If one mole of an ideal gas at ( P _ 1 , V _ 1 ) is allowed to...

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 $\left(B \to C\right) .$ 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:

If one mole of an ideal gas at ( P _ 1 , V _ 1 ) is allowed to expand reversibly and isothermally ( A to B ) its

Options

A $0$
B $RTln\left(2\right)$
C- \dfrac{RT}{2 $\left(\gamma - 1\right)}$
D $RT \left[\ln \left(2\right) - \dfrac{1}{2\left(\gamma - 1\right)}\right]$

Correct answer

D. $RT \left[\ln \left(2\right) - \dfrac{1}{2\left(\gamma - 1\right)}\right]$

Step-by-step solution

$A - B =$ isothermal process

${W}_{AB} = {P}_{1}{V}_{1}\ln \left[\dfrac{2{V}_{1}}{{V}_{1}}\right] = {P}_{1}{V}_{1}\ln \left(2\right)$

$B - C \to$ Isochoric process

${W}_{BC} = 0$

$C - A \to$ Adiabatic process

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

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

$= {P}_{1}{V}_{1}\ln \left(2\right) + 0 + \dfrac{{P}_{1}{V}_{1}}{2\left(1 - \gamma\right)}$

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

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