The entropy change for the expansion of 2 , t | vedclass

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(D) The formula for entropy change in an isothermal expansion is $\Delta S = nR \ln \left( \frac{V_2}{V_1} ight) = 2.303 	imes nR \log \left( \frac

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$9.2$

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(D) The formula for entropy change in an isothermal expansion is $\Delta S = nR \ln \left( \frac{V_2}{V_1} ight) = 2.303 	imes nR \log \left( \frac{V_2}{V_1} ight)$.Given: $n = 2 \, 	ext{mol}$,$R = 2 \, 	ext{cal/mol K}$,$V_1 = 2 \, 	ext{L}$,$V_2 = 20 \, 	ext{L}$.Substituting the values: $\Delta S = 2.303 	imes 2 	imes 2 	imes \log \left( \frac{20}{2} ight)$.$\Delta S = 2.303 	imes 4 	imes \log(10)$.Since $\log(10) = 1$,$\Delta S = 2.303 	imes 4 = 9.212 \, 	ext{cal/K}$.

The entropy change for the expansion of $2 \, \text{mol}$ of an ideal gas from $2 \, \text{L}$ to $20 \, \text{L}$ at a temperature of $27 \, ^oC$ is ...... $(R = 2 \, \text{cal/mol K})$

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