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, then $γ$ (speciﬁc heat) satisﬁes the equation

1 - 2^{1 - γ} = (γ - 1) l n 2

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1 - 2^{γ - 1} = (1 - γ) l n 2

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2^{γ - 1} = l n 2

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1 - 2^{1 - γ} = γ l n 2

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Solution

The correct option is A $1 - 2^{1 - γ} = (γ - 1) l n 2$
Work done $= n R T_{2} l_{n} (\frac{V_{2}}{V_{1}})$ for isothermal process.
Work done $= n R (\frac{T_{2} - T_{1}}{γ - 1})$ for Adiabatic process.
from question,
$n R T_{2} l n (\frac{V_{2}}{V_{1}}) = n R (\frac{T_{2} - T_{1}}{γ - 1})$
$∵ \frac{T_{2}}{T_{1}} = {(\frac{V_{2}}{V_{1}})}^{y - 1}$
$∴ l n 2 (γ - 1) = (1 - \frac{T_{1}}{T_{2}})$
$∴ l n 2 (γ - 1) = 1 - \frac{1}{2^{γ - 1}}$
$∴ l n 2 (γ - 1) = 1 - 2^{1 - γ}$

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The correct option is A 1−21−γ=(γ−1)ln2Work done =nRT2ln(V2V1) for isothermal process. Work done =nR(T2−T1γ−1) for Adiabatic process. from question, nRT2ln(V2V1)=nR(T2−T1γ−1) ∵T2T1=(V2V1)y−1 ∴ln2(γ−1)=(1−T1T2) ∴ln2(γ−1)=1−12γ−1 ∴ln2(γ−1)=1−21−γ