1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Standard Entropy calculation

  1. May 22, 2014 #1
    1. The problem statement, all variables and given/known data

    vapour pressures of a liquid have been measured and fit to the following equation:
    Log10 (mmHg) = -3571/T + 6.124
    The melting point has been determined to be 392.7 K.
    A Cp value given for the liquid is 250 J/mol K
    and theΔSvap is 117 J/mol K

    2. Relevant equations



    3. The attempt at a solution

    1.
    Calculate the standard entropy of the liquid at the melting point.


    2. Relevant equations



    3. The attempt at a solution

    T1= 392.7 K P1 = 0.8045 mmHg (from the equation, giving the vapour pressure of the LIQUID at the melting point).
    T2= 298.15 K P2= 1.052E-3 mmHg

    From the Clausius Clapeyron:

    ln(0.8045/1.052E-3)= ΔH/R*(1/298.15 K - 1/392.7 K)

    ΔH/R = 8221.89
    ΔH = 68361 J/mol

    ΔH/T = ΔS = 68361/392.7 = 174.07 J/mol K

    However the answer is about twice this value
     
  2. jcsd
  3. May 22, 2014 #2
    I'm getting different values for P1 and P2, based on the equation Log10 (mmHg) = -3571/T + 6.124.

    I find that equation particularly annoying because strictly speaking it makes no sense. What is the logarithm of a unit?
     
  4. May 22, 2014 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It's OK. The value 6.124 is specific to the assumption that the pressure is measured in mmHg. Different units, different constant. Likewise, the 3571 assume degrees Kelvin.
     
  5. May 23, 2014 #4
    Yes, thank you haruspex, it is Log of the Pressure in mmHG and assuming Kelvin for temperature units
     
  6. May 23, 2014 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Good, but I agree with dauto that there's something wrong with your calculation of P1 and P2. If you can't find an error, please post the details of your working.
     
  7. May 23, 2014 #6
    Just noticed an error in the formula I gave. The +6.124 is for the pressure in bar, when I gave it in mmHg.
    I will correct it to the given data in mmHg (+8.999)
     
  8. May 23, 2014 #7
    for some reason I cannot edit the formula above, it should be:

    Log10 P (mmHg) = -3571/T + 8.999
     
  9. May 23, 2014 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    OK, but I don't get your value for ΔH/R from that using your equations. I may well have made a mistake, but perhaps you could check yours.
     
  10. May 23, 2014 #9
    I rechecked it, I get 8222.53 instead of what is listed above, which doesn't change my final answer drastically
     
  11. May 23, 2014 #10
    Yes, I know that. Doesn't keep me from feeling annoyed. It's very poor form.
     
  12. May 23, 2014 #11

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    That is for a reason. It is discouraged in this forum to edit your posts after you received replies. This for the simple reason that the other posts will look strange and off-topic if you edited out mistakes from your original post. We prefer that you just post the corrections in a new post and leave the original post as it is.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Standard Entropy calculation
  1. Entropy calculation (Replies: 16)

  2. Entropy calculation (Replies: 9)

  3. Entropy calculation (Replies: 5)

Loading...