n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes -
$\mathrm{n}$ mole of a perfect gas undergoes a cyclic process $\mathrm{ABCA}$ (see figure) consisting of the following processes -
$\mathrm{A} \rightarrow \mathrm{B}$ : Isothermal expansion at temperature $\mathrm{T}$ so that the volume is doubled from $\mathrm{V}_{1}$ to $\mathrm{V}_{2}=2 \mathrm{~V}_{1}$ and pressure changes from $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$. $\mathrm{B} \rightarrow \mathrm{C}$ : Isobaric compression at pressure $\mathrm{P}_{2}$ to initial volume $\mathrm{V}_{1}$. $\mathrm{C} \rightarrow \mathrm{A}$ : Isochoric change leading to change of pressure from $\mathrm{P}_{2}$ to $\mathrm{P}_{1}$. Total workdone in the complete cycle $\mathrm{ABCA}$ is -
Correct Option: 4,
$A \rightarrow B=$ isotheraml process
$\mathrm{B} \rightarrow \mathrm{C}=$ isobaric process
$C \rightarrow A=$ isochoric process
also, $V_{2}=2 V_{1}$
work done by gas in the complete cycle $\mathrm{ABCA}$ is
$\Rightarrow \mathrm{W}=\mathrm{W}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{CA}} \quad \ldots(1)$
$\Rightarrow \mathrm{W}_{\mathrm{CA}}=0$, as isochoric process
$\Rightarrow \mathrm{W}_{\mathrm{AB}}=2 \mathrm{P}_{1} \mathrm{~V}_{1} \ln \left(\frac{\mathrm{v}_{2}}{\mathrm{~V}_{1}}\right)=2 \mathrm{nRT} \ln (2)$
$\Rightarrow W_{B C}=P_{2}\left(V_{1}-V_{2}\right)=P_{2}\left(V_{1}-2 V_{1}\right)=-P_{2} V_{1}=-n R T$
$\Rightarrow$ Now put the value of $w_{A B}, w_{B C}$ and $w_{C A}$ in equation, we get
$\Rightarrow w=2 n R T \ln (2)-n R T+0$
$\Rightarrow w=\mathrm{nRT}[2 \ln (2)-1]$
$\Rightarrow w=\mathrm{nRT}\left[\ln (2)-\frac{1}{2}\right]$