The work done in doubling the volume of 5 mol | vedclass

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(B) For an isothermal process,the work done $W$ is given by the formula: $W = nRT \ln\left(\frac{V_f}{V_i}ight)$.Given: $n = 5 \, 	ext{moles}$,$T =

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$14400$

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(B) For an isothermal process,the work done $W$ is given by the formula: $W = nRT \ln\left(\frac{V_f}{V_i}ight)$.Given: $n = 5 \, 	ext{moles}$,$T = 500 \, K$,$V_f = 2V_i$,and $R \approx 8.314 \, J/(mol \cdot K)$.Substituting the values: $W = 5 	imes 8.314 	imes 500 	imes \ln(2)$.Since $\ln(2) \approx 0.693$,we have $W = 5 	imes 8.314 	imes 500 	imes 0.693$.$W = 2500 	imes 8.314 	imes 0.693 \approx 14412 \, J$.Rounding to the nearest provided option,the work done is approximately $14400 \, J$.

The work done in doubling the volume of $5$ moles of an ideal gas at a constant temperature of $500 \, K$ is ....... $J$.

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