Calculate the work done when 1 text{ mole} of | vedclass

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Question

(A) For a reversible isothermal expansion of an ideal gas, the work done $w$ is given by the formula:$w = -nRT \ln\left(\frac{P_1}{P_2}\right)$Given:$

Vedclass · Accepted Answer

$-5.744$

Vedclass · Answer

(A) For a reversible isothermal expansion of an ideal gas, the work done $w$ is given by the formula:$w = -nRT \ln\left(\frac{P_1}{P_2}ight)$Given:$n = 1 	ext{ mol}$$R = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$$T = 300 	ext{ K}$$P_1 = 10 	ext{ bar}$$P_2 = 1 	ext{ bar}$Substituting the values:$w = -(1 	ext{ mol}) 	imes (8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}) 	imes (300 	ext{ K}) 	imes \ln\left(\frac{10}{1}ight)$$w = -2494.2 	imes \ln(10)$Since $\ln(10) \approx 2.303$:$w = -2494.2 	imes 2.303 \approx -5744.14 	ext{ J}$Converting to kilojoules:$w \approx -5.744 	ext{ kJ}$Thus, the correct option is $A$.

Reaction	$\Delta H \ (kJ \ mol^{-1}) / \Delta S \ (J \ K^{-1} \ mol^{-1})$
$I$	$+98.0, +14.8$
$II$	$-55.5, -84.6$
$III$	$+28.3, -17.0$
$IV$	$-40.5, +24.6$
$V$	$+34.7, 0.0$

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