Question

If one mole of an ideal gas at $(P_1,V_1,T)$ is allowed to expand reversibly and isothermally $(A$ to$B)$ its pressure is reduced to $\frac{1}{2}$ of original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of initial value $(B\to C)$. Then it is restored to its initial state by a reversible adiabatic compression $(C$ to $A)$. The net work done by the gas is equal to :

• A. $0$
• B. $-\frac{RT}{2(\gamma -1)}$
• C. $RT[\ln 2-\frac{1}{2(\gamma -1)}]$
• D. $RT\ln 2$

Answer 1

The correct option is C $RT[\ln 2-\frac{1}{2(\gamma -1)}]$

$AB\to$ Isothermal process,

$\therefore W_{\text{AB}}=nRT\ln 2=RT\ln 2$

$BC\to$ Isochoric process,
$\therefore W_{\text{BC}}=0$

$CA\to$ Adiabatic compression,

$\therefore W_{\text{CA}}=\frac{P_1{V}_1-\frac{P_1}{4}\times 2V_1}{1-\gamma }$

$=\frac{P_1{V}_1}{2(1-\gamma )}=\frac{RT}{2(1-\gamma )}$

So, total work done,
$W_{\text{ABCA}}=RT\ln 2+\frac{RT}{2(1-\gamma )}$
$=RT[\ln 2-\frac{1}{2(\gamma -1)}]$