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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 $( B \rightarrow C ) .$ 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 ...... .
$RT \left(\ln 2-\frac{1}{2(\gamma-1)}\right)$
$-\frac{ RT }{2(\gamma-1)}$
$0$
$RT \ln 2$
Solution
$A-B=$ isothermal process
$W _{ AB }= P _{1} V _{1} \ln \left[\frac{2 V _{1}}{ V _{1}}\right]= P _{1} V _{1} \ln (2)$
$B – C \rightarrow$ Isochoric process
$W _{ BC }=0$
$C – A \rightarrow$ Adiabatic process
$W _{ CA }=\frac{ P _{1} V _{1}-\frac{ P _{1}}{4} \times 2 V _{1}}{1-\gamma}=\frac{ P _{1} V _{1}\left[1-\frac{1}{2}\right]}{1-\gamma}=\frac{ P _{1} V _{1}}{2(1-\gamma)}$
$W _{ net }= W _{ AB }+ W _{ BC }+ W _{ CA } \quad\left\{ P _{1} V _{1}= RT \right\}$
$= P _{1} V _{1} \ln (2)+0+\frac{ P _{1} V _{1}}{2(1-\gamma)}$
$W _{\text {net }}= RT \left[\ln (2)-\frac{1}{2(\gamma-1)}\right]$
