# Proof that PV^gamma is constant in a reversable adiabatic condition.

1. Feb 28, 2008

### Jennifer Lyn

I've been working on this proof, and I just can't get it backwards or forwards, so I must be going about it wrong or missing something. I'm on a re-do for the homework, because my first attempt was completely wrong, so here I go with the second.

Any guidance would be appreciated.

1. The problem statement, all variables and given/known data
The relationship between pressure and volume of an ideal gas is expressed as pv=constant in a reversable isothermal condition. Show that the relationship between pressure and volume of the same gas is expressed as pV^gamma=constant in a reversible adiabatic condition where gamma=Cp,m/Cv,m.

2. Relevant equations
gamma=Cp,m/Cv,m.
pv=constant, rev. ideal

3. The attempt at a solution

p1v1^(Cpm/Cvm)=p2v2^(Cpm/Cvm) take ln and mult both sides by the Cvm/Cpm
ln p1v1 = ln p2v2 e to both sides
p1v1=p2v2
p1v1/p2v2 = 1
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 28, 2008

### Mapes

Hi Jennifer, welcome to PF. I can see a problem with your first solution step:

$$PV^\gamma\neq(PV)^\gamma$$

and

$$\ln(PV^\gamma)\neq\gamma\ln(PV)$$

Here is a suggestion for the proof: divide the adiabatic process into two processes, one isobaric (constant pressure) and one isochoric (constant volume). You can do this because you're dealing with state variables, and it doesn't matter what path you take from one state to another.

In the constant-volume process, the enthalpy increases by $V\,dp$. But because you're dealing with an ideal gas, you can also say that the enthalpy increases by $c_p\,dT$ (you should know or verify these relations).

In the constant-pressure process, the energy increases by $-p\,dV$. For an ideal gas, it also increases by $c_V\,dT$.

You should be able to integrate the two processes separately and combine the results to show what you want to show.

3. Feb 28, 2008

### Lojzek

You can also use equipartition theorem: every degree of freedom in the system contributes
kT/2 to average energy of the system.

So energy of ideal gas is E=x*N*kT/2=x*nRT/2, where x is the number of degrees of freedom per molecule, N is number of molecules in the gas and n=N/Avogadro number.

This formula allows you to calculate increase of temperature because of added work:

x*nRdT=-p*dV (you can also eliminate p with ideal gas equation)

Now you have both dV and dT you can calculate dp from ideal gas equation:

p=nRT/V -> dp=nRdT/V-nRTdV/V^2

Finally you integrate the relation between dV and dp.

You can also express the result with cp/cv using relation cp/cv=(x+2)/x

Last edited: Feb 28, 2008