Question

what is the delta u when 1.00mol of liquid water vaporizes @ 100C? The heat of the vaporization, delta h, of water @ 100C is 40.66kj/mol

Answer 1

heat= moleswater*Hv = 1*40.66kj/mol

Answer 2

the answer in the back of the book says 37.56 KJ

Answer 3

Pretty sure Bob's wrong here, Yesenia's right.
The equation for finding delta u: delta u = q + w
q = heat, w = work done by system.
The question says q = 40.66, so we need to find the value of w.
w = -P(delta V)
P = pressure, delta V = change in volume
You'd have to assume STP (standard temperature and pressure), which means P = 1.01*10^5 Pa.
You'd also be able to find delta V with this assumption, because according to Avogadro's Law, 1 mol of gas at STP has a volume of 22.41 L. This is applicable, because the liquid water vaporizes into a gas.
But the question says this experiment is done at 100 C (373.15 K), which is not standard temperature (273.15 K).
So you'll have to pull in Charles' Law, which states (Vi/Ti) = (Vf/Tf); meaning the initial volume*initial temperature = final volume*final temperature
Initial volume = 22.41 L (you got this from assuming STP and Avogadro's Law), Initial temperature = 273.15 K (standard temperature), Final volume = Vf (unknown), and Final temperature = 373.15 K (the temperature at which the experiment was conducted)
(22.41 L)(273.15 K) = (Vf)(373.15 K), Vf = 30.6 L
Now you can go back to: w = -P(delta V)
*You also have to convert the liters unit to m^3, which is done by: 30.6 L/1000 = 0.0306 m^3
Substitue values: w = -(1.01*10^5)(0.0306 m^3), w = -3092 J
Now the original equation: delta U = q + w
*q's value is given in kJ, so w has to be converted as well: -3092 J/1000 = -3.09 kJ
Substitute values: delta U = 40.66 + (-3.09), delta U = 37.6 kJ