Reversible adiabatic expansion using Van der Waals' equation

[Kweh-chan]

Homework Statement

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 = C_vT - A/V

where the molar heat capacity at constant volume, C_v , 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 C_v and R .

Homework Equations

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

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 = C_vdT + (A/V²)dV

0 = C_vdT + (P + A/V²)dV = C_vdT + RT/(V-B)
∫R/(V-B) dV = -∫C_v(dT/T)
Rln(V-B) + C_vln(T) = const.
ln(V-B)^R + ln(T)^C_V = const
ln[(V-B)^R(T)^C_V] = const.
(V-B)^R(T)^C_V = 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)^C_V = 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!

 

[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 p₁v₁^γ = constant.

I think.

 

[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.

 

[Chestermiller]

(V-B)^R(T)^C_V = 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)^C_V = 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!

If you have

(V-B)^R(T)^C_V = const.

and you multiply both sides of the equation by R^C_V, you get


(V-B)^R(RT)^C_V = (const.)(R^C_V) = New constant

Chet