A kettle containing 1kg of boiling water is heated until complete evaporation. Calculate q, w, "delta"U and "delta"H for this process. Suppose that the water vapor is an ideal gas and molar "delta"H of vaporisation at 373K = 40.6kJ/mol.
H = U + PV
"delta"U = W + q
The Attempt at a Solution
number of moles = 1000g/(18.02g/mol) = 55.49mol
q = (molar "delta"H of vaporisation) x number of moles = 40.6x10^3J/mol x 55.49mol = 2.25x10^6J
U = H - PV
"delta"U = "delta"H - [("delta"P)(V) + ("delta"V)(P)] = "delta"H - [0 + nR"delta"T] = "delta"H
(since there is no pressure or temperature variation, "delta"U = "delta"H? I know pressure is constant, but is temperature constant too during this phase change?)
Therefore, "delta"U = 2.25x10^6J
W = "delta"U - q = 0J
I feel like I'm missing something in the calculation of "delta"U. Any help is greatly appreciated. Thanks.