Question

A sample of 1.00 mol H2O(g) is condensed isothermally and reversibly to liquid water at 100ºC and 1 atm.. Find w, q, ΔU, and ΔH for this process. (The latent heat of vaporization, i.e., standard enthalpy of vaporization, of water at 100ºC is 40.7 kJ/mol at 1 atm.)

Answer 1

H = ΔH(cond) = -ΔH(vaP) = -1 mole*40.7kj/mole = -40.7 kj
since the condensation is done isothermally and external pressure is constant at 1 atm.
Hence q =q(at constant pressure) = ΔH = -40.7kj
w = -p(external)dv
where dv = V(liquid) - V(vapour) which almost equal to -V(vapour) since V(L)<<<V((vpour)
on the assumption that H2O(g) is a perfect gas V(VPour) = nRT/P
P= P(external)
W = -p(external)*nRT/p(external) = nRT = 1 *8.314*373 = 3.1*10^3 j =3.1Ki