reversible adiabatic expansion

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.

nickjer

If [tex]PV^\gamma[/tex] 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

Apply the first law: [itex]\Delta Q = \Delta U + W[/itex].

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

AM

tensus2000

still lost

Andrew Mason

[tex]\Delta Q = \Delta U + W = 0[/tex]

(1) [tex]\therefore W = - \Delta U[/tex]


Now:

(2) [tex]\Delta U = nC_v\Delta T[/tex] and

[tex]C_v = C_p/\gamma = (C_v+R)/\gamma[/tex] so:

(3) [tex]C_v = R/(\gamma-1)[/tex]

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