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(B) Since the process occurs very slowly and the system remains in thermal equilibrium with the surroundings,it is an isothermal process.For an isothe

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$345$

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(B) Since the process occurs very slowly and the system remains in thermal equilibrium with the surroundings,it is an isothermal process.For an isothermal process,the work done $W$ is given by the formula:$W = \mu RT \ln\left(\frac{V_2}{V_1}ight)$Given:Number of moles $\mu = 0.2 \, mol$Temperature $T = 27^{\circ}C = 27 + 273 = 300 \, K$Universal gas constant $R = 8.3 \, J/(mol \cdot K)$Final volume $V_2 = 2V_1$,so $\frac{V_2}{V_1} = 2$Substituting the values:$W = 0.2 	imes 8.3 	imes 300 	imes \ln(2)$$W = 0.2 	imes 8.3 	imes 300 	imes 0.693$$W = 498 	imes 0.693 \approx 345 \, J$Thus,the work done is approximately $345 \, J$.

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