The entropy change involved in the isothermal | vedclass

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Question

(A) The entropy change for an isothermal reversible process is given by the formula: $\Delta S = n R \ln \frac{V_2}{V_1}$ Substituting the given value

Vedclass · Accepted Answer

$38.3$

Vedclass · Answer

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

The entropy change involved in the isothermal reversible expansion of $2 \, \text{mole}$ of an ideal gas from a volume of $10 \, dm^3$ to a volume of $100 \, dm^3$ at $27 \, ^oC$ is : .............. $J \, K^{-1}$

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