NCERT Solutions · Class 11 Physics · NCERT Physics Part I and II Part II · Chapter 11
Chapter 11: Thermodynamics (Thermodynamics)
Step-by-step solutions to all 8 exercises of Chapter 11, Thermodynamics (NCERT Class 11 Physics Part II, 2026-27): fuel use of a geyser, heat at constant pressure, why bodies need not reach the mean temperature, adiabatic compression, the first law along two paths, free expansion, rate of change of internal energy and work from a P–V diagram. All 8 questions are answered, with the key answer highlighted.
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First law: ΔQ=ΔU+ΔW (ΔW = work done by the system). Work W=∫PdV = area under the P–V curve. At constant pressure Q=nCpΔT; diatomic gas Cp=27R. Adiabatic: PVγ = constant, γ = 1.4 for diatomic gases. In free expansion Q = 0, W = 0, so ΔU = 0.
A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °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 × 10⁴ J/g?
Solution
Mass of water per minute =3000 g. Heat needed per minute =mcΔT=3000×4.2×50=6.3×105 J.
What amount of heat must be supplied to 2.0 × 10⁻² kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N₂ = 28; R = 8.3 J mol⁻¹ K⁻¹.)
Solution
n=2820=0.714 mol. Nitrogen is diatomic: Cp=27R=29.05J mol−1K−1.
Explain why (a) Two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have 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.
Solution
(a) Heat lost by one body equals heat gained by the other: m1c1(T1−T)=m2c2(T−T2). The final T is the mean only if the heat capacities m1c1 and m2c2 are equal.
(b) A liquid with high specific heat absorbs a large amount of heat for a small rise in its own temperature, so it carries heat away efficiently.
(c) While driving, friction with the road and flexing of the tyre heat the air inside. At (nearly) constant volume, pressure rises with temperature.
(d) A harbour town is next to the sea. Water has a high specific heat, and sea breezes and land breezes keep the town from getting very hot by day or very cold at night. A desert has no such moderating effect.
(a) Heat capacities may differ (b) high specific heat absorbs more heat (c) air heats up at constant volume (d) the sea moderates temperature
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?
Solution
The cylinder is insulated, so the compression is adiabatic: P1V1γ=P2V2γ, with γ = 1.4 for hydrogen.
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)
Solution
Adiabatic path: Q = 0, W (by system) = −22.3 J, so ΔU=Q−W=+22.3 J. ΔU depends only on the states A and B, so it is the same on the second path.
Second path: Q=9.35×4.19=39.2 J; W=Q−ΔU=39.2−22.3.
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?
Solution
(a) The gas doubles its volume at the same temperature (see (c)), so the pressure halves: 0.5 atm.
(b) Insulated: Q = 0. The gas expands into vacuum: W = 0. So ΔU = 0.
(c) For an ideal gas U depends only on temperature, so the temperature does not change.
(d) No. Free expansion is rapid and uncontrolled; the intermediate states are not equilibrium states and do not satisfy the gas equation. Only the final state lies on the P-V-T surface.
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13). Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F. [D: (2.0 m³, 600 N m⁻²), E: (5.0 m³, 300 N m⁻²), F: (2.0 m³, 300 N m⁻²).]
Solution
D → E (expansion): work = area of trapezium =2600+300×(5.0−2.0)=1350 J.
E → F (compression at 300 N m⁻²): W=300×(2.0−5.0)=−900 J.
Total =1350−900=450 J, the area of triangle DEF =21×3.0×300.
450 J
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