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

Vedclass · Accepted Answer

$1729$

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(A) For an isothermal reversible compression 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}$$P_1 = x 	ext{ kPa}$$P_2 = 2x 	ext{ kPa}$Substituting the values:$W = -(1 	ext{ mol}) 	imes (8.314 	ext{ J K}^{-1} 	ext{mol}^{-1}) 	imes (300 	ext{ K}) 	imes \ln\left(\frac{x}{2x}ight)$$W = -2494.2 	imes \ln(0.5)$$W = -2494.2 	imes (-0.693)$$W \approx 1729 	ext{ J}$Since the gas is compressed, work is done on the system, hence the value is positive.

$1 \text{ mole}$ of an ideal gas is compressed isothermally and reversibly from an initial pressure $x \text{ kPa}$ to a final pressure $2x \text{ kPa}$ at $300 \text{ K}$. Find the work done $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$. (in $\text{ J}$)

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