ThermodynamicsHard

Question

An ideal gas with the adiabatic exponent $\gamma$goes through a process P = Po – αV, where Po and α are positive constants and V is the volume. At what volume will the gas entropy have the maximum value?

Options

A.$\frac{\gamma P_{o}}{\alpha(\gamma - 1)}$
B.$\frac{\gamma P_{o}}{\alpha(\gamma + 1)}$
C.$\frac{\alpha P_{o}}{\gamma + 1}$
D.$\frac{\alpha P_{o}}{\gamma - 1}$

Solution

$dS = \frac{n.C_{V,m}.dT + P.dV}{T}$

For maximum entropy, $\frac{dS}{dV} = 0$

$\text{Or, }n.C_{V,m}\frac{dT}{dV} + P = 0(1) $$$\text{Now, }P = \frac{RT}{V} = P_{0} - \alpha V \Rightarrow \frac{dT}{dV} = \frac{1}{R}\left( P_{0} - 2\alpha V \right)(2)$$

From (1) and (2),

$1 \times \frac{R}{\gamma - 1} \times \frac{1}{R}\left( P_{0} - 2\alpha V \right) + \left( P_{0} - \alpha V \right) = 0 $$$\therefore V = \frac{\gamma.P_{0}}{\alpha(\gamma + 1)}$$

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