# Thermodynamic Proof

1. Nov 14, 2012

### cmmcnamara

1. The problem statement, all variables and given/known data

Show that for an isentropic compression/expansion process that $Pv^k=constant$

2. Relevant equations

The usual thermodynamic potentials, maxwell relations

3. The attempt at a solution

The solution I am arriving at follows that of the solutions manual, but there is a substitution they use in the proof that is non-intuitive to me.

$$s=s(P,v)$$
$$ds=\left(\frac{∂s}{∂P}\right)_vdP+\left(\frac{∂s}{∂v}\right)_Pdv$$

Since the process is isentropic, ds=0, and using maxwell substitutions we arrive at:

$$dP-\left(\frac{∂P}{∂v}\right)_sdv=0$$

This is where our proofs are the same, they then diverge by making the substitution, and dividing by pressure:

$$k=-\frac{v}{P}\left(\frac{∂P}{∂v}\right)_s$$
$$\frac{dP}{P}+k\frac{dv}{v}=0$$
$$lnP+klnv=constant$$
$$Pv^k=constant$$

I don't find their substitution very intuitive. They describe it as the isentropic expansion coefficient. I do not understand how they can treat it as a constant however in the differential equation. Likewise, I feel that by their solution, if k can be treated as a constant then I would expect the partial differential component could be treated as such as well and the portion I had completed could be used to arrive at an algebraically different equation that could be rearranged but I don't see it. Can anyone help me understand the reasoning behind all of this?

NOTE: Later as a part of the problem it asks to prove that k reduces to the ideal gas ratio, for an ideal gas. I also tried working it from that angle but to no avail.

2. Nov 14, 2012

### Yuu Suzumi

Are you sure they are claiming this more generally than for an ideal gas?
I remember that even for Van der Vaals Gases this expression has to be modified in a way similar to the equation of state!

3. Nov 14, 2012

### cmmcnamara

Yes I am sure it is. The latter part of the proof is shown to be proven through further maxwell relations coupled with the cyclic properties of partial derivatives. I just don't understand the choice made for k and setting it constant. It just seems a bit arbitrary to me.

4. Nov 15, 2012

### Yuu Suzumi

It has been a while since I last did thermodynamics. But why may I not write this:

$k \equiv -\frac{\frac{dp}{p}}{\frac{dV}{V}} = \frac{Vdp}{pdV}$

and then, using that I have an isentropic process:

$k = \frac{dH}{dU} = \frac{C_p}{C_v}$.

If this is still valid, then $k$ can not be constant in a real polyatomic gas. With varying temperature the number of "frozen" degrees of freedom in the gas will change!

5. Nov 19, 2012

### cmmcnamara

Sorry for my absence! Although I am not familiar with your first expression, the second I am familiar with as a definition for k. That ratio is constant for an ideal gas which is what the reduction should go to, but if that is the case I cannot see how they would allow it to be considered constant in the first portion of the proof.

6. Nov 19, 2012

Exactly.