Thermal - derive a work equation

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Homework Help Overview

The discussion revolves around deriving the work equation W = (P2V2 - P1V1) / (γ - 1) using the relationship PV^γ = constant and the integral for work done during a process. The subject area is thermodynamics, specifically focusing on adiabatic processes and the work done by gases.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the integral form of work and question how to apply the relationship PV^γ = constant in the context of the problem. Some express uncertainty about the implications of γ and its role in the equations. Others discuss the need for substitutions and the nature of variables involved in the integral.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning their understanding of the relationships between pressure, volume, and temperature. Some guidance has been offered regarding the substitution of variables and the interpretation of the equations, but no consensus has been reached on the approach to take.

Contextual Notes

Participants note that the problem may not require complex methods such as partial differential equations, and there is some confusion regarding the assumptions about constants and variables in the equations being discussed.

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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?
 
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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 = ____ ?​
 

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