Let an object of mass \( m \) and specific heat capacity \( s \) be at temperature \( T \). If the surroundings are at temperature \( T_s \), and the object's temperature drops by a small amount \( dT \) in time \( dt \), the heat lost is \( dQ = -msdT \). The negative sign implies a decrease in heat.
From the law, \( \frac { dQ }{ dt } \propto (T - T_s) \). Therefore, \( \frac { dQ }{ dt } = -k(T - T_s) \), where \( k \) is a positive constant. Combining these, we get: \( -ms\frac { dT }{ dt } = -k(T - T_s) \)
Rearranging to separate variables: \( \frac{dT}{T - T_s} = \frac{k}{ms} dt \) Let \( A = \frac{k}{ms} \). Integrating both sides: \( \int \frac{dT}{T - T_s} = \int A dt \) \( \ln(T - T_s) = A t + B \) (where B is the integration constant) Exponentiating both sides: \( T - T_s = e^{At+B} = e^B e^{At} \) Let \( C = e^B \). \( T = T_s + C e^{At} \) This exponential decay formula shows that the temperature difference decreases exponentially over time. It is crucial for understanding how objects cool down in various situations.
In simple words: Newton's law of cooling says that hot things cool faster if there's a big difference between their temperature and the air around them. As the difference gets smaller, they cool slower. This is why a cup of hot tea cools down quickly at first, then more slowly.

🎯 Exam Tip: When describing Newton's law of cooling, clearly state the direct proportionality between the rate of heat loss and the temperature difference. The mathematical expression and the exponential decay curve are essential for a complete answer.

Question 8. Explain Wien's law and why our eyes are sensitive only to visible rays?
Answer: Wien's displacement law states that the wavelength at which a black body emits the maximum intensity of radiation is inversely proportional to its absolute temperature. This means that hotter objects emit most of their radiation at shorter wavelengths (like blue light or UV), while cooler objects emit at longer wavelengths (like red light or infrared).
Mathematically, Wien's law is given by: \( \lambda_m T = b \) or \( \lambda_m = \frac { b }{ T } \) where \( \lambda_m \) is the peak wavelength, \( T \) is the absolute temperature in Kelvin, and \( b \) is Wien's displacement constant (approximately \( 2.898 \times 10^{-3} \text{ m K} \)).
If a body's temperature increases, the peak wavelength shifts towards shorter wavelengths, meaning higher frequencies and more energetic radiation. For example, a heating iron bar first glows red, then orange, then yellow, and if heated enough, it would appear white or blue-white.
**Wien's law and Human Vision:** Our eyes are sensitive only to visible light, which spans wavelengths from about 400 nm to 700 nm. This range is not arbitrary; it's a result of evolution. The Sun, which is the primary source of light for life on Earth, acts approximately like a black body with a surface temperature of around 5700 K.
Using Wien's law for the Sun: \( \lambda_m = \frac { 2.898 \times 10^{-3} \text{ m K} }{ 5700 \text{ K} } \approx 508 \times 10^{-9} \text{ m} = 508 \text{ nm} \) This calculated peak wavelength (508 nm) falls right in the middle of the visible light spectrum (green-blue light). Humans evolved under the Sun's radiation, so our visual systems naturally became most sensitive to the wavelengths that the Sun emits most intensely. This adaptation ensures we can efficiently perceive our surroundings using the most abundant light available.



In simple words: Wien's law says that hotter objects glow with shorter wavelengths (like blue), while cooler objects glow with longer wavelengths (like red). Our eyes see only visible light because the Sun, our main light source, emits most of its light in this range, so our eyes evolved to best use that light.

🎯 Exam Tip: When explaining Wien's law, define it and provide the formula. For human eye sensitivity, connect it directly to the Sun's peak emission wavelength and evolutionary adaptation.

Question 9. Discuss the,
(i) Thermal equilibrium
(ii) Mechanical equilibrium
(iii) Chemical equilibrium
(iv) Thermodynamic equilibrium
Answer:
(i) **Thermal Equilibrium:** Two systems are in thermal equilibrium if they are at the same temperature and there is no net flow of heat between them. Their temperatures will not change over time.
(ii) **Mechanical Equilibrium:** A system is in mechanical equilibrium if there are no unbalanced forces acting on it or on its surroundings. This means the system is either at rest or moving at a constant velocity.
(iii) **Chemical Equilibrium:** Chemical equilibrium occurs when there is no net chemical reaction taking place within the system or between systems in contact. The concentrations of reactants and products remain constant over time.
(iv) **Thermodynamic Equilibrium:** A system is in thermodynamic equilibrium if it is simultaneously in thermal, mechanical, and chemical equilibrium with itself and its surroundings. In this state, all macroscopic properties like pressure, volume, and temperature have fixed, unchanging values.
In simple words: Equilibrium means a balanced state. Thermal equilibrium means everything is the same temperature. Mechanical means no forces are pushing or pulling it. Chemical means no new chemicals are being made. Thermodynamic equilibrium means all three of these are balanced at the same time.

🎯 Exam Tip: For each type of equilibrium, clearly state the condition that must be met (e.g., same temperature for thermal, no unbalanced forces for mechanical). Emphasize that thermodynamic equilibrium is a combination of all three.

Question 10. Explain Joule's Experiment of the mechanical equivalent of heat.
Answer: James Prescott Joule's experiment showed that mechanical energy can be directly converted into internal energy (heat). In his setup, two masses were tied to a rope that turned a paddle wheel immersed in a container of water. The container was insulated to prevent heat loss to the surroundings.
When the masses fell due to gravity, they lost potential energy (equal to \( 2mgh \), where \( m \) is the mass, \( g \) is gravity, and \( h \) is the fall height). This mechanical energy made the paddle wheel turn inside the water. The friction between the paddle wheel and the water caused the water's temperature to rise.

Joule found that a specific amount of mechanical work always produced the same amount of heat, causing a predictable rise in water temperature. For example, he determined that 4.186 J of mechanical energy is equivalent to 1 calorie of heat energy (the energy needed to raise the temperature of 1 gram of water by 1°C). This constant is known as the mechanical equivalent of heat. Joule's experiment confirmed the principle of conservation of energy by showing that different forms of energy (mechanical and thermal) are interchangeable.
In simple words: Joule's experiment showed that rubbing or moving things creates heat. He used falling weights to turn paddles in water, making the water warmer. This proved that mechanical energy can be changed directly into heat energy, and there's a fixed amount of work that makes a fixed amount of heat.

🎯 Exam Tip: Describe Joule's experimental setup and the energy transformations involved. State the mechanical equivalent of heat (1 calorie = 4.186 J) and emphasize how it demonstrates the conservation of energy.

Question 11. Derive the expression for the work done in a volume change in a thermodynamic system.
Answer: Let's consider an ideal gas inside a cylinder fitted with a movable, frictionless piston. When the gas expands quasi-statically (meaning very slowly, so it always stays in equilibrium), it pushes the piston by a small distance \( dx \). During this process, the pressure, temperature, and internal energy of the gas remain well-defined at every instant.
The force exerted by the gas on the piston is \( F = P A \), where \( P \) is the pressure of the gas and \( A \) is the area of the piston.
The small amount of work (\( dW \)) done by the gas on the piston as it moves \( dx \) is given by: \( dW = F \, dx \) Substituting \( F = PA \): \( dW = PA \, dx \) Since \( A \, dx \) is the small change in volume (\( dV \)) of the gas (area multiplied by displacement), we can write: \( dW = P \, dV \)
If the gas expands from an initial volume \( V_i \) to a final volume \( V_f \), the total work done by the gas is found by integrating this expression: \( W = \int_{V_i}^{V_f} P \, dV \)

If the volume increases (\( dV > 0 \)), the gas does positive work. If the gas is compressed (\( dV < 0 \), meaning \( V_i > V_f \)), then work is done *on* the system, and \( W \) is negative. This integral is very important for analyzing thermodynamic processes. The area under the pressure-volume (PV) curve represents the work done.
In simple words: When a gas pushes a piston and expands, it does work. If the gas shrinks, work is done on it. We find the total work by adding up all the small amounts of work done as the volume changes. The formula for this is \( W = P \times \text{change in volume} \).

🎯 Exam Tip: Clearly define work done as force multiplied by displacement, then relate it to pressure and volume change. Remember that positive work means expansion (gas does work), and negative work means compression (work done on gas).

Question 12. Derive Meyer's relation for an ideal gas.
Answer: Meyer's relation links the molar specific heat capacities of an ideal gas at constant pressure (\( C_P \)) and constant volume (\( C_V \)). We can derive this using the first law of thermodynamics.
Consider one mole of an ideal gas.
**Case 1: Gas heated at constant volume.** If the gas is heated at a constant volume, no work is done (\( W = 0 \)). According to the first law of thermodynamics (\( Q = \Delta U + W \)), the heat supplied (\( Q \)) directly increases the internal energy (\( \Delta U \)) of the gas. If the temperature increases by \( dT \), the change in internal energy \( dU \) is given by: \( dU = \mu C_V dT \) ... (1) Here, \( \mu \) is the number of moles.
**Case 2: Gas heated at constant pressure.** If the gas is heated at a constant pressure, its temperature increases by \( dT \), and its volume also expands by \( dV \). The heat supplied (\( Q \)) is: \( Q = \mu C_P dT \) ... (2) The work done by the gas during expansion is: \( W = P \, dV \) ... (3)
Now, apply the first law of thermodynamics \( Q = dU + W \): Substitute (1), (2), and (3) into this equation: \( \mu C_P dT = \mu C_V dT + P \, dV \) ... (4)
For an ideal gas, the equation of state is \( PV = \mu RT \). If we differentiate this equation with respect to temperature at constant pressure: \( P \, dV = \mu R \, dT \) ... (5)
Substitute \( P \, dV \) from (5) into (4): \( \mu C_P dT = \mu C_V dT + \mu R \, dT \) Divide the entire equation by \( \mu dT \) (assuming \( \mu \ne 0 \) and \( dT \ne 0 \)): \( C_P = C_V + R \) Rearranging this gives Meyer's relation: \( C_P - C_V = R \) ... (6) This relation is fundamental for ideal gases and shows that \( C_P \) is always greater than \( C_V \) because extra energy is needed to do work during expansion at constant pressure.
In simple words: Meyer's relation shows that the specific heat of a gas at constant pressure is always bigger than its specific heat at constant volume. This is because when a gas heats up and expands while keeping pressure steady, it uses some energy to push things away (do work), not just to get hotter. The difference between these two specific heats equals the ideal gas constant R.

🎯 Exam Tip: When deriving Meyer's relation, clearly state the first law of thermodynamics. Break down the process into constant volume and constant pressure heating, correctly applying the work done equation and the ideal gas law. Remember to explain why \( C_P \) is greater than \( C_V \).

Question 13. Explain in detail the isothermal process.
Answer: An isothermal process is a thermodynamic process where the temperature of the system remains constant throughout. Even though the pressure and volume of the gas change, the temperature stays fixed. For an ideal gas, the equation of state is \( PV = \mu RT \). Since \( \mu \), \( R \), and \( T \) are constant in an isothermal process, it means that \( PV = \text{constant} \). This relationship is often called Boyle's Law.
When we plot pressure (P) against volume (V) for an isothermal process on a PV diagram, the graph forms a hyperbola. This curve is also known as an isotherm.
An important consequence of constant temperature in an ideal gas is that its internal energy (\( \Delta U \)) also remains constant, meaning \( \Delta U = 0 \). This is because the internal energy of an ideal gas depends only on its temperature.
According to the first law of thermodynamics, \( Q = \Delta U + W \). Since \( \Delta U = 0 \) for an isothermal process, the first law simplifies to \( Q = W \). This means that any heat added to the system is completely converted into work done by the system, and vice versa. For example, if a gas expands isothermally, it absorbs heat from its surroundings and uses that energy to do work.

In simple words: An isothermal process means the temperature stays exactly the same. For a gas, this means its pressure and volume change in a way that \( P \times V \) always stays constant. If heat goes in, it's used to do work, and the gas's internal energy doesn't change.

🎯 Exam Tip: When explaining an isothermal process, emphasize that temperature remains constant and that internal energy change (\( \Delta U \)) is zero for an ideal gas. Remember to mention the \( PV = \text{constant} \) relationship and its hyperbolic graph on a PV diagram.

Question 14. Derive the work done in an isothermal process.
Answer: Let's derive the expression for the work done when an ideal gas undergoes an isothermal process. In an isothermal process, the temperature (\( T \)) remains constant.
We know that the work done by a gas during a volume change from \( V_i \) to \( V_f \) is given by: \( W = \int_{V_i}^{V_f} P \, dV \) ... (1)
For an ideal gas, the equation of state is \( PV = \mu RT \), where \( \mu \) is the number of moles and \( R \) is the ideal gas constant. Since \( T \) is constant in an isothermal process, we can express pressure \( P \) as: \( P = \frac { \mu RT }{ V } \) ... (2)
Substitute this expression for \( P \) into the work done integral (1): \( W = \int_{V_i}^{V_f} \frac { \mu RT }{ V } \, dV \) Since \( \mu \), \( R \), and \( T \) are all constants in an isothermal process, we can take them out of the integral: \( W = \mu RT \int_{V_i}^{V_f} \frac { 1 }{ V } \, dV \)
The integral of \( \frac { 1 }{ V } \) with respect to \( V \) is \( \ln(V) \). So, evaluating the definite integral: \( W = \mu RT [\ln(V)]_{V_i}^{V_f} \) \( W = \mu RT (\ln(V_f) - \ln(V_i)) \) Using the logarithm property \( \ln(a) - \ln(b) = \ln(\frac{a}{b}) \): \( W = \mu RT \ln \left( \frac { V_f }{ V_i } \right) \) ... (3)
This is the expression for the work done in an isothermal process.
If the gas expands (\( V_f > V_i \)), then \( \frac{V_f}{V_i} > 1 \), so \( \ln \left( \frac{V_f}{V_i} \right) \) is positive, and the work done \( W \) is positive. This means the gas does work on its surroundings.
If the gas is compressed (\( V_f < V_i \)), then \( \frac{V_f}{V_i} < 1 \), so \( \ln \left( \frac{V_f}{V_i} \right) \) is negative, and the work done \( W \) is negative. This means work is done *on* the gas by the surroundings.
The work done is also represented by the area under the PV curve on a graph.

In simple words: To find the work done when a gas changes volume at a steady temperature, we use the ideal gas law to link pressure and volume. Then we use calculus to add up all the tiny bits of work. The final formula tells us the work depends on the number of gas particles, the temperature, and how much the volume changes.

🎯 Exam Tip: Start with the basic work integral \( W = \int P \, dV \). Correctly substitute \( P \) from the ideal gas law, and perform the integration carefully. Remember to discuss the sign of work for both expansion and compression.

Question 15. Explain in detail an adiabatic process.
Answer: An adiabatic process is a thermodynamic process where no heat energy is exchanged between the system and its surroundings. This means \( Q = 0 \). Even though there's no heat transfer, the pressure, volume, and temperature of the system can all change.
According to the first law of thermodynamics (\( Q = \Delta U + W \)), if \( Q = 0 \), then \( \Delta U = -W \). This implies that any work done by the gas comes from its own internal energy (causing its temperature to drop), or work done *on* the gas increases its internal energy (causing its temperature to rise).
An adiabatic process can happen in two main ways:
(i) **Thermal Insulation:** The system is perfectly insulated from its surroundings, preventing any heat from entering or leaving. For example, a gas inside a thermos flask that is quickly compressed or expanded would undergo an adiabatic process because the insulation prevents heat flow.
(ii) **Rapid Process:** The process occurs so quickly that there isn't enough time for heat to transfer. For instance, when a bicycle tire bursts, the air inside expands extremely fast. This rapid expansion is essentially adiabatic, causing the air to cool down noticeably even though the tire is not insulated.
The relationship between pressure and volume in an adiabatic process is given by \( PV^\gamma = \text{constant} \), where \( \gamma \) (gamma) is the adiabatic exponent, a ratio of specific heats \( C_P/C_V \). This constant is important for understanding how the gas behaves during adiabatic changes.


In adiabatic compression, work is done *on* the gas, increasing its internal energy and causing its temperature and pressure to rise. In adiabatic expansion, the gas does work *on* its surroundings, using its own internal energy, which causes its temperature and pressure to drop. This is why when a spray can is used, the gas escaping feels cold-it's undergoing rapid adiabatic expansion.
In simple words: An adiabatic process is when no heat goes in or out of a system. If you push on a gas quickly (compression), it gets hotter. If a gas expands quickly, it gets colder. This is because all energy changes come from the gas's own internal energy, not from outside heat.

🎯 Exam Tip: Define an adiabatic process as one with no heat exchange (\( Q=0 \)). Explain its impact on internal energy and work (\( \Delta U = -W \)). Provide examples for both thermally insulated and rapid processes, and mention the relationship \( PV^\gamma = \text{constant} \).

Question 16. Derive the work done in an adiabatic process.
Answer: Let us think about a gas with 'p' moles inside a cylinder. The cylinder has walls and a base that do not let heat pass through. A piston, which is also frictionless and insulated, is fitted into the cylinder, as shown in the picture. The piston's top surface has an area 'A'. When this system goes from an initial state (with pressure \( P_i \), volume \( V_i \), and temperature \( T_i \)) to a final state (with \( P_f \), \( V_f \), \( T_f \)) without any heat exchange, this is called an adiabatic process.

The work done, W, in this adiabatic process when the volume changes from \( V_i \) to \( V_f \) is given by:
\( W = \int_{V_{i}}^{V_{f}} P d V \) ... (1)
Since the adiabatic process happens very slowly (quasi-statically), the ideal gas law still holds true at every point. The adiabatic equation of state is \( PV^\gamma = \text{constant} \). We can write pressure as \( P = \frac{\text{constant}}{V^\gamma} \). If we put this into equation (1), we get:
\( W_{adia} = \int_{V_{i}}^{V_{f}} \text{constant} \frac{dV}{V^\gamma} \)
\( W_{adia} = \text{constant} \int_{V_{i}}^{V_{f}} V^{-\gamma} dV \)
\( W_{adia} = \text{constant} \left[ \frac{V^{-\gamma+1}}{-\gamma+1} \right]_{V_{i}}^{V_{f}} \)
\( W_{adia} = \frac{\text{constant}}{1-\gamma} [V_f^{1-\gamma} - V_i^{1-\gamma}] \)
Since \( PV^\gamma = \text{constant} \), we can write the constant as \( P_f V_f^\gamma \) or \( P_i V_i^\gamma \). So, we can substitute this constant back into the equation:
\( W_{adia} = \frac{1}{1-\gamma} [P_f V_f^\gamma V_f^{1-\gamma} - P_i V_i^\gamma V_i^{1-\gamma}] \)
\( W_{adia} = \frac{1}{1-\gamma} [P_f V_f - P_i V_i] \) ... (2)
From the ideal gas law, we know that \( P_i V_i = \mu R T_i \) and \( P_f V_f = \mu R T_f \). Substituting these into equation (2), we get:
\( W_{adia} = \frac{1}{1-\gamma} [\mu R T_f - \mu R T_i] \)
\( W_{adia} = \frac{\mu R}{1-\gamma} [T_f - T_i] \)
\( W_{adia} = \frac{\mu R}{\gamma-1} [T_i - T_f] \) ... (3)
In an adiabatic expansion, the gas does work, so \( W_{adia} \) is positive. This means \( T_i > T_f \), and the gas cools down. In an adiabatic compression, work is done *on* the gas, so \( W_{adia} \) is negative. This means \( T_i < T_f \), and the gas temperature increases.
In simple words: When a gas expands without exchanging heat, it uses its own internal energy to do work, which makes it cooler. If you push the piston and compress the gas quickly, you do work on it, and it gets hotter because its internal energy increases.

🎯 Exam Tip: Remember that in an adiabatic process, there is no heat exchange (\( Q=0 \)). The work done comes directly from the change in internal energy, which affects the temperature.

Question 17. Explain the isobaric process and derive the work done in this process.
Answer: An isobaric process is a type of thermodynamic process where the pressure of a system stays constant. Even though the pressure is consistent, the temperature, volume, and internal energy of the system can change. According to the ideal gas equation, \( V = \frac{\mu R}{P} T \). Here, \( \frac{\mu R}{P} \) is a constant, so \( V \propto T \). This means that in an isobaric process, the volume is directly proportional to the temperature. The V-T graph for an isobaric process is a straight line that passes through the origin.
The work done by the gas when its volume increases from \( V_i \) to \( V_f \) is given by:
\( W = \int_{V_{i}}^{V_{f}} P d V \)
Since the pressure (P) is constant in an isobaric process, we can take it out of the integral:
\( W = P \int_{V_{i}}^{V_{f}} d V \)
\( W = P [V]_{V_i}^{V_f} \)
\( W = P[V_f - V_i] \)
\( W = P \Delta V \)
If the change in volume \( \Delta V \) is positive (expansion), the work done by the gas is positive. If \( \Delta V \) is negative (compression), the work done by the gas is negative (work is done on the gas). The area under the isobaric curve on a PV diagram represents the work done by the gas.
The first law of thermodynamics for an isobaric process is \( \Delta U = Q - P \Delta V \). Here, some heat supplied (Q) increases the internal energy (\( \Delta U \)) and the rest is used to do work (\( P \Delta V \)). This means that the internal energy does not stay constant.
In simple words: An isobaric process is when the pressure of a gas stays the same. The work done by the gas in this process is simply the pressure multiplied by how much the volume changes. When the gas expands, it does positive work; when it gets compressed, negative work is done on it.

🎯 Exam Tip: Remember that the key feature of an isobaric process is constant pressure. This simplifies the work done calculation significantly to \( P \Delta V \).

Question 18. Explain in detail the isochoric process.
Answer: An isochoric process is a thermodynamic process where the volume of the system stays constant. Even though the volume does not change, the pressure, temperature, and internal energy can still vary. The pressure-volume (PV) graph for an isochoric process is a vertical line, parallel to the pressure axis, as shown in the figures.

In the figures, (a) shows an increase in pressure, and (b) shows a decrease in pressure at constant volume.
The equation of state for an isochoric process is \( P = (\frac{\mu R}{V}) T \). Since \( \frac{\mu R}{V} \) is a constant, we have \( P \propto T \). This means that in an isochoric process, the pressure is directly proportional to the temperature. Therefore, the P-T graph for an isochoric process is a straight line passing through the origin.
For an isochoric process, the change in volume \( \Delta V = 0 \). Because there is no change in volume, no work is done by or on the gas (\( W=0 \)). According to the first law of thermodynamics, \( \Delta U = Q - W \). Since \( W=0 \), the equation becomes \( \Delta U = Q \). This means that all the heat supplied to the system is used only to increase its internal energy, which in turn raises the temperature and pressure of the gas. This is a very efficient way to raise the internal energy.
In simple words: An isochoric process is one where the volume of a gas never changes. Because the gas cannot expand or shrink, no work is done. All the heat you add to the gas simply makes it hotter and increases its pressure.

🎯 Exam Tip: The key takeaway for an isochoric process is "constant volume" and "no work done" (\(W=0\)), meaning all heat added goes directly into internal energy and temperature increase.

Question 19. What are the limitations of the first law of thermodynamics?
Answer: The first law of thermodynamics is very good at explaining how heat and work can be converted into each other. However, it does not tell us the direction in which a process will naturally occur. For example:
(i) If a hot object touches a cold object, heat will always flow from the hot object to the cold object. The first law allows for heat to flow in either direction, but in nature, it only flows from hotter to colder. It never spontaneously flows from a cold object to a hot object.
(ii) When you apply brakes in a car, the car stops due to friction, and the kinetic energy is turned into heat. However, this heat energy cannot be completely converted back into the car's kinetic energy to make it move again. The first law does not explain why this reverse conversion is not possible or why there is always some loss when energy changes form. This limitation shows that the first law is not enough to explain many natural processes. It does not account for the quality or usability of energy.
In simple words: The first law tells us that energy is conserved, but it doesn't say *which way* things will happen. It doesn't explain why heat only goes from hot to cold, or why you can't get all the energy back from a process like friction.

🎯 Exam Tip: The first law is about energy conservation, but it lacks information about the direction of processes and the concept of entropy, which is addressed by the second law.

Question 20. Explain the heat engine and obtain its efficiency.
Answer: A heat engine is a device that takes heat energy as input and converts some of it into mechanical work by following a cyclic process. It typically consists of three main parts:
(i) **Hot reservoir (or Source):** This is a high-temperature source (\( T_H \)) that supplies heat to the engine. It can provide a large amount of heat without its temperature changing significantly. This is where the engine gets its input energy.
(ii) **Working substance:** This is the material inside the engine (like a gas or water vapor) that absorbs heat from the hot reservoir, expands, and does work.
(iii) **Cold reservoir (or Sink):** This is a low-temperature sink (\( T_L \)) where the engine releases the unused heat after doing work. It can absorb heat without its temperature changing significantly.
The diagram below shows a schematic of a heat engine:

The heat engine operates in a cycle, meaning after one full process, it returns to its original state. Because it returns to its initial state, the change in its internal energy over one cycle is zero. The efficiency of a heat engine (\( \eta \)) is defined as the ratio of the useful work done (output) to the heat absorbed from the hot reservoir (input) during one cycle.
\( \eta = \frac{\text{Work done}}{\text{Heat extracted}} = \frac{W}{Q_H} \) ... (1)
According to the first law of thermodynamics, the heat absorbed (\( Q_H \)) is used for both the work done (W) and the heat rejected to the cold reservoir (\( Q_L \)). So, \( Q_H = W + Q_L \), which means \( W = Q_H - Q_L \).
Substituting W into the efficiency equation:
\( \eta = \frac{Q_H - Q_L}{Q_H} = 1 - \frac{Q_L}{Q_H} \) ... (2)
This formula shows that the efficiency of a heat engine depends on the ratio of heat rejected to heat absorbed. For maximum efficiency, \( Q_L \) should be as small as possible, ideally zero, which would give \( \eta = 1 \) (100% efficiency). However, in practice, some heat must always be rejected to the cold reservoir.
In simple words: A heat engine takes heat from a hot place, turns some of it into useful work, and then dumps the rest of the heat into a cold place. Its efficiency is how much useful work it does compared to the total heat it took in.

🎯 Exam Tip: The efficiency of a heat engine is always less than 1 (or 100%) because some heat must always be rejected to the cold reservoir, as stated by the second law of thermodynamics.

Question 21. Explain in detail Carnot heat engine.
Answer: A Carnot engine is a theoretical reversible heat engine that operates in a specific cycle between two temperatures. It is the most efficient possible heat engine. A Carnot engine has four main parts:
1. **Source:** This is a hot reservoir maintained at a high temperature \( T_H \). It can provide any amount of heat (\( Q_H \)) without changing its temperature.
2. **Sink:** This is a cold reservoir maintained at a constant low temperature \( T_L \). It can absorb any amount of heat (\( Q_L \)) without changing its temperature.
3. **Insulating Stand:** This stand is made of perfectly non-conducting material, meaning no heat can pass through it. It helps to isolate the working substance when needed.
4. **Working Substance:** This is typically an ideal gas enclosed in a cylinder. The cylinder has perfectly non-conducting walls and a perfectly conducting bottom. A non-conducting and frictionless piston is fitted in it. The working substance undergoes a series of processes to convert heat into work.
These four parts are shown in the following figure:

**Carnot's Cycle:** The Carnot cycle consists of four reversible processes:
1. **Isothermal Expansion (A → B):** The cylinder is placed on the hot source (\( T_H \)). The gas absorbs heat \( Q_H \) and expands isothermally (at constant temperature). The piston moves outward, and work is done by the gas.
Work done: \( W_{A \to B} = \int_{V_1}^{V_2} P dV = \mu R T_H \ln(\frac{V_2}{V_1}) \)
2. **Adiabatic Expansion (B → C):** The cylinder is moved to the insulating stand. The gas continues to expand, but now adiabatically (no heat exchange). The temperature drops from \( T_H \) to \( T_L \). Work is done by the gas.
Work done: \( W_{B \to C} = \frac{\mu R}{\gamma-1} (T_H - T_L) \)
3. **Isothermal Compression (C → D):** The cylinder is placed on the cold sink (\( T_L \)). The gas is compressed isothermally, releasing heat \( Q_L \) to the sink. Work is done *on* the gas.
Work done: \( W_{C \to D} = \int_{V_3}^{V_4} P dV = \mu R T_L \ln(\frac{V_4}{V_3}) \)
4. **Adiabatic Compression (D → A):** The cylinder is again moved to the insulating stand. The gas is compressed adiabatically, and its temperature rises from \( T_L \) back to \( T_H \). Work is done *on* the gas, and the system returns to its initial state.
Work done: \( W_{D \to A} = \frac{\mu R}{\gamma-1} (T_L - T_H) \)
The PV diagram for these processes is shown below:

The net work done by the working substance over one complete cycle is the area enclosed by the cycle on the PV diagram (area ABCD). It is the sum of the work done in each process:
\( W_{net} = W_{A \to B} + W_{B \to C} + W_{C \to D} + W_{D \to A} \)
The adiabatic work terms cancel out because \( W_{B \to C} = -W_{D \to A} \).
So, \( W_{net} = W_{A \to B} + W_{C \to D} \)
This net work done is positive for the cycle shown, meaning the engine does useful work. This is an important way to analyze how heat engines operate and how much work they can produce.
In simple words: A Carnot engine is a perfect, imaginary engine that works in four steps: two steps where temperature stays the same (isothermal) and two steps where no heat moves in or out (adiabatic). It's the best possible engine because it gets the most work out of the heat difference between hot and cold places.

🎯 Exam Tip: The Carnot cycle is a theoretical benchmark. Understanding its four reversible processes is crucial for grasping the maximum possible efficiency of any heat engine operating between two given temperatures.

Question 22. Derive the expression for Carnot engine efficiency.
Answer: The efficiency \( \eta \) of any heat engine is defined as the ratio of the net work done (W) to the heat absorbed from the hot source (\( Q_H \)).
\( \eta = \frac{W}{Q_H} \) ... (1)
From the first law of thermodynamics, for one complete cycle, the net work done is the difference between the heat absorbed and the heat rejected: \( W = Q_H - Q_L \).
Substituting this into the efficiency equation, we get:
\( \eta = \frac{Q_H - Q_L}{Q_H} = 1 - \frac{Q_L}{Q_H} \) ... (2)
For the isothermal expansion (A to B) in the Carnot cycle, the heat absorbed is \( Q_H = \mu R T_H \ln(\frac{V_2}{V_1}) \).
For the isothermal compression (C to D), the heat rejected is \( Q_L = \mu R T_L \ln(\frac{V_3}{V_4}) \). (The negative sign for work done is omitted here as we're interested in the magnitude of heat ejected).
So, we have:
\( \frac{Q_L}{Q_H} = \frac{\mu R T_L \ln(\frac{V_3}{V_4})}{\mu R T_H \ln(\frac{V_2}{V_1})} = \frac{T_L}{T_H} \frac{\ln(\frac{V_3}{V_4})}{\ln(\frac{V_2}{V_1})} \) ... (3)
For the adiabatic processes in the Carnot cycle, we know that \( T_H V_2^{\gamma-1} = T_L V_3^{\gamma-1} \) and \( T_H V_1^{\gamma-1} = T_L V_4^{\gamma-1} \).
Dividing these two equations, we get:
\( \frac{T_H V_2^{\gamma-1}}{T_H V_1^{\gamma-1}} = \frac{T_L V_3^{\gamma-1}}{T_L V_4^{\gamma-1}} \)
\( (\frac{V_2}{V_1})^{\gamma-1} = (\frac{V_3}{V_4})^{\gamma-1} \)
This implies that \( \frac{V_2}{V_1} = \frac{V_3}{V_4} \). So, the logarithmic terms in equation (3) cancel out: \( \ln(\frac{V_3}{V_4}) = \ln(\frac{V_2}{V_1}) \).
Therefore, equation (3) simplifies to:
\( \frac{Q_L}{Q_H} = \frac{T_L}{T_H} \) ... (4)
Now, substituting this ratio back into the efficiency formula (2):
\( \eta = 1 - \frac{T_L}{T_H} \)
This is the expression for the efficiency of a Carnot engine. It shows that the efficiency only depends on the absolute temperatures of the hot and cold reservoirs. To achieve higher efficiency, the temperature difference between the source and sink should be as large as possible. An engine is more efficient when the hot reservoir is very hot or the cold reservoir is very cold. It highlights that no heat engine can be 100% efficient unless the cold reservoir is at absolute zero temperature, which is practically impossible.
In simple words: The efficiency of a Carnot engine depends only on how hot the hot part is and how cold the cold part is. The bigger the temperature difference, the more efficient the engine can be. This means to get more work, the heat needs to come from a much hotter place or go to a much colder place.

🎯 Exam Tip: The Carnot efficiency formula \( \eta = 1 - \frac{T_L}{T_H} \) is fundamental. Always use absolute temperatures (Kelvin) for \( T_L \) and \( T_H \) in this formula to get the correct efficiency.

Question 23. Explain the second law of thermodynamics in terms of entropy.
Answer: The second law of thermodynamics can be explained using the concept of entropy. Entropy is a measure of the disorder or randomness of a system. For a reversible process, the change in entropy is defined as \( \Delta S = \frac{Q}{T} \), where Q is the heat transferred and T is the absolute temperature.
For a Carnot engine, we found that \( \frac{Q_L}{Q_H} = \frac{T_L}{T_H} \), which can be rearranged as \( \frac{Q_H}{T_H} = \frac{Q_L}{T_L} \).
Here, \( \frac{Q_H}{T_H} \) represents the entropy received from the hot reservoir, and \( \frac{Q_L}{T_L} \) represents the entropy given out to the cold reservoir. For a reversible Carnot engine, the total change in entropy over one cycle is zero, meaning the entropy received equals the entropy given out.
However, the second law of thermodynamics in terms of entropy states a more general principle for all processes in nature:
"For all natural processes (which are irreversible), the total entropy of an isolated system always increases. For reversible processes, the total entropy remains unchanged."
This means that in any real-world process, the universe or an isolated system tends to become more disordered. Entropy determines the direction in which natural processes occur. For example, heat flows from hot to cold because that process increases the total entropy of the universe. If heat were to flow from cold to hot, the total entropy would decrease, which is forbidden by the second law. This law implies that systems spontaneously move towards states of greater probability and greater disorder. This is a very important idea that explains why things tend to spread out, break down, and become less organized over time.
In simple words: The second law of thermodynamics, when we talk about entropy, means that everything in the universe tends to get more messy and disordered over time. Heat will always move in a way that makes the total messiness (entropy) of everything bigger.

🎯 Exam Tip: When explaining the second law in terms of entropy, clearly state that entropy always increases for irreversible processes and remains constant for reversible ones, emphasizing the concept of increasing disorder.

Question 24. Explain in detail the working of a refrigerator.
Answer: A refrigerator is essentially a Carnot engine operating in reverse. Its main purpose is to transfer heat from a cold region (the inside of the refrigerator) to a hotter region (the surroundings, like the kitchen air). This process requires external work to be done on the system. The working of a refrigerator can be understood with the following components and diagram:

1. **Cold Reservoir (Sink):** The working substance (refrigerant gas) absorbs a quantity of heat \( Q_L \) from the cold interior of the refrigerator, which is at a lower temperature \( T_L \). This makes the inside of the refrigerator even colder.
2. **Work Done:** External work (W) is done on the working substance, typically by a compressor. This work is necessary to move the heat from the colder region to the warmer region, as heat does not naturally flow this way.
3. **Hot Reservoir (Source):** The working substance, now at a higher temperature and pressure due to compression, rejects a larger quantity of heat \( Q_H \) to the hot surroundings (like the kitchen air, at temperature \( T_H \)). This is why the air near the back of a refrigerator feels warm.
According to the first law of thermodynamics, for one complete cycle, the heat rejected to the hot reservoir is equal to the heat absorbed from the cold reservoir plus the work done on the system: \( Q_H = Q_L + W \).
By continuously performing this cycle, the refrigerator keeps its contents cold and transfers the heat to the warmer environment. This process highlights that transferring heat against its natural flow (from cold to hot) requires an input of energy.
In simple words: A refrigerator works like an engine in reverse. It uses energy (electricity) to pull heat from inside its cold space and push that heat out into the warmer room, keeping your food fresh.

🎯 Exam Tip: Remember that a refrigerator works by doing work to move heat from a cold region to a hot region, which is the opposite of a heat engine. The coefficient of performance (COP) is used to measure its efficiency, not the thermal efficiency.

IV. Numerical Problems:

Question 1. Calculate the number of moles of air in the inflated balloon at room temperature as shown in the figure.
Answer:
We are given the room temperature and the radius of the balloon. We need to find the number of moles of air.
Room temperature \( T = 273 + 30 = 303 \text{ K} \)
Radius of the balloon \( R = 10 \times 10^{-2} \text{ m} \)
Pressure inside the balloon \( P = 1.8 \times 10^5 \text{ Pa} \)
Volume of the balloon, which is a sphere, is given by \( V = \frac{4}{3} \pi R^3 \).
\( V = \frac{4}{3} \times 3.14 \times (10 \times 10^{-2})^3 \)
\( V = \frac{4 \times 3.14 \times 10^3 \times 10^{-6}}{3} \)
\( V = \frac{4 \times 3.14 \times 10^{-3}}{3} \)
We use the ideal gas law \( PV = \mu RT \), where \( \mu \) is the number of moles and \( R \) is the ideal gas constant (8.314 J/mol·K).
So, number of moles of air \( \mu = \frac{PV}{RT} \)
\( \mu = \frac{ (4 \times 3.14 \times 10^{-3} / 3) \times (1.8 \times 10^5) }{ 8.314 \times 303 } \)
\( \mu = \frac{ (4 \times 3.14 \times 10^{-3}) \times (1.8 \times 10^5) }{ 3 \times 8.314 \times 303 } \)
\( \mu \approx 0.3 \text{ moles} \)
In simple words: First, we find the balloon's volume using its radius. Then, we use the ideal gas law, which connects pressure, volume, temperature, and the amount of gas, to calculate how many moles of air are inside.

🎯 Exam Tip: Remember the ideal gas law formula \( PV = \mu RT \) and the formula for the volume of a sphere. Ensure consistent units (SI units) for all quantities.

Question 2. In the planet Mars, the average temperature is around -53°C and atmospheric pressure is 0.9 kPa. Calculate the number of moles of the molecules in unit volume in the planet Mars? Is this greater than that in earth?
Answer:
First, we convert the given temperature to Kelvin and pressure to Pascals.
Average temperature of Mars \( T_{mars} = -53^\circ \text{C} = -53 + 273 = 220 \text{ K} \)
Atmospheric pressure on Mars \( P_{mars} = 0.9 \text{ kPa} = 0.9 \times 10^3 \text{ Pa} \)
We need to find the number of moles per unit volume, which is \( \frac{\mu}{V} \). We use the ideal gas law: \( PV = \mu RT \).
Rearranging for moles per unit volume: \( \frac{\mu}{V} = \frac{P}{RT} \)
Substituting the values for Mars:
\( \frac{\mu}{V}_{mars} = \frac{0.9 \times 10^3}{8.314 \times 220} \approx 0.4921 \text{ mol/m}^3 \)
For Earth, at room temperature (let's assume \( 30^\circ \text{C} \) or \( 303 \text{ K} \)) and standard atmospheric pressure (1 atm = \( 1.01325 \times 10^5 \text{ Pa} \)):
\( \frac{\mu}{V}_{earth} = \frac{1.01325 \times 10^5}{8.314 \times 303} \approx 40.2 \text{ mol/m}^3 \)
Comparing the values:
Number of moles in unit volume on Mars \( \approx 0.49 \text{ mol/m}^3 \)
Number of moles in unit volume on Earth \( \approx 40.2 \text{ mol/m}^3 \)
The number of moles of molecules in a unit volume on Mars is much smaller than on Earth. The lower pressure and different temperature on Mars lead to a less dense atmosphere in terms of molecular count.
In simple words: We calculated how many gas molecules are in a small box on Mars and compared it to a similar box on Earth. Mars has fewer molecules in the same space because of its lower pressure and colder temperature.

🎯 Exam Tip: Remember to convert temperatures to Kelvin for gas law calculations. Pay attention to the units given (kPa vs Pa) and ensure consistency. The ideal gas law is key for these types of problems.

Question 3. An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁ and contains ideal gas at pressure P₁ and temperature T₁. The other chamber has volume V₁ and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.
Answer:
Let's assume the first chamber has initial conditions \( (P_1, V_1, T_1) \) and contains \( n_1 \) moles of gas. The second chamber has initial conditions \( (P_2, V_2, T_2) \) and contains \( n_2 \) moles of gas.
According to the ideal gas law, \( P_1 V_1 = n_1 R T_1 \) and \( P_2 V_2 = n_2 R T_2 \).
From these, we can find the number of moles in each chamber:
\( n_1 = \frac{P_1 V_1}{R T_1} \) and \( n_2 = \frac{P_2 V_2}{R T_2} \)
When the partition is removed, the total volume becomes \( V_{total} = V_1 + V_2 \) and the total number of moles is \( n_{total} = n_1 + n_2 \). Since no work is done and the container is insulated, the total internal energy of the system remains constant.
The internal energy of an ideal gas is \( U = n C_V T \), where \( C_V \) is the molar specific heat at constant volume.
So, the initial total internal energy is \( U_{initial} = n_1 C_V T_1 + n_2 C_V T_2 \).
The final total internal energy is \( U_{final} = (n_1 + n_2) C_V T_f \), where \( T_f \) is the final equilibrium temperature.
Since \( U_{initial} = U_{final} \):
\( n_1 C_V T_1 + n_2 C_V T_2 = (n_1 + n_2) C_V T_f \)
Dividing by \( C_V \) (assuming it's constant for both gases):
\( n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_f \)
Therefore, the final equilibrium temperature is:
\( T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \)
Now, substitute the expressions for \( n_1 \) and \( n_2 \):
\( T_f = \frac{(\frac{P_1 V_1}{R T_1}) T_1 + (\frac{P_2 V_2}{R T_2}) T_2}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}} \)
\( T_f = \frac{\frac{P_1 V_1}{R} + \frac{P_2 V_2}{R}}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}} \)
Multiply numerator and denominator by \( R \):
\( T_f = \frac{P_1 V_1 + P_2 V_2}{\frac{P_1 V_1}{T_1} + \frac{P_2 V_2}{T_2}} \)
To simplify the denominator, find a common denominator:
\( T_f = \frac{P_1 V_1 + P_2 V_2}{\frac{P_1 V_1 T_2 + P_2 V_2 T_1}{T_1 T_2}} \)
Finally, the expression for the final equilibrium temperature is:
\( T_f = \frac{(P_1 V_1 + P_2 V_2) T_1 T_2}{P_1 V_1 T_2 + P_2 V_2 T_1} \)
In simple words: When two gases mix in an insulated container, their total energy stays the same. We use the ideal gas law to find out how much gas is in each part, and then we use the idea that the total internal energy is conserved to calculate the new, single temperature after they mix.

🎯 Exam Tip: This problem relies on the conservation of internal energy for an insulated system and the ideal gas law. Remember to express the number of moles in terms of P, V, and T before substituting into the energy conservation equation.

Question 4. The temperature of a uniform rod of length L having a coefficient of linear expansion \( \alpha_L \) is changed by \( \Delta T \). Calculate the new moment of inertia of the uniform rod about axis passing through its centre and perpendicular to an axis of the rod.
Answer:
The initial moment of inertia of a uniform rod of mass M and length L about its center, perpendicular to its axis, is given by \( I = \frac{ML^2}{12} \).
When the temperature of the rod changes by \( \Delta T \), its length changes. The increase in length \( \Delta L \) is given by:
\( \Delta L = L \alpha_L \Delta T \)
The new length of the rod, \( L' \), will be:
\( L' = L + \Delta L = L (1 + \alpha_L \Delta T) \)
The mass M of the rod remains constant. So, the new moment of inertia \( I' \) about the same axis will be:
\( I' = \frac{M(L')^2}{12} \)
Substitute the expression for \( L' \):
\( I' = \frac{M [L (1 + \alpha_L \Delta T)]^2}{12} \)
\( I' = \frac{M L^2 (1 + \alpha_L \Delta T)^2}{12} \)
We know that \( I = \frac{ML^2}{12} \), so we can write the new moment of inertia in terms of the initial moment of inertia:
\( I' = I (1 + \alpha_L \Delta T)^2 \)
Expanding the term \( (1 + \alpha_L \Delta T)^2 \), we get \( 1 + 2\alpha_L \Delta T + (\alpha_L \Delta T)^2 \).
Since \( \alpha_L \Delta T \) is usually very small, the term \( (\alpha_L \Delta T)^2 \) is even smaller and can often be ignored.
Therefore, for small changes, the new moment of inertia is approximately:
\( I' \approx I (1 + 2\alpha_L \Delta T) \)
In simple words: When you heat a rod, it gets a little longer. Since the moment of inertia depends on the length, the moment of inertia also changes. We use the linear expansion formula to find the new length and then plug it into the moment of inertia formula to get the new value.

🎯 Exam Tip: Remember the basic formula for moment of inertia of a rod and the linear thermal expansion formula. Be careful with squaring the length term in the moment of inertia calculation. Often, approximations are valid for small temperature changes.

Question 5. Draw the TP diagram (P-x axis, T-y axis), VT(T-x axis, V-y axis) diagram for (a) Isochoric process (b) Isothermal process (c) isobaric process.
Answer:
(a) For an isochoric process (constant volume, \( V = V_0 \)), pressure is directly proportional to temperature (\( P \propto T \)).
The TP diagram will show temperature (T) on the x-axis and pressure (P) on the y-axis. The relationship is \( P = (\frac{\mu R}{V_0}) T \), which is a straight line passing through the origin.
The VT diagram will show temperature (T) on the x-axis and volume (V) on the y-axis. Since volume is constant, this will be a horizontal line at \( V = V_0 \).
(b) For an isothermal process (constant temperature, \( T = T_0 \)), pressure is inversely proportional to volume (\( P \propto \frac{1}{V} \)).
The TP diagram will show temperature (T) on the x-axis and pressure (P) on the y-axis. Since temperature is constant, this will be a vertical line at \( T = T_0 \).
The VT diagram will show temperature (T) on the x-axis and volume (V) on the y-axis. Since temperature is constant, this will be a horizontal line at \( T = T_0 \).
(c) For an isobaric process (constant pressure, \( P = P_0 \)), volume is directly proportional to temperature (\( V \propto T \)).
The TP diagram will show temperature (T) on the x-axis and pressure (P) on the y-axis. Since pressure is constant, this will be a horizontal line at \( P = P_0 \).
The VT diagram will show temperature (T) on the x-axis and volume (V) on the y-axis. The relationship is \( V = (\frac{\mu R}{P_0}) T \), which is a straight line passing through the origin.
In simple words: These diagrams show how pressure, volume, and temperature change together during different kinds of gas processes. For a constant volume process, pressure changes directly with temperature. For constant temperature, pressure and volume change opposite to each other. For constant pressure, volume changes directly with temperature.

🎯 Exam Tip: Clearly label axes and identify the type of curve for each process. Remember the ideal gas law \( PV = \mu RT \) and how constant variables affect the proportionality relationships to sketch these diagrams accurately.

Question 6. A man starts bicycling in the morning at a temperature around 25°C, he checked the pressure of tire which is equal to be 500 kPa. Afternoon he found that the absolute pressure in the tyre is increased to 520 kPa. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?
Answer:
Given initial conditions (morning):
Initial temperature \( T_1 = 25^\circ \text{C} = 25 + 273 = 298 \text{ K} \)
Initial pressure \( P_1 = 500 \text{ kPa} \)
Given final conditions (afternoon):
Final pressure \( P_2 = 520 \text{ kPa} \)
We need to find the final temperature \( T_2 \).
Since the expansion of the tyre is negligible, the volume of the air inside the tyre remains constant. This is an isochoric process.
For an isochoric process, Charles's Law states that pressure is directly proportional to absolute temperature: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)
Rearranging to solve for \( T_2 \):
\( T_2 = \frac{P_2 T_1}{P_1} \)
Substitute the given values:
\( T_2 = \frac{520 \text{ kPa} \times 298 \text{ K}}{500 \text{ kPa}} \)
\( T_2 = \frac{154960}{500} \)
\( T_2 = 309.92 \text{ K} \)
Now, convert this temperature back to Celsius:
\( T_2 = 309.92 - 273 = 36.92^\circ \text{C} \)
Therefore, the temperature of the tyre in the afternoon is approximately \( 36.92^\circ \text{C} \). This shows how warmer temperatures can increase tire pressure.
In simple words: Since the tire volume doesn't change much, we use a gas law that connects pressure and temperature. We found that as the pressure in the tire went up, the temperature of the air inside also increased, which is why tires often feel harder in hot weather.

🎯 Exam Tip: Always convert temperatures to Kelvin when using gas laws. Recognize an isochoric process when volume is stated as negligible or constant. Ensure you apply the correct gas law relationship.

Question 7. The temperature of a normal human body is 98.6°F. During high fever if the temperature increases to 104°F, what is the change in peak wavelength that emitted by our body? (Assume human body is a black body)
Answer:
We need to use Wien's Displacement Law, which states \( \lambda_{max} T = b \), where \( b \) is Wien's displacement constant (\( 2.898 \times 10^{-3} \text{ m K} \)). First, convert temperatures from Fahrenheit to Kelvin.
Normal body temperature \( T_{normal} = 98.6^\circ \text{F} \).
Conversion to Celsius: \( C = \frac{F-32}{1.8} \)
\( C_{normal} = \frac{98.6-32}{1.8} = \frac{66.6}{1.8} = 37^\circ \text{C} \)
Conversion to Kelvin: \( T_{K, normal} = 37 + 273 = 310 \text{ K} \)
Peak wavelength at normal temperature:
\( \lambda_{max, normal} = \frac{b}{T_{K, normal}} = \frac{2.898 \times 10^{-3} \text{ m K}}{310 \text{ K}} \)
\( \lambda_{max, normal} = 0.009348 \times 10^{-3} \text{ m} = 9348 \times 10^{-9} \text{ m} = 9348 \text{ nm} \)
Fever temperature \( T_{fever} = 104^\circ \text{F} \).
Conversion to Celsius: \( C_{fever} = \frac{104-32}{1.8} = \frac{72}{1.8} = 40^\circ \text{C} \)
Conversion to Kelvin: \( T_{K, fever} = 40 + 273 = 313 \text{ K} \)
Peak wavelength at fever temperature:
\( \lambda_{max, fever} = \frac{b}{T_{K, fever}} = \frac{2.898 \times 10^{-3} \text{ m K}}{313 \text{ K}} \)
\( \lambda_{max, fever} = 0.009258 \times 10^{-3} \text{ m} = 9258 \times 10^{-9} \text{ m} = 9258 \text{ nm} \)
Change in peak wavelength \( = \lambda_{max, normal} - \lambda_{max, fever} = 9348 \text{ nm} - 9258 \text{ nm} = 90 \text{ nm} \).
A higher temperature leads to a shorter peak wavelength of emitted radiation.
In simple words: We used Wien's Law to find the main wavelength of heat radiation from the body at normal and fever temperatures. Since the body gets hotter during a fever, it emits heat waves with slightly shorter wavelengths.

🎯 Exam Tip: Remember to convert temperatures to Kelvin for Wien's Law. Pay attention to the units for Wien's constant and ensure your final wavelength unit is consistent (e.g., nm).

Question 8. In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure? (For air \( \gamma \) = 1.4)
Answer:
For an adiabatic process, the relationship between pressure and volume is given by \( PV^\gamma = \text{Constant} \).
Taking the derivative of this equation with respect to P and V:
\( P \gamma V^{\gamma-1} dV + V^\gamma dP = 0 \)
Divide by \( V^\gamma \):
\( P \gamma V^{-1} dV + dP = 0 \)
Rearrange to find the relationship between percentage changes:
\( \frac{dP}{P} = -\gamma \frac{dV}{V} \)
We are given that the volume is increased by 4%, so the percentage change in volume is \( \frac{\Delta V}{V} \times 100\% = 4\% \).
Therefore, \( \frac{\Delta V}{V} = 0.04 \).
Also given, for air \( \gamma = 1.4 \).
Now, we can find the percentage change in pressure:
\( \frac{\Delta P}{P} \times 100\% = -\gamma \left(\frac{\Delta V}{V} \times 100\%\right) \)
\( \frac{\Delta P}{P} \times 100\% = -1.4 \times (4\%) \)
\( \frac{\Delta P}{P} \times 100\% = -5.6\% \)
The negative sign indicates that the pressure decreases. So, the percentage change in pressure is a 5.6% decrease. This is a common effect in adiabatic processes, where expansion leads to a drop in pressure and temperature.
In simple words: When air expands without heat being added or taken away (adiabatic process), its pressure drops. For a 4% increase in volume, the pressure goes down by 5.6% because of a special number for air called gamma.

🎯 Exam Tip: Remember the adiabatic relation \( PV^\gamma = \text{constant} \) and its differential form \( \frac{dP}{P} = -\gamma \frac{dV}{V} \). Ensure you include the negative sign to show a decrease in pressure during expansion.

Question 9. In a petrol engine, (internal combustion engine) air at atmospheric pressure and temperature of 20°C is compressed in the cylinder by the piston to 1/8 of its original volume. Calculate the temperature of the compressed air. (For air \( \gamma \) = 1.4)
Answer:
Given initial conditions:
Initial temperature \( T_1 = 20^\circ \text{C} = 20 + 273 = 293 \text{ K} \)
Initial volume \( V_1 = V \)
Given final conditions:
Final volume \( V_2 = \frac{1}{8} V \)
For air, \( \gamma = 1.4 \).
We need to find the final temperature \( T_2 \).
For an adiabatic process (like compression in an engine, which happens very quickly), the relationship between temperature and volume is given by \( T V^{\gamma-1} = \text{Constant} \).
So, \( T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \)
Rearranging to solve for \( T_2 \):
\( T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1} \)
Substitute the given values:
\( T_2 = 293 \text{ K} \left( \frac{V}{\frac{1}{8}V} \right)^{1.4-1} \)
\( T_2 = 293 \text{ K} (8)^{0.4} \)
Calculate \( 8^{0.4} \):
\( 8^{0.4} \approx 2.297 \)
\( T_2 = 293 \text{ K} \times 2.297 \)
\( T_2 \approx 673.1 \text{ K} \)
Convert this temperature back to Celsius:
\( T_2 = 673.1 - 273 = 400.1^\circ \text{C} \)
Therefore, the temperature of the compressed air is approximately \( 400^\circ \text{C} \). This rise in temperature is why engine cylinders get very hot during operation.
In simple words: When air in a petrol engine is squeezed quickly to a much smaller space, its temperature goes up a lot. We used a special formula for fast squeezing (adiabatic compression) to find that the air gets very hot, reaching around 400 degrees Celsius.

🎯 Exam Tip: Identify adiabatic processes in engine compression problems. Use the correct adiabatic relation \( T V^{\gamma-1} = \text{constant} \). Ensure temperature is in Kelvin and perform power calculations accurately.

Question 10. Draw the same cyclic process qualitatively in the V-T diagram where T is taken along x direction and V is taken along y-direction. Analyze the nature of heat exchange in each process.
Answer:
The provided figure (from the source, usually a P-V diagram with states 1, 2, 3) shows a cyclic process. Let's describe the processes and their heat exchange:
**Process 1 to 2:** This is an isothermal expansion (temperature constant, volume increases). During isothermal expansion, the gas absorbs heat from the surroundings and uses it to do work, keeping its internal energy unchanged.
**Process 2 to 3:** This is an isochoric heating process (volume constant, temperature increases). Since the volume is constant, no work is done by or on the gas. Heat is added to the system, which increases its internal energy and thus its temperature.
**Process 3 to 1:** This is an isobaric compression (pressure constant, volume decreases, temperature decreases). Work is done on the gas by the surroundings, and heat flows out of the system. This leads to a decrease in internal energy and temperature.
Now, let's describe how to draw this on a V-T diagram:
- **Process 1 to 2 (Isothermal Expansion):** Since temperature (T) is constant, this path will be a vertical line on the V-T diagram, going upwards (increasing V) at a fixed T.
- **Process 2 to 3 (Isochoric Heating):** Since volume (V) is constant, this path will be a horizontal line on the V-T diagram, moving to the right (increasing T) at a fixed V.
- **Process 3 to 1 (Isobaric Compression):** Since pressure (P) is constant, volume (V) is directly proportional to temperature (T) (\( V \propto T \)) for an ideal gas. This means the path will be a straight line passing through the origin (if extrapolated). On the V-T diagram, it will be a diagonal line going from a higher V and T (state 3) to a lower V and T (state 1).
In simple words: The process starts with the gas expanding at a steady temperature, where it takes in heat to do work. Next, the gas is heated at a constant volume, so all the heat increases its internal energy. Finally, the gas is squeezed at a constant pressure, releasing heat and getting cooler.

🎯 Exam Tip: Understand the relationship between P, V, T for isothermal, isochoric, and isobaric processes. Clearly identify heat exchange directions (absorbed/released) and work done (by/on the gas) for each step of the cycle.

Question 11. An ideal gas is taken in a cyclic process as shown in the figure. Calculate (a) work done by the gas. (b) work done on the gas. (c) Net work done in the process
Answer:
The figure shows a P-V diagram for a cyclic process with points A, B, C.
(a) Work done by the gas along AB (isobaric expansion):
From the figure, pressure for A-B is \( P = 600 \text{ Pa} \).
Change in volume \( \Delta V = V_B - V_A = 6 \text{ m}^3 - 3 \text{ m}^3 = 3 \text{ m}^3 \).
Work done \( W_{AB} = P \Delta V = 600 \text{ Pa} \times 3 \text{ m}^3 = 1800 \text{ J} \). (Work done *by* the gas, so it's positive)
(b) Work done by the gas along BC (isochoric process):
From the figure, volume is constant along B-C (\( V = 6 \text{ m}^3 \)).
For an isochoric process, \( \Delta V = 0 \), so work done \( W_{BC} = 0 \text{ J} \).
Work done by the gas along CA (isobaric compression):
From the figure, pressure for C-A is \( P = 400 \text{ Pa} \).
Change in volume \( \Delta V = V_A - V_C = 3 \text{ m}^3 - 6 \text{ m}^3 = -3 \text{ m}^3 \).
Work done \( W_{CA} = P \Delta V = 400 \text{ Pa} \times (-3 \text{ m}^3) = -1200 \text{ J} \). (Work done *by* the gas, negative because volume decreases).
(c) Net work done in the cyclic process is the sum of work done in each part:
\( W_{net} = W_{AB} + W_{BC} + W_{CA} \)
\( W_{net} = 1800 \text{ J} + 0 \text{ J} + (-1200 \text{ J}) \)
\( W_{net} = 600 \text{ J} \)
Alternatively, net work done is the area enclosed by the cycle on the P-V diagram. This is a rectangle:
Width = \( (6-3) \text{ m}^3 = 3 \text{ m}^3 \)
Height = \( (600-400) \text{ Pa} = 200 \text{ Pa} \)
Area = \( \text{Width} \times \text{Height} = 3 \text{ m}^3 \times 200 \text{ Pa} = 600 \text{ J} \). Since the cycle is clockwise, the net work is positive (work done by the gas).
In simple words: We calculated the work done during each step of the gas cycle shown in the diagram. We added up the work from expanding (positive work) and compressing (negative work) to find the total work done by the gas in one full cycle.

🎯 Exam Tip: Work done is the area under the P-V curve. For a cyclic process, net work is the area enclosed by the cycle. Clockwise cycles mean net work is done *by* the gas (positive), counter-clockwise means net work is done *on* the gas (negative).

Question 12. For a given ideal gas \( 6 \times 10^5 \) J heat energy is supplied and the volume of gas is increased from \( 4 \text{ m}^3 \) to \( 6 \text{ m}^3 \) at atmospheric pressure. Calculate (a) the work done by the gas (b) change in internal energy of the gas (c) graph this process in PV and TV diagram.
Answer:
Given:
Heat supplied \( Q = 6 \times 10^5 \text{ J} \)
Initial volume \( V_i = 4 \text{ m}^3 \)
Final volume \( V_f = 6 \text{ m}^3 \)
Atmospheric pressure \( P = 1.01325 \times 10^5 \text{ Pa} \) (standard atmospheric pressure, as not specified otherwise).
(a) Work done by the gas \( W \):
Since pressure is constant (atmospheric pressure), this is an isobaric process.
\( W = P (V_f - V_i) \)
\( W = 1.01325 \times 10^5 \text{ Pa} \times (6 \text{ m}^3 - 4 \text{ m}^3) \)
\( W = 1.01325 \times 10^5 \times 2 \text{ J} \)
\( W = 2.0265 \times 10^5 \text{ J} \)
(b) Change in internal energy of the gas \( \Delta U \):
According to the First Law of Thermodynamics, \( \Delta U = Q - W \).
\( \Delta U = 6 \times 10^5 \text{ J} - 2.0265 \times 10^5 \text{ J} \)
\( \Delta U = (6 - 2.0265) \times 10^5 \text{ J} \)
\( \Delta U = 3.9735 \times 10^5 \text{ J} \)
(c) Graph this process in P-V and T-V diagrams:
**P-V Diagram:**
- Pressure (P) on y-axis, Volume (V) on x-axis.
- Since pressure is constant, the process will be a horizontal line starting from \( V_i = 4 \text{ m}^3 \) to \( V_f = 6 \text{ m}^3 \) at \( P = 1 \text{ atm} \).
**T-V Diagram:**
- Temperature (T) on y-axis, Volume (V) on x-axis.
- Since \( P \) is constant and \( V \) increases, according to the ideal gas law \( PV = \mu RT \), if \( P \) is constant and \( V \) increases, then \( T \) must also increase. So, it will be a line sloping upwards to the right, from \( V_i \) to \( V_f \) with increasing T.
In simple words: We calculated how much work the gas did as it expanded under constant pressure after receiving heat. Then, using the first law of thermodynamics, we found how much of that heat actually went into increasing the gas's internal energy. The process would look like a straight horizontal line on a pressure-volume graph and a sloped line on a temperature-volume graph.

🎯 Exam Tip: For isobaric processes, work done is simply \( P \Delta V \). Always remember the First Law of Thermodynamics \( \Delta U = Q - W \). When sketching diagrams, correctly represent constant quantities as horizontal or vertical lines and proportional quantities as sloped lines.

Question 13. Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase the efficiency. (a) By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant (b) By increasing the temperature of the hot reservoir from 300°C tc 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?
Answer:
The efficiency of a reversible heat engine (Carnot engine) is given by \( \eta = 1 - \frac{T_L}{T_H} \), where \( T_L \) is the cold reservoir temperature and \( T_H \) is the hot reservoir temperature, both in Kelvin.
Initial conditions:
Cold reservoir temperature \( T_L = 100^\circ \text{C} = 100 + 273 = 373 \text{ K} \)
Hot reservoir temperature \( T_H = 300^\circ \text{C} = 300 + 273 = 573 \text{ K} \)
Initial efficiency:
\( \eta_{initial} = 1 - \frac{373}{573} = 1 - 0.6509 = 0.3491 \approx 34.91\% \)
(a) Method 1: Decrease cold reservoir temperature to 50°C, keep hot reservoir constant.
New cold reservoir temperature \( T_L' = 50^\circ \text{C} = 50 + 273 = 323 \text{ K} \)
Hot reservoir temperature \( T_H = 573 \text{ K} \)
New efficiency for method (a):
\( \eta_a = 1 - \frac{323}{573} = 1 - 0.5637 = 0.4363 \approx 43.63\% \)
(b) Method 2: Increase hot reservoir temperature to 350°C, keep cold reservoir constant.
Cold reservoir temperature \( T_L = 373 \text{ K} \)
New hot reservoir temperature \( T_H' = 350^\circ \text{C} = 350 + 273 = 623 \text{ K} \)
New efficiency for method (b):
\( \eta_b = 1 - \frac{373}{623} = 1 - 0.5987 = 0.4013 \approx 40.13\% \)
Comparing the efficiencies: \( \eta_a \approx 43.63\% \) and \( \eta_b \approx 40.13\% \).
Method (a) results in higher efficiency (43.63%) compared to method (b) (40.13%). Therefore, decreasing the cold reservoir temperature is the more suitable method to increase the efficiency of the heat engine in this scenario.
In simple words: To make a heat engine work better, you can either make the hot part hotter or the cold part colder. By calculating the efficiency for both options, we found that making the cold part colder gave a bigger boost to how well the engine worked.

🎯 Exam Tip: Always convert temperatures to Kelvin for efficiency calculations. To maximize Carnot efficiency, the temperature difference between the hot and cold reservoirs should be as large as possible. Decreasing \( T_L \) generally has a greater impact than increasing \( T_H \) by the same amount.

Question 14. An engine whose efficiency is 45% takes heat from a source maintained at a temperature of 327°C. To have an engine of efficiency 60% what must be the intake temperature for the same exhaust (sink) temperature?
Answer:
First, let's find the exhaust (sink) temperature using the initial efficiency.
Initial efficiency \( \eta_1 = 45\% = 0.45 \)
Source temperature \( T_{H1} = 327^\circ \text{C} = 327 + 273 = 600 \text{ K} \)
Using the efficiency formula \( \eta = 1 - \frac{T_L}{T_H} \):
\( 0.45 = 1 - \frac{T_L}{600} \)
\( \frac{T_L}{600} = 1 - 0.45 \)
\( \frac{T_L}{600} = 0.55 \)
\( T_L = 0.55 \times 600 = 330 \text{ K} \)
So, the exhaust (sink) temperature is 330 K.
Now, we want to achieve a new efficiency \( \eta_2 = 60\% = 0.60 \) with the *same* exhaust temperature \( T_L = 330 \text{ K} \). We need to find the new intake (source) temperature \( T_{H2} \).
\( \eta_2 = 1 - \frac{T_L}{T_{H2}} \)
\( 0.60 = 1 - \frac{330}{T_{H2}} \)
\( \frac{330}{T_{H2}} = 1 - 0.60 \)
\( \frac{330}{T_{H2}} = 0.40 \)
\( T_{H2} = \frac{330}{0.40} \)
\( T_{H2} = 825 \text{ K} \)
Convert this new intake temperature back to Celsius:
\( T_{H2, \text{C}} = 825 - 273 = 552^\circ \text{C} \)
Therefore, to achieve 60% efficiency with the same exhaust temperature, the intake temperature must be \( 552^\circ \text{C} \). This demonstrates that a higher operating temperature range improves engine performance.
In simple words: We first used the engine's current efficiency and hot temperature to figure out its cold exhaust temperature. Then, to get a higher efficiency of 60% with that same exhaust temperature, we calculated that the engine's hot intake temperature needs to be increased significantly.

🎯 Exam Tip: This problem involves two stages of efficiency calculation. It is crucial to correctly identify which temperature (source or sink) is kept constant between the two scenarios. Remember to always convert temperatures to Kelvin.

Question 15. An ideal refrigerator keeps its content at 0°C while the room temperature is 27°C. Calculate its coefficient of performance.
Answer: First, convert the temperatures to Kelvin. The cold reservoir temperature \( T_L \) is \( 0^\circ C + 273 = 273 \, K \). The hot reservoir temperature \( T_H \) is \( 27^\circ C + 273 = 300 \, K \). The coefficient of performance (\( \beta \)) for an ideal refrigerator is calculated using the formula: \( \beta = \frac{T_L}{T_H - T_L} \).
\( \beta = \frac{273}{300 - 273} \)
\( \beta = \frac{273}{27} \)
\( \beta = 10.11 \)
So, the coefficient of performance for this refrigerator is 10.11. This means the refrigerator is quite efficient at moving heat.
In simple words: We find out how well the refrigerator works by using a formula that compares the cold temperature inside to the temperature difference between inside and outside. After converting temperatures to Kelvin, we divide the cold temperature by the difference between hot and cold temperatures.

🎯 Exam Tip: Always remember to convert Celsius temperatures to Kelvin for thermodynamic calculations by adding 273 to the Celsius value. This is a common point of error.