# Thermodynamic question involving work

1. Sep 13, 2011

### Liquidxlax

1. The problem statement, all variables and given/known data
Show that the work done by a reversible function from P1V1 to P2V2 is

W = (P1V1 -P2V2)/ (x - 1)

using the relation PVx = C where X and C are constants

2. Relevant equations

dW = PdV

3. The attempt at a solution

I'm going to have to update this more when i get home because i forgot my textbook and notebook

Basically what I did was draw a graph from P1V1 to P2V2 and the whole point is to find the area underneath the curve

My problem is i'm not sure how to include the change in pressure and the change in volume in this integral.

I'm basically getting (x-y)/xy where x and y are different volumes

Like i said i will update my work when i get home, but any help for now will be appreciated

2. Sep 13, 2011

### pabloenigma

dw =PdV,there you are.
So to compute work,all you need to do is compute the definite integral of PdV from V1 to V2.
Express P in terms of V,by using the equation PVx=constant(say k),ie P=kV-x

Now Integrate KV-x from V1 toV2,and you are done,provided you put the limits nicely(substitute k=P1V1x in the 2nd term and k=P2V2x in the first term)

3. Sep 13, 2011

### Liquidxlax

funny thing is that if you are right i did that, and i guess i must have messed up my algebra. Curse the few first days of school

4. Sep 14, 2011

### Liquidxlax

I did do that, but do you justify that k=P2V2x=P1V1x?

5. Sep 15, 2011

### pabloenigma

In your question, PVx=constant is the equation of the process,ie this equation characterizes and defines the process the system is made to go through.This defines the way P and V of the system changes. So, P and V of the system always saties this equation. If P1V1, P2V2,P3V3 , etc are pressure and volume of the system in different points of time, then for all thes states, PiVi=k is satisfied.

So from the initial and final states we get

P1V1= P2V[/SUB]2[/SUB]=k.
Got it?