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!

The Pressure of a liquid in equilibrium?

  1. Jul 9, 2016 #1
    • Homework posted in wrong forum, so no template
    The pressure of a liquid in equilibrium, is equal to the pressure of it's vapour, or to the sum of vapour pressure plus atmospheric pressure?
    My doubt starts from this problem:

    At 293 K and 1 atm, the vapour pressure of water is 565.8 Pa. Calculate the vapour pressure of water, when the total pressure is 2 atm, knowing that the molar volume of water is 1.8*10^-5 m^3/mol.

    The formula that I'm supposed to use is the one relative to the chemical potential at equilibrium between to phases, at constant temperature:

    V(liq)dP= RT d(lnp) in which P stands for the pressure on the liquid, and p the vapour pressure.
    Integrating the two quantities the resulting formula is ln(p'/p) = V(liq)*(P-p)/(RT) (p' stands for the new vapour pressure)

    Now the solution to the problem on the book is:

    ln(p'/565.8) = [1.8*10^-5*(202650 - 565.8)]/(R*T)

    My question is, why there is the subtraction between the final total pressure on the liquid and the initial vapour pressure, instead doing 202650 - (101325 + 565.8), in which (101325 + 565.8) is the initial total pressure? What am I getting wrong?
    Thank you in advance ;)
     
  2. jcsd
  3. Jul 9, 2016 #2
    This problem, and the solution in the book, is strange at many levels. If you look up the equilibrium vapor pressure of pure water at 293 C, you find that the value is 2340 Pa. If the system contains air, so that the total pressure is 1 atm., it is difficult to see how the partial pressure of the water vapor in the gas would be only 565.8 Pa. The partial pressure should have been very close to 2340 Pa.

    I also agree with your comment about how the effect of the total pressure on the fugacity of the liquid water should have been calculated. In fact, the 568 should not even have been included, and it should just have been 202650-101325.
     
  4. Jul 9, 2016 #3
    In fact I don't know what to think about this exercise. Anyway regarding the concept, I have another question about your answer, if you don't mind. If the right answer is with the subtraction 202650-101325, why 101325 is taken as the initial total pressure? If the problem doesn't say clearly for example that " the total pressure is 1 atm" but instead says the vapour pressure at 1 atm is 2340 Pa, Shouldn't the total initial pressure on the liquid be the sum of the vapour pressure and the atmospheric pressure? Or it is usually intended as a partial pressure, the vapour pressure?
     
  5. Jul 9, 2016 #4
    I would interpret it as the total pressure being 1 atm and the partial pressure of water vapor being 2340 Pa.

    In my mind another issue is this: if they are applying a non-ideality correction to the liquid, why aren't they also applying a non-ideality correction to the vapor.

    Chet
     
  6. Jul 9, 2016 #5
    I think that's my fault, I didn't mention that the book before introducing the formula expresses clearly that it is going to consider the vapor as an ideal gas.
    Anyway, thank you very much for your help!
     
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: The Pressure of a liquid in equilibrium?
  1. Equilibrium Pressure (Replies: 2)

  2. Pressure Equilibrium (Replies: 14)

Loading...