The work done in ergs for the reversible expa | vedclass

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Question

(B) For a reversible isothermal expansion of an ideal gas,the work done is given by the formula: $W = -2.303 nRT \log(V_2/V_1)$.Here,$n = 1 \ mol$,$T

Vedclass · Accepted Answer

$-298 	imes 10^7 	imes 8.31 	imes 2.303 	imes \log 2$

Vedclass · Answer

(B) For a reversible isothermal expansion of an ideal gas,the work done is given by the formula: $W = -2.303 nRT \log(V_2/V_1)$.Here,$n = 1 \ mol$,$T = 25^{\circ} C = 298 \ K$,$V_1 = 10 \ L$,and $V_2 = 20 \ L$.The gas constant $R$ in $erg \ K^{-1} \ mol^{-1}$ is $8.314 	imes 10^7 \ erg \ K^{-1} \ mol^{-1}$.Substituting the values: $W = -2.303 	imes 1 	imes (8.314 	imes 10^7) 	imes 298 	imes \log(20/10)$.$W = -2.303 	imes 8.314 	imes 10^7 	imes 298 	imes \log 2$.This matches the expression in option $B$.

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 $25^{\circ} C$ is :

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