Question

Calculate the total work done by the gas in the process. (given $ln2=0.693)$

Answer 1

Total work done by gas, $W_{Total}=W_{AB}+W_{BC}+W_{CA}$
$W_{AB}=nRTln\frac{4V}{V}=2nRTln2=2PVln2$.
Also $P_A{V}_A=P_B{V}_B$ (As AB is an isothermal process)
or, $P_B=\frac{P_A{V}_A}{V_B}=\frac{PV}{4V}=\frac{P}{4}$.
In the step BC, the pressure remains constant. Hence the work done is,
$W_{BC}=\frac{P}{4}(V-4V)=-\frac{3PV}{4}$.
In the step CA, the volume remains constant and so the work done is zero. The net work done by the gas in the cyclic process is
$W=W_{AB}+W_{BC}+W_{CA}$
$=2PVln2-\frac{3PV}{4}+0$
Hence, the work done by the gas is $0.636PV$.