Thermal - derive a work equation

[accountkiller]

Homework Statement

Show how W= (P²V² - P¹V¹) / [itex]\gamma[/itex] -1

can be derived using relations between PV^gamma = constant, and W(1 to 2) = -[itex]\int[/itex] P(T,V) dV (from v1 to v2).

Homework Equations

I think we can use R = C_v ([itex]\gamma[/itex] - 1)

The Attempt at a Solution

Not sure how to start. The integral would be W = [itex]\int[/itex] 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?

 

[Redbelly98]

Actually, the formula for work is [itex]W = \int P \ dV[/itex]. You can use [itex]\ P \ V^{\gamma} = \mbox{constant} \ [/itex] in that integral.

 

[accountkiller]

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

 

[Redbelly98]

What exactly does it mean for PV^{[itex]\gamma[/itex]} to equal a constant? It means that P and V are inversely proportional, right?

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
[tex] P \ V^{\ \gamma} = \mbox{constant}[/tex]
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.

I'm not sure what the [itex]\gamma[/itex] as an exponent of the V being constant means though.

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.

 

[accountkiller]

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

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

 

[accountkiller]

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

 

[Redbelly98]

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

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

P is not constant. [itex] PV^{\ \gamma} [/itex] is a constant -- call it k, if you wish, and solve for P:
[tex]PV^{\ \gamma} = k[/tex]
Therefore,

P = ____ ?​