Finding area from work, pressure and volume

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Homework Help Overview

The discussion revolves around a problem involving the calculation of area from work, pressure, and volume, specifically using the integral of pressure with respect to volume. Participants are exploring the relationships between pressure and volume in a physical context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the integral and the relationships between different pressures and volumes. There are attempts to solve for an unknown variable, A, and questions arise regarding the mixing of pressure terms and the implications of having multiple unknowns.

Discussion Status

Some participants have provided guidance on the relationships between the pressures and volumes, suggesting that there are equations that can be used to eliminate certain variables. There is an ongoing exploration of the implications of these relationships and the correct interpretation of the problem statement.

Contextual Notes

There is a mention of potential confusion regarding the definitions of pressure at different states and the need to clarify the initial conditions. Participants are also reflecting on the algebraic relationships that govern the problem.

JoeyBob
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Homework Statement
see attached
Relevant Equations
W=integral(PdV)
So I basically took the integral and ended up with W=PVf-A(Vf^3)/3-PVi+A(Vf^3)/3

so 65.7=72*5.3-A(5.3)^3/3-72(2.4)+A(2.4)^3/3

But when I solve for A I get the wrong answer of 3.179 when the answer is suppose to be 5.05. I've checked my calculation with an algebra calculator too...
 

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JoeyBob said:
Homework Statement:: see attached
Relevant Equations:: W=integral(PdV)

So I basically took the integral and ended up with W=PVf-A(Vf^3)/3-PVi+A(Vf^3)/3

so 65.7=72*5.3-A(5.3)^3/3-72(2.4)+A(2.4)^3/3

But when I solve for A I get the wrong answer of 3.179 when the answer is suppose to be 5.05. I've checked my calculation with an algebra calculator too...
You seem to have mixed up P0 and P1.
 
haruspex said:
You seem to have mixed up P0 and P1.
Wait wouldn't tha mean there's two unknowns then? How would I find P0?
 
JoeyBob said:
Wait wouldn't tha mean there's two unknowns then? How would I find P0?
You have two equations relating P0, P1, P2, V1, V2 and A.
 
haruspex said:
You have two equations relating P0, P1, P2, V1, V2 and A.

Wait so is it P1=P0+AV and then the integral eqn?
 
JoeyBob said:
Wait so is it P1=P0+AV and then the integral eqn?
No.
You are given P=P0-AV2 as a general fact. In particular this will be true at the initial and final states. Plug in the values/variables for those states to get two equations.
 
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You have $$P_1=P_0-AV_1^2$$ so $$P_0=P_1+AV_1^2$$So you can eliminate ##P_0##
 
Chestermiller said:
Well the problem statement says: $$P=P_0-AV^2$$so at the initial condition, $$P_1=P_0-AV_1^2$$So my algebra tells me that $$P_0=P_1+AV_1^2$$
What am I missing?
Sorry, I read what I was expecting, not what you wrote.
I read your second line as $$P_2=P_0+AV_2^2$$, reinforced by your comment about eliminating P0. So what I should have written is $$P_2=P_0-AV_2^2$$.
 
haruspex said:
Sorry, I read what I was expecting, not what you wrote.
I read your second line as $$P_2=P_0+AV_2^2$$, reinforced by your comment about eliminating P0. So what I should have written is $$P_2=P_0-AV_2^2$$.
What do you need to know P2 for?
 
  • #10
Chestermiller said:
What do you need to know P2 for?
I'm not saying you do. I'm explaining what I thought you had written and what I thought you had intended.
 

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