Class XI · Chapter 11
Thermodynamics — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering the first and second laws, heat engines, and Carnot efficiency.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Why is the change in internal energy, ΔU, always zero over a complete cyclic thermodynamic process?
Answer
Internal energy is a state function — it depends only on the system’s current state, not on the path taken to reach it. Since a cyclic process returns the system to its exact starting state, ΔU over the full cycle is always zero.
Internal energy is a state function — it depends only on the system’s current state, not on the path taken to reach it. Since a cyclic process returns the system to its exact starting state, ΔU over the full cycle is always zero.
Derive Mayer’s relation, Cp − Cv = R, for one mole of an ideal gas.
Answer
At constant volume, all added heat raises internal energy: dQ = Cv dT = dU (no expansion work).At constant pressure, added heat both raises internal energy and does expansion work: dQ = Cp dT = dU + PdV. For one mole of ideal gas, PV = RT → PdV = RdT (at constant P).So Cp dT = Cv dT + RdT → Cp − Cv = R.
At constant volume, all added heat raises internal energy: dQ = Cv dT = dU (no expansion work).At constant pressure, added heat both raises internal energy and does expansion work: dQ = Cp dT = dU + PdV. For one mole of ideal gas, PV = RT → PdV = RdT (at constant P).So Cp dT = Cv dT + RdT → Cp − Cv = R.
Q3. (a) State the Kelvin-Planck and Clausius statements of the second law of thermodynamics. (b) A Carnot engine operates between reservoirs at 600 K and 300 K, absorbing 800 J of heat from the hot reservoir. Find the work done and the engine’s efficiency.
Answer
(a) Kelvin-Planck: No engine operating in a cycle can convert heat completely into work — some heat must always be rejected. Clausius: Heat cannot spontaneously flow from a colder body to a hotter one without external work being done.(b) η = 1 − Tc/Th = 1 − 300/600 = 0.5 = 50%.W = η × Qh = 0.5 × 800 = 400 J.
(a) Kelvin-Planck: No engine operating in a cycle can convert heat completely into work — some heat must always be rejected. Clausius: Heat cannot spontaneously flow from a colder body to a hotter one without external work being done.(b) η = 1 − Tc/Th = 1 − 300/600 = 0.5 = 50%.W = η × Qh = 0.5 × 800 = 400 J.
2NEET Style (Single-Correct MCQ)
Q4. In an isothermal process for an ideal gas, the change in internal energy is:
Solution
Internal energy of an ideal gas depends only on temperature; with temperature constant (isothermal), ΔU = 0, so Q = W entirely.
Internal energy of an ideal gas depends only on temperature; with temperature constant (isothermal), ΔU = 0, so Q = W entirely.
Q5. In an adiabatic process:
Solution
By definition, an adiabatic process has no heat exchange with the surroundings (Q = 0), so ΔU = −W entirely.
By definition, an adiabatic process has no heat exchange with the surroundings (Q = 0), so ΔU = −W entirely.
Q6. The efficiency of any real heat engine is always:
Solution
The second law guarantees some heat must always be rejected, so η = 1 − Qc/Qh is always strictly less than 1.
The second law guarantees some heat must always be rejected, so η = 1 − Qc/Qh is always strictly less than 1.
Q7. The coefficient of performance (COP) of a refrigerator can be:
Solution
Unlike engine efficiency, COP = Qc/W can exceed 1 — a refrigerator can move more heat energy than the work it consumes, which is why it’s more efficient than direct electrical heating.
Unlike engine efficiency, COP = Qc/W can exceed 1 — a refrigerator can move more heat energy than the work it consumes, which is why it’s more efficient than direct electrical heating.
3JEE Main Style
Q8. A gas absorbs 500 J of heat while doing 300 J of work on its surroundings. Find the change in its internal energy.
Solution
ΔU = Q − W = 500 − 300 = 200 J.
ΔU = Q − W = 500 − 300 = 200 J.
Q9. An isochoric (constant volume) process has:
Solution
W = ∫PdV = 0 when volume doesn’t change — so all heat added goes directly into internal energy: ΔU = Q.
W = ∫PdV = 0 when volume doesn’t change — so all heat added goes directly into internal energy: ΔU = Q.
Q10. A Carnot engine has an efficiency of 40%, operating with a hot reservoir at 500 K. Find the temperature of the cold reservoir.
Solution
0.4 = 1 − Tc/500 → Tc/500 = 0.6 → Tc = 300 K.
0.4 = 1 − Tc/500 → Tc/500 = 0.6 → Tc = 300 K.
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