Calculate the entropy change in JK^{-1} for t | vedclass

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(D) The formula for entropy change in a reversible isothermal expansion is: $\Delta S = 2.303 	imes n 	imes R 	imes \log(\frac{V_2}{V_1})$Given: $n

Vedclass · Accepted Answer

$95.73$

Vedclass · Answer

(D) The formula for entropy change in a reversible isothermal expansion is: $\Delta S = 2.303 	imes n 	imes R 	imes \log(\frac{V_2}{V_1})$Given: $n = 5 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$V_1 = 8 \ dm^3$,$V_2 = 80 \ dm^3$.Substituting the values: $\Delta S = 2.303 	imes 5 	imes 8.314 	imes \log(\frac{80}{8})$$\Delta S = 2.303 	imes 5 	imes 8.314 	imes \log(10)$Since $\log(10) = 1$,$\Delta S = 2.303 	imes 5 	imes 8.314 = 95.73 \ J \ K^{-1}$.

Calculate the entropy change in $JK^{-1}$ for the reversible expansion of $5 \ mol$ of an ideal gas at $27 \ ^oC$ from $8 \ dm^3$ to $80 \ dm^3$.

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