Two samples A and B, of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 2^1−γ= (γ − 1) ln2.

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Solution

Let,
Initial pressure of the gas = P₁
Initial volume of the gas = V₁
Final pressure of the gas= P₂
Final volume of the gas = V₂
Given, V₂ = 2 V₁, for each case.
In an isothermal expansion process,
work done $= n R T_{1} \ln \frac{V_{2}}{V_{1}}$
Adiabatic work done,
$W = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
It is given that same work is done in both cases.
So,
$n R T_{1} \ln (\frac{V_{2}}{V_{1}}) = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} . . . (1)$
In an adiabatic process,
$P_{2} = P_{1} {(\frac{V_{1}}{V_{2}})}^{γ} = P_{1} {(\frac{1}{2})}^{γ}$
From eq (1),
$n R T_{1} \ln 2 = \frac{P_{1} V_{1} (1 - \frac{1}{2^{γ}} \times 2)}{γ - 1}$
and nRT₁ = P₁V₁
$So, \ln 2 = \frac{1 - \frac{1}{2^{γ}} . 2}{γ - 1}$
Or (γ − 1) ln 2 = 1 − 2^1−γ

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