Posted by Linda on Monday, December 8, 2008 at 3:14pm.
Have you tried using the Clausius-Clapeyron equation to find delta H vap? You have P1 and P2, and T1 and T2. Or is that how you found 192 J/g?
And I don't see a starting point for temperature. If you are to determine total heat required from some T to vaporization, you must have both delta H vap and a starting T.
I found 32.54 kJ/mol for delta H vap (on the Internet).
thank you!!!
Secondly, how is it that you solve this type of question, I have no clue in finding a way to start the problem.
thanks
Do you have a temperature from which to start? Said another way, are you to determine the total heat from some temperature up to and including the vaporization OR are you to determine only the heat from the vaporization?
I am to find how much heat is required to vaporize 3.50L of at its normal boiling point?
such that the normal boiling point: 77 Degrees Celsius
and the following data are provided,
Normal melting point, -23 Degrees Celsius
density of liquid, 1.59g/mL
Delta H (FUSION): 3.28 kJ mol^-1
vapor pressure at 25 Degrees Celsius, 110 Torr
Thanks.
If you need only the heat required to vaporize it, then most of the data provided is not useful (heat fusion, etc).
First, convert 3.50 L to grams using the density listed.
I looked up the heat of vap on the web and found 32.54 kJ/mol. Change that to J/gram
32.54 kJ/mol x (1 mol/molar mass CCl4) x (1000 J/kJ) = heat vap in J/g.
Now
mass CCl4 x heat vap in J/g = ??
Check my thinking. Check my work.
Yes!!!
Thank you very much for guiding through this question!
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