Thermodynamics

JEE Main level, single correct, medium

165 of 190

Question 165 of 190

The equation of state for a gas is given by PV=nRT+αV\text{PV}=\text{nRT}+\alpha \text{V}, where n is the number of moles and α\alpha is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are T0 and P0 respectively. The work done by the gas when its temperature doubles isobarically will be :

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Answer: (C) P0T0RP0−α\dfrac{{\text{P}}_{\text{0}}{T}_{\text{0}}R}{{\text{P}}_{\text{0}}-\alpha }

PV=nRT+αV\text{PV}=\text{nRT}+\alpha \text{V}

P0dVdT=R+αdVdT\dfrac{{\text{P}}_{0}\text{dV}}{\text{dT}}=\text{R}+\alpha \dfrac{\text{dV}}{\text{dT}}

(P0−α)dVdT=R\left({\text{P}}_{0}-\alpha \right)\dfrac{\text{dV}}{\text{dT}}=\text{R}

∴dV=RdTP0−α∴\text{dV}=\dfrac{\text{RdT}}{{\text{P}}_{0}-\alpha }

∴W=∫P0dV=∫T02T0P0RP0−αdT∴\text{W}=\int {P}_{0}\text{dV}=\int _{{\text{T}}_{0}}^{2{\text{T}}_{0}}\dfrac{{\text{P}}_{0}\text{R}}{{\text{P}}_{0}-\alpha }\text{dT}

=P0T0RP0−α=\dfrac{{\text{P}}_{0}{\text{T}}_{0}\text{R}}{{\text{P}}_{0}-\alpha }

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