Isothermal Process

Question

Maintained in a water bath at constant temperature, two moles of an ideal gas expand slowly until the volume doubles at 300 K. Taking $R=8.31	ext{ J mol}^{-1}	ext{K}^{-1}$, what work does the gas do?

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An isothermal expansion keeps temperature fixed, so for an ideal gas the work done is $W = nRT\ln\!\left(\dfrac{V_f}{V_i}ight)$, derived by integrating $P = nRT/V$ over volume at constant $T$. Substituting $n = 2$, $R = 8.31$, $T = 300	ext{ K}$, and $V_f/V_i = 2$ gives $W = 2(8.31)(300)\ln 2 = 4986 	imes 0.693 \approx 3454	ext{ J}$. 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, $\Delta \cup = 0$ for the ideal gas, so by the First Law the heat absorbed equals this work, $Q = 3454	ext{ J}$. 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.

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