The entropy change involved in the isothermal reversible expansion of $2$ moles 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:

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

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Solution

The correct option is D $38.3 J {m o l}^{- 1} K^{- 1}$
Given,
A mole of an ideal gas $= 2$

Initial volume $(V_{1}) = 10 d m^{3}$

Final volume $(V_{2}) = 100 d m^{3}$

To find entropy change in the isothermal process.

$Δ S = n R l n (\frac{V_{2}}{V_{1}})$ or $Δ S = n R l n (\frac{P_{1}}{P_{2}})$

$Δ S = 2 \times 8.314 \times l n (\frac{100}{10})$

$Δ S = 38.28 \sim 38.3 \frac{J}{m o l K}$

$Δ S = 38.3 \frac{J}{m o l K}$

Hence, the correct option is $D$

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

The correct option is D 38.3Jmol−1K−1Given, A mole of an ideal gas =2Initial volume (V1)=10dm3Final volume (V2)=100dm3To find entropy change in the isothermal process.ΔS=nRln(V2V1) or ΔS=nRln(P1P2)ΔS=2×8.314×ln(10010)ΔS=38.28∼38.3Jmol KΔS=38.3JmolKHence, the correct option is D

The entropy change involved in the isothermal reversible expansion of $2$ moles 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: