Finding area from work, pressure and volume

AI Thread Summary
The discussion revolves around solving for the area (A) in the equation derived from the work-energy principle, where the user initially calculated an incorrect value of 3.179 instead of the expected 5.05. Participants point out potential confusion regarding the variables P0 and P1, suggesting that the user may have mixed them up. They clarify that two equations can be formed from the relationships between pressure and volume, allowing for the elimination of P0. The conversation emphasizes the importance of correctly interpreting the equations and variables involved in the calculations. Accurate identification of the initial and final states is crucial for solving the problem correctly.
JoeyBob
Messages
256
Reaction score
29
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...
 

Attachments

  • question.PNG
    question.PNG
    5.4 KB · Views: 143
Physics news on Phys.org
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.
 
  • Like
Likes Delta2
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.
 
Back
Top