How is Work Calculated in Reversible Adiabatic Expansion?

[tensus2000]

Homework Statement

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

Homework Equations

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 haven't 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 don't know how to do this.

 

[nickjer]

If $$PV^\gamma$$ is a constant, then finding the difference in it would be 0, and not equal to the work. For more help on this, please check the wikipedia page:

http://en.wikipedia.org/wiki/Adiabatic_process

 

[Andrew Mason]

Homework Statement

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

Homework Equations

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 haven't 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 don't know how to do this.

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

 

[tensus2000]

still lost

 

[Andrew Mason]

still lost

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

(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

 

[tensus2000]

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