Thermodynamics problem: Piston in cylinder

Click For Summary
SUMMARY

The discussion centers on a thermodynamics problem involving a piston in a horizontal cylinder with specific dimensions and initial conditions. The initial pressure is set at p0=10^5 N/m², and the piston moves 0.4m, leading to new pressures P1 and P2 calculated as P1= P0L/(L-2h) and P2=P0L/(L+2h). The force required to hold the piston in position is determined using the formula F=(P2-P1)*S, resulting in a force of 888N. The conversation emphasizes the importance of the ideal gas law and the assumption of isothermal conditions for accurate calculations.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of pressure and volume relationships in thermodynamics
  • Familiarity with force calculations (F=PA)
  • Basic concepts of isothermal and adiabatic processes
NEXT STEPS
  • Study the ideal gas law and its applications in thermodynamics
  • Learn about isothermal and adiabatic processes in detail
  • Explore the concept of energy conservation in thermodynamic systems
  • Investigate the effects of temperature changes on gas behavior
USEFUL FOR

Students and professionals in engineering, physics, and thermodynamics who are looking to deepen their understanding of gas laws and piston mechanics in thermodynamic systems.

Adrian379
Messages
6
Reaction score
0

Homework Statement


[/B]
A horizontal cylinder with length L=1m and section S=2*10^(-3) is divided in two parts by a piston. In theese compartments is air at pressure p0=10^5N/m^2 and same temperature.
The piston move with h=0,4m from initial position. Which is force to hold piston in this position ?

Homework Equations


[/B]
My Tries: just this figure and same irrelevant ecuations like PV=nRT and i think P0V0=P1V1, but i don't know why. Thank you very much, i am at beginning with thermodynamics.

bUybp1r.png
 
Physics news on Phys.org
I'll give some hints. The first step is going to be finding the new pressures on both sides in the piston after you move the piston. For that, you'll be using P0V0=P1V1.

After you know the pressure on both sides, you need to find the net force on the dividing wall between the two sides. If you know the pressure, and you know the area, how do you find the force exerted by either side? Look at the units for pressure and area.
 
Yeah but can you explain me why p0v0=p1v1 ? Thanky you for response.
 
I'm not an expert here so you might want to double check this, but in my understanding it is a consequence of the law of conservation of energy.

Pressure times volume is a form of energy. So if you change the volume, you must also change the pressure. Energy can be neither created nor destroyed. However, this is only valid if you don't add energy from an outside source.
 
Adrian379 said:
Yeah but can you explain me why p0v0=p1v1 ? Thanky you for response.
It follows from the ideal gas law, assuming that the temperature and number of moles don't change.$$nRT_0=P_0V_0=P_1V_1$$
 
Chestermiller said:
It follows from the ideal gas law, assuming that the temperature and number of moles don't change.$$nRT_0=P_0V_0=P_1V_1$$
Yes but, if the piston move, the temperature from compartments don't change ?

I've found P1 & P2:
P1= P0L/L-2h

P2=P0L/L+2h

Now, i must to found the force.

F=P*S but what P i must multiply to S to found force ?

I solve it ! F=(p2-p1)*S = 888N
Thank you very much !
 
Last edited:
Adrian379 said:
Yes but, if the piston move, the temperature from compartments don't change ?
Apparently they expect you to assume that the cylinder and piston are not insulated. So, when the system equilibrates, the temperature will be unchanged.
 
Am I right in thinking that without the assumption that the process is Isothermal, there is no way to find the pressure or temperature? It's been a while since I solved this sort of thing.
 
John Morrell said:
Am I right in thinking that without the assumption that the process is Isothermal, there is no way to find the pressure or temperature? It's been a while since I solved this sort of thing.
It could also be done if the two chambers and the piston were adiabatic, and the piston were moved very slowly.
 

Similar threads

Replies
8
Views
2K
Replies
49
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 5 ·
Replies
5
Views
11K
Replies
13
Views
2K
Replies
1
Views
2K