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Thermal Physics problem: Van der Waals

  1. Nov 22, 2013 #1
    1. Find an expression that relates T and V in a process in which the entropy S is constant, by doing the following:

    a) From thermodynamic identity dS=(1/t)dU + (p/v)dV
    find the expression for dS as a function of T and V (and dT and dV).

    b) Integrate the expression found in (a) along the line of constant S.


    2. Relevant equations
    I tried to use the equation of state for a Van der Waals gas

    p=(NkT)/V-a(N^2/V^2)

    a is constant
    its energy as a function of temperature and volume is given by
    U(T,V)=2NkT-a(N^2/V)

    If you could give any hints that would be great!
     
    Last edited: Nov 22, 2013
  2. jcsd
  3. Nov 23, 2013 #2

    rude man

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    That equation is dimensionally inconsistent and therefore wrong.

    But it's just one letter aaway from right so is probably a typo. Anyway, fix it.

    Are you allowed to assume an ideal gas?
     
  4. Nov 24, 2013 #3
    I took a a snapshot of the problem to be more specific
     

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  5. Nov 24, 2013 #4

    TSny

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    As noted by rude man, the equation given for dS is incorrect. Since the equation is just a simple rearrangement of the first law (with dQ = T dS), you can see how to fix it.

    To proceed with the problem, you will need to use the expression for U and find it's differential dU in terms of dT and dV.
     
  6. Nov 24, 2013 #5

    rude man

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    OK.
    You're given U(T,V) and dS = 1/T dU + p/V dV
    so taking dU is just basic calculus. N, k and a are constants.

    Warning: I've carried this thru to dS = 0 in terms of T and V but I wind up with an expression that I can't integrate (tried separation of parts, couldn't do it). Someone else help!
     
    Last edited: Nov 24, 2013
  7. Nov 24, 2013 #6
    ok so thermodynamic identity --> dS=(1/T)dU + (p/T)dV
     
  8. Nov 24, 2013 #7

    TSny

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    Good.

    What does this expression for dS become after substituting for dU and simplifying?
     
  9. Nov 24, 2013 #8

    rude man

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    Hey, how come everyone except I caught the misstatement in the problem? Now I gotta start over ... :surprised:

    I hope at least U(T,V) is given correctly. Someone else want to check dimensions?

    Still can't seem to separate variables though ...
     
  10. Nov 24, 2013 #9
    ok dU=(∂U/∂T)dT + (∂U/∂V)dV --> dU = 2NKdT + aN^2(1/V^2)dV

    I plug in P and dU into dS = (1/T)dU + (P/T)dV

    I get...

    dS = (1/T)[2NkdT + a(N^2)/(V^2)dV] + (NKT/V - (aN^2)/(V^2))/T)dV

    please correct me if im wrong but since S is constant, then dS equals 0,
    therefore the lefthand side is 0, we seperate the varibales for dT and dV and simplify
    i get
    -2dT=(T/V)dV
    integrate both parts and solve for Vf --> Vf=Vi/16

    Thanks so much guys!!!
     
  11. Nov 24, 2013 #10

    rude man

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    Congartulations! Would you (or TSny or whoever) mind posting how you did the separation of variables? I couldn't do it. :blushing:

    EDIT: never mind, I hadn't noticed two terms canceling each other, leaving only 2 terms to be separated.

    Good show!!
     
    Last edited: Nov 24, 2013
  12. Nov 24, 2013 #11
    i forgot a step, -2dT = (T/V)dV --> (-2/T)dT=(1/V)dV
    but i seperated varibles by putting all T's and V's on opposite sides of the equation

    -(1/T)[2NkdT] = (1/T)[a(N^2)/(V^2) + NKT/V - (aN^2)/(V^2))/T)dV

    simplify a bit...
    2NkdT = (NKT/V)dV

    2dT = (T/V)dV

    (2/T)dT = (1/v)dV then integrate both sides ---> 2ln(Tf/Ti) = ln(Vf/Vi) , where Tf = 4Ti , 2ln(4) = ln(Vf/Vi)
    then solve for Vf
    =)
     
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