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Question

The equation of state for a gas is given by $P V = n R T + α V$ , where $n$ is the number of moles and $α$ is a positive constant. The initial pressure and temperature of 1 mole of the gas contained in a cylinder are $P_{0}$ and $T_{0}$ respectively. The work done by the gas when its temperature doubles isobarically is

W = \frac{P_{0} R T_{0}}{P_{0} - α}

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W = \frac{R T_{0}}{P_{0} - α}

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W = \frac{2 P_{0} R T_{0}}{P_{0} - α}

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W = \frac{P_{0} R T_{0}}{2 (P - α)}

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Solution

The correct option is A $W = \frac{P_{0} R T_{0}}{P_{0} - α}$
Work done for isobaric process,
$Δ W = P Δ V . . . . . . (I)$
From question,
$P V = n R T + α V$
$\Rightarrow P Δ V = n R Δ T + α Δ V$
[Here, $P, n, R and α$ are constants]
$\Rightarrow P Δ V - α Δ V = n R Δ T$
$\Rightarrow Δ V = \frac{n R Δ T}{P - α}$
Putting the value of $Δ V$ in Eq.(I),
$\Rightarrow Δ W = P Δ V = \frac{P n R Δ T}{P - α}$
[Given $n = 1 mole, P_{i} = P_{0}, T_{i} = T_{0}$ ]
$Δ T = 2 T_{0} - T_{0} = T_{0}$
$\Rightarrow Δ W = P Δ V = \frac{P_{0} R T_{0}}{P_{0} - α}$
Hence, option (a) is the correct answer.

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Standard XII Physics

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The equation of state for a gas is given by PV=nRT+αV, where n is the number of moles and α is a positive constant. The initial pressure and temperature of 1 mole of the gas contained in a cylinder are P0 and T0 respectively. The work done by the gas when its temperature doubles isobarically is