Q.Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 500 ml to a volume of 2 L at $25^\circ$C and normal pressure.

Step 1. Reversible isothermal work: $w_{rev}=-nRT\ln(V_f/V_i)$ (Section ~7.2.6, eq. 7.6); $n=2$ mol, $T=25+273=298$ K, $V_i=500$ mL $=0.5$ L, $V_f=2$ L.

Step 2. $V_f/V_i=2/0.5=4$; $\ln4=1.3863$.

Step 3. $w_{rev}=-2\times8.314\times298\times1.3863=-2\times8.314\times298\times1.3863$. Computing: $8.314\times298=2477.6$; $\times2=4955.3$; $\times1.3863\approx6870$ J. …

Identify this as a reversible isothermal expansion and apply $w_{rev}=-nRT\ln(V_f/V_i)$, conve …

• Forgetting to convert 500 mL to 0.5 L before taking the ratio $V_f/V_i$ …