A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volume $V_{0}$ , in which an ideal gas is contained under the same pressure $P_{0}$ and is always at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas $η$ times compared to that of the other by slowly moving the piston?

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Solution

Let the agent move the piston as shown.
In equilibrium position,
$P_{1} A + F_{a g e n t} = P_{2} A$
$F_{a g e n t} = (P_{2} - P_{1}) A$
Work done by the agent
$F_{a g e n t} d x = (P_{2} - P_{1}) A \times d x = (P_{2} - P_{1}) d V$
Applying $P V =$ constant for two parts, we have
$P_{1} (V_{0} + A x) = P_{0} V_{0}$ and $P_{2} (V_{0} - A x) = P_{0} V_{0}$
$P_{1} = \frac{P_{0} V_{0}}{(V_{0} + A x)}$ and $P_{2} = \frac{P_{0} V_{0}}{(V_{0} - A x)}$
$∴ P_{2} - P_{1} = \frac{P_{0} V_{0} (2 A x)}{V_{0}^{2} - A^{2} x^{2}} = \frac{2 P_{0} V_{0} V}{V_{0}^{2} - V^{2}}$
When the volume of the left end is $η$ times the volume of right end,
we have
$(V_{0} + V) = η (V_{0} - V)$
$V = (\frac{η - 1}{η + 1}) V_{0} . . . . (2)$
The work done is given by
$W = \int_{0}^{V} (P_{2} - P_{1}) d V = \int_{0}^{V} \frac{2 P_{0} V_{0} V}{(V_{0}^{2} - V^{2})} d V = - P_{0} V_{0} [l n (V_{0}^{2} - V^{2})]_{0}^{V}$
$= - P_{0} V_{0} [l n (V_{0}^{2} - V^{2}) - l n V_{0}^{2}] = - P_{0} V_{0} [l n {V_{0}^{2} - {(\frac{η - 1}{η + 1})}^{2} V_{0}^{2}} - l n V_{0}^{2}]$
$= - P_{0} V_{0} [l n {4 η / (η + 1)^{2}}] = P_{0} V_{0} l n [\frac{(η + 1)^{2}}{4 η}]$ .

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