Finding area from work, pressure and volume

In summary: I was mistaken.In summary, the conversation is about solving for the value of A in the equation W=PVf-A(Vf^3)/3-PVi+A(Vf^3)/3=65.7=72*5.3-A(5.3)^3/3-72(2.4)+A(2.4)^3/3, where the correct answer is supposed to be 5.05. The problem arises from mixing up P0 and P1 in the equation P=P0-AV^2, leading to two equations and two unknowns. The correct equation should be P2=P0-AV2^2.
  • #1
JoeyBob
256
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: 117
Physics news on Phys.org
  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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?
 
  • #6
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
  • #7
You have $$P_1=P_0-AV_1^2$$ so $$P_0=P_1+AV_1^2$$So you can eliminate ##P_0##
 
  • #8
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 $$P_2=P_0-AV_2^2$$.
 
  • #9
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.
 

FAQ: Finding area from work, pressure and volume

How do you calculate area from work, pressure, and volume?

To calculate the area from work, pressure, and volume, you can use the formula: Area = Work/Pressure x Volume. This formula is derived from the ideal gas law, which states that the pressure and volume of a gas are inversely proportional at a constant temperature. By rearranging the formula, you can solve for area.

What units are used to measure area in this context?

In this context, area is typically measured in square meters (m^2). However, depending on the specific situation, other units such as square centimeters (cm^2) or square feet (ft^2) may also be used.

Can you use this method to find the area of any shape?

No, this method is specifically used for finding the area of a gas in a container at a constant temperature. It cannot be applied to finding the area of a 2D shape, such as a circle or triangle.

What is the significance of finding area from work, pressure, and volume?

Finding the area from work, pressure, and volume is important in understanding the behavior of gases. It allows scientists to calculate the amount of space that a gas occupies in a container and how it changes under different conditions. This information is useful in various fields such as chemistry, physics, and engineering.

Are there any limitations to using this method?

Yes, there are some limitations to using this method. It assumes that the gas is ideal, meaning that it follows the ideal gas law perfectly. In reality, most gases deviate from this law at high pressures or low temperatures. Additionally, this method may not be accurate for gases that undergo chemical reactions or phase changes.

Back
Top