1. Not finding help here? Sign up for a free 30min 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!

Van de Waals fluid in Free energy, Enthalpy representations

  1. Feb 10, 2013 #1
    Compute the coefficient of expansion α in terms of P and V....

    1. The problem statement, all variables and given/known data

    Compute the coefficient of expansion α in terms of P and V for an ideal Van der Waals
    gas

    2. Relevant equations

    (p+a/v^2)(v-b)=RT

    3. The attempt at a solution

    Is this as simple as solving for a? How would I go about eliminating T? I believe I have to take a derivative.
     
    Last edited: Feb 10, 2013
  2. jcsd
  3. Feb 11, 2013 #2
    Re: Compute the coefficient of expansion α in terms of P and V....

    The coefficient of thermal expansion is defined as:

    [tex]\alpha=\frac{1}{v}(\frac{\partial v}{\partial T})_p[/tex]
     
    Last edited: Feb 11, 2013
  4. Feb 11, 2013 #3
    Re: Compute the coefficient of expansion α in terms of P and V....

    ah yes, I should have looked that up myself I assumed it was the a in the formula. Since it seems I cannot isolate v in this equation I cannot explicitly find [tex](\frac{\partial v}{\partial T})_p[/tex] I tried looking for some nifty maxwell's relations but I cannot find any that would be useful.. any helpful hints? and thank you for the reply!
     
  5. Feb 11, 2013 #4
    Who says you have to do it explicitly?
     
  6. Feb 11, 2013 #5
    Ok bare with me I'm not the brightest. So are you implying I should compute

    ∂/∂T(pv-pb+a/v-ba/v^2=RT)

    and get something like p∂v/∂T-a/v^2(∂v/∂T)+ba/v^3(∂v/∂T)=R∂T/∂T

    Then factor and get ∂v/∂T= R/(p-a/v^2+ba/v^3)

    then


    α=(1/v)(∂v∂T)p = (1/v)R/(p-a/v^2+ba/v^3)

    it's in terms of p,v at least.. is this correct?
     
  7. Feb 11, 2013 #6
    Looks OK, except for the omission of a factor of 2 in the ba term. If I were you, I would try playing with the final equation a little bit to see if I could combine it with the original equation in some way to manipulate it into a simpler form. If you don't feel like doing this, that's OK. Your answer is fine as it is. Nice job.
     
  8. Feb 11, 2013 #7
    Pretty awesome stuff Chestermiller I appreciate it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Van de Waals fluid in Free energy, Enthalpy representations
  1. Van Der-Waals forces. (Replies: 0)

Loading...