Q.Two moles of an ideal gas at a temperature of $400\ \text{K}$ expand isothermally and reversibly from a volume of $5\ \text{L}$ to a volume of $15\ \text{L}$. Calculate the work done by the gas. (Take $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$, $\ln 3 = 1.099$.)

An isothermal process is one in which the temperature of a thermodynamic system stays constant while its pressure and volume are free to change. Since $PV=\mu RT$ and $T$ is fixed, the equation of state becomes $PV = \text{constant}$, so if the gas moves from $(P_1,V_1)$ to $(P_2,V_2)$, then $P_1V_1=P_2V_2$; on a P-V diagram this traces a hyperbola called an isotherm, with a curve closer to the origin corresponding to a lower temperature. Since $U$ for an ideal gas depends only on $T$, and $T$ is constant, $\Delta U = 0$ throughout an isothermal process, so the first law reduces to $Q=W$: all the heat supplied is used entirely to do external work (isothermal expansion, $Q,W>0$), or all the work done on the gas during compression flows straight out as heat ($Q,W<0$). The work done during a quasi-static isothermal expansion from $V_i$ to $V_f$ is $W = \mu RT\ln\left(\dfrac{V_f}{V_i}\right),$ positive for expansion and negative for compression. Everyday examples include boiling/melting at a fixed temperature until the phase change is complete, and biological processes occurring at the body's constant 37°C.