The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of 10 $d m^{3}$ to a volume of 100 $d m^{3}$ at $27^{o} C$ is:

38.3 J m o l^{- 1} K^{- 1}

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35.8 J m o l^{- 1} K^{- 1}

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32.3 J m o l^{- 1} K^{- 1}

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42.3 J m o l^{- 1} K^{- 1}

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Solution

The correct option is A $38.3 J m o l^{- 1} K^{- 1}$
$Δ S = \frac{Q_{r e v}}{T}$
$Q_{r e v} = 2.303 n R T log \frac{V_{f}}{V_{i}}$
$Δ S = 2.303 n R log \frac{V_{f}}{V_{i}}$
$Δ S = 2.303 \times 2 \times 8.314 log \frac{100}{10}$
$Δ S = 38.294 \approx 38.3 J m o l^{- 1} K^{- 1}$

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The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of $10 d m^{3}$ to a volume of $100 d m^{3}$ at $27^{\circ} C$ is :

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The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27oC is:

The correct option is A 38.3 Jmol−1K−1ΔS=QrevTQrev=2.303nRTlogVfViΔS=2.303nRlogVfViΔS=2.303×2×8.314log10010ΔS=38.294≈38.3 Jmol−1K−1

The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of 10 $d m^{3}$ to a volume of 100 $d m^{3}$ at $27^{o} C$ is:

The correct option is A $38.3 J m o l^{- 1} K^{- 1}$
$Δ S = \frac{Q_{r e v}}{T}$
$Q_{r e v} = 2.303 n R T log \frac{V_{f}}{V_{i}}$
$Δ S = 2.303 n R log \frac{V_{f}}{V_{i}}$
$Δ S = 2.303 \times 2 \times 8.314 log \frac{100}{10}$
$Δ S = 38.294 \approx 38.3 J m o l^{- 1} K^{- 1}$