Calculate the entropy change in J , mol^{-1} | vedclass

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Question

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

Vedclass · Accepted Answer

$38.3$

Vedclass · Answer

(C) The formula for entropy change in an isothermal reversible process 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: $n = 2 \, mol$,$V_1 = 10 \, dm^3$,$V_2 = 100 \, dm^3$,$R = 8.314 \, J \, mol^{-1} K^{-1}$.Substituting the values: $\Delta S = 2.303 	imes 2 	imes 8.314 	imes \log \left( \frac{100}{10} ight)$.$\Delta S = 2.303 	imes 2 	imes 8.314 	imes \log(10)$.Since $\log(10) = 1$,$\Delta S = 2.303 	imes 2 	imes 8.314 = 38.29 \approx 38.3 \, J \, mol^{-1} K^{-1}$.

Calculate the entropy change in $J \, mol^{-1} K^{-1}$ for the isothermal reversible expansion of $2 \, mol$ of an ideal gas from a volume of $10 \, dm^3$ to $100 \, dm^3$ at $27 \, ^oC$.

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