Calculate the maximum work obtained when 0.75 mol of an ideal gas expands isothermally and reversible at 27°C from a volume of 15 L to 25 L.
For an isothermal reversible expansion of an ideal gas w = – nRT log V2 V1 = – 2.303 nRT log V2 V1 Putting n = 0.75 mol; V1 = 15 L; V2 = 25 L, T = 27 + 273 = 300 K R = 8.314 JK-1 mol-1. w = – 2.303 × 0.75 × 8.314 × 300 log 2515 = – 955.5 J.
