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One mole of an ideal gas having initial volume $V$, pressure $2P$ and temperature $T$ undergoes a cyclic process $ABCDA$ as shown below : The net work done in the complete cycle is
zero
$\frac{1}{2} RT\,ln\,2$
$RT\,ln\,2$
$\frac{3}{2} RT\,ln\,2$
Solution
Wrok done, $\Delta \mathrm{W}=\mathrm{P} \Delta \mathrm{V}$
At constant pressure,
$\Delta \mathrm{W}=\mathrm{P}\left(\mathrm{V}_{\mathrm{t}}-\mathrm{V}_{\mathrm{i}}\right)=\mathrm{nR}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right)$
At constant temperature,
$\Delta \mathrm{W}=\mathrm{nRT} \ln \left(\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right)=\mathrm{nRT} \ln \left(\frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{V}_{\mathrm{f}}}\right)$
Therefore, work done for path $\mathrm{AB}, \mathrm{BC}, \mathrm{CD}$ and
$DA$ respectively will be.
$\Delta \mathrm{W}_{\mathrm{AB}}=1 \times \mathrm{R} \times(2 \mathrm{T}-\mathrm{T})=\mathrm{RT}$
$\Delta \mathrm{W}_{\mathrm{BC}}=1 \times \mathrm{R} \times 2 \mathrm{T} \ln \left(\frac{2 \mathrm{P}}{\mathrm{P}}\right)=2 \mathrm{RT} \ln 2$
$\Delta \mathrm{W}_{\mathrm{CD}}=1 \times \mathrm{R} \times(\mathrm{T}-2 \mathrm{T})=-\mathrm{RT}$
$\Delta W_{D A}=1 \times R \times T \ln \left(\frac{P}{2 P}\right)=R T \ln \left(\frac{1}{2}\right)$
Net work done in the complete cycle is, $\Delta \mathrm{W}=\Delta \mathrm{W}_{\mathrm{AB}}+\Delta \mathrm{W}_{\mathrm{BC}}+\Delta \mathrm{W}_{\mathrm{CD}}+\Delta \mathrm{W}_{\mathrm{DA}}$
$=\mathrm{RT}+2 \mathrm{RT} \ln 2-\mathrm{RT}+\mathrm{RT} \ln \left(\frac{1}{2}\right)$
$=2 \mathrm{RT} \ln 2+\mathrm{RT} \ln 1-\mathrm{RT} \ln 2$
$=2 \mathrm{RT} \ln 2-\mathrm{RT} \ln 2$
$(\because \ln 1=0)$
$=\mathrm{RT} \ln 2$
