Calculating Pressure & Work in a Piston-Cylinder Assembly

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Homework Help Overview

The problem involves calculating the initial and final pressures and the work done in a piston-cylinder assembly as warm air cools and the volume changes. The scenario includes a spring force that varies linearly, with given atmospheric pressure and piston area.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate initial and final pressures based on the relationship between pressure, area, and force. They express uncertainty about their pressure calculations and how to integrate pressure to find work.
  • Some participants suggest using average force to calculate work instead of integration, questioning the necessity of complex calculations.
  • There is a discussion about finding the distance traveled by the piston using volume and area.

Discussion Status

Participants are exploring different methods to approach the problem, with some providing guidance on simplifying calculations. There is a lack of explicit consensus, but some participants affirm the correctness of the original poster's calculations.

Contextual Notes

Participants note the importance of understanding the problem before performing calculations and discuss the implications of varying forces on work done. There is an emphasis on the assumptions made regarding the linear variation of force and neglecting friction.

krnhseya
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Homework Statement



Warm air is contained in a piston-cylinder assembly oriented horizontally. The air cools slowly from an initial volume of 0.003m^3 to a final volume of 0.002m^3. During the process, the spring exerts a force that varies linearly from an initial value of 900N to a final value of zero. The atmospheric pressure is 100kPa, and the area of the piston face is 0.018m^2. Friction is neglected between piston and cylinder. For the air, determine the initial and final pressures and work.

Homework Equations



Pressure x Area = Force

Work = [tex]\int[/tex] p dV

The Attempt at a Solution



Initial Pressure of Air x Area = Atmospheric Pressure x Area + Spring Force
Initial Pressure of Air x 0.018m^2 = (100x10^3Pa x 0.018m^2) + 900N
Initial Pressure of Air = 150kPa

Final Pressure of Air x Area = Atmospheric Pressure x Area
Final Pressure of Air = Atmospheric Pressure
Final Pressure of Air = 100kPa

Work = [tex]\int[/tex] p dV

Volume is changing from 0.003m^3 to 0.002m^3, which can be used as an interval of this integration. I am not sure about how I should use pressure to calculate the work.

Also, I am not really sure if I got those pressures correctly.
Thank you.
 
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Physics tip - don't do the maths until you understand the problem.
If you can solve the problem with simple maths do so.

Pressure - at the start the pressure is atmopshere+spring, at the end it is just atmosphere
Work - work is force*distance. Since the force is varying linearly you don't need to integrate just use the average/mid point force
 
mgb_phys said:
Physics tip - don't do the maths until you understand the problem.
If you can solve the problem with simple maths do so.

Pressure - at the start the pressure is atmopshere+spring, at the end it is just atmosphere
Work - work is force*distance. Since the force is varying linearly you don't need to integrate just use the average/mid point force

Did I do it wrong? I thought those calculations for pressure were quite simple...?

Well for the work part, here's what I did.

Since the area for both initial and final are constant, find the distance traveled by dividing volume by area. (Initial Volume/Area and Final Volume/Area)
Once I get those, subtrace one to another to final the distance travelled.
Then like you mentioned earlier, average force (450N) times the distance that I got to find the work done, which is 25J or 0.03kJ)
 
Sounds correct
 
mgb_phys said:
Sounds correct

Thank you very much. Have a great day! :)
 

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