NEET · 16 September 2026
Isothermal expansion: which statements are true?
An ideal gas is enclosed in a cylinder fitted with a frictionless piston and kept in contact with a large heat reservoir at temperature T. The gas expands quasi-statically and isothermally from volume V to 2V. Read the following statements and choose the correct answer from the options given below.
Statements A–E
A. Throughout the expansion the gas remains in thermal equilibrium with the reservoir at temperature T.
B. The internal energy of the gas increases because the gas does positive work on the piston.
C. The work done by the gas during this expansion equals nRT ln 2.
D. Since Δ U = 0, the heat absorbed by the gas from the reservoir equals the work done by the gas.
E. If the same expansion from V to 2V occurred adiabatically, the reservoir would still supply the same amount of heat.
Answer options A–D
A. A, C and D only
B. A, B and C only
C. B, D and E only
D. A, C, D and E only
Reveal answer and explanation
Correct answer: option A — statements A, C and D only.
A is true in the ideal quasi-static reservoir setup stated here: the gas stays at the reservoir temperature T in the equilibrium limit. Isothermal alone means constant temperature; it does not by itself require reservoir contact.
B is false. For a fixed amount of ideal gas, internal energy depends only on temperature. Here T is constant, so ΔU = 0 despite positive work by the gas.
C is true. For this quasi-static expansion, Wby = p dV′ = nRT ln(2V/V) = nRT ln 2 > 0. Here n is the number of moles, R the gas constant and T the absolute temperature. This is not the work formula for every irreversible isothermal expansion.
D is true. Taking work done by the gas as positive, ΔU = Q − Wby, hence Q = Wby. This describes one expansion, not a complete cyclic heat engine.
E is false. Adiabatic means Q = 0, so the reservoir supplies no heat. A quasi-static adiabatic expansion doing positive work cools an ideal gas. In contrast, insulated free expansion of an ideal gas has Q = W = 0 and unchanged temperature.
Takeaway: Here isothermal gives ΔU = 0 and Q = W; adiabatic gives Q = 0.
This revisits the September 15 practice question with a clarified explanation. It does not mean the original explanation was correct or has been updated in the app.
Sources: OpenStax: First law · OpenStax: Thermodynamic processes · OpenStax: Adiabatic processes and free expansion
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