The work done in ergs for the reversible expa | vedclass

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(B) For a reversible isothermal expansion,the work done is given by $W = 2.303 \, nRT \log \frac{V_2}{V_1}$.Given: $n = 1 \, 	ext{mole}$,$T = 25 + 27

Vedclass · Accepted Answer

$298 	imes 10^7 	imes 8.314 	imes 2.303 \log \, 2$

Vedclass · Answer

(B) For a reversible isothermal expansion,the work done is given by $W = 2.303 \, nRT \log \frac{V_2}{V_1}$.Given: $n = 1 \, 	ext{mole}$,$T = 25 + 273 = 298 \, K$,$V_1 = 10 \, L$,$V_2 = 20 \, L$.Since $R = 8.314 \, J \, K^{-1} \, mol^{-1}$ and $1 \, J = 10^7 \, 	ext{ergs}$,$R = 8.314 	imes 10^7 \, 	ext{ergs} \, K^{-1} \, mol^{-1}$.Substituting the values: $W = 2.303 	imes 1 	imes (8.314 	imes 10^7) 	imes 298 	imes \log \frac{20}{10}$.$W = 298 	imes 10^7 	imes 8.314 	imes 2.303 \log \, 2 \, 	ext{ergs}$.

The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of $10 \ L$ to $20 \ L$ at $25 \, ^oC$ is

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