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(C) For an isothermal reversible expansion,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \ \log\left(\frac{P_1}{P_2}\right)$.Given va

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$-5.74$

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(C) For an isothermal reversible expansion,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \ \log\left(\frac{P_1}{P_2}ight)$.Given values are: $n = 1 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$P_1 = 10 \ bar$,and $P_2 = 1 \ bar$.Substituting these values into the equation:$w = -2.303 	imes 1 	imes 8.314 	imes 300 	imes \log\left(\frac{10}{1}ight)$.Since $\log(10) = 1$,we have:$w = -2.303 	imes 8.314 	imes 300 	imes 1$.$w = -5744.14 \ J$.Converting to $kJ$,we get $w = -5.744 \ kJ \approx -5.74 \ kJ$.

Calculate work done in isothermal reversible expansion of $1 \ mol$ of an ideal gas from an initial pressure of $10 \ bar$ to a final pressure of $1 \ bar$ at $300 \ K$ $\left(R = 8.314 \ J \ K^{-1} \ mol^{-1}\right)$. (in $kJ$)

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