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Finding ΔHvap of water from graph of ln (p of H2O in atm) versus 1/T 
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#1
Dec1211, 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 yaxis is ln (p of H2O in atm) and the xaxis 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! 


#2
Dec1311, 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 xcoordinates 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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