Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Dec 12, 2011 #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!
  2. jcsd
  3. Dec 13, 2011 #2
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook