Is PV^gamma Constant in Adiabatic Processes?

[stratz]

Homework Statement

Proving PV^gamma = constant In adiabatic expansion. Q = 0

Homework Equations

N/A

The Attempt at a Solution

ΔE_int = 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.

 

[TSny]

ΔE_int = W

nCvdT = PdV = nRT / V dV

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

 

[stratz]

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

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

 

[TSny]

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

 

[stratz]

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

 

[TSny]

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

Yes.

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

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

Either way, dE = dQ - PdV.

 

[stratz]

Alright, I got this to work correctly, I had to use the Cv + R identity which canceled 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?

 

[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 = P₁V₁^γ.

 

[stratz]

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 = P₁V₁^γ.

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

Thanks for the help

 

[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.