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Reversible adiabatic expansion using Van der Waals' equation

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data
    A real gas obeys Van der Waals‟ equation, which for one mole of gas is

    (p + A/V2)(V-B) = RT

    and its internal energy is given by

    U = CvT - A/V

    where the molar heat capacity at constant volume, Cv , is independent of the
    temperature and pressure. Show that the relation between the pressure p and
    the volume V of the Van der Waals‟ gas during a reversible adiabatic
    expansion can be written as

    (p + A/V2)(V-B)[itex]\gamma[/itex] = const.

    and find the expression for the parameter [itex]\gamma[/itex] in terms of Cv and R .

    2. Relevant equations

    (p + A/V2)(V-B) = RT
    U = CvT - A/V
    Q= U + W

    3. The attempt at a solution

    There is already a given solution and method for this equation. I worked through this much:

    0 = U + W
    0 = dU + PdV

    dU = (dU/dT)dT - (dU/dV)dV = CvdT + (A/V2)dV

    0 = CvdT + (P + A/V2)dV = CvdT + RT/(V-B)
    ∫R/(V-B) dV = -∫Cv(dT/T)
    Rln(V-B) + Cvln(T) = const.
    ln(V-B)R + ln(T)CV = const
    ln[(V-B)R(T)CV] = const.
    (V-B)R(T)CV = const.


    I got stuck here and checked the method. My process was right, but according to it, the next line of work is:

    (V-B)R(RT)CV = const.

    I don't understand where this mystery R comes from. I've tried rearranging the ideal gas equation, and the first given equation to no avail. Could someone please explain how I get this R in the process?

    Thanks!
     
  2. jcsd
  3. May 6, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    R is the universal molargas constant and is part of the definition of a van der walal gas, just as it is for an ideal gas.

    I haven't worked this out completely, but I would:

    let p1 = p + A/v2
    v1 = v - B
    then p1v1 = RT.
    By your du = -dw and dT = (1/R)(p1dv1 + v1dp1) you can eliminate T, integrate by parts to get p1v1γ = constant.

    I think.
     
  4. Apr 29, 2014 #3
    I was too thinking the same.

    All I can think of, for a simple solution, is that you multiply both sides by R^Cv (which in itself is constant since Cv=1.5R when n=1)

    This is valid since the right side remains constant = constant times R^Cv

    The left side is now what we require.
     
    Last edited: Apr 29, 2014
  5. Apr 29, 2014 #4
    If you have

    (V-B)R(T)CV = const.

    and you multiply both sides of the equation by RCV, you get


    (V-B)R(RT)CV = (const.)(RCV) = New constant

    Chet
     
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