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?

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Solution

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

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Q. 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?

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Q. The entropy change involved in the isothermal reversible expansion of

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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?

The correct option is B Δs=nRlnv2v1=2.303nRlogv2v1=2.303×5mol×(8.314JK−1mol−1)log100L10L=95.73JK−1

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