The change in entropy for a reversible isothermal expansion of $n$ moles of an ideal gas from volume $V_1$ to $V_2$ at temperature $T$ is:

1. A $nR\ln(V_2/V_1)$ Correct answer
2. B $nR(V_2-V_1)$
3. C $nR/T$
4. D $nRT$

Solution

$\Delta S = \int dQ_{rev}/T = nR\ln(V_2/V_1)$ for isothermal expansion.

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Second Law and Entropy

This question is part of PrepWiser's free JEE Main question bank. 15 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.

More Thermodynamics questions

• Easy MCQ

  The work done by a gas in an isobaric expansion from $V_1$ to $V_2$ at pressure $P$ is:

• Medium MCQ

  For an adiabatic process, $PV^{\gamma} = $ const. For a monatomic ideal gas, $\gamma$ equals:

• Easy MCQ

  In an isothermal process for an ideal gas:

• Medium MCQ

  According to the Stefan-Boltzmann law, the power radiated by a black body is proportional to:

• Easy MCQ

  Most good thermal conductors are also: