RBSE Solutions Class 11 Physics Chapter 13 Thermodynamics

RBSE Class 11 Physics Chapter 13 Very Short Answer Type Questions

Question 1. If two systems A and B are in thermal equilibrium with C, then will A and B also be in thermal equilibrium?
Answer: Yes, if system A is in thermal balance with system C, and system B is also in thermal balance with system C, then A and B will also be in thermal balance with each other. This is explained by the Zeroth Law of Thermodynamics.
In simple words: Yes, if two things are balanced in heat with a third thing, they are also balanced in heat with each other.

🎯 Exam Tip: Always state the law or principle when directly asked to justify a thermodynamic statement. The Zeroth Law is fundamental for defining temperature.

Question 2. On which law is the first law of thermodynamics based on?
Answer: The First Law of Thermodynamics is based on the principle that energy cannot be created or destroyed, only changed from one form to another. This is known as the Law of Conservation of Energy.
In simple words: The first law of thermodynamics is all about how energy never goes away; it just changes its form.

🎯 Exam Tip: Remember that the first law is a direct application of the conservation of energy principle to thermodynamic systems.

Question 4. Write the P-V relation for adiabatic expansion of ideal gas.
Answer: For an ideal gas expanding without heat transfer (adiabatically), the relationship between its pressure (P) and volume (V) is given by \( PV^{\gamma} = \text{constant} \). Here, \( \gamma \) is a special ratio of specific heats (\( \frac{C_p}{C_v} \)) and is also called the atomicity of the gas. This shows how pressure and volume change together.
In simple words: For a gas that expands without gaining or losing heat, its pressure multiplied by its volume raised to the power of gamma always stays the same. Gamma is a special number for each gas.

🎯 Exam Tip: Make sure to correctly write the exponent \( \gamma \) for volume in the adiabatic relation, as it is a common point of error.

Question 5. What are the dimensions of efficiency of heat engine?
Answer: The efficiency (\( \eta \)) of a heat engine is calculated as the ratio of the useful work done (W) to the heat supplied from the source (\( Q_1 \)), so \( \eta = \frac{W}{Q_1} \). Since both work and heat are forms of energy and have the same units, their ratio (efficiency) has no units or dimensions. So, its dimensional formula is \( [M^0L^0T^0] \), meaning it has no mass, length, or time components.
In simple words: A heat engine's efficiency tells us how much work it does compared to the heat it uses. Since both work and heat are measured in the same way, efficiency doesn't have any units or size like meters or seconds; it's just a number.

🎯 Exam Tip: Remember that any ratio of quantities with the same units will be dimensionless. Efficiency is a classic example of this.

Question 6. What is the effect on pressure on change of state in a system in an isobaric process?
Answer: In an isobaric process, the pressure within a system does not change, even if the state of the system changes (like volume or temperature). Therefore, the change in pressure (\( \Delta P \)) during such a process is always zero.
In simple words: In an isobaric process, the pressure always stays the same. So, there is no change in pressure at all.

🎯 Exam Tip: "Isobaric" literally means "constant pressure." Clearly stating this definition is key to answering questions about it.

Question 7. Can heat be supplied without increasing the temperature of gas?
Answer: Yes, it is possible to supply heat to a gas without making its temperature go up. This happens when the gas uses that heat energy to expand and do work on its surroundings. In such cases, the added heat is converted into work instead of raising the temperature. A good example is during an isothermal expansion where temperature remains constant.
In simple words: Yes, you can add heat to a gas and its temperature won't go up if the gas uses that heat to push outwards and do work.

🎯 Exam Tip: Relate this concept to the First Law of Thermodynamics, where heat can be converted into work instead of increasing internal energy (temperature).

Question 9. What is the specific heat of a gas in isothermal and adiabatic process?
Answer: The specific heat (\( s \)) of a substance is defined by the formula \( s = \frac{Q}{m\Delta T} \), where \( Q \) is the heat added, \( m \) is the mass, and \( \Delta T \) is the change in temperature.
In an isothermal process, the temperature change \( \Delta T \) is zero. If you put \( \Delta T = 0 \) in the formula, the specific heat becomes infinitely large (\( s = \infty \)). This means a gas can absorb or release a lot of heat without changing its temperature.
In an adiabatic process, no heat (\( Q \)) is transferred. If you put \( Q = 0 \) in the formula, the specific heat becomes zero (\( s = 0 \)). This means that even a small amount of heat transfer can cause a large temperature change.
In simple words: Specific heat shows how much heat is needed to change a substance's temperature. For an isothermal process (constant temperature), specific heat is infinite because you add heat but the temperature doesn't change. For an adiabatic process (no heat added or removed), specific heat is zero because even a tiny heat transfer would cause a huge temperature change.

🎯 Exam Tip: Clearly state the condition (\( \Delta T = 0 \) or \( Q = 0 \)) for each process and how it affects the specific heat formula.

Question 10. What type of process is a Carnot cyclic?
Answer: A Carnot cycle is a type of theoretical thermodynamic process that is considered perfectly reversible. This means it can go forward and backward without any energy loss, and it returns to its starting state.
In simple words: A Carnot cycle is a perfect process that can go forwards and backwards smoothly, always returning to where it started.

🎯 Exam Tip: Emphasize "theoretical" and "reversible" when describing a Carnot cycle, as these are its defining characteristics.

Question 11. On what factors does the efficiency of a Carnot engine depend?
Answer: The efficiency (\( \eta \)) of a Carnot heat engine depends only on the absolute temperatures of the heat source (\( T_1 \)) and the heat sink (\( T_2 \)). The formula is \( \eta = 1 - \frac{T_2}{T_1} \). This means its efficiency is solely determined by how hot the source is and how cold the sink is, not by the type of working substance used.
In simple words: The efficiency of a Carnot engine depends only on the temperatures of the hot part (source) and the cold part (sink) it works between. The bigger the temperature difference, the more efficient it is.

🎯 Exam Tip: Remember to use absolute temperatures (Kelvin) for all efficiency calculations involving heat engines.

RBSE Class 11 Physics Chapter 13 Short Answer Type Questions

Question 1. Reversibility is the basis of an ideal engine. Briefly explain the statement.
Answer: A reversible process is one where both the system and its surroundings can go back to their exact starting conditions by reversing the process. An ideal engine, like the theoretical Carnot engine, is based on this idea of reversibility. The Carnot engine shows the highest possible efficiency for converting heat into work because it operates through a series of reversible steps between two temperature reservoirs. This is why reversibility is fundamental to understanding how an ideal engine works.
In simple words: An ideal engine works perfectly because it's reversible. This means if you run it forwards and then backwards, everything goes back to how it was before, with no energy lost. The best engine, called a Carnot engine, uses this idea to be as efficient as possible.

🎯 Exam Tip: Explain that "ideal" in thermodynamics often implies "reversible" and "no energy losses," leading to maximum theoretical efficiency.

Question 3. What is Thermodynamics? Explain the Zeroth law of Thermodynamics and give its importance.
Answer: Thermodynamics is the branch of physics that studies how heat relates to other forms of energy and work. It's all about how energy transforms. The Zeroth Law of Thermodynamics is based on the concept of temperature. To explain it, imagine two systems, A and B, which are separated from each other by a wall that prevents heat transfer (adiabatic wall). Both A and B are connected to a third system C by walls that allow heat to flow (conducting walls). Over time, A will reach thermal equilibrium with C, and B will also reach thermal equilibrium with C. This means that A and B will eventually have the same temperature as C. Now, if we replace the adiabatic wall between A and B with a conducting wall and insulate C from A and B, we find that A and B are already in thermal equilibrium with each other. This state, where the temperatures are equal for two systems in thermal equilibrium, is called temperature (T). So, if \( T_A = T_C \) and \( T_B = T_C \), it naturally means \( T_A = T_B \). Therefore, systems A and B are in thermal equilibrium. This law is important because it defines temperature and provides a way to compare the thermal states of different systems.
In simple words: Thermodynamics is the study of heat and how it turns into other types of energy or work. The Zeroth Law of Thermodynamics is about temperature. It says that if two things (A and B) are both separately in heat balance with a third thing (C), then A and B are also in heat balance with each other. This law helps us understand what temperature really is.

🎯 Exam Tip: When explaining the Zeroth Law, use clear examples or analogies of systems in thermal contact to illustrate the concept of thermal equilibrium.

Question 4. Explain heat, work and internal energy of a system.
Answer: Heat, work, and internal energy are fundamental concepts in thermodynamics.
Heat: Heat is energy that moves from a hotter place to a colder place because of a temperature difference. It's a non-mechanical way of transferring energy. When an object loses heat to its surroundings (like the atmosphere), we consider that heat energy as negative. When it gains heat from the surroundings, it's positive.
Work: In thermodynamics, work is done when a force causes movement, often involving a change in volume for a gas. For example, if a gas expands against external pressure, it performs work. The work done during an isothermal expansion (where temperature stays constant) for \( \mu \) moles of an ideal gas from initial volume \( V_1 \) to final volume \( V_2 \) can be calculated. We know the ideal gas equation is \( PV = \mu RT \), so \( P = \frac{\mu RT}{V} \). The total work done (\( W_{isothermal} \)) is given by:
\[ W_{isothermal} = \int_{V_1}^{V_2} P dV \] Now, substitute P:
\[ W_{isothermal} = \int_{V_1}^{V_2} \frac{\mu RT}{V} dV \] Since \( \mu \) and \( T \) are constant for the process:
\[ W_{isothermal} = \mu RT \int_{V_1}^{V_2} \frac{1}{V} dV \]
\[ W_{isothermal} = \mu RT [\log_e V]_{V_1}^{V_2} \]
\[ W_{isothermal} = \mu RT (\log_e V_2 - \log_e V_1) \]
\[ W_{isothermal} = \mu RT \log_e \frac{V_2}{V_1} \] This can also be written using base-10 logarithm:
\[ W_{isothermal} = 2.303 \mu RT \log_{10} \frac{V_2}{V_1} \] The relationship between pressure and volume in an isothermal process is \( PV = \text{constant} \). From this, we can find the slope of the P-V curve:
\( PdV + VdP = 0 \)
\( \implies VdP = -PdV \)
\( \implies \frac{dP}{dV} = \frac{-P}{V} \) This means the slope of the P-V curve for an isothermal process is negative. It shows that as volume increases, pressure decreases. This formula helps us find out how much work a gas does when it expands at a steady temperature.
Internal Energy: Internal energy is all the energy stored inside a system due to the random motion and arrangement of its atoms and molecules. It includes both kinetic energy (from movement) and potential energy (from forces between molecules). For an ideal gas, internal energy mainly depends on its temperature. If a system's internal energy changes, it means its temperature or molecular state has changed.
In simple words:
Heat: Heat is energy moving from something warm to something cool because of temperature difference.
Work: Work is energy used when a gas expands or gets squeezed. For a gas expanding at a steady temperature, we can use a special math formula with pressure and volume to find the work done.
Internal Energy: This is all the energy hidden inside a gas from its tiny moving parts. It mostly depends on how hot the gas is.

🎯 Exam Tip: Define each term clearly and concisely, using simple examples. For work, differentiate between work done *by* the system (expansion) and work done *on* the system (compression).

Question 6. Explain the First Law of Thermo-dynamics.
Answer: The First Law of Thermodynamics is essentially the law of conservation of energy applied to heat systems. It states that the change in a system's internal energy (\( \Delta U \)) is equal to the heat added to the system (\( Q \)) minus the work done by the system (\( W \)). The formula is \( \Delta U = Q - W \).
When a gas expands, it does work. If a gas inside a cylinder pushes a piston, the work done (\( dW \)) is given by \( P dV \), where \( P \) is pressure and \( dV \) is the change in volume.
For an adiabatic process, where no heat is exchanged (\( Q = 0 \)), the work done is directly related to the change in internal energy and temperature. The ratio of specific heats is represented by \( \gamma \). The work done in an adiabatic process (\( W_{adiabatic} \)) for \( \mu \) moles of an ideal gas changing from state \( (P_1, V_1, T_1) \) to \( (P_2, V_2, T_2) \) is derived as:
\[ W_{adiabatic} = \int_{V_1}^{V_2} P dV \] Using the adiabatic relation \( PV^\gamma = K \) (constant), so \( P = K V^{-\gamma} \):
\[ W_{adiabatic} = \int_{V_1}^{V_2} K V^{-\gamma} dV \]
\[ W_{adiabatic} = K \left[ \frac{V^{-\gamma+1}}{-\gamma+1} \right]_{V_1}^{V_2} \]
\[ W_{adiabatic} = \frac{K}{1-\gamma} [V_2^{1-\gamma} - V_1^{1-\gamma}] \] Since \( K = P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \), we can substitute and simplify to:
\[ W_{adiabatic} = \frac{1}{1-\gamma} [P_2 V_2 - P_1 V_1] \] This can also be written as:
\[ W_{adiabatic} = \frac{1}{\gamma-1} [P_1 V_1 - P_2 V_2] \] Using the ideal gas law \( PV = \mu RT \):
\[ W_{adiabatic} = \frac{1}{\gamma-1} [\mu RT_1 - \mu RT_2] \]
\[ W_{adiabatic} = \frac{\mu R}{\gamma-1} (T_1 - T_2) \] This formula shows that if the gas does work (expands), \( W_{adiabatic} \) is positive, which means \( T_1 > T_2 \) (temperature decreases). If work is done on the gas (compressed), \( W_{adiabatic} \) is negative, meaning \( T_1 < T_2 \) (temperature increases).
In simple words: The First Law of Thermodynamics says that energy is conserved. It means the heat you add to a system (like a gas) is used either to increase the gas's internal energy (make it hotter) or to do work (make it expand). If a gas expands without heat entering or leaving it (adiabatic process), the work it does will cause its temperature to drop. If you compress it without heat transfer, its temperature will rise.

🎯 Exam Tip: The First Law is a restatement of energy conservation. Remember the signs: heat *added* is positive, work *done by* the system is positive.

Question 7. Derive the Mayer's relation.
Answer: Mayer's relation connects the specific heat of a gas at constant pressure (\( C_p \)) and specific heat at constant volume (\( C_v \)). To derive this, let's imagine one mole of an ideal gas.
The First Law of Thermodynamics states:
\[ dQ = dU + PdV \] where \( dQ \) is the heat added, \( dU \) is the change in internal energy, and \( PdV \) is the work done.
For one mole of an ideal gas, the change in internal energy (\( dU \)) when temperature changes by \( dT \) at constant volume is:
\[ dU = C_v dT \] When heat is supplied at constant pressure, the heat added is:
\[ dQ = C_p dT \] Now, substitute these into the First Law of Thermodynamics:
\[ C_p dT = C_v dT + PdV \] For one mole of an ideal gas, the ideal gas equation is \( PV = RT \).
If we consider a process at constant pressure, differentiating the ideal gas equation gives:
\[ PdV = RdT \] Substitute \( PdV = RdT \) into the modified First Law equation:
\[ C_p dT = C_v dT + RdT \] Divide all terms by \( dT \):
\[ C_p = C_v + R \]
\[ \implies C_p - C_v = R \] This is Mayer's relation. \( R \) is the universal gas constant, which is approximately \( 8.31 \text{ J} \cdot \text{mol}^{-1} \text{K}^{-1} \). This relation shows a fundamental connection between how a gas absorbs heat at constant pressure versus constant volume.
In simple words: Mayer's relation is a simple formula that tells us the difference between how much heat a gas needs to get hotter when its pressure stays the same, versus when its volume stays the same. The difference between these two values is equal to a special constant number called the gas constant. It's like saying \( \text{Heat at steady pressure} - \text{Heat at steady volume} = \text{A fixed number} \).

🎯 Exam Tip: Clearly state the assumptions made (ideal gas, one mole) and show each substitution step from the First Law and ideal gas equation to arrive at Mayer's relation.

Question 8. Differentiate between reversible and irreversible process.
Answer: Let's differentiate between reversible and irreversible processes in thermodynamics:
Reversible Process:
1. These processes happen extremely slowly, step-by-step, allowing the system to remain in equilibrium at every point.
2. The system is always very close to a state of balance, and no significant disturbance occurs.
3. They would ideally take an infinite amount of time to complete because each step is so gradual.
4. The work obtained from a reversible process is always the maximum possible work that can be extracted from a system.
Irreversible Process:
1. These processes happen quickly and are not in equilibrium during the change.
2. They involve factors like friction, rapid expansion, or heat transfer across a large temperature difference, which make them impossible to perfectly reverse.
3. They complete in a finite amount of time and usually lead to an increase in the entropy (disorder) of the universe.
4. The work obtained from an irreversible process is always less than the maximum possible work.
In simple words: A reversible process is like a perfect movie that can play forwards and backwards exactly the same, giving the most work possible. An irreversible process is like real life; it only goes one way, always loses some energy (like friction), and gives less work.

🎯 Exam Tip: Focus on key differences like speed, equilibrium, and work output when comparing reversible and irreversible processes.

Question 9. Explain the work done by gas in an adiabatic expansion.
Answer: An adiabatic expansion is a process where a gas expands, doing work, but no heat is exchanged between the gas and its surroundings. This means the system is completely insulated, or the process happens so quickly that heat doesn't have time to flow in or out. In an adiabatic expansion, the internal energy of the gas decreases because it's doing work, which also causes its temperature to drop.
The essential conditions for an adiabatic process are:
• The process (expansion or compression) must happen very quickly, so there's no time for heat to transfer.
• The container holding the gas must be perfectly insulated to prevent any heat exchange with the outside.
In such a process, all energy changes are due to work being done, not heat transfer.
In simple words: An adiabatic expansion is when a gas pushes outwards and does work, but no heat goes in or out of it. This can happen if it's very fast or in a perfectly insulated container. When the gas expands this way, it uses its own inside energy, so it gets colder.

🎯 Exam Tip: Highlight the two main conditions for an adiabatic process: no heat exchange and often a rapid process or perfect insulation.

Question 10. Derive the formula for efficiency of Carnot engine.
Answer: The Carnot engine is an ideal, theoretical heat engine proposed by Sadi Carnot in 1824. He aimed to find the maximum possible efficiency for an engine. It operates through a special sequence of four reversible thermodynamic processes, known as the Carnot cycle, which always brings the working substance back to its initial state. A Carnot engine has several key components:
(i) Working substance: This is the material (often an ideal gas) inside the engine that undergoes the cycle and performs work. It is typically contained in a cylinder with a frictionless piston.
(ii) Source: This is a high-temperature reservoir (at temperature \( T_1 \)) from which the engine absorbs heat. It has an infinitely large heat capacity, so its temperature remains constant even when heat is drawn from it.
(iii) Insulating Stand: This is a non-conducting platform used for adiabatic processes, meaning no heat can enter or leave the working substance when placed on it.
(iv) Sink: This is a low-temperature reservoir (at temperature \( T_2 \)) to which the engine expels excess heat. It also has an infinitely large heat capacity, so its temperature remains constant even when heat is rejected to it.
The efficiency (\( \eta \)) of a Carnot engine is defined as the ratio of the net work done by the engine to the heat absorbed from the source. The formula is given by:
\[ \eta = \frac{\text{Useful work}}{\text{Heat taken from source}} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} \] For a reversible Carnot engine, the ratio of heat exchanged is equal to the ratio of absolute temperatures: \( \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \). Substituting this into the efficiency formula:
\[ \eta = 1 - \frac{T_2}{T_1} \] This formula shows that the efficiency of a Carnot engine depends only on the absolute temperatures of the heat source (\( T_1 \)) and the heat sink (\( T_2 \)). It can also be expressed as a percentage:
\[ \eta = \left( 1 - \frac{T_2}{T_1} \right) \times 100\% \] The maximum possible efficiency occurs when \( T_2 \) is very low (ideally 0 K) or \( T_1 \) is very high, but it can never be 100% unless \( T_2 \) is absolute zero, which is practically impossible.
In simple words: The Carnot engine is a perfect, imaginary engine. It works in a special cycle to get the most energy possible from heat. It has four main parts: the working gas, a hot place (source), a cold place (sink), and an insulating stand. The efficiency of this engine tells us how much useful work it does compared to the heat it takes in. The formula shows it only depends on how hot the "hot place" is and how cold the "cold place" is.

🎯 Exam Tip: When deriving, clearly state the definition of efficiency and the relationship between heat and temperatures for a reversible engine. Remember to use absolute temperatures.

Question 11. Derive the formula for the efficiency of carnot heat engine and explain it.
Answer: The efficiency of a heat engine (\( \eta \)) is defined as the ratio of the useful work done to the total heat absorbed from the source. For a Carnot engine, which is an ideal reversible heat engine, its efficiency is determined by the temperatures of the heat source (\( T_1 \)) and the heat sink (\( T_2 \)).
The efficiency formula is:
\[ \eta = \frac{\text{Useful work}}{\text{Heat taken from source}} = \frac{Q_1 - Q_2}{Q_1} \] where \( Q_1 \) is the heat absorbed from the source and \( Q_2 \) is the heat rejected to the sink.
For an ideal Carnot cycle, the ratio of heat transferred is proportional to the absolute temperatures:
\[ \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \] Substituting this into the efficiency formula, we get:
\[ \eta = 1 - \frac{Q_2}{Q_1} \]
\[ \implies \eta = 1 - \frac{T_2}{T_1} \] This formula is derived from the properties of the Carnot cycle.
Percentage efficiency:
\[ \eta = \left[ 1 - \frac{T_2}{T_1} \right] \times 100\% \] From this relation, it is clear that the efficiency of the engine depends only on the temperatures of the source (\( T_1 \)) and sink (\( T_2 \)).
• If \( T_1 = T_2 \), then \( \eta = 0 \), meaning no heat can be converted into work if the source and sink are at the same temperature.
• For efficiency to be 100%, either \( T_2 \) must be 0 K (absolute zero) or \( T_1 \) must be infinitely high. Both these conditions are practically impossible. Therefore, the efficiency of a heat engine can never be 100%. The higher the temperature difference between the source and the sink, the greater the efficiency.
In simple words: The efficiency of a Carnot heat engine tells us how well it turns heat into useful work. The formula for it is \( \eta = 1 - \frac{T_2}{T_1} \), which means it only depends on the temperatures of the hot part (\( T_1 \)) and the cold part (\( T_2 \)). If the hot and cold parts are at the same temperature, the engine won't work at all. It can never be 100% efficient, because that would mean the cold part is at absolute zero, which isn't possible.

🎯 Exam Tip: Clearly define the terms \( T_1 \) (source temperature) and \( T_2 \) (sink temperature) and explain why 100% efficiency is impossible in practice.

RBSE Class 11 Physics Chapter 13 Long Answer Type Questions

Question 1. Explain briefly the zeroth, first and second law of Thermodynamics.
Answer:
Zeroth Law of Thermodynamics: This law is about temperature. It states that if two systems (like A and B) are each in thermal equilibrium with a third system (C) separately, then the first two systems (A and B) are also in thermal equilibrium with each other. Simply put, if A has the same temperature as C, and B also has the same temperature as C, then A and B must have the same temperature. This law defines the concept of temperature and is the basis for thermometers.
First Law of Thermodynamics: This law is based on the principle of conservation of energy. It states that energy cannot be created or destroyed, only transformed. For a thermodynamic system, the change in its internal energy (\( \Delta U \)) is equal to the heat supplied to the system (\( \Delta Q \)) minus the work done by the system (\( \Delta W \)). The mathematical form is:
\[ \Delta Q = \Delta U + \Delta W \] Here, internal energy (\( U \)) depends only on the initial and final states, not the path taken. Work done by the system against a constant pressure \( P \) is \( \Delta W = P\Delta V \). So, the first law can also be written as:
\[ \Delta Q = \Delta U + P\Delta V \] This means any heat energy supplied to a system is used partly to increase the system's internal energy and partly to do work on its surroundings.
Second Law of Thermodynamics: This law focuses on the direction of natural processes and the concept of entropy (disorder). It states that the total entropy of an isolated system either increases over time or remains constant in ideal, reversible processes. It means that natural processes are irreversible; for example, heat always flows from a hot object to a cold object on its own, and never the other way around, unless external work is done. It sets limits on the efficiency of heat engines and explains why some processes occur spontaneously and others do not.
In simple words:
Zeroth Law: If two things are both at the same temperature as a third thing, then those two things are also at the same temperature as each other. This is how we know what temperature means.
First Law: This law says energy is always conserved. It means the heat you add to a system either makes its inside energy go up or makes it do work, but the total energy stays the same.
Second Law: This law tells us that things naturally move towards more disorder (entropy). Heat always flows from hot to cold, never the other way naturally. It also tells us there's a limit to how efficient engines can be.

🎯 Exam Tip: Clearly distinguish between the focus of each law: Zeroth defines temperature, First deals with energy conservation, and Second with the direction of processes and entropy.

Question 2. Explain the various processes of thermodynamics and work done in these processes.
Answer: Every balanced state of a thermodynamic system is fully described by specific values of some observable variables. If variables like temperature (T), pressure (P), and volume (V) change over time in a thermodynamic system, the process is called a thermodynamic process. There are several types of thermodynamic processes:
(i) Isothermal process:
An isothermal process is a system change where the temperature stays constant, meaning \( \Delta T = 0 \). This usually happens when a system is in contact with an outside heat reservoir, and the change occurs slowly, allowing the system to continuously adjust its temperature through heat exchange.
An isothermal process can happen in any system that controls temperature. In the thermodynamic analysis of chemical reactions, the first step is to analyze what happens under isothermal conditions. In an isothermal process, the internal energy of an ideal gas remains constant. This is because there are no intermolecular forces in an ideal gas. During isothermal compression of a gas, work is done on the system to reduce its volume and increase its pressure. Doing work on the gas can increase internal energy and temperature. For the temperature to stay constant, energy must leave the system. For an ideal gas, the amount of energy entering the environment is equal to the work done on the gas, as its internal energy does not change.
(ii) Isochoric process:
If the pressure (P) and temperature change in a physical change of a thermodynamic system, but the volume stays constant, it is called an isochoric process. In this process, no work is done by the gas because the volume (V) is constant, meaning \( dV = 0 \).
(iii) Isobaric process:
If the temperature (T) and volume (V) change in a physical change of a thermodynamic system, and the pressure stays constant, it is called an isobaric process.
(iv) Adiabatic process:
An adiabatic process is one where temperature (T), pressure (P), and volume (V) can change, but no heat is transferred between the thermodynamic system and its surroundings. In this process, energy is transferred only as work. This process provides a strong theoretical basis for explaining the first law of Thermodynamics.
A process where no heat is transferred in the system ( \( Q = 0 \) ) is called an adiabatic process. For example, when a gas inside an engine's cylinder is compressed very quickly, little energy can be transferred as heat out of the system during the compression time. Essential conditions for an adiabatic process are that the expansion or compression should be sudden so that heat does not have time to be exchanged with the surroundings, and the container walls must be perfectly insulated to prevent any heat exchange between the gas and its surroundings.
(v) Cyclic process:
A cyclic process in thermodynamics is when a system undergoes various changes and then returns to its initial state. In this process, the change in internal energy \( \Delta U = 0 \), because internal energy is a function of the state. According to the first law of Thermodynamics, \( dQ = dU + dW \). Since \( dU = 0 \), it means \( dQ = dW \). This indicates that the total heat absorbed by the system is equal to the work done by the system.
In simple words: Thermodynamic processes describe how a system changes, like temperature, pressure, or volume. Isothermal means constant temperature, isochoric means constant volume, and isobaric means constant pressure. Adiabatic means no heat transfer, and cyclic means the system returns to its original state.

🎯 Exam Tip: When explaining thermodynamic processes, clearly define each process by its constant variable (e.g., constant temperature for isothermal) and briefly describe its main characteristic and implications for heat transfer and work.

Question 3. Differentiate between the isothermal and adiabatic processes and calculate the work done in these process.
Answer: An isothermal process is a change in a system where the temperature remains constant, so \( \Delta T = 0 \). This often happens when a system is in contact with an outside heat reservoir and changes slowly, allowing it to adjust to the reservoir's temperature through heat exchange. In an isothermal process, the internal energy of an ideal gas stays constant.
An adiabatic process is one where the temperature, pressure, and volume can change, but no heat is transferred between the system and its surroundings, so \( Q = 0 \). In this process, energy is transferred only as work. This provides a strong basis for explaining the first law of Thermodynamics.
Work done in an isothermal process:
Consider a cylinder with \( \mu \) moles of an ideal gas undergoing isothermal expansion. The system moves from an initial state \( (P_1, V_1) \) to a final state \( (P_2, V_2) \). If at some point the pressure is \( P \) and the volume changes from \( V \) to \( V + dV \), the total work done is given by:
\[ W_{isothermal} = \int_{V_1}^{V_2} P dV \]
From the ideal gas equation for \( \mu \) moles of a gas, \( PV = \mu RT \), so \( P = \frac{\mu RT}{V} \).
Substituting \( P \) in the work equation:
\[ W_{isothermal} = \int_{V_1}^{V_2} \frac{\mu RT}{V} dV \]
Since \( T \) is constant in an isothermal process:
\[ W_{isothermal} = \mu RT \int_{V_1}^{V_2} \frac{1}{V} dV \]
\[ W_{isothermal} = \mu RT [\log_e V]_{V_1}^{V_2} \]
\[ W_{isothermal} = \mu RT (\log_e V_2 - \log_e V_1) \]
\[ W_{isothermal} = \mu RT \log_e \left(\frac{V_2}{V_1}\right) \]
We can also write this using base-10 logarithm:
\[ W_{isothermal} = 2.303 \mu RT \log_{10} \left(\frac{V_2}{V_1}\right) \]
The slope of the P-V curve for an isothermal process is negative.
Work done in an adiabatic process:
Consider a cylinder with \( n \) moles of ideal gas made of completely insulated material with a frictionless insulated piston. Let the gas pressure be \( P \) and the piston's cross-section area be \( A \). If the piston moves upwards by a distance \( dx \), the work done by the gas due to expansion is:
\[ dW = P A dx \]
Since \( dV = A dx \) is the increase in volume, then:
\[ dW = P dV \]
If the gas expands adiabatically from initial state \( (P_1, V_1, T_1) \) to final state \( (P_2, V_2, T_2) \), the total work done by the gas is derived from the relation \( PV^\gamma = \text{constant} \). The work done is:
\[ W_{adiabatic} = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \]
Since \( P_1 V_1 = \mu R T_1 \) and \( P_2 V_2 = \mu R T_2 \) for \( \mu \) moles of an ideal gas:
\[ W_{adiabatic} = \frac{\mu R T_1 - \mu R T_2}{\gamma - 1} \]
\[ W_{adiabatic} = \frac{\mu R (T_1 - T_2)}{\gamma - 1} \]
This expression shows the work done for \( \mu \) moles of an ideal gas in an adiabatic process. If work is done by the gas, \( W_{adiabatic} \) is positive, which happens when \( T_1 > T_2 \). If the gas is compressed, work is done on the gas, so \( W_{adiabatic} \) is negative, which happens when \( T_1 < T_2 \), leading to an increase in gas temperature.
In simple words: Isothermal processes keep temperature constant, while adiabatic processes prevent heat transfer. Work done in isothermal expansion uses a log formula with constant temperature, while work done in adiabatic expansion uses a formula based on initial and final pressures and volumes (or temperatures) and a constant \( \gamma \).

🎯 Exam Tip: Remember that for an ideal gas, internal energy is constant in an isothermal process, while in an adiabatic process, the temperature changes due to work being done, affecting the internal energy.

Question 4. Write the working principle of Carnot's reversible engine and plot the P-V curve for work in each process, and obtain the formula for efficiency.
Answer: In 1824, scientist Sadi Carnot imagined a theoretical engine. He wanted to find out if the maximum efficiency of an engine could be 100% when heat from a high-temperature source is completely changed into useful work. He concluded that an ideal reversible heat engine working between two temperatures is called a Carnot engine.
A Carnot Cycle: In Carnot's engine, the working material starts from a definite state, goes through different reversible processes, and then returns to its initial position. This is called the Carnot cycle. A Carnot engine has the following parts:
(i) Working substance: This is the material that does the work. In a Carnot engine, an ideal gas in a cylinder with a frictionless piston is used as the working substance.
(ii) Source: This is a high-temperature object at a constant temperature \( T_1K \). It should have infinite heat capacity so that its temperature does not change when heat is taken from it.
(iii) Insulating Stand: This is a completely non-conducting object on which the working substance is placed during the adiabatic process.
(iv) Sink: This is a low-temperature object at a constant temperature \( T_2K \). It should also have infinite heat capacity so that its temperature does not change when heat is given to it.

The efficiency (\( \eta \)) of a Carnot engine is defined as the ratio of the work done by the engine to the heat absorbed from the source. It depends only on the temperatures of the source (\( T_1 \)) and the sink (\( T_2 \)).
The formula for efficiency of a Carnot heat engine is:
\[ \eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1} \]
where \( W \) is the useful work done, \( Q_1 \) is the heat taken from the source, and \( Q_2 \) is the heat rejected to the sink.
For a reversible Carnot engine, the ratio of heat exchanged is equal to the ratio of absolute temperatures:
\[ \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \]
Substituting this into the efficiency formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
The percentage efficiency is \( \eta = \left(1 - \frac{T_2}{T_1}\right) \times 100\% \)
From this relation, it is clear that the engine's efficiency depends on the temperatures of the source (\( T_1 \)) and the sink (\( T_2 \)). The efficiency is always less than one, meaning less than 100%. If \( T_1 = T_2 \), then \( \eta = 0 \), which means no heat energy can be converted to mechanical energy if the source and sink are at the same temperature. An efficiency of 100% would require either \( T_1 = \infty \) or \( T_2 = 0 \), both of which are not possible situations. Therefore, efficiency can never be 100%.
In simple words: A Carnot engine is a perfect engine that converts heat into work using a cycle. Its efficiency depends only on the temperatures of the hot heat source and the cold heat sink. The formula for its efficiency is \( \eta = 1 - \frac{T_2}{T_1} \), where \( T_1 \) is the hot temperature and \( T_2 \) is the cold temperature.

🎯 Exam Tip: When deriving the Carnot engine efficiency, clearly state the four parts of the engine and the key assumption that the process is reversible, which allows the use of the temperature ratio.

Question 5. Write the statements of Kelvin-Planck and Clausius of second law of Thermodynamics and show the equivalence of the two statements.
Answer: The second law of Thermodynamics states that the total entropy of an isolated system always increases over time, or stays the same in ideal cases (like reversible processes or steady states). The first law of Thermodynamics defines internal energy and states the law of conservation of energy. The second law of Thermodynamics deals with the direction of natural processes. It states that a natural process is not reversible. For example, heat flows from a hot region to a cold region, but never the reverse, unless some external work is done on the system.
1. Statement of Kelvin-Planck: According to this statement, "It is impossible to make an engine in which the working substance takes heat from a single source and converts it completely into work in a complete cycle, without any other changes to itself or its surroundings." This means that it is not possible to get work continuously from a single heat source. A heat sink is always necessary to release some heat. This statement is based on the principle of the Carnot heat engine.
2. Statement of Clausius: According to this statement, "It is an impossible process for a working substance in a cyclic process to transfer heat directly from a low-temperature object to a high-temperature object without external work." This means that heat transfer from a low-temperature object to a high-temperature object cannot happen on its own. External work must be done on the working substance. This statement is based on the principle of a refrigerator.
Equivalence of Clausius and Kelvin Statements:
Let's imagine an engine that violates the Kelvin-Planck statement. This engine would take heat from a source and convert it entirely into work in a cycle, with no other effects. Now, let's connect this engine with a reverse Carnot engine (a refrigerator).

The combined system (the engine violating Kelvin-Planck and the refrigerator) transfers heat \( \Delta Q = Q \left(\frac{1}{\eta} - 1\right) \) from the colder region (sink) to the hotter region (source). This transfer of heat from cold to hot without external work directly violates the Clausius statement. Thus, a violation of the Kelvin-Planck statement means a violation of the Clausius statement, implying that the Clausius statement includes the Kelvin-Planck statement.
Similarly, we can prove that the Kelvin-Planck statement implies the Clausius statement, and therefore, both statements are equivalent.
In simple words: The Kelvin-Planck statement says you can't build an engine that turns all heat into work without any waste. The Clausius statement says heat can't naturally flow from a cold place to a hot place without outside help. Both statements are different ways of saying the same thing: you can't get something for nothing in terms of heat and work, and heat always flows from hot to cold unless forced.

🎯 Exam Tip: To show the equivalence, assume one statement is violated and then demonstrate that this violation leads to a contradiction of the other statement, proving they are logically linked.

Question 6. Write and obtain the Carnot's theorem.
Answer: Carnot's theorem states:
(a) All heat engines operating between two given heat reservoirs are less efficient than a Carnot heat engine operating between the same reservoirs. The efficiency of a Carnot (reversible) engine is the maximum possible.
(b) The efficiency of all Carnot heat engines working between the same pair of heat reservoirs is equal, regardless of the working substance used.
Proof:
Consider two heat reservoirs at different temperatures. Let engine 'I' (irreversible) and engine 'R' (reversible) operate between these two reservoirs.

Assume that the irreversible engine (I) is more efficient than the reversible engine (R). Let \( \eta_I \) be the efficiency of the irreversible engine and \( \eta_R \) be the efficiency of the reversible engine. So, we assume \( \eta_I > \eta_R \).
Let the irreversible engine (I) take heat \( Q_1 \) from the source at \( T_1 \), perform work \( W \), and reject heat \( Q_{1I} - W \) to the sink at \( T_2 \). The efficiency is \( \eta_I = \frac{W}{Q_1} \).
Now, let the reversible engine (R) operate in reverse (as a refrigerator). It takes heat \( Q_2 \) from the sink at \( T_2 \), has work \( W \) done on it, and rejects heat \( Q_2 + W \) to the source at \( T_1 \). Its efficiency operating as an engine would be \( \eta_R = \frac{W}{Q_2+W} \).
If we couple the irreversible engine (I) with the reversible engine (R) acting as a refrigerator, such that the work produced by engine I (\( W \)) drives the refrigerator R, then:
Engine I takes \( Q_1 \) from \( T_1 \), produces \( W \), rejects \( Q_1 - W \) to \( T_2 \).
Refrigerator R takes \( Q_2 \) from \( T_2 \), absorbs \( W \), rejects \( Q_2 + W \) to \( T_1 \).
Total heat taken from source \( T_1 \) by Engine I: \( Q_1 \).
Total heat given to source \( T_1 \) by Refrigerator R: \( Q_2 + W \).
Net heat flow to source \( T_1 \): \( (Q_2 + W) - Q_1 \).
Total heat taken from sink \( T_2 \) by Refrigerator R: \( Q_2 \).
Total heat given to sink \( T_2 \) by Engine I: \( Q_1 - W \).
Net heat flow from sink \( T_2 \): \( Q_2 - (Q_1 - W) = Q_2 - Q_1 + W \).
Since \( W = \eta_I Q_1 \), we can substitute \( W \) into the net heat flows.
If \( \eta_I > \eta_R \), it implies that the irreversible engine is doing more work for the same heat input or needing less heat input for the same work compared to a reversible engine. If we assume \( \eta_I \) is higher, then \( Q_1 \) (heat from source to irreversible engine) will be smaller for the same \( W \). If \( W \) from I drives R, then the combined system would be effectively transferring heat from the colder reservoir to the hotter reservoir without any external work input, which violates the Clausius statement of the second law of Thermodynamics.
This contradiction proves that our initial assumption (\( \eta_I > \eta_R \)) must be false. Therefore, no engine can be more efficient than a reversible Carnot engine operating between the same two temperatures.
For the second statement of Carnot's theorem, we can consider two reversible engines (R and R') working between the same two temperatures. If we assume \( \eta_R > \eta_{R'} \), then by using the same logic as above, we would again violate the second law. This means that \( \eta_R \) cannot be greater than \( \eta_{R'} \). Similarly, \( \eta_{R'} \) cannot be greater than \( \eta_R \). Therefore, \( \eta_R = \eta_{R'} \), which proves that all reversible engines working between the same two temperatures have the same efficiency.
In simple words: Carnot's theorem states that no engine can be more efficient than a perfect, ideal engine (a Carnot engine) working between the same two temperatures. It also says that all perfect engines working between the same two temperatures will have the exact same efficiency, no matter what they are made of. This means there's a limit to how good a heat engine can be.

🎯 Exam Tip: When explaining Carnot's theorem, clearly state both parts of the theorem. For the proof, focus on demonstrating how assuming a more efficient engine leads to a violation of the second law of thermodynamics.

RBSE Class 11 Physics Chapter 13 Numerical Questions

Question 1. The efficiency of an ideal engine is 75% and it emits heat 2 × 10³ W to sink at 283 K, then calculate :
(i) Temperature of source.
(ii) Work done per minute by the engine.
(iii) Absorbed heat from the source in a cycle.
Answer:
Given:
Efficiency \( \eta = 75\% = \frac{75}{100} = \frac{3}{4} \)
Heat emitted to sink \( Q_2 = 2 \times 10^3 W \)
Sink temperature \( T_2 = 283 K \)
(i) Temperature of source \( T_1 \):
The efficiency of an ideal engine is given by:
\( \eta = 1 - \frac{T_2}{T_1} \)
So, \( \frac{T_2}{T_1} = 1 - \eta \)
\( \frac{283}{T_1} = 1 - \frac{3}{4} \)
\( \frac{283}{T_1} = \frac{4 - 3}{4} \)
\( \frac{283}{T_1} = \frac{1}{4} \)
\( T_1 = 4 \times 283 \)
\( T_1 = 1132 K \)
(ii) Work done per minute by the engine \( W \):
We know that \( \eta = \frac{W}{Q_1} \) and \( \frac{Q_2}{Q_1} = 1 - \eta \)
From \( \frac{Q_2}{Q_1} = 1 - \eta \)
\( \frac{2 \times 10^3}{Q_1} = 1 - \frac{3}{4} \)
\( \frac{2 \times 10^3}{Q_1} = \frac{1}{4} \)
\( Q_1 = 4 \times 2 \times 10^3 \)
\( Q_1 = 8 \times 10^3 J/s \)
Work done \( W = Q_1 - Q_2 \)
\( W = 8 \times 10^3 J/s - 2 \times 10^3 J/s \)
\( W = 6 \times 10^3 J/s \)
Work done per minute:
\( W_{per\,minute} = W \times 60 \text{ seconds} \)
\( W_{per\,minute} = 6 \times 10^3 J/s \times 60 s \)
\( W_{per\,minute} = 360 \times 10^3 J \)
\( W_{per\,minute} = 3.6 \times 10^5 J \)
(iii) Absorbed heat from the source in a cycle:
The heat absorbed from the source in one cycle is \( Q_1 \).
From part (ii), we already calculated:
\( Q_1 = 8 \times 10^3 J/s \)
This is the heat absorbed per second. For "a cycle," typically we refer to per unit time if power is given.
In simple words: First, we found the hot source temperature using the engine's efficiency and cold temperature. Then, we calculated the total heat taken from the source and used it to find the work done by the engine each minute. Finally, the absorbed heat from the source is the total heat input for the engine.

🎯 Exam Tip: Always convert percentage efficiency to a decimal or fraction before using it in calculations. Pay attention to units (Joules per second for power, Joules for energy, Kelvin for temperature) and timeframes (per minute vs. per second).

Question 2. Calculate the efficiency of a Carnot engine working between freezing point and boiling point.
Answer:
Given:
Freezing point \( T_2 = 0^\circ C \)
Boiling point \( T_1 = 100^\circ C \)
We need to convert temperatures to Kelvin:
\( T_2 = 0 + 273 = 273 K \)
\( T_1 = 100 + 273 = 373 K \)
The efficiency of a Carnot engine is given by:
\( \eta = 1 - \frac{T_2}{T_1} \)
Substitute the values:
\( \eta = 1 - \frac{273}{373} \)
To simplify, find a common denominator or perform the subtraction:
\( \eta = \frac{373 - 273}{373} \)
\( \eta = \frac{100}{373} \)
\( \eta \approx 0.26809 \)
As a percentage:
\( \eta \approx 0.26809 \times 100\% \)
\( \eta \approx 26.8\% \)
Rounding to the nearest whole number for simplicity, as shown in source:
\( \eta \approx 27\% \)
In simple words: We converted the freezing and boiling points to Kelvin. Then, we used the Carnot efficiency formula, which compares the hot and cold temperatures, to find that the engine is about 27% efficient.

🎯 Exam Tip: Always convert Celsius temperatures to Kelvin when calculating the efficiency of a Carnot engine, as the formula uses absolute temperatures. A common mistake is using Celsius directly, which leads to incorrect results.

Question 4. A Carnot refrigerator is operated between 260 K and 400 K. It takes 600 cal heat from sink at low temperature, then calculate the heat given to source at high temperature and the work in every cycle.
Answer:
Given:
Lower temperature \( T_2 = 260 K \)
Higher temperature \( T_1 = 400 K \)
Heat taken from sink \( Q_2 = 600 \text{ cal} \)
For a Carnot refrigerator, the coefficient of performance (\( \beta \)) is given by:
\( \beta = \frac{T_2}{T_1 - T_2} \)
Substitute the given temperatures:
\( \beta = \frac{260}{400 - 260} \)
\( \beta = \frac{260}{140} \)
\( \beta = \frac{26}{14} = \frac{13}{7} \)
Also, the coefficient of performance is defined as the ratio of heat extracted from the cold reservoir to the work done on the refrigerator:
\( \beta = \frac{Q_2}{W} \)
So, \( \frac{13}{7} = \frac{600}{W} \)
Solving for \( W \):
\( W = \frac{600 \times 7}{13} \)
\( W = \frac{4200}{13} \)
\( W \approx 323.0769 \text{ cal} \)
Therefore, the work done in every cycle is approximately \( 323.1 \text{ cal} \).
Now, to find the heat given to the source at high temperature (\( Q_1 \)):
For a refrigerator, \( Q_1 = Q_2 + W \)
\( Q_1 = 600 \text{ cal} + 323.1 \text{ cal} \)
\( Q_1 = 923.1 \text{ cal} \)
The work done in calories can be converted to Joules using the conversion factor 1 cal \( \approx \) 4.2 J:
\( W_{Joules} = 323.1 \text{ cal} \times 4.2 J/\text{cal} \)
\( W_{Joules} = 1357.02 J \)
So, the heat given to the source is \( 923.1 \text{ cal} \) and the work done is \( 1357.02 J \).
In simple words: We calculated how well the refrigerator works using its hot and cold temperatures. Then, using the heat it removes from the cold place, we found out how much work is needed to run it and how much heat it releases to the warmer surroundings.

🎯 Exam Tip: For refrigerator problems, remember that the coefficient of performance \( \beta = \frac{T_2}{T_1 - T_2} \) and also \( \beta = \frac{Q_2}{W} \). The relationship \( Q_1 = Q_2 + W \) is crucial for finding the heat rejected to the hot reservoir.

Question 5. The efficiency of a Carnot engine at temperature 100 K and T K; and 180 K and 900 K is same, then calculate T.
Answer:
Given:
First case: Sink temperature \( T_{2A} = 100 K \), Source temperature \( T_{1A} = T K \)
Second case: Sink temperature \( T_{2B} = 180 K \), Source temperature \( T_{1B} = 900 K \)
The efficiency \( \eta \) for a Carnot engine is given by \( \eta = 1 - \frac{T_{sink}}{T_{source}} \).
For the first case, efficiency \( \eta_A \):
\( \eta_A = 1 - \frac{100}{T} \)
For the second case, efficiency \( \eta_B \):
\( \eta_B = 1 - \frac{180}{900} \)
\( \eta_B = 1 - \frac{1}{5} \)
\( \eta_B = \frac{5 - 1}{5} \)
\( \eta_B = \frac{4}{5} \)
Given that the efficiencies are the same, \( \eta_A = \eta_B \):
\( 1 - \frac{100}{T} = \frac{4}{5} \)
Rearrange the equation to solve for \( T \):
\( \frac{100}{T} = 1 - \frac{4}{5} \)
\( \frac{100}{T} = \frac{5 - 4}{5} \)
\( \frac{100}{T} = \frac{1}{5} \)
\( T = 100 \times 5 \)
\( T = 500 K \)
The temperature \( T \) is \( 500 K \).
In simple words: We set the efficiency formulas for two Carnot engines equal to each other because their efficiencies are the same. By knowing the temperatures for one engine and one temperature for the other, we solved for the unknown temperature, which came out to be 500 K.

🎯 Exam Tip: When efficiencies are equal, equate the efficiency formulas. Be careful with algebraic manipulation when solving for an unknown temperature, especially when dealing with fractions.

Question 6. Two Carnot engines A and B are working in series. First engine A takes heat at 900 K and gives heat to the sink at T K. Another engine B, takes the heat from A and emits it to sink at 400 K. Calculate the temperature T in following conditions :
(i) when same work is done by both the engines.
(ii) when efficiency of both the engines is equal.
Answer:
Given:
Engine A: Source temperature \( T_{1A} = 900 K \), Sink temperature \( T_{2A} = T K \)
Engine B: Source temperature \( T_{1B} = T K \), Sink temperature \( T_{2B} = 400 K \)

(i) When the same work is done by both the engines ( \( W_A = W_B \) ):
Let \( Q_1 \) be the heat absorbed by engine A from the source at 900 K.
Let \( Q_2 \) be the heat rejected by engine A to the sink at T K. This heat \( Q_2 \) is then absorbed by engine B.
Let \( Q_3 \) be the heat rejected by engine B to the sink at 400 K.
Work done by engine A: \( W_A = Q_1 - Q_2 \)
Work done by engine B: \( W_B = Q_2 - Q_3 \)
We know the relation \( \frac{Q_{hot}}{Q_{cold}} = \frac{T_{hot}}{T_{cold}} \). So, for engine A: \( \frac{Q_1}{Q_2} = \frac{900}{T} \)
For engine B: \( \frac{Q_2}{Q_3} = \frac{T}{400} \)
From the first relation, \( Q_1 = Q_2 \frac{900}{T} \). So, \( W_A = Q_2 \frac{900}{T} - Q_2 = Q_2 \left(\frac{900}{T} - 1\right) \).
From the second relation, \( Q_3 = Q_2 \frac{400}{T} \). So, \( W_B = Q_2 - Q_2 \frac{400}{T} = Q_2 \left(1 - \frac{400}{T}\right) \).
Given \( W_A = W_B \):
\( Q_2 \left(\frac{900}{T} - 1\right) = Q_2 \left(1 - \frac{400}{T}\right) \)
Divide by \( Q_2 \) (assuming \( Q_2 \neq 0 \)):
\( \frac{900}{T} - 1 = 1 - \frac{400}{T} \)
Combine terms with \( T \):
\( \frac{900}{T} + \frac{400}{T} = 1 + 1 \)
\( \frac{900 + 400}{T} = 2 \)
\( \frac{1300}{T} = 2 \)
\( T = \frac{1300}{2} \)
\( T = 650 K \)

(ii) When the efficiency of both engines is equal ( \( \eta_A = \eta_B \) ):
Efficiency of engine A: \( \eta_A = 1 - \frac{T_{2A}}{T_{1A}} = 1 - \frac{T}{900} \)
Efficiency of engine B: \( \eta_B = 1 - \frac{T_{2B}}{T_{1B}} = 1 - \frac{400}{T} \)
Given \( \eta_A = \eta_B \):
\( 1 - \frac{T}{900} = 1 - \frac{400}{T} \)
\( \frac{T}{900} = \frac{400}{T} \)
\( T^2 = 900 \times 400 \)
\( T = \sqrt{900 \times 400} \)
\( T = \sqrt{360000} \)
\( T = 600 K \)
In simple words: When two Carnot engines work in a row, we found the middle temperature T under two conditions. If both engines do the same amount of work, T is 650 K. If both engines have the same efficiency, T is 600 K. We used the formulas for work and efficiency of Carnot engines for each part.

🎯 Exam Tip: For engines in series, the heat rejected by the first engine is absorbed by the second. Remember to use the correct temperature limits for each engine when calculating efficiency or work done. When \( W_A = W_B \), set the work expressions equal. When \( \eta_A = \eta_B \), set the efficiency expressions equal and solve for T.

Question 8. If a gas (y = 1.5) is compressed at pressure 27 times the initial pressure, then calculate the change in temperature if the initial temperature is 27°C.
Answer:
Given:
Ratio of specific heats \( \gamma = 1.5 \)
Initial pressure \( P_1 = P \)
Final pressure \( P_2 = 27P \)
Initial temperature \( T_1 = 27^\circ C \)
Convert initial temperature to Kelvin:
\( T_1 = 27 + 273 = 300 K \)
We need to find the change in temperature, \( \Delta T = T_2 - T_1 \). First, we find \( T_2 \).
For an adiabatic process, the relationship between pressure and temperature is:
\( \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}} \)
Substitute the given values:
\( \frac{T_2}{300} = \left(\frac{27P}{P}\right)^{\frac{1.5 - 1}{1.5}} \)
\( \frac{T_2}{300} = (27)^{\frac{0.5}{1.5}} \)
\( \frac{T_2}{300} = (27)^{\frac{1}{3}} \)
We know that \( 27 = 3^3 \), so \( (27)^{\frac{1}{3}} = (3^3)^{\frac{1}{3}} = 3 \).
\( \frac{T_2}{300} = 3 \)
\( T_2 = 3 \times 300 \)
\( T_2 = 900 K \)
Now, calculate the change in temperature \( \Delta T \):
\( \Delta T = T_2 - T_1 \)
\( \Delta T = 900 K - 300 K \)
\( \Delta T = 600 K \)
If we want the change in Celsius, it's the same:
\( T_2 = 900 - 273 = 627^\circ C \)
\( \Delta T = 627^\circ C - 27^\circ C = 600^\circ C \)
So, the change in temperature is \( 600 K \) or \( 600^\circ C \).
In simple words: We used the adiabatic process formula that links pressure and temperature. By plugging in the initial temperature, the pressure change, and the given \( \gamma \) value, we found the new temperature. Then, we subtracted the initial temperature to get the total change, which is 600 K.

🎯 Exam Tip: Always remember to convert Celsius temperatures to Kelvin when working with gas laws, especially in adiabatic processes. Be careful with fractional exponents when calculating, as \( (X^a)^{b} = X^{ab} \).

Question 9. A refrigerator transfers 200 J average heat per second from temperature -10°C to 27°C. Taking the cycle to be reversible, calculate the average power when there is no loss of heat.
Answer: First, convert the given temperatures to Kelvin:
Temperature of the cold reservoir (sink), \( T_2 = -10^\circ C + 273 = 263 K \).
Temperature of the hot reservoir (source), \( T_1 = 27^\circ C + 273 = 300 K \).
Heat transferred from the sink, \( Q_2 = 200 J \cdot s^{-1} \).

The coefficient of performance (\( \beta \)) for a reversible refrigerator is given by:
\( \beta = \frac{T_2}{T_1 - T_2} \)
Substitute the temperature values:
\( \beta = \frac{263}{300 - 263} = \frac{263}{37} \)

We also know that the coefficient of performance is defined as the ratio of heat removed from the cold reservoir to the work done by the refrigerator:
\( \beta = \frac{Q_2}{W} \)
Where W is the work done per second, which is the average power.
Rearranging for W:
\( W = \frac{Q_2}{\beta} \)
Now, substitute the value of \( Q_2 \) and \( \beta \):
\( W = \frac{200 J \cdot s^{-1}}{263/37} \)
\( W = \frac{200 \times 37}{263} \)
\( W = \frac{7400}{263} \approx 28.136 J \cdot s^{-1} \)
The average power of the refrigerator is approximately \( 28.1 W \).
In simple words: First, change the temperatures to Kelvin. Then, use these temperatures to find the refrigerator's performance factor. After that, divide the heat removed from the cold part by this factor to get the power needed to run the refrigerator.

🎯 Exam Tip: Remember to always convert temperatures to Kelvin for thermodynamic calculations. Clearly show each step of the formula application and calculation to avoid losing marks.

Question 11. A Carnot engine operates between 373K and 283 K. Then, calculate the efficiency and deduce that when would its efficiency be 100%.
Answer: Given temperatures:
Temperature of the hot reservoir (source), \( T_1 = 373 K \).
Temperature of the cold reservoir (sink), \( T_2 = 283 K \).

The efficiency (\( \eta \)) of a Carnot engine is calculated using the formula:
\( \eta = 1 - \frac{T_2}{T_1} \)
Substitute the given temperature values into the formula:
\( \eta = 1 - \frac{283}{373} \)
To simplify, find a common denominator:
\( \eta = \frac{373 - 283}{373} = \frac{90}{373} \)
Now, calculate the decimal value and convert to a percentage:
\( \eta \approx 0.24128 \)
\( \eta \approx 0.241 \times 100\% \)
So, the efficiency of the Carnot engine is approximately \( 24.1\% \).

For the efficiency of a Carnot engine to be \( 100\% \) (or 1 in decimal form), we set \( \eta = 1 \):
\( 1 = 1 - \frac{T_2}{T_1} \)
This means:
\( \frac{T_2}{T_1} = 0 \)
For this ratio to be zero, the temperature of the cold reservoir (\( T_2 \)) must be \( 0 K \).
Therefore, a Carnot engine would have 100% efficiency only if the temperature of the sink (\( T_2 \)) were absolute zero (0 K). This is practically impossible to achieve.
In simple words: To find the efficiency, subtract the cold temperature divided by the hot temperature from 1. For a machine to be 100% efficient, the cold temperature must be zero Kelvin, which is not possible in real life.

🎯 Exam Tip: Always express efficiency as a percentage. Remember that 100% efficiency for a heat engine requires an absolute zero temperature sink, which is a theoretical limit and not achievable in practice.