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Entropy for a Reversible Process

1 mole of an ...

Question

1 mole of an ideal gas is allowed to expand isothermally at $27^{o} C$ until its volume is tripled. Calculate $△ S_{s y s}$ and $△ S_{u n i v}$ under the reversible expansion.

△ S_{s y s} = 9.1 J K^{- 1} m o l^{- 1}

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△ S_{u n i v} = △ S_{s y s} = 9.1 J K^{- 1} m o l^{- 1}

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△ S_{s u r r} = - 9.1 J K^{- 1} m o l^{- 1}

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△ S_{u n i v} = 0

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Solution

The correct option is D $△ S_{u n i v} = 0$
(i) In isothermal reversible process:
$△ S = \frac{q_{r e v}}{T}$
$q = - W = 2.303 R T l o g \frac{V_{2}}{V_{1}}$
$= 2.303 \times 8.314 \times 300 l o g 3 = 2740.6 J m o l^{- 1}$
$△ S_{s y s t e m} = \frac{q_{r e v}}{T} = \frac{2740.6}{300} = 9.135 J K^{- 1} m o l^{- 1}$
$△_{s u r r} = - △ S_{s y s t e m} = - 9.135 J K^{- 1} m o l^{- 1}$
$△ S_{u n i v} = 0$

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