Two samples A and B of the same gas have equal volumes and pressuresa. THe gas in sample A is expanded isotherm ally to duble ist volume and the gas in B is expanded adiabat ically to double its volume. If the work done by the gas is the same for the two cases, show that $γ$ satisfies the equation $1 - 2^{1 - γ}$ = $(γ - 1)$ In2

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Solution

$P_{1}$ = Initial pressure,

$V_{1}$ = Initial volume

$P_{2}$ = Final pressure,

$V_{2}$ = Final volume

Given, $V_{2} = 2 V_{1}$

Isothermal work done

= $n R T_{1} I n \frac{V_{2}}{V_{1}}$

Adiabatic work done

= $\frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$

Given that work done at both cases are same.

Hence,

$n R T_{1} I n = \frac{V_{2}}{V_{1}}$

= $\frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$ ....(1)

An adiabatic process,

$P_{2} = P_{1} {(\frac{V_{2}}{V_{1}})}^{γ} = P_{1} {(\frac{1}{2})}^{γ}$

From the equation (1)

$n R T_{1} I n_{2} = \frac{P_{1} V_{1} (1 - \frac{1}{2^{γ}} .2)}{γ - 1}$

and $n R T_{1} = P_{1} V_{1}$

So In 2 = $\frac{1 - \frac{1}{2^{γ}} .2}{γ - 1}$

or $(γ - 1) I n 2 = 1 - 2^{1 - γ}$

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