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 ...... .

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Correct Answer

(A) \(RT \left(\ln 2-\frac{1}{2(\gamma-1)}\right)\)

Step-by-step Solution

Detailed explanation

\(A-B=\) 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

\(W _{ CA }=\frac{ P _{1} V _{1}-\frac{ P _{1}}{4} \times 2 V _{1}}{1-\gamma}=\frac{ P _{1} V _{1}\left[1-\frac{1}{2}\right]}{1-\gamma}=\frac{ P _{1} V _{1}}{2(1-\gamma)}\)

\(W _{ net }= W _{ AB }+ W _{ BC }+ W _{ CA } \quad\left\{ P _{1} V _{1}= RT \right\}\)

\(= P _{1} V _{1} \ln (2)+0+\frac{ P _{1} V _{1}}{2(1-\gamma)}\)

\(W _{\text {net }}= RT \left[\ln (2)-\frac{1}{2(\gamma-1)}\right]\)

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