Reversible adiabatic expansion proof

The relationship is: T^(Cp,m/R)/P = constantWhere R is the gas constant and Cp,m is the molar heat capacity.
  • #1
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Homework Statement


Prove the relationship between the pressure, P, and the temperature, T, for an ideal gas with a reversible adiabatic expansion. Base the proof on the first law of thermodynamics and the ideal gas law.

The relationship is: T^(Cp,m/R)/P = constant

Where R is the gas constant and Cp,m is the molar heat capacity.

Homework Equations


ΔU=Q-W
pV=nRT

Possibly relevant:
Wrev=-pdV
γ=Cp,m/Cv,m where γ is the heat capacity ratio.
PVγ= constant

The Attempt at a Solution


I don't even know where to start explaining what I have tried... I have been kicking around a lot of things. I can't seem to manage to manipulate things so that temperature is to the power of anything, although I have burnt a lot of time trying to turn PVγ into some relation for T to the power of some derivative of gamma. I assume since gamma has Cp,m in it, that it is going to play in, and especially since it is already a power for volume, this flagged me in that direction, but I am just going around in circles.

Any help in the right direction would be GREATLY appreciated.
 
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  • #2
Latsabb said:
The relationship is: T^(Cp,m/R)/P = constant

Where R is the gas constant and Cp,m is the molar heat capacity.

Homework Equations


ΔU=Q-W
pV=nRT

Possibly relevant:
Wrev=-pdV
γ=Cp,m/Cv,m where γ is the heat capacity ratio.
PVγ= constant

The Attempt at a Solution


I don't even know where to start explaining what I have tried... I have been kicking around a lot of things. I can't seem to manage to manipulate things so that temperature is to the power of anything, although I have burnt a lot of time trying to turn PVγ into some relation for T to the power of some derivative of gamma. I assume since gamma has Cp,m in it, that it is going to play in, and especially since it is already a power for volume, this flagged me in that direction, but I am just going around in circles.
Start with PVγ= constant and substitute nRT/P for V.

AM
 
  • #3
Ok, so I end up with:
P*(nRT/P)γ=c

Which I then turn into:
P*(nRT)γ/Pγ=c

And then:
(nRT)γ/Pγ-1=c

Am I on the right track? Because I have been playing around with it, and I can't seem to extract nR out, nor am I having luck manipulating the powers to what I need.
 
  • #4
If you are supposed to use the first law, then you are expected to start with

##dU=nCvdT=-PdV##
 
  • #5
I finally got it nailed down. Chester was correct, the missing piece here was that dU=Cv,mdT=-PdV. An integration of that, and some minor modifying of terms produced what I was after. Thank you!
 

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