Pressure on Piston: Conservation of Energy Explained

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Homework Help Overview

The discussion revolves around the concept of pressure in the context of a piston exerting force on water in a tank. The original poster attempts to reconcile their understanding of pressure as defined by the equation P = F/S with their observations about summing pressures from multiple molecules in contact with the piston.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the nature of pressure as an intensive quantity, questioning the validity of summing pressures from individual molecules. They discuss the implications of pressure in relation to forces and equilibrium in systems with multiple pistons.

Discussion Status

Some participants have offered clarifications about the nature of pressure and its intensive properties, while others are questioning the original poster's understanding of these concepts. The discussion is ongoing, with various interpretations being explored.

Contextual Notes

There is an underlying assumption that the original poster's understanding of pressure may be flawed, leading to confusion regarding conservation of energy and the behavior of forces in equilibrium. The discussion also touches on the implications of hydrostatic pressure in liquid systems.

CGandC
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Homework Statement


Suppose there is a tank filled with water and a piston of area S exerts a force F on the water.
upload_2018-6-11_23-15-46.png

Suppose I divide the water boundary touching the piston to - N small equal " square " molecules.
Then , the force on the upper face of each molecule is F/N .
Also, the area of the upper face of each molecule is of length S/N .
Thus , the pressure on each molecule touching the piston is : ## P = \frac{F/N}{S/N} = \frac{F}{S} ##
And this is the same pressure that each molecule exerts on the piston

Question: If ## P = \frac{F}{S} ## is also the pressure that each molecule ( touching the piston ) exerts on the piston, then , if I sum the pressures exerted by each such molecule on the piston, I get:

## P = \frac{F}{S} + \frac{F}{S} + ... + \frac{F}{S} = \frac{N*F}{S} ##

which contradicts conservation of energy and also because it just seems wrong because I was told the pressure on the piston is just : ## P = \frac{F}{S} ## and not ## P = \frac{N*F}{S} ##

Why is there contradiction of conservation of energy and why the pressure on the piston is not ## P = \frac{N*F}{S} ##?

Homework Equations

The Attempt at a Solution

 

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Pressure is an intensive quantity, not an extensive one. You sum forces, not pressures.
 
Orodruin said:
You sum forces, not pressures.
Well, you can sum pressures if they all act on the same patch of surface, but then we would have pressure NF/S acting on area S/N for a total force of F and an average pressure over S of F/S.
 
Orodruin said:
Pressure is an intensive quantity, not an extensive one. You sum forces, not pressures.

So if for example, I have a trapezoid block of mass and two pistons are acting on either side of it, each exerting force F1 and F2 upon the faces L1 and L2 respectively, as shown in the figure below:

upload_2018-6-12_10-18-11.png


Then, I think the total pressure on the trapezoid should be:
## P = \frac{F_1}{L_1} + \frac{F_2}{L_2} ##
But according to what you said, pressure is an intensive property so summation like this cannot be done.

So my question is: what should be the total pressure acting on the trapezoid?
 

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You seem to have a fundamental misunderstanding of what pressure means. It is not the property of an object as a whole, it is a local property that describes force per unit area over an intersection. In your trapezoid example, the pistons need to apply the same pressure for the system to be in equilibrium - the hydrostatic pressure of the liquid. Otherwise you cannot apply hydrostatics, which is also what let's you assume that the pressure is constant throughout the liquid.
 
Orodruin said:
a fundamental misunderstanding of what pressure means
And maybe one regarding compression and tension.
Compression is not a force, exactly, but more a pair of equal and opposite forces. Even just looking at the forces rather than pressure, the compression is not F1+F2. Assuming equilibrium, the compression is F1=F2.
 

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