Question:

Which of the following statements/relationships is not correct in thermodynamic changes ?

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Use the sign convention in which work done on the system is positive; an expanding gas does work, so w is negative.
Updated On: Oct 31, 2026
  • \(\Delta U = 0\) (isothermal reversible expansion of a gas)
  • \(w = -nRT \ln \frac{V_2}{V_1}\) (isothermal reversible expansion of an ideal gas)
  • \(w = nRT \ln \frac{V_2}{V_1}\) (isothermal reversible expansion of an ideal gas)
  • For a system of constant volume heat involved directly changes to internal energy.
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The Correct Option is C

Solution and Explanation

For an isothermal process of an ideal gas \(\Delta U = 0\) (U depends only on T), so statement (1) is acceptable.
Reversible isothermal expansion: \(w = -\int_{V_1}^{V_2} P\,dV = -nRT\ln\frac{V_2}{V_1}\) (work done on the system, IUPAC convention). Hence (2) is correct.
Since \(V_2 \gt V_1\), \(w\) must be negative for expansion, so \(w = +nRT\ln\frac{V_2}{V_1}\) in (3) has the wrong sign.
At constant volume \(w = 0\), so \(\Delta U = q_V\): statement (4) is correct.
Therefore the incorrect statement is (3).
Answer: (3) \(w = nRT \ln \frac{V_2}{V_1}\) (isothermal reversible expansion of an ideal gas)
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