Thermodynamics

JEE Main level, single correct, medium

49 of 190

Question 49 of 190

If one mole of an ideal gas at (P1,V1)\left({P}_{1},{V}_{1}\right) is allowed to expand reversibly and isothermally (AA to BB) 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→C).\left(B\rightarrow C\right). Then it is restored to its initial state by a reversible adiabatic compression (CC to AA). The net workdone by the gas is equal to:

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Answer: (D) RT[ln⁡(2)−12(γ−1)]RT\left[\ln \left(2\right)-\dfrac{1}{2\left(\gamma -1\right)}\right]

A−B=A-B= isothermal process
WAB=P1V1ln⁡[2V1V1]=P1V1ln⁡(2){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→B-C\rightarrow Isochoric process
WBC=0{W}_{BC}=0
C−A→C-A\rightarrowAdiabatic process
WCA=P1V1−P14×2V11−γ=P1V1[1−12]1−γ=P1V12(1−γ){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)}
Wnet=WAB+WBC+WCA{P1V1=RT}{W}_{net}={W}_{AB}+{W}_{BC}+{W}_{CA}\left\{{P}_{1}{V}_{1}=RT\right\}
=P1V1ln⁡(2)+0+P1V12(1−γ)={P}_{1}{V}_{1}\ln \left(2\right)+0+\dfrac{{P}_{1}{V}_{1}}{2\left(1-\gamma \right)}
Wnet=RT[ln⁡(2)−12(γ−1)]{W}_{net}=RT\left[\ln \left(2\right)-\dfrac{1}{2\left(\gamma -1\right)}\right]

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