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(D) For an isothermal process,the work done $BY$ the gas is given by $W_{by} = \int_{V_i}^{V_f} P \, dV$. Since $PV = nRT$,we have $P = nRT/V$. Integr

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$W = -nRT \ln(P_f/P_i)$

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(D) For an isothermal process,the work done $BY$ the gas is given by $W_{by} = \int_{V_i}^{V_f} P \, dV$. Since $PV = nRT$,we have $P = nRT/V$. Integrating this,$W_{by} = nRT \int_{V_i}^{V_f} \frac{1}{V} dV = nRT \ln(V_f/V_i)$. By Boyle's Law,$P_i V_i = P_f V_f$,so $V_f/V_i = P_i/P_f$. Thus,$W_{by} = nRT \ln(P_i/P_f) = -nRT \ln(P_f/P_i)$. The work done $ON$ the gas is the negative of the work done by the gas: $W_{on} = -W_{by} = -nRT \ln(V_f/V_i) = nRT \ln(P_f/P_i)$.

An ideal gas is subjected to an isothermal expansion such that its volume changes from $V_i$ to $V_f$ and pressure from $P_i$ to $P_f$. The work done on the gas is:

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