The entropy change in the isothermal reversib | vedclass

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Question

(C) For an isothermal reversible expansion,the entropy change is given by the formula: $\Delta S = n R \ln \frac{V_2}{V_1} = 2.303 n R \log \frac{V_2}

Vedclass · Accepted Answer

$38.3$

Vedclass · Answer

(C) For an isothermal reversible expansion,the entropy change is given by the formula: $\Delta S = n R \ln \frac{V_2}{V_1} = 2.303 n R \log \frac{V_2}{V_1}$.Given values: $n = 2 \, 	ext{mol}$,$V_1 = 10 \, L$,$V_2 = 100 \, L$,and $R = 8.314 \, J \, K^{-1} \, 	ext{mol}^{-1}$.Substituting the values: $\Delta S = 2.303 	imes 2 	imes 8.314 	imes \log \frac{100}{10}$.$\Delta S = 2.303 	imes 2 	imes 8.314 	imes \log(10)$.Since $\log(10) = 1$,we get $\Delta S = 2.303 	imes 2 	imes 8.314 	imes 1 = 38.297 \approx 38.3 \, J \, K^{-1}$.

The entropy change in the isothermal reversible expansion of $2 \, \text{moles}$ of an ideal gas from $10 \, L$ to $100 \, L$ at $300 \, K$ is $..... \, J \, K^{-1}$.

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