Three moles of an ideal gas expand isothermally and reversibly from $90$ to $300 L$ at $300 K$ . Calculate the entropy change for the process.

10.01 J m o l^{1} K^{1}

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20.01 J m o l^{1} K^{1}

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30.01 J m o l^{1} K^{1}

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None of the above

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Solution

The correct option is A $10.01 J m o l^{1} K^{1}$
For an isothermal reversible process, the work is,
$W = - n R T ln \frac{V_{2}}{V_{1}}$
Therefore, the work per mole is
$W = - R T ln \frac{V_{2}}{V_{1}} = - 8.316 \times 300 ln \frac{300}{90} = - 3003 J m o l^{- 1}$
From the first law, $Δ U = Q + W$ or in terms of molar quantities, $Δ U_{m} = Q_{m} + W_{m}$ . Since $Δ U_{m} = 0$ for this
isothermal process, that means the heat per mole is
$Q_{m} = - W_{m} = 3003 J K^{- 1}$
Finally, since $Δ S = \frac{d Q_{r e v}}{T}$
for an isothermal process, the molar entropy change is
$Δ S = \frac{d Q_{m}}{T} = \frac{3003}{300} = 10.01 J m o l^{- 1} K^{- 1}$

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