Finding area from work, pressure and volume

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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?
 
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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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 $$P_2=P_0-AV_2^2$$.