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(A) For an isothermal reversible compression,the work done $w$ is given by the formula: $w = -nRT \ln\left(\frac{P_2}{P_1}\right)$.Given values: $n =

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$1.729$

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(A) For an isothermal reversible compression,the work done $w$ is given by the formula: $w = -nRT \ln\left(\frac{P_2}{P_1}ight)$.Given values: $n = 1 \ mol$,$T = 300 \ K$,$P_1 = x \ bar$,$P_2 = 2x \ bar$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Substituting these values into the formula:$w = -(1 \ mol) 	imes (8.314 \ J \ K^{-1} \ mol^{-1}) 	imes (300 \ K) 	imes \ln\left(\frac{2x}{x}ight)$.$w = -8.314 	imes 300 	imes \ln(2)$.Using $\ln(2) \approx 0.693$:$w = -2494.2 	imes 0.693 \approx -1728.5 \ J$.Converting to $kJ$: $w \approx -1.729 \ kJ$.The magnitude of work done on the system is $1.729 \ kJ$.

Calculate the work done if $1 \ mole$ of a certain gas is compressed isothermally and reversibly at $300 \ K$ from an initial pressure $x \ bar$ to a final pressure $2x \ bar$ $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$. (in $kJ$)

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