Maharashtra Board Class 12 Physics Chapter 4 Thermodynamics Exercise Solutions

1. Choose the correct option.

(i) A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in the internal energy of the system?
(a) 10J
(b) 4J
(c) -10J
(d) – 4J
Answer: (b) 4J
In simple words: When you add 10 Joules of heat, the gas uses some energy to lift the lid (3 Newtons × 2 meters = 6 Joules of work). The remaining energy (10 - 6 = 4 Joules) stays inside as a change in internal energy.

📝 Teacher's Note: This is a direct application of the First Law of Thermodynamics (\( Q = \Delta U + W \)). Remind students that work done by the system must be subtracted from the total heat added to find the change in internal energy.

🎯 Exam Tip: Always calculate the work done first using \( W = \text{Force} \times \text{Displacement} \) if it is not given directly. Ensure all units are in Joules.

(ii) Which of the following is an example of the first law of thermodynamics?
(a) The specific heat of an object explains how easily it changes temperatures.
(b) While melting, an ice cube remains at the same temperature.
(c) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(d) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.
Answer: (b) While melting, an ice cube remains at the same temperature. [Here, \( \Delta u = 0 \), \( W = Q \)]
(c) When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
(d) After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.
In simple words: The First Law is basically energy conservation; it says energy can change forms (like heat into work) but isn't lost. In melting ice, the heat energy added is perfectly balanced by the energy used to break the solid bonds, keeping the temperature constant.

📝 Teacher's Note: Option (d) is also a statement of energy conservation, but in the context of specific textbook exercises, option (b) is highlighted for its thermodynamic process (\( Q = \Delta U + W \)). Focus on the relationship between heat and state change.

🎯 Exam Tip: Look for scenarios where heat, work, and internal energy are interrelated. Melting at constant temperature is a classic example where added heat equals the work done in changing the phase.

(iii) Efficiency of a Carnot engine is large when
(a) \( T_H \) is large
(b) \( T_C \) is low
(c) \( T_H - T_C \) is large
(d) \( T_H - T_C \) is small
Answer: (c) \( T_H - T_C \) is large
In simple words: A heat engine works better when there is a huge temperature difference between the hot source and the cold sink, just like a waterfall is more powerful if it's very tall.

📝 Teacher's Note: Use the formula \( \eta = 1 - \frac{T_C}{T_H} \). Show that as the numerator (\( T_C \)) decreases or the denominator (\( T_H \)) increases, the overall fraction gets smaller, making the efficiency larger.

🎯 Exam Tip: Remember that "large efficiency" implies the efficiency is closer to 1 (or 100%). This happens when the temperature of the sink is much lower than the source.

(iv) The second law of thermodynamics deals with transfer of:
(a) work done
(b) energy
(c) momentum
(d) mass
Answer: (b) energy
In simple words: The second law tells us the direction in which energy likes to move — usually from a hot place to a cold place naturally.

📝 Teacher's Note: While the first law is about the quantity of energy, the second law is about the quality and direction of energy transfer (entropy). It dictates that heat cannot move from cold to hot without external work.

🎯 Exam Tip: Identify the Second Law with concepts like "direction of heat flow" and "efficiency limits."

(v) During refrigeration cycle, heat is rejected by the refrigerant in the :
(a) condenser
(b) cold chamber
(c) evaporator
(d) hot chamber
Answer: closed tube[See the textbook]
In simple words: In a fridge, the coils at the back (or bottom) get hot because they are "throwing away" the heat collected from inside the fridge into the room.

📝 Teacher's Note: The condenser is specifically designed to reject heat to the surroundings. The refrigerant changes from vapor to liquid here by releasing its latent heat.

🎯 Exam Tip: Be careful with the components: Evaporator absorbs heat, Condenser rejects heat.

2. Answer in brief.

(i) A gas contained in a cylinder surrounded by a thick layer of insulating material is quickly compressed.
(a) Has there been a transfer of heat?
(b) Has work been done?
Answer:
(a) There is no transfer of heat.
(b) The work is done on the gas.
In simple words: Because the cylinder is insulated and the compression is "quick," heat doesn't have time to escape (this is an adiabatic process). Since the gas was squeezed, someone had to do work "on" it.

📝 Teacher's Note: Explain that "insulating material" plus "quickly" are keywords for an adiabatic process (\( Q = 0 \)). Compressing a gas increases its internal energy through work done on it.

🎯 Exam Tip: In adiabatic compression, the temperature of the gas always rises because the work done on it is converted into internal energy with no heat loss.

(ii) Give an example of some familiar process in which no heat is added to or removed form a system, but the temperature of the system changes.
Answer:
Hot water in a container cools after sometime. Its temperature goes on decreasing with time and after sometime it attains room temperature.
[Note : Here, we do not provide heat to the water or remove heat from the water. The water cools on exchange of heat with the surroundings. Recall the portion covered in chapter 3.]
In simple words: Think of a hot cup of tea left on a table; you aren't actively taking heat out with a machine, but its temperature drops because it naturally shares its energy with the cooler air around it.

📝 Teacher's Note: This refers to spontaneous heat transfer due to a temperature gradient. Another classic example is the sudden expansion of air from a tire valve (adiabatic expansion), which feels cold.

🎯 Exam Tip: For processes where temperature changes without "active" heat addition, mention "exchange with surroundings" or "adiabatic processes."

(iii) Give an example of some familiar process in which heat is added to an object, without changing its temperature.
Answer:
1. Melting of ice
2. Boiling of water.
In simple words: When ice melts or water boils, the heat you add is used entirely to change the state (solid to liquid or liquid to gas) rather than making it hotter.

📝 Teacher's Note: This is the concept of Latent Heat. During a phase change, the temperature remains constant until the entire mass has changed state.

🎯 Exam Tip: Always use "Phase Change" as the keyword for heat addition at constant temperature.

Question 2. (iv) What sets the limits on efficiency of a heat engine?
Answer:
The temperature of the cold reservoir sets the limit on the efficiency of a heat engine.
[Notes : (1) \( \eta = 1 - \frac{T_C}{T_H} \)
This formula shows that for maximum efficiency, \( T_C \) should be as low as possible and \( T_H \) should be as high as possible.
(2) For a Carnot engine, efficiency
\( \eta = 1 - \frac{T_C}{T_H} \). \( \eta \rightarrow 1 \) as \( T_C \rightarrow 0 \).]
In simple words: The efficiency is limited by how cold you can make the "exhaust" part. If the cold part was at absolute zero, the engine would be perfectly efficient, but that's impossible in real life.

📝 Teacher's Note: Emphasize the Kelvin-Planck statement of the second law: no engine can be 100% efficient because some heat must always be rejected to a cold reservoir.

🎯 Exam Tip: Mention both the source (\( T_H \)) and sink (\( T_C \)) temperatures in your answer, as the ratio between them determines the limit.

Question 2. (v) Why should a Carnot cycle have two isothermal two adiabatic processes?
Answer:
With two isothermal and two adiabatic processes, all reversible, the efficiency of the Carnot engine depends only on the temperatures of the hot and cold reservoirs.
[Note : This is not so in the Otto cycle and Diesel cycle.]
In simple words: These specific steps allow the engine to be perfectly "reversible," meaning it doesn't waste any energy on friction or messy heat leaks, making it the most efficient theoretical engine possible.

📝 Teacher's Note: Explain that isothermal processes allow heat exchange at constant temperature, while adiabatic processes change the temperature of the working substance without losing heat to the surroundings.

🎯 Exam Tip: Use the term "reversibility" as the core reason for choosing these four processes.

3. Answer the following questions.

(i) A mixture of hydrogen and oxygen is enclosed in a rigid insulting cylinder. It is ignited by a spark. The temperature and the pressure both increase considerably. Assume that the energy supplied by the spark is negligible, what conclusions may be drawn by application of the first law of thermodynamics?
Answer:
The internal energy of a system is the sum of potential energy and kinetic energy of all the constituents of the system. In the example stated above, conversion of potential energy into kinetic energy is responsible for a considerable rise in pressure and temperature of the mixture of hydrogen and oxygen ignited by the spark.
In simple words: Even though no outside heat was added (the cylinder is insulated), a chemical reaction turned "hidden" chemical energy into heat, making the molecules move faster and bump harder against the walls.

📝 Teacher's Note: In this case, \( Q = 0 \) and \( W = 0 \) (rigid cylinder), so \( \Delta U = 0 \). However, the internal energy is redistributed from chemical potential energy to thermal kinetic energy.

🎯 Exam Tip: Note that for a "rigid" cylinder, the work done (\( W \)) is zero. Focus the answer on the transformation of internal energy forms.

(ii) A resistor held in running water carries electric current. Treat the resistor as the system
(a) Does heat flow into the resistor?
(b) Is there a flow of heat into the water?
(c) Is any work done?
(d) Assuming the state of resistance to remain unchanged, apply the first law of thermodynamics to this process.
Answer:
(a) Heat is generated into the resistor due to the passage of electric current. In the usual notation, heat generated = \( I^2Rt \).
(b) Yes. Water receives heat from the resistor.
(c) \( I^2Rt = MC\Delta T + P\Delta V \)
\( (Q) \) \(( \Delta U )\) \( (W) \)

Here, I = current through the resistor, R = resistance of the resistor, t = time for which the current is passed through the resistor, M = mass of the water, S = specific heat of water, T = rise in the temperature of water, P = pressure against which the work is done by the water, \( \Delta u \)= increase in the volume of the water.
In simple words: Electricity makes the resistor hot, and that heat moves into the water. This heat makes the water warmer and also causes it to expand slightly (doing work).

📝 Teacher's Note: Clarify that in thermodynamic terms, the electrical energy entering the resistor is considered "work" or "heat input" depending on the system boundary. In this specific textbook context, it's treated as heat generation (\( Q \)).

🎯 Exam Tip: When applying the first law, identify the "heat" term with Joule heating (\( I^2Rt \)).

(iii) A mixture of fuel and oxygen is burned in a constant-volume chamber surrounded by a water bath. It was noticed that the temperature of water is increased during the process. Treating the mixture of fuel and oxygen as the system,
(a) Has heat been transferred ?
(b) Has work been done?
(c) What is the sign of \( \Delta u \) ?
Answer:
(a) Heat has been transferred from the chamber to the water bath.
(b) No work is done by the system (the mixture of fuel and oxygen) as there is no change in its volume.
(c) There is increase in the temperature of water. Therefore, \( \Delta u \) is positive for water. For the system (the mixture of fuel and oxygen), \( \Delta u \) is negative.
In simple words: The fire in the box heats up the water outside. Since the box is rigid, it doesn't move or expand (no work). The fuel loses internal energy (negative \( \Delta u \)) because it's giving it away to the water.

📝 Teacher's Note: Emphasize that "constant-volume" strictly implies \( W = 0 \). The sign of \( \Delta U \) depends on whether you are looking at the fuel (losing energy) or the water (gaining energy).

🎯 Exam Tip: Always define the "system" clearly. If the question asks for the sign of \( \Delta U \), specify whether it's for the system or the surroundings.

(iv) Draw a p-V diagram and explain the concept of positive and negative work. Give one example each.
Answer:
Consider some quantity of an ideal gas enclosed in a cylinder fitted with a movable, massless and frictionless piston.
Suppose the gas is allowed to expand by moving the piston outward extremely slowly. There is decrease in pressure of the gas as the volume of the gas increases. Below figure shows the corresponding P-V diagram.In this case, the work done by the gas on its surroundings, \( W = \int_{V_i}^{V_f} PdV \) (= area under the curve) is positive as the volume of the gas has increased from \( V_i \) to \( V_f \).

Let us now suppose that starting from the same initial condition, the piston is moved inward extremely slowly so that the gas is compressed. There is increase in pressure of the gas as the volume of the gas decreases. Figure shows the corresponding P-V diagram.In this case, the work done by the gas on its surroundings, \( W = \int_{V_i}^{V_f} PdV \) (= area under the curve) is negative as the volume of the gas has decreased from \( V_i \) to \( V_f \).
In simple words: When a gas pushes a piston out, it does positive work (like you pushing a door open). When an outside force pushes the piston in, the gas does negative work (the door is being pushed against you).

📝 Teacher's Note: Use the area-under-the-curve concept to explain magnitude. For the sign, focus on the direction of volume change: expansion means positive work by the gas, compression means negative work by the gas.

🎯 Exam Tip: Always draw the arrows on the p-V curve to show the direction of the process. Label the axes clearly and shade the area to represent work.

(v) A solar cooker and a pressure cooker both are used to cook food. Treating them as thermodynamic systems, discuss the similarities and differences between them.
Answer:
Similarities :
1. Heat is added to the system.
2. There is increase in the internal energy of the system.
3. Work is done by the system on its environment.
Differences : In a solar cooker, heat is supplied in the form of solar radiation. The rate of supply of heat is relatively low.
In a pressure cooker, usually LPG is used (burned) to provide heat. The rate of supply of heat is relatively high. As a result, it takes very long time for cooking when a solar cooker is used. With a pressure cooker, it does not take very long time for cooking.
[Note : A solar cooker can be used only when enough solar radiation is available.]
In simple words: Both cookers use heat to warm up the food. The solar cooker uses sunlight and is slow, while the pressure cooker uses a stove flame and is very fast because it can supply much more heat quickly.

📝 Teacher's Note: Explain that the pressure cooker also works at a higher pressure, which raises the boiling point of water, further speeding up the cooking process compared to a solar cooker at atmospheric pressure.

🎯 Exam Tip: Focus on "Rate of Heat Transfer" as the primary scientific difference between the two systems.

Question 4. A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system. [Ans: -106.5 J]
Answer:
Data : \( P = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \), \( V_1 = 5 \text{ litres} = 5 \times 10^{-3} \text{ m}^3 \), \( V_2 = 10 \text{ litres} = 10 \times 10^{-3} \text{ m}^3 \), \( Q = 400\text{J} \).
The work done by the system (gas in this case) on its surroundings,
\( W = P(V_2 - V_1) \)
\( = (1.013 \times 10^5 \text{ Pa}) (10 \times 10^{-3} \text{ m}^3 - 5 \times 10^{-3} \text{ m}^3) \)
\( = 1.013 (5 \times 10^2)\text{J} = 5.065 \times 10^2 \text{J} = 506.5 \text{J} \)
The change in the internal energy of the system, \( \Delta u = Q - W = 400\text{J} - 506.5\text{J} = -106.5\text{J} \)
The minus sign shows that there is a decrease in the internal energy of the system.
In simple words: The gas took in 400 Joules of heat but did 506.5 Joules of work pushing the piston. Since it did more work than the energy it received, its internal "savings" (internal energy) dropped by 106.5 Joules.

📝 Teacher's Note: Remind students to convert Liters to cubic meters (\( 1 \text{ L} = 10^{-3} \text{ m}^3 \)) and Atmospheres to Pascals (\( 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \)) before solving.

🎯 Exam Tip: A negative \( \Delta U \) indicates the temperature of the gas decreased. This is a common point for conceptual sub-questions.

Question 5. A system releases 130 kJ of heat while 109 kJ of work is done on the system. Calculate the change in internal energy. [Ans: \( \Delta U = 21 \text{ kJ} \)]
Answer:
Data : \( Q = -130\text{kj} \), \( W= - 109\text{kJ} \)
\( \Delta u = Q - W = - 130\text{kJ} - ( - 109\text{kJ}) \)
\( = (-130 + 109) \text{ kJ} = - 21 \text{ kj} \).
This is the change (decrease) in the internal energy.
In simple words: The system lost 130 units of heat but someone helped it by pushing in 109 units of work. Overall, it still ended up losing 21 units of its internal energy.

📝 Teacher's Note: Sign conventions are critical here: heat released is negative (\( -Q \)), work done *on* the system is negative (\( -W \) in the IUPAC convention, though many textbooks use \( Q = \Delta U + W \) where \( W \) is work *by* the system). Follow the convention used in the textbook solution provided.

🎯 Exam Tip: Watch the signs! "Heat released" is \( - \), "Work done on the system" is \( - \). Subtracting a negative number becomes adding.

Question 6. Efficiency of a Carnot cycle is 75%. If temperature of the hot reservoir is 727ºC, calculate the temperature of the cold reservoir. [Ans: 23ºC]
Answer:
Data : \( \eta = 75\% = 0.75 \), \( T_H = (273 + 727) \text{ K} = 1000 \text{ K} \)
\( \eta = 1 - \frac{T_C}{T_H} \)

\( \implies \frac{T_C}{T_H} = 1 - \eta \)

\( \implies T_C = T_H(1 - \eta) = 1000 \text{ K} (1 - 0.75) \)
\( = 250\text{K} = (250 - 273)^\circ\text{C} = -23 \text{ } ^\circ\text{C} \)
This is the temperature of the cold reservoir.
In simple words: To get 75% efficiency from a furnace that is at 1000 Kelvin, the "cooling" end of the engine needs to be as cold as -23 degrees Celsius.

📝 Teacher's Note: Students often forget to convert Celsius to Kelvin. Always add 273. The answer key in the source says 23ºC, but the calculation correctly leads to -23ºC. Point out this distinction in class.

🎯 Exam Tip: For efficiency problems, the temperature must always be in Kelvin. Don't forget to convert the final answer back to Celsius if the question asks for it.

Question 7. A Carnot refrigerator operates between 250K and 300K. Calculate its coefficient of performance. [Ans: 5]
Answer:
Data : \( T_C = 250 \text{ K} \), \( T_H = 300 \text{ K} \)
\( K = \frac{T_C}{T_H - T_C} = \frac{250 \text{ K}}{300 \text{ K} - 250 \text{ K}} = \frac{250}{50} = 5 \)
This is the coefficient of performance of the refrigerator.
In simple words: The "Coefficient of Performance" (COP) tells you how well a fridge works. A COP of 5 means for every 1 unit of electricity you use, you remove 5 units of heat from inside the fridge.

📝 Teacher's Note: Contrast this with efficiency (\( \eta \)). Efficiency is always less than 1, but COP can be much greater than 1 because it represents a transfer ratio, not energy creation.

🎯 Exam Tip: Note that for COP, the cold temperature (\( T_C \)) is in the numerator, unlike efficiency where it's part of the subtraction.

Question 8. An ideal gas is taken through an isothermal process. If it does 2000 J of work on its environment, how much heat is added to it? [Ans: 2000J]
Answer:
Data : \( W = 2000 \text{ J} \), isothermal process
In this case, the change in the internal energy of the gas, \( \Delta u \), is zero as the gas is taken through an isothermal process.
Hence, the heat added to it,
\( Q = \Delta u + W = 0 + W = 2000\text{J} \)
In simple words: In an isothermal process, the temperature stays the same, so the gas's internal energy doesn't change. This means every bit of heat you give the gas is immediately used by it to do work.

📝 Teacher's Note: Isothermal implies \( \Delta T = 0 \), which for an ideal gas means \( \Delta U = 0 \). This is a key conceptual link to the first law.

🎯 Exam Tip: When you see "isothermal," immediately write down \( \Delta U = 0 \) to simplify the First Law equation to \( Q = W \).

Question 9. An ideal monatomic gas is adiabatically compressed so that its final temperature is twice its initial temperature. What is the ratio of the final pressure to its initial pressure? [Ans: 5.656]
Answer:
Data : \( T_f = 2T_i \), monatomic gas \( \implies \gamma = 5/3 \)

\( P_i V_i^\gamma = P_f V_f^\gamma \) in an adiabatic process
Now, \( PV = nRT \)

\( \implies V = \frac{nRT}{P} \)

\( \therefore V_i = \frac{nRT_i}{P_i} \) and \( V_f = \frac{nRT_f}{P_f} \)

\( \therefore P_i \left( \frac{nRT_i}{P_i} \right)^\gamma = P_f \left( \frac{nRT_f}{P_f} \right)^\gamma \)

\( \therefore P_i^{1-\gamma} T_i^\gamma = P_f^{1-\gamma} T_f^\gamma \)

\( \therefore \left( \frac{T_f}{T_i} \right)^\gamma = \left( \frac{P_f}{P_i} \right)^{\gamma - 1} \)

\( \therefore \left( \frac{T_f}{T_i} \right)^{\frac{\gamma}{\gamma - 1}} = \frac{P_f}{P_i} \)

\( \therefore 2^{\frac{5/3}{5/3-1}} = \left( \frac{P_f}{P_i} \right) = 2^{5/2} \)

\( \therefore \frac{P_f}{P_i} = 2^{2.5} = 5.656 \)
This is the ratio of the final pressure (\( P_f \)) to the initial pressure (\( P_i \)).
In simple words: If you compress a monatomic gas so it gets twice as hot without losing any heat, the pressure will jump up to about 5.6 times what it was at the start.

📝 Teacher's Note: The relationship between P and T in an adiabatic process is \( P^{1-\gamma} T^\gamma = \text{constant} \). Remind students that for monatomic gases, \( \gamma = 1.67 \) (or \( 5/3 \)).

🎯 Exam Tip: Practice log calculations or root calculations for \( 2^{2.5} \) as it is a common step in adiabatic problems. \( 2^{2.5} = \sqrt{2^5} = \sqrt{32} \approx 5.66 \).

Question 10. A hypothetical thermodynamic cycle is shown in the figure. Calculate the work done in 25 cycles.
[Ans: \( 7.855 \times 10^4 \text{ J} \)]
Answer:
Data : \( a = \frac{\Delta V_{max}}{2} = \left( \frac{6-2}{2} \right) \times 10^{-3} \text{ m}^3 = 2 \times 10^{-3} \text{ m}^3 \),
\( b = \left( \frac{\Delta P_{max}}{2} \right) = \left( \frac{11-1}{2} \right) \times 10^5 \text{ Pa} = 5 \times 10^5 \text{ Pa} \),
25 cycles
The work done in one cycle, \( \oint PdV \)
\( = \pi ab = (3.142) (2 \times 10^{-3} \text{ m}^3) (5 \times 10^5 \text{ Pa}) \)
\( = 3.142 \times 10^3 \text{J} \)
Hence, the work done in 25 cycles
\( = (25) (3.142 \times 10^3 \text{ J}) = 7.855 \times 10^4 \text{ J} \)
In simple words: The "area" inside the loop on the graph represents the work done in one full turn. By calculating the area of this ellipse-like shape and multiplying it by 25 turns, we find the total energy spent.

📝 Teacher's Note: Explain that for a circular/elliptical P-V diagram, the work done per cycle is the area of the figure. Use the formula \( \text{Area} = \pi \times \text{semi-major axis} \times \text{semi-minor axis} \).

🎯 Exam Tip: Be very careful with the powers of 10 on the graph axes. Always multiply the numerical result by the scale factor given on the axis label.

Question 11. The figure shows the V-T diagram for one cycle of a hypothetical heat engine which uses the ideal gas. Draw the p-V diagram and P-V diagram of the system.
Ans: (a)]
[Ans: (b)]
Answer:
(a) P-V diagram (Schematic)
ab: isobaric process,
bc : isothermal process,
cd : isobaric process,
da : isothermal process
\( \frac{P_a V_a}{T_a} = \frac{P_b V_b}{T_b} = \frac{P_c V_c}{T_c} = \frac{P_d V_d}{T_d} = nR \)
In simple words: This graph translates how the gas's volume and pressure change together during different steps of the engine's cycle, like a map of its movement.

📝 Teacher's Note: Mapping processes from one diagram type (V-T) to another (P-V) requires identifying the nature of each leg. For example, a straight line through the origin in V-T is an isobaric process (\( V \propto T \)).

🎯 Exam Tip: Remember: Isotherms are curves on a P-V diagram (\( P \propto 1/V \)), and Isobaric processes are horizontal lines on a P-V diagram.

Question 12. A system is taken to its final state from initial state in hypothetical paths as shown figure. Calculate the work done in each case.
[Ans: \( AB = 2.4 \times 10^6 \text{ J} \), \( CD = -8 \times 10^5 \text{ J} \), BC and DA zero, because constant volume change]
Answer:
Data: \( P_A = P_B = 6 \times 10^5 \text{ Pa} \), \( P_C = P_D = 2 \times 10^5 \text{ Pa} \), \( V_A = V_D = 2 \text{ L} \), \( V_B = V_C = 6 \text{ L} \), \( 1 \text{ L} = 10^{-3} \text{ m}^3 \)
(i) The work done along the path A → B (isobaric process), \( W_{AB} = P_A (V_B - V_A) = (6 \times 10^5 \text{ Pa})(6 - 2)(10^{-3} \text{ m}^3) = 2.4 \times 10^3 \text{ J} \)
(ii) \( W_{BC} \) = zero as the process is isochoric (\( V = \text{constant} \)).
(iii) The work done along the path C → D (isobaric process), \( W_{CD} = P_C (V_D - V_C) = (2 \times 10^5 \text{ Pa}) (2 - 6) (10^{-3} \text{ m}^3) = -8 \times 10^2 \text{ J} \)
(iv) \( W_{DA} \) = zero as \( V = \text{constant} \).
In simple words: Work only happens when the box expands or shrinks. From A to B, it's expanding, so it does work. From B to C, it stays the same size, so work is zero. From C to D, it's shrinking, so work is negative.

📝 Teacher's Note: The work done during any isochoric (vertical line on P-V) process is always zero because \( \Delta V = 0 \). For isobaric (horizontal) processes, use \( W = P\Delta V \).

🎯 Exam Tip: Note the units! If the volume is in Liters and Pressure is in Pascals, your raw calculation might have an error unless you convert Liters to \( \text{m}^3 \) using the \( 10^{-3} \) factor.

Can you tell? (Textbook Page No. 76)

Question 1. Why is it that different objects kept on a table at room temperature do not exchange heat with the table ?
Answer:
The objects do exchange heat with the table but there is no net transfer of energy (heat) as the objects and the table are at the same temperature.
In simple words: Energy is actually jumping back and forth between the table and the objects, but since they are at the same temperature, the amount going in equals the amount going out — so nothing changes overall.

📝 Teacher's Note: This is a key point about Thermal Equilibrium. Heat transfer is driven by temperature difference. When \( \Delta T = 0 \), we reach dynamic equilibrium.

🎯 Exam Tip: Use the phrase "thermal equilibrium" and "no net heat transfer" to answer this correctly.

Can you tell? (Textbook Page No. 77)

Question 1. Why is it necessary to make a physical contact between a thermocouple and the object for measuring its temperature ?
Answer:
For heat transfer to develop thermoemf.
In simple words: To measure how hot something is, the thermocouple needs to actually "feel" the heat so it can turn that thermal energy into a tiny bit of electricity.

📝 Teacher's Note: A thermocouple relies on the Seebeck effect. Without physical contact, conduction cannot occur efficiently to equalize the temperature of the junction with the object.

🎯 Exam Tip: Focus on "conduction" as the primary mode of heat transfer required for a thermocouple to work.

Can you tell? (Textbook Page No. 81)

Question 1. Can you explain the thermodynamics involved in cooking food using a pressure cooker ?
Answer:
Basically, heat is supplied by the burning fuel causing increase in the internal energy of the food (system), and the system does some work on its surroundings. In the absence of any data about the components of food and their thermal and chemical properties, we cannot evaluate changes in internal energy and work done.
In simple words: The stove gives energy to the food, which gets hot and pressurized. Some of that energy tries to push the lid or the weight up (work), while the rest cooks the food faster.

📝 Teacher's Note: Mention that the closed environment prevents steam from escaping, which builds pressure. Higher pressure raises the boiling point of water, allowing the food to cook at temperatures higher than 100°C.

🎯 Exam Tip: Connect heat input to increased "Internal Energy" and "Pressure" for a complete thermodynamic explanation.

Use your brain power (Textbook Page No. 85)

Question 1. Verify that the area under the P-V curve has dimensions of work.
Answer:
Area under the P-V curve is \( \int_{V_i}^{V_f} PdV \), where P is the pressure and V is the volume.
\( [\text{pressure}] = \frac{[\text{force}]}{[\text{area}]} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}] \)
\( [\text{Volume}] = [M^0L^3T^0] \)
\( \therefore [\text{Pressure} \times \text{Volume}] = \frac{[MLT^{-2}]}{[L^2]} \cdot [L^3] = [MLT^{-2}][L] = [\text{force}][\text{distance}] = [\text{work}] \).
In simple words: Pressure is like pushing force on a surface, and Volume is like moving that surface through space. When you multiply force by distance, you get work!

📝 Teacher's Note: Dimensional analysis is a powerful way to verify physical formulas. Showing that \( \text{Nm}^{-2} \times \text{m}^3 = \text{Nm} = \text{Joule} \) helps students visualize work in thermodynamics.

🎯 Exam Tip: For dimensional verification, always break down units into fundamental M, L, and T components.

Use your brain power (Textbook Page No. 91)

Question 1. Show that the isothermal work may also be expressed as \( W = nRT \ln \left( \frac{P_i}{P_f} \right) \),
Answer:
In the usual notation,
\( W = nRT \ln \frac{V_f}{V_i} \) and \( P_i V_i = P_f V_f = nRT \) in an isothermal process
\( \therefore \frac{V_f}{V_i} = \frac{P_i}{P_f} \) and \( W = nRT \ln \frac{P_i}{P_f} \)
In simple words: Since pressure and volume are opposites in an isothermal process (if one goes up, the other goes down by the same ratio), we can swap Volume in the formula for Pressure just by flipping the starting and ending values.

📝 Teacher's Note: This derivation uses Boyle's Law (\( P_1V_1 = P_2V_2 \)). It's useful when you know the initial and final pressures but don't have measurements for volume expansion.

🎯 Exam Tip: Note that for expansion, \( V_f > V_i \) but \( P_i > P_f \). This ensures the log term remains positive in both versions of the formula.

Use your brain power (Textbook Page No. 94)

Question 1. Why is the P-V curve for an adiabatic process steeper than that for an isothermal process ?
Answer:
Adiabatic process : \( PV^\gamma = \text{constant} \)
\( \therefore V^\gamma dP + \gamma PV^{\gamma-1} dV = 0 \)
\( \therefore \frac{dP}{dV} = -\gamma \frac{P}{V} \)
Isothermal process : \( PV = \text{constant} \)
\( \therefore pdV + VdP = 0 \implies \frac{dP}{dV} = \frac{P}{-V} \)
Now, \( \gamma > 1 \)
\( \therefore \left| \left( \frac{dP}{dV} \right)_{adiabatic} \right| > \left| \left( \frac{dP}{dV} \right)_{isotherm} \right| \)
\( \therefore \frac{dP}{dV} \) is the slope of the P - V curve.
In simple words: In an adiabatic squeeze, the gas gets hotter because heat can't escape, which makes it push back even harder. This causes the pressure to rise much faster and more steeply than if the gas stayed at the same temperature.

📝 Teacher's Note: The factor \( \gamma \) (ratio of specific heats) is always greater than 1, so the "slope" for adiabatic curves is always steeper by that factor compared to isothermal curves.

🎯 Exam Tip: Draw both curves starting from the same point on a P-V diagram to clearly show the difference in steepness. Mention \( \gamma > 1 \).

Question 2. Explain formation of clouds at high altitude.
Answer:
As the temperature of the earth increases due to absorption of solar radiation, water from rivers, lakes, oceans, etc. evaporates and rises to high altitude. Water vapour forms clouds as water molecules come together under appropriate conditions. Clouds are condensed water vapour and are of various type, names as cumulus clouds, nimbostratus clouds, stratus clouds and high-flying cirrus clouds.
In simple words: The sun heats up water into invisible steam that floats up. High in the sky, it's colder, so the steam turns back into tiny water droplets or ice bits that clump together to form the fluffy clouds we see.

📝 Teacher's Note: This is an application of adiabatic cooling. As moist air rises, it expands due to lower pressure. This expansion does work and, with no heat added, the air temperature drops, leading to condensation.

🎯 Exam Tip: Include "adiabatic expansion" and "cooling" in your explanation of cloud formation for better marks.

Can you tell? (Textbook Page No. 95)

Question 1. How would you interpret Eq. 4.21 (Q = W) for a cyclic process ?
Answer:
It means \( \Delta u = 0 \) for a cyclic process as the system returns to its initial state.
In simple words: In a loop, the engine ends up exactly where it started. Since its energy state is back to normal, all the heat you gave it during the trip must have been turned into work.

📝 Teacher's Note: In any cycle, because state variables (P, V, T) return to their original values, the total change in internal energy must be zero.

🎯 Exam Tip: For a cyclic process, always assume \( \Delta U = 0 \) as the starting point for your calculations.

Question 2. An engine works at 5000 rpm, and it performs 1000 J of work in one cycle. If the engine runs for 10 min, how much total work is done by the engine ?
Answer:
The total work done by the engine = (1000 J/cycle) (5000 cycles/min) (10 min) = \( 5 \times 10^7 \text{ J} \).
In simple words: If the engine does a bit of work every time it turns, you just multiply the work per turn by how many times it turns in a minute, and then by the total minutes.

📝 Teacher's Note: This problem connects mechanical power and thermodynamic cycles. Remind students to check that the units of time (minutes in rpm and minutes in duration) cancel out.

🎯 Exam Tip: Be careful with the units "per minute" vs "per second" (Hertz) if the question provides frequency in different units.

Do you know? (Textbook Page No. 101)

Question 1. The capacity of an air conditioner is expressed in a tonne. Do you know why?
Answer:
Before refrigerators and AC were invented, cooling was done by using blocks of ice. When cooling machines were invented, their capacity was expressed in terms of the equivalent amount of ice melted in a day (24 hours). The same term is used even today.
(Note : 1 tonne = 1000 kg = 2204.6 pounds, 1 ton (British) = 2240 pounds = 1016.046909 kg, 1 ton (US) = 2000 pounds = 907.184 kg.]
In simple words: People used to buy actual tons of ice to keep things cool. So, when machines were made, they were rated by how many tons of ice they could "replace" in a single day.

📝 Teacher's Note: This is a historical unit of power, not mass. One ton of refrigeration is equivalent to the cooling provided by melting 2000 lbs of ice in 24 hours (roughly 3.5 kW or 12,000 BTU/hr).

🎯 Exam Tip: Understanding unit origins helps in conceptual questions. Remember that "tonne" in AC ratings refers to the "cooling power."

Use your brainpower (Textbook Page No. 105)

Question 1. Suggest a practical way to increase the efficiency of a heat engine.
Answer:
The efficiency of a heat engine can be increased by choosing the hot reservoir at very high temperature and cold reservoir at very low temperature.
In simple words: Make the furnace extra hot and the radiator extra cold. The bigger the gap between "hot" and "cold," the more work the engine can squeeze out of the heat.

📝 Teacher's Note: This reflects the formula \( \eta = 1 - \frac{T_C}{T_H} \). Practically, this is limited by the melting points of materials used to build the engine (for \( T_H \)) and the surrounding ambient temperature (for \( T_C \)).

🎯 Exam Tip: Mention both increasing \( T_H \) and decreasing \( T_C \) to provide a complete answer.