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Thermodynamics — Previous Year Questions (Class 11 Physics)

Thermodynamics explores heat, internal energy, and work. Master the first law and heat engines to excel in board examinations.

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TL;DR: Thermodynamics explores heat, internal energy, and work. Master the first law and heat engines to excel in board examinations.

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Aug 5, 2026

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Thermodynamics explores heat, internal energy, and work. Master the first law and heat engines to excel in board examinations.

Thermodynamics — Previous Year Questions with Solutions

Q (2023, 2 marks): A system absorbs 500 J of heat and performs 200 J of work. Calculate the change in internal energy of the system.

Answer: Using the first law of thermodynamics:
ΔU = Q - W
where Q is heat absorbed and W is work done by the system.
ΔU = 500 - 200 = 300 J
Final Answer: ΔU = 300 J

Q (2022, 3 marks): An ideal gas undergoes an isothermal process. Prove that the work done by the gas equals the heat absorbed.

Answer: For an isothermal process, temperature is constant, so ΔT = 0.
For an ideal gas, internal energy depends only on temperature: ΔU = nC_v ΔT
Since ΔT = 0, we have ΔU = 0.
Using the first law of thermodynamics:
Q = ΔU + W
Q = 0 + W
Q = W
Therefore, in an isothermal process, heat absorbed equals work done by the gas.
Final Answer: Proved that Q = W in isothermal process

Q (2023, 3 marks): Calculate the work done by an ideal gas during isobaric expansion from 1 L to 5 L at a constant pressure of 2 atm.

Answer: Given: V₁ = 1 L, V₂ = 5 L, P = 2 atm (constant)
For isobaric process: W = P(V₂ - V₁)
W = 2 × (5 - 1) = 2 × 4 = 8 atm·L
Converting to Joules: 1 atm·L = 101.325 J
W = 8 × 101.325 ≈ 810.6 J
Alternatively: W = 8 × 100 = 800 J (using approximation 1 atm·L ≈ 100 J)
Final Answer: W ≈ 810.6 J or 800 J

Q (2021, 5 marks): Define specific heat capacity and explain why the specific heat of a gas at constant pressure (C_p) is greater than at constant volume (C_v).

Answer: Specific heat capacity is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K.
For a gas at constant volume (C_v): All heat goes into increasing internal energy, W = 0, so Q = ΔU.
For a gas at constant pressure (C_p): Heat increases both internal energy and performs work against external pressure.
C_p = C_v + R (for ideal gas)
Since R > 0, we have C_p > C_v.
Physical reason: At constant pressure, the gas expands during heating, so work is done by the gas, requiring more heat input for the same temperature rise.
Final Answer: C_p > C_v because additional heat is needed for expansion work at constant pressure

Q (2022, 3 marks): A Carnot engine operates between two temperature reservoirs at 500 K and 300 K. Calculate its efficiency.

Answer: For a Carnot engine, efficiency is:
η = 1 - (T_cold/T_hot)
where T_hot = 500 K and T_cold = 300 K
η = 1 - (300/500)
η = 1 - 0.6 = 0.4 = 40%
Final Answer: η = 40% or 0.4

Q (2023, 5 marks): A heat engine absorbs 1000 J of heat from a hot reservoir and rejects 600 J to a cold reservoir. Calculate (i) work output, (ii) efficiency, (iii) coefficient of performance if it operates as a heat pump.

Answer: (i) Work output:
W = Q_in - Q_out = 1000 - 600 = 400 J
(ii) Efficiency (as heat engine):
η = W/Q_in = 400/1000 = 0.4 = 40%
(iii) Coefficient of performance (as heat pump):
COP = Q_in/W = 1000/400 = 2.5
(This means 2.5 units of heat are pumped for every unit of work input)
Final Answer: Work = 400 J, Efficiency = 40%, COP = 2.5

Frequently Asked Questions

What is the second law of thermodynamics?

The entropy of an isolated system always increases (or remains constant for reversible processes). Alternatively: Heat flows spontaneously from a hotter to a colder body, not vice versa. Or: No heat engine can operate at 100% efficiency.

Can the internal energy of an ideal gas change in an isothermal process?

No. In an isothermal process, temperature remains constant. Since internal energy of an ideal gas depends only on temperature (U = nC_v T), internal energy remains constant. Any heat absorbed is entirely converted to work done by the gas.

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