# Decrease in pressure due to adiabatic expansion derivation?

1. Oct 23, 2013

### monnapomona

1. The problem statement, all variables and given/known data
Show that $\frac{dp}{p}$ =$\frac{\gamma}{\gamma-1}$$\frac{dT}{T}$ if the decrease in pressure is due to an adiabatic expansion.

2. Relevant equations
Poisson equations:
Pv$^{\gamma}$
Tv$^{\gamma - 1}$

Ideal Gas Law:
Pv=R$_{d}$T, where R$_{d}$ is the dry air gas constant.

Hydrostatic Equation:
$\frac{dp}{dz}$ = - ρg

3. The attempt at a solution
I tried making those two equations equal to each other since they are equivalent (i think) and differentiating on both sides but i ended up with $\frac{\gamma - 1}{\gamma}$ in the final result...

EDIT: I was looking up adiabatic atmosphere and found this. I'm wondering where the formula p$^{\gamma - 1}$T$^{\gamma}$ = constant comes from...

Last edited: Oct 24, 2013
2. Oct 24, 2013

### ehild

How is γ related to Cp and Cv?

You can consider air as ideal gas. Use the Ideal Gas Law to eliminate V.

ehild

Last edited: Oct 24, 2013
3. Oct 24, 2013

### monnapomona

I think $\gamma$ = Cp/Cv and R = Cp - Cv.
I used an entropy equation and made it equal to 0:
0 = Cp*ln(T2/T1) - R*ln(p2/p1) and got (R/Cp)*ln(p2/p1) = ln(T2/T1). So I solved for R/Cv = $\gamma$ -1 / $\gamma$... and the final result was
(P2/P1)^(($\gamma$ -1) / $\gamma$) = T2/T1

Am I on the right track with this?

4. Oct 24, 2013

### ehild

You are on the right track, but write the equation with the differentials.

ehild

5. Oct 24, 2013

### monnapomona

This is where I'm stuck... could i say P2/P1 = P and T2/T1 = T, then take the ln of the equation to bring the exponent down?

Last edited: Oct 25, 2013
6. Oct 25, 2013

### ehild

Assume that P1=Po and T1=To are fixed, and P2=Po+dP, T2=To+dT, dT and dP very small. How are dP and dT related? Have you studied differentials?

ehild

7. Oct 26, 2013

### monnapomona

Is dt and dp directly proportional?

And I have calculus background from a few years ago so I'm a bit rusty on some concepts... like differentials.

8. Oct 26, 2013

### ehild

Try to refresh your Calculus knowledge.

You have the equation (P2/P1)^((γ -1) / γ) = T2/T1.

Write P2/P1=(Po+dP)/Po=1+dP/Po and T2/T1=1+dT/To. dP/Po << and dT/To <<1. The original equation becomes
(1+dP/Po)(γ-1)/γ=1+dT/To.
Use the approximation (1+δ)n ≈1+nδ,valid for δ<<1.
You can check it, what is (1+0.001)3, for example?

What do you get?

ehild

9. Oct 29, 2013

### monnapomona

You would get 1.003003 so its approximately 1. Curious question, why does this approximation matter to the proof or derivation of the original question?

10. Oct 29, 2013

### ehild

You get the desired formula with that approximation.

ehild