Question

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

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

Answer 1

Answer: (A)

$A-B=$ isothermal process
$W_{AB}=P_1{V}_1\ln \left[\frac{2V_1}{V_1}\right]=P_1{V}_1\ln (2)$
$B-C\to$ Isochoric process
$W_{BC}=0$
$C-A\to$ Adiabatic process
$W_{CA}=\frac{P_1{V}_1-\frac{P_1}{4}\times 2V_1}{1-\gamma }$
$=\frac{P_1{V}_1\left[1-\frac{1}{2}\right]}{1-\gamma }=\frac{P_1{V}_1}{2(1-\gamma )}$
$W_{net}=W_{AB}+W_{BC}+W_{CA}$
$\left\{P_1{V}_1=RT\right\}$
$=P_1{V}_1\ln (2)+0+\frac{P_1{V}_1}{2(1-\gamma )}$
$W_{net}=RT\left[\ln (2)-\frac{1}{2(\gamma -1)}\right]$