1. The problem statement, all variables and given/known data A sample of 1.00 mol H20(g) is condensed isothermally and reversibly to liquid water at 100°C. The standard enthalpy of vaporization of water at 100°C is 40.656 kJ mol-1. Find w, q, change in internal energy, and change in enthalpy for this process. 2. Relevant equations 3. The attempt at a solution So since this is an isothermal reversible compression, change in temperature is equal to 0, so change in internal energy should also be equal to 0. Because deltaH = deltaU + delta(pV), and for an ideal gas pV = nRT so deltaH = 0 + delta(nRT) = 0, change in enthalpy should also be equal to 0. HOWEVER the solutions were quite different. Apparently deltaH = -standard enthalpy of vapourization, q = deltaH, w = -(external pressure) x deltaV, and deltaU = q + w. What am I missing here? Why are the changes in internal energy and enthalpy not zero if the change in temperature is zero? I know this is probably a really basic thermodynamics question but I just don't get it!