What is the entropy change involved in the isothermal expansion of 5 mol of ideal gas from a volume of 10L to 100L at 300 K?

- 95.7 J K^{- 1}

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+ 95.7 k J K^{- 1}

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28.72 k J K^{- 1}

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- 28.72 k J K^{- 1}

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Solution

The correct option is B $+ 95.7 k J K^{- 1}$
$Δ s = n R l n \frac{v_{2}}{v_{1}} = 2.303 n R l o g \frac{v_{2}}{v_{1}} = 2.303 \times 5 m o l \times (8.314 J K^{- 1} m o l^{- 1}) l o g \frac{100 L}{10 L} = 95.73 J K^{- 1}$

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