# Reversible adiabatic expansion

1. Apr 15, 2010

### tensus2000

1. The problem statement, all variables and given/known data

I'm in a rutt for a tutorial question:

The question is basically to show that the work done during a reversible adiabatic expansion of an ideal gas is

W = (P1V1 – P2V2)/(1 ‐ γ) ........ Y is gamma

2. Relevant equations

3. The attempt at a solution

I've got so far as to get W= (P1V1^γ – P2V2^γ)
due to P1V1^γ being constant for a reversible adiabat
also that W= -PdV

But i havent a clue how to get the 1-Y at the bottom, me thinks its intergrating for V to get this but my maths is very bad so i dont know how to do this.

2. Apr 15, 2010

### nickjer

3. Apr 16, 2010

### Andrew Mason

Apply the first law: $\Delta Q = \Delta U + W$.

Since $\Delta Q = 0$ and $\Delta U = nC_v\Delta T$ where $C_v = R/(\gamma - 1)$ you should be able to work it out quickly. (Hint: apply the ideal gas law: PV=nRT).

AM

4. Apr 16, 2010

### tensus2000

still lost

5. Apr 16, 2010

### Andrew Mason

$$\Delta Q = \Delta U + W = 0$$

(1) $$\therefore W = - \Delta U$$

Now:

(2) $$\Delta U = nC_v\Delta T$$ and

$$C_v = C_p/\gamma = (C_v+R)/\gamma$$ so:

(3) $$C_v = R/(\gamma-1)$$

Substitute (3) into (2) and then just substitute PV for nRT

AM

6. Apr 17, 2010

### tensus2000

Cheers, i got it
The way you showed was much easier then the way i was going about it