Thermal - derive a work equation

AI Thread Summary
The discussion focuses on deriving the work equation W = (P2V2 - P1V1) / (γ - 1) using the relationship PV^γ = constant and the work integral W(1 to 2) = -∫ P(T,V) dV. Participants express confusion about starting the derivation, particularly regarding the implications of γ as an exponent and how it affects the relationship between pressure and volume. The integral for work is clarified as W = ∫ P dV, with suggestions to substitute for P using the constant relationship. Ultimately, the conversation emphasizes the need to express P in terms of V and constants to facilitate the integration process.
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Homework Statement


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).


Homework Equations


I think we can use R = Cv (\gamma - 1)


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?
 
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Actually, the formula for work is W = \int P \ dV. You can use \ P \ V^{\gamma} = \mbox{constant} \ in that integral.
 
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.
 
mbradar2 said:
What exactly does it mean for PV\gamma 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
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.

I'm not sure what the \gamma 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:
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?
 
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:
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?
P is not constant. PV^{\ \gamma} is a constant -- call it k, if you wish, and solve for P:
PV^{\ \gamma} = k
Therefore,
P = ____ ?​
 

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