# Reversible adiabatic expansion using Van der Waals' equation

1. May 6, 2012

### Kweh-chan

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)$\gamma$ = const.

and find the expression for the parameter $\gamma$ 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. May 6, 2012

### rude man

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.

3. Apr 29, 2014

### pjhaynes

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
4. Apr 29, 2014

### Staff: Mentor

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