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The equation of state for a gas is given by PV = nTR + $α$ V where $n$ is the no. of moles and $α$ a positive constant. The initial pressure and temperature of one mole of the gas contained in a cylinder is $P_{0} and T_{0}$ respectively. The work done by the gas when its temperature doubles
isobarically will be

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$\frac{P_{o} T_{o} R}{P_{o} + α}$

none of these

$\frac{P_{0} T_{0} R}{P_{0} - α}$

$P_{0} T_{0} R \ln 2$

answer is A.

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Detailed Solution

$∆ W = P ∆ V [PV = μRT + αV]$
$P_{o} [\frac{μR ∆ T}{P_{o} - α}] = ∴ P ∆ V = μRT + α ∆ T$

or $∆ V = \frac{(μR ∆ T)}{(P - α)}$

i.e., $∆ W = \frac{P_{0} {RT}_{0}}{P_{0} - α}$ [ $μ = 1$ ]

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