Thermodynamics

JEE Main level, single correct, medium

92 of 190

Question 92 of 190

The pressure and volume of an ideal gas are related as PV32=KP{V}^{\dfrac{3}{2}}=K (Constant). The work done when the gas is taken from state A(P1,V1,T1)A\left({P}_{1},{V}_{1},{T}_{1}\right) to state B(P2,V2,T2)B\left({P}_{2},{V}_{2},{T}_{2}\right) is :

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Answer: (A) 2(P1V1−P2V2)2({P}_{1}{V}_{1}-{P}_{2}{V}_{2})

For PVx=P{V}^{x}= constant, work done by the gas is given by,
W=nRΔT1−x=P2V2−P1V11−xW=\dfrac{nR\Delta T}{1-x}=\dfrac{{P}_{2}{V}_{2}-{P}_{1}{V}_{1}}{1-x}.
Here x=32x=\dfrac{3}{2}, therefore we can write
∴W=P2V2−P1V1−12∴W=\dfrac{{P}_{2}{V}_{2}-{P}_{1}{V}_{1}}{-\dfrac{1}{2}}
=2(P1V1−P2V2)=2\left({P}_{1}{V}_{1}-{P}_{2}{V}_{2}\right)

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