For a reversible expansion of 5 , text{mol} o | vedclass

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(C) The formula for the change in entropy $(\Delta S)$ for a reversible isothermal expansion of an ideal gas is given by: $\Delta S = nR \ln(\frac{V_2

Vedclass · Accepted Answer

$95.73$

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(C) The formula for the change in entropy $(\Delta S)$ for a reversible isothermal expansion of an ideal gas is given by: $\Delta S = nR \ln(\frac{V_2}{V_1})$.Given values are: $n = 5 \, 	ext{mol}$,$R = 8.314 \, J \, K^{-1} \, 	ext{mol}^{-1}$,$V_1 = 8 \, 	ext{dm}^3$,$V_2 = 80 \, 	ext{dm}^3$.Substituting the values: $\Delta S = 5 	imes 8.314 	imes \ln(\frac{80}{8})$.$\Delta S = 5 	imes 8.314 	imes \ln(10)$.Since $\ln(10) \approx 2.303$,we have: $\Delta S = 5 	imes 8.314 	imes 2.303$.$\Delta S \approx 95.73 \, J \, K^{-1}$.

For a reversible expansion of $5 \, \text{mol}$ of an ideal gas at $27 \, ^\circ\text{C}$ from a volume of $8 \, \text{dm}^3$ to $80 \, \text{dm}^3$,the change in entropy is ...... $J \, K^{-1}$.

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