Question

$n$ mole of a perfect gas undergoes a cyclic process $ABCA$ (see figure) consisting of the following processes –

$A\to B$ : Isothermal expansion at temperature $T$ so that the volume is doubled from $V_1$ to $V_2=2V_1$ and pressure changes from $P_1$ to $P_2$.

$B\to C$ : Isobaric compression at pressure $P_2$ to initial volume $V_1$.

$C\to A$ : Isochoric change leading to change of pressure from $P_2$ to $P_1$.

Total work done in the complete cycle $ABCA$ is –

• A. $0$
• B. $nRT$$(\ln 2-\frac{1}{2})$
• C. $nRT$ $(ln2+\frac{1}{2})$
• D. $nRT\ln 2$

Answer 1

The correct option is B $nRT$$(\ln 2-\frac{1}{2})$



$A\to B=$ isothermal process

$B\to C=$ isobaric process

$C\to A=$ isochoric process

also, $V_2=2V_1$

Work done by gas in the complete cycle $ABCA$ is

$W=W_{AB}+W_{BC}+W_{CA}\dots \dots \dots .\left(1\right)$

$A\to B$ is an isothermal process, therefore work done is

$W_{AB}=nRT\ln \left(\frac{V_2}{V_1}\right)=nRT\ln 2$

$B\to C$ is an isobaric process, therefore work done is

$W_{BC}=P(V_2–V_1)$

$=P_2(V_1–2V_1)=-P_2{V}_1$

From gas equation, $PV=nRT\Rightarrow P=\frac{nRT}{V}$

Therefore $W_{BC}=-P_2{V}_1=-\frac{nRT}{V_2}{V}_1=-\frac{nRT}{2V_1}V1=-\frac{nRT}{2}$

$C\to A$ is an isochoric process, where work done is zero,

$\Rightarrow W_{CA}=0$

Substituting in equation $\left(1\right)$

$W=nRT\ln \left(2\right)-\frac{nRT}{2}+0$

$W=nRT\left(\ln \right(2)-\frac{1}{2})$

Hence , option $\left(D\right)$ is correct.