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!

Homework Help: 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?

  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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted