Thermal - derive a work equation

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accountkiller
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Homework Statement


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

can be derived using relations between PVgamma = 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 = Cv ([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?
 
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.
 
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.
 
mbradar2 said:
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.
 
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?
 
Actually, I'm sure I can substitute something into T in terms of V. I'll be back.
 
mbradar2 said:
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 = ____ ?​