note that H is the extensive (total, not per mol) enthalpy
Q =
H = n Hwhere H is the intensive (per mol) enthalpy
One kilogram of liquid water at 100° is vaporized to steam at 100° in a piston. (Equilibrium can be assumed). What procedure is needed to determine how much heat would be required and how much work is done in this process? If Q does not equal W, what accounts for the difference?
At constant pressure: dQ = dH
note that H is the extensive (total, not per mol) enthalpy
Q = H = n H
where H is the intensive (per mol) enthalpy
H = Hsensible + Hlatent
Hsensible = CpT
Since T=0, Hsensible=0
nH = nHlatent = heat of vaporization
In plain terms, the water is going from liquid water at 100° to gaseous water at 100°. There is no change in temperature, so all the change in heat must be due to the change in state from liquid to gas. This is the heat of vaporization.
Calculations:
nHvap = (1000g x 1 mol/18g) x
(40.6 kJ/mol)
= 2256 kJ
Q = 2256 kJ
This value of Q (2256 kJ) is greater than the value W (169.4 kJ).
The difference in Q and W is used to break intermolecular forces used to hold the water molecules together in the liquid.
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