The entropy change in the isothermal reversible expansion of 2 moles of an ideal gas from $10$ to $100$ L at $300$ K is:

42.3 J K^{- 1}

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35.8 J K^{- 1}

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38.3 J K^{- 1}

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32.3 J K^{- 1}

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Solution

The correct option is A $42.3 J K^{- 1}$
For an isothermal process $Δ U = 0$

According to the first law of thermodynamics

$Δ U = q + w$

$\Rightarrow q = - w = + 2.303 n R T l o g \frac{V_{f}}{V_{i}}$

$= + 2.303 \times 2 \times 8.314 l o g (\frac{100}{10}) T$

$= 38.29 T$

$Δ S = \frac{q}{T} = 38.29 J / K$

Hence option $C$ is correct.

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The entropy change in the isothermal reversible expansion of 2 moles of an ideal gas from 10 to 100 L at 300 K is:

The correct option is A 42.3 JK−1For an isothermal process ΔU=0According to the first law of thermodynamicsΔU=q+w⇒q=−w=+2.303 nRT logVfVi=+2.303×2×8.314 log(10010)T =38.29 TΔS=qT=38.29 J/KHence option C is correct.

The entropy change in the isothermal reversible expansion of 2 moles of an ideal gas from $10$ to $100$ L at $300$ K is: