Isothermal Process

An isothermal expansion keeps temperature fixed, so for an ideal gas the work done is $W=nRT\ln \left(\frac{V_f}{V_i}\right)$, derived by integrating $P=nRT/V$ over volume at constant $T$. Substituting $n=2$, $R=8.31$, $T=300\text{ K}$, and $V_f/V_i=2$ gives $W=2(8.31)(300)\ln 2=4986\times 0.693\approx 3454\text{ J}$. The value 4986 J is the prefactor $nRT$ alone, forgetting the logarithmic factor. The value 2494 J corresponds to one mole instead of two, halving the answer incorrectly. The value 1727 J results from using $\ln 2$ on the one-mole figure, compounding the same error. The logarithmic form arises because pressure falls continuously as the volume grows, so the work cannot be a simple product but must be integrated across the changing pressure. Because temperature is constant, $$\begin{gathered} \Delta \\ cup=0 \end{gathered}$$ for the ideal gas, so by the First Law the heat absorbed equals this work, $Q=3454\text{ J}$. Physically the surrounding water bath feeds in exactly enough heat to replace the energy leaving as expansion work, keeping the temperature pinned at 300 K. As a magnitude check, the logarithm of two is about 0.69, so the work is roughly 69 percent of $nRT$, which matches 3454 J being a bit under 4986 J.