Harry S.

asked • 07/28/19

What is Delta U when 1.0 mol of liquid water vaporizes at 100 degrees C? The heat of vaporization of water at 100 degrees C is 44.6 kJ/mol.

the book says the answer is 37.56 KJ. I've tried approaching the problem in more than one way I haven't gotten this answer. Thank you for any help you can give.

J.R. S.

tutor
It would be useful if you included the various approaches you have taken. Did you use ∆U = w + P∆V? Did you include the pressure-volume work being done by the system on the surroundings?
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07/29/19

Harry S.

I tried the approach you mentioned. This is the only information the book gives
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07/30/19

1 Expert Answer

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Mark O. answered • 07/29/19

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Hi Harry,


I think your book is wrong. The heat of vaporization of water is Lv = 2256 X 103 J./kg. The mass of a mole of water is m = 18.0153 X 10-3 kg.


Q = mLv = (18.0153 X 10-3 kg)(2256 X 103 J./kg) = 4.064 X 104 J = 40.64 kJ

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Tella G.

Hey Mark, I actually think you're wrong here. The textbook answer is correct. Here's the process: 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 All of the equations I've used in my explanation are included within the textbook. I'm learning chemistry along with the textbook, and I haven't ever seen the equation that you've listed there Mark, which may be the reason why you get a different answer.
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08/14/21

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