Finding ΔHvap of water from graph of ln (p of H2O in atm) versus 1/T


by icecream1023
Tags: Δhvap, 1 or t, graph, versus, water
icecream1023
icecream1023 is offline
#1
Dec12-11, 03:52 PM
P: 1
1. From a lab experiment, I measured the volume of trapped air at various temperatures. I know I did calculations correctly up to the point that I drew the linear relationship between ln (p of H2O in atm) to 1/T (in Kelvin). My textbook says the slope of this graph is supposed to equal -ΔHvap/R, but the first thing that's confusing me is that I don't know what the units are and the second thing that's confusing me is just that the numbers don't work out. The slope of the line in this graph is -5158.73, but I don't understand how units fit into this. As I've said, the y-axis is ln (p of H2O in atm) and the x-axis is 1/T (in Kelvin).



2. I don't...know what to put in this section.



3. If the slope is equal to -ΔHvap/R, then...
-5158.73 = -ΔHvap/R
ΔHvap = 5158.73 * 0.08206 (Again, I have no idea where my units are!!)
ΔHvap = 423.3 *something*


I appreciate any help; thank you!
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cheme101
cheme101 is offline
#2
Dec13-11, 12:47 AM
P: 7
You're on the right track, you just need to get your units right. Your slope (-ΔH/R) has to be in K because your x-coordinates are in 1/K; their product needs to be dimensionless to agree with ln(P).

With consistent units:

lnP = -(H/R)*T
[unitless]= -([J/mol] / [J/mol K]) * (1/K)

So if you use R=8.314 J / mol K,
you should get -ΔHvap = 8.314 * -5158.73 J/mol


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