# Homework Help: Proving PV^gamma = Constant

1. Oct 29, 2016

### stratz

1. The problem statement, all variables and given/known data
Proving PV^gamma = constant In adiabatic expansion. Q = 0

2. Relevant equations
N/A

3. The attempt at a solution

ΔEint = W

nCvdT = PdV = nRT / V dV

∫Cv/T dT = ∫R/V dV

Cv ln(T2/T1) = R ln(V2/V1)

ln(T2/T1) = (R/Cv) ln(V2/V1)

T2/T1 = (V2/V1)R⋅gamma / Cp

P2V2 / P1V1 = (V2/V1)R⋅gamma / Cp

I did some extra rearranging after this point, but it appears that the equation in this form cannot get PV^gamma = constant or for P1V1^gamma = P2V2^gamma. Is there a mistake somewhere?

Thanks.

Last edited: Oct 29, 2016
2. Oct 29, 2016

### TSny

Watch the signs. Does W stand for the work done by the gas or the work done on the gas?

3. Oct 29, 2016

### stratz

Well it is the differentiated form of work, but I guess that doesn't matter since the rate of change of work done on the gas is negative right?

I'll try adding a negative sign to that and see if it works. Thanks

4. Oct 29, 2016

### TSny

To see if dT and dV have the same sign, think about what happens to the temperature during an adiabatic expansion.

5. Oct 29, 2016

### stratz

Okay, so temperature will decrease as energy is lost in the form of work on the environment due to expansion of volume right?

6. Oct 29, 2016

### TSny

Yes.

Some people write the first law for an infinitesimal reversible step as dE = dQ - dWby where dWby is the work done by the gas: dWby = +PdV.

Others write it as dE = dQ + dWon where dWon is the work done on the gas: dWon = - PdV

Either way, dE = dQ - PdV.

7. Oct 29, 2016

### stratz

Alright, I got this to work correctly, I had to use the Cv + R identity which cancelled out the Cp in the denominator. However this gave me P1V1^gamma = P2V2^gamma. Would it be possible to rearrange this somehow to make just one side equal to a constant?

8. Oct 29, 2016

### TSny

State 2 can be thought of as any state during the adiabatic process. So for any state (P, V) during the process, you have P V γ = P1V1γ.

9. Oct 29, 2016

### stratz

Yeah, it's just that I saw a proof somewhere where they got PVgamma = econstant which ended up being PVgamma = constant. Anyways, it's pretty much the same result.

Thanks for the help

10. Oct 29, 2016

### TSny

Instead of doing definite integrals from state 1 to state 2, you could do indefinite integrals. This will lead to P V γ = const where the const is related to the constants of integration of the indefinite integrals.