Question

Two samples A and B of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, then $\gamma$ (specific heat) satisfies the equation

• A. $1-2^{1-\gamma }=(\gamma -1)\text{ln}2$
• B. $1-2^{\gamma -1}=(1-\gamma )\text{ln}2$
• C. $2^{\gamma -1}=\text{ln}2$
• D. $1-2^{1-\gamma }=\gamma \text{ln}2$

Answer 1

The correct option is A $1-2^{1-\gamma }=(\gamma -1)\text{ln}2$

Work done $=nR{T}_1\text{ln}\left(\frac{V_2}{V_1}\right)$ for isothermal process,
work done $=nR\left(\frac{T_1-T_2}{\gamma -1}\right)$ for adiabatic change.
From question,
$nR{T}_1\text{ln}\left(\frac{V_2}{V_1}\right)=nR\left(\frac{T_1-T_2}{\gamma -1}\right)$
$\because \frac{T_1}{T_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma -1}$
$\therefore \text{ln}2(\gamma -1)=(1-\frac{T_2}{T_1})$
$\therefore \text{ln}2(\gamma -1)=1-\frac{1}{2^{\gamma -1}}$
$\therefore \text{ln}2(\gamma -1)=1-2^{1-\gamma }$

Why this question? Problems comparing isothermal and adiabatic work done are generally asked in JEE.