# Thermal - derive a work equation

1. Sep 11, 2011

### accountkiller

1. The problem statement, all variables and given/known data
Show how W= (P2V2 - P1V1) / $\gamma$ -1

can be derived using relations between PVgamma = constant, and W(1 to 2) = -$\int$ P(T,V) dV (from v1 to v2).

2. Relevant equations
I think we can use R = Cv ($\gamma$ - 1)

3. The attempt at a solution
Not sure how to start. The integral would be W = $\int$ P2V2 - P1V1 dV, but that means we'd have to do partial differential equations, and the problem is not meant to be so difficult.

Any suggestions?

2. Sep 11, 2011

### Redbelly98

Staff Emeritus
Actually, the formula for work is $W = \int P \ dV$. You can use $\ P \ V^{\gamma} = \mbox{constant} \$ in that integral.

3. Sep 12, 2011

### accountkiller

What exactly does it mean for PV$\gamma$ to equal a constant? It means that P and V are inversely proportional, right? I'm not sure what the $\gamma$ as an exponent of the V being constant means though.

4. Sep 12, 2011

### Redbelly98

Staff Emeritus
Only if γ is equal to 1. Inversely proportional would mean that PV=PV1=constant.

But since γ is not 1, P and V are not inversely proportional. So we have to leave it as
$$P \ V^{\ \gamma} = \mbox{constant}$$
You can use that relation to substitude for P in the integral used to calculate W. If you do that substitution, then the integrand will be in terms of V and constants, and can be integrated.

γ is just an exponent, the relation involving P and V γ is just how the relation between P and V is expressed for an ideal gas undergoing an adiabatic process.

5. Sep 13, 2011

### accountkiller

All right, by taking P to be constant and substituting PV = nRT, here's where I'm at:
$W = \int P dV = \int \frac{nRT}{V} dV = nR \int \frac{T}{V} dV$

T is in the equation, so it's not just V that I can take an integral of.
Did I have a wrong step?

6. Sep 14, 2011

### accountkiller

Actually, I'm sure I can substitute something into T in terms of V. I'll be back.

7. Sep 14, 2011

### Redbelly98

Staff Emeritus
P is not constant. $PV^{\ \gamma}$ is a constant -- call it k, if you wish, and solve for P:
$$PV^{\ \gamma} = k$$
Therefore,
P = ____ ?​