Exercises — Thermodynamics

1. 11.13 marksNCERT Cl-11 Physics Part II, Ch11 Exercises, Q11.1

   A geyser heats water flowing at the rate of 3.0 litres per minute from 27 degrees C to 77 degrees C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 x 10^4 J/g?

   Hint. 1 litre of water has a mass of 1 kg, so the mass flow rate directly gives the heat needed per minute via Q = m s delta-T. Dividing that by the heat of combustion per gram gives the fuel consumption rate.

   Step 1 — Find the mass flow rate and temperature rise. Mass flow rate = 3.0 kg/min (since 1 L of water has mass 1 kg). Temperature rise delta-T = 77 - 27 = 50 degrees C.

   Step 2 — Find the heat required per minute. Q = m s delta-T = 3.0 x 4186 x 50 = 627900 J/min, using the specific heat of water s = 4186 J/kg/K.

   Step 3 — Convert to a fuel consumption rate. Fuel needed per minute = Q / (heat of combustion) = 627900 / (4.0 x 10^4) = 15.7 g/min.

   ✦ The burner must consume fuel at about 15.7 g/min, since that is the mass of fuel whose combustion releases exactly the heat the flowing water absorbs every minute.

2. 11.23 marksNCERT Cl-11 Physics Part II, Ch11 Exercises, Q11.2

   What amount of heat must be supplied to 2.0 x 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45 degrees C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J/mol/K.)

   Hint. Nitrogen is diatomic, so at constant pressure use Cp = 7R/2, not Cp = 5R/2. Convert the given mass to moles first using the molecular mass.

   Step 1 — Convert mass to moles. mu = mass / molecular mass = (2.0 x 10^-2 x 1000 g) / 28 = 20/28 = 0.7143 mol.

   Step 2 — Find Cp for a diatomic gas. N2 is diatomic, so Cp = 7R/2 = 7 x 8.3 / 2 = 29.05 J/mol/K.

   Step 3 — Apply Q = mu Cp delta-T. Q = 0.7143 x 29.05 x 45 = 933.8 J.

   ✦ About 933.8 J of heat must be supplied, because the constant-pressure heat capacity of a diatomic gas is 7R/2, not the monatomic value 5R/2 or 3R/2 for solids — using the wrong Cp is the most common way this problem goes wrong.

3. 11.34 marksNCERT Cl-11 Physics Part II, Ch11 Exercises, Q11.3

   Explain why: (a) two bodies at different temperatures T1 and T2, if brought into thermal contact, do not necessarily settle to the mean temperature (T1 + T2)/2. (b) the coolant in a chemical or nuclear plant should have a high specific heat. (c) air pressure in a car tyre increases during driving. (d) the climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

   Hint. For (a), think about what determines the final temperature when heat capacities differ. For (c) and (d), both come down to what happens to the temperature of a fixed volume, or a slowly-varying reservoir, when heat is added.

   (a) The common final temperature reached depends on how much heat each body must gain or lose per degree, i.e. on their heat capacities (mass x specific heat), not on temperature alone. Heat lost by the hotter body equals heat gained by the cooler one, so the final temperature is a heat-capacity-weighted average of T1 and T2. It equals the simple arithmetic mean (T1 + T2)/2 only in the special case where the two heat capacities happen to be equal.

   (b) A coolant's job is to carry away large amounts of heat from a plant without its own temperature rising dangerously, and without needing an impractically large volume of it to circulate. A high specific heat means a given mass of coolant absorbs a large quantity of heat for only a small rise in its own temperature, which is exactly the property that makes it effective and safe to use.

   (c) Friction between the tyre and the road, and repeated flexing of the tyre wall, convert mechanical energy into heat, raising the temperature of the air trapped inside. Since the tyre's volume stays essentially fixed, Gay-Lussac's law (P proportional to T at constant V) means this temperature rise shows up directly as a pressure rise.

   (d) Water has a much higher specific heat than land. The sea near a harbour town absorbs heat slowly and releases it slowly, keeping nearby air temperatures moderate through the day-night and seasonal cycle. A desert town, surrounded by low-specific-heat land, heats up and cools down quickly, producing much larger temperature swings at the same latitude.

   ✦ All four cases trace back to the same idea: how much a body's temperature changes for a given heat exchange depends on its heat capacity, not on the heat exchanged alone.

4. 11.43 marksNCERT Cl-11 Physics Part II, Ch11 Exercises, Q11.4

   A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

   Hint. Insulated walls and an insulated piston mean no heat can enter or leave — this is an adiabatic process, so use PV^gamma = constant with gamma for a diatomic gas (hydrogen).

   Step 1 — Identify the process. Both the cylinder walls and the piston are insulated, so no heat is exchanged: this is an adiabatic compression.

   Step 2 — Use the adiabatic relation. For an adiabatic process, P1 V1^gamma = P2 V2^gamma, so P2/P1 = (V1/V2)^gamma. Hydrogen is diatomic, so gamma = 7/5 = 1.4.

   Step 3 — Substitute V1/V2 = 2 (volume halved). P2/P1 = 2^1.4 = 2.64.

   ✦ The pressure increases by a factor of about 2.64, since compressing an insulated gas adiabatically raises both its temperature and its pressure more sharply than an isothermal compression to the same volume would.

5. 11.53 marksNCERT Cl-11 Physics Part II, Ch11 Exercises, Q11.5

   In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J.)

   Hint. Internal energy is a state variable, so delta-U between A and B is the same regardless of which path is used to get there. Find delta-U from the adiabatic path first, then apply the first law to the second path.

   Step 1 — Find delta-U from the adiabatic path. Work done ON the system is 22.3 J, so work done BY the system is delta-W = -22.3 J. Since the process is adiabatic, delta-Q = 0, so delta-U = delta-Q - delta-W = 0 - (-22.3) = 22.3 J.

   Step 2 — Carry delta-U over to the second path. Because U is a state variable, delta-U depends only on states A and B, not on the path — so delta-U = 22.3 J for the second path too.

   Step 3 — Apply the first law to the second path. delta-Q = 9.35 cal = 9.35 x 4.19 = 39.18 J. From delta-Q = delta-U + delta-W, delta-W = delta-Q - delta-U = 39.18 - 22.3 = 16.88 J.

   ✦ The net work done by the system in the second process is about 16.9 J, because although Q and W separately depend on the path taken between A and B, their combination delta-Q - delta-W does not — that is exactly what makes delta-U path-independent.

6. 11.64 marksNCERT Cl-11 Physics Part II, Ch11 Exercises, Q11.6

   Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following: (a) What is the final pressure of the gas in A and B? (b) What is the change in internal energy of the gas? (c) What is the change in the temperature of the gas? (d) Do the intermediate states of the system, before settling to the final equilibrium state, lie on its P-V-T surface?

   Hint. This is free expansion into a vacuum — no heat is exchanged (insulated) and no work is done (there is nothing on the other side of the stopcock to push against), which fixes both delta-Q and delta-W at zero.

   (a) The gas expands to fill both cylinders, doubling its volume at the same temperature (justified in part (c) below). By Boyle's law at fixed T, P_initial x V = P_final x 2V, so P_final = P_initial / 2 — the pressure halves.

   (b) The system is insulated, so delta-Q = 0. The gas expands into a vacuum with nothing to push against, so delta-W = 0 as well. By the first law, delta-U = delta-Q - delta-W = 0.

   (c) For an ideal gas, internal energy depends only on temperature. Since delta-U = 0, the temperature does not change: delta-T = 0.

   (d) No. During the expansion the gas is not in equilibrium — pressure is not uniform throughout the combined volume while the gas rushes to fill the vacant cylinder. The P-V-T surface represents only equilibrium states, so these non-equilibrium intermediate states cannot be plotted on it at all.

   ✦ Free expansion is the standard example of an irreversible process: energy is conserved throughout (parts b and c), but the process passes through states with no well-defined pressure, which is exactly why part (d)'s answer is no.

7. 11.72 marksNCERT Cl-11 Physics Part II, Ch11 Exercises, Q11.7

   An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing?

   Hint. Work directly with rates by dividing the first law through by time: the rate of change of internal energy equals the rate of heat supply minus the rate at which the system does work.

   Step 1 — Write the first law as a rate equation. dU/dt = dQ/dt - dW/dt.

   Step 2 — Substitute the given rates. dU/dt = 100 - 75 = 25 W.

   ✦ The internal energy increases at 25 J/s, since only the part of the supplied heat not converted into output work goes toward raising the system's internal energy.

8. 11.83 marksNCERT Cl-11 Physics Part II, Ch11 Exercises, Q11.8

   A thermodynamic system is taken from an original state D to an intermediate state E by a linear process in which pressure falls from 600 N/m^2 at V = 2.0 m^3 (point D) to 300 N/m^2 at V = 5.0 m^3 (point E). Its volume is then reduced to the original value of 2.0 m^3 from E to F by an isobaric process at 300 N/m^2. Calculate the total work done by the gas from D to E to F.

   Hint. Split the path into its two legs and compute the area under each on the P-V diagram separately — D to E is a trapezoid (a linear P-V segment, not a named process), and E to F is a simple rectangle since pressure is constant.

   Step 1 — Work done D to E (linear segment, trapezoidal area). W_DE = average pressure x change in volume = [(600 + 300)/2] x (5.0 - 2.0) = 450 x 3.0 = 1350 J.

   Step 2 — Work done E to F (isobaric compression). W_EF = P x delta-V = 300 x (2.0 - 5.0) = 300 x (-3.0) = -900 J.

   Step 3 — Add the two legs. W_total = W_DE + W_EF = 1350 + (-900) = 450 J.

   ✦ The gas does a net 450 J of work over the full D-E-F path, since the expansion leg D-E contributes positive work while the compression leg E-F, though smaller in magnitude here, subtracts from it rather than adding.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part II textbook, Reprint 2026-27 (keph204.pdf, 18 pages, "CHAPTER ELEVEN"), cross-checked against the official CBSE curriculum 2026-27, Subject Code 042, Unit VIII. One end-of-chapter Exercises set (8 questions, 11.1-11.8) — the rationalised chapter dropped the longer exercise sets of earlier NCERT editions. CBSE's Unit VIII syllabus line matches the chapter content directly; no orphaned-exercise gap was found, unusually for this book.. Questions are referenced from the NCERT textbook for identification.