Calculating Pressure & Work in a Piston-Cylinder Assembly

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SUMMARY

The discussion centers on calculating pressure and work in a piston-cylinder assembly containing warm air. The initial pressure of the air is determined to be 150 kPa, while the final pressure is equal to the atmospheric pressure of 100 kPa. The work done during the process is calculated using the average force of 450 N and the distance traveled by the piston, resulting in a total work of 25 J. Key equations utilized include the relationship between pressure, area, and force, as well as the work integral.

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  • Knowledge of integration for work calculations
  • Ability to apply the concept of average force in varying force scenarios
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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 = \int 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 = \int 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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