One mole of an ideal gas is expanded isotherm | vedclass

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(B) For an isothermal process,the change in entropy $\Delta S$ is given by the formula: $\Delta S = nR \ \ln \ (P_1 / P_2)$.Given that the gas is expa

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$5.76$

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(B) For an isothermal process,the change in entropy $\Delta S$ is given by the formula: $\Delta S = nR \ \ln \ (P_1 / P_2)$.Given that the gas is expanded to half of its initial pressure,$P_2 = P_1 / 2$,so $P_1 / P_2 = 2$.For $n = 1 \ mol$,the expression becomes: $\Delta S = 1 	imes R 	imes \ln \ (2)$.Substituting the values: $\Delta S = 8.314 	imes 0.693 = 5.76 \ J \ K^{-1} \ mol^{-1}$.

One mole of an ideal gas is expanded isothermally and reversibly to half of its initial pressure. $\Delta S$ for the process in $J \ K^{-1} \ mol^{-1}$ is $[\ln \ 2 = 0.693$ and $R = 8.314 \ J \ mol^{-1} \ K^{-1}]$

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