Work done on the system through a piston?

In summary, the work done on a system is positive when the volume is expanded and negative when the volume is compressed.
  • #1
betelgeuse91
32
0
I am stuck on like the first page of thermal physics. It seems like the signs of the work done on the system are opposite when the volume is expanded and compressed. But when I imagine myself pushing or pulling the piston, I get confused from [tex]W = \textbf{F}\cdot \textbf{d}[/tex]
This work will be positive as the direction of the force and the direction of the displacement of the piston will be the same regardless of the change in volume.

What am I missing?
yiwBXz2.png

(I got the image from http://www.splung.com/content/sid/6/page/work)
 

Attachments

  • yiwBXz2.png
    yiwBXz2.png
    11.6 KB · Views: 988
Physics news on Phys.org
  • #2
betelgeuse91 said:
What am I missing?
In the one case it's the fluid that's doing the positive work and in the other it's you, the pusher. So then the fluid does negative work: it exerts a force outwards but the displacement is inward.
 
  • #3
BvU said:
In the one case it's the fluid that's doing the positive work and in the other it's you, the pusher. So then the fluid does negative work: it exerts a force outwards but the displacement is inward.

If I consider a piston in equilibrium with atmosphere such that it won't move unless I push it or pull it. This case, whichever way I do, won't I be doing positive work? From above formula for work.
 
  • #4
Correct.
 
  • #5
betelgeuse91 said:
If I consider a piston in equilibrium with atmosphere such that it won't move unless I push it or pull it. This case, whichever way I do, won't I be doing positive work? From above formula for work.
BvU said:
Correct.

Then either case I am doing positive work. But the above and other text says when volume expands, external work done on the system is negative and vice versa (so the signs of work I do for compression and expansion of gas are opposite). How does this work?
 
  • #6
betelgeuse91 said:
I am stuck on like the first page of thermal physics. It seems like the signs of the work done on the system are opposite when the volume is expanded and compressed. But when I imagine myself pushing or pulling the piston, I get confused from [tex]W = \textbf{F}\cdot \textbf{d}[/tex]
This work will be positive as the direction of the force and the direction of the displacement of the piston will be the same regardless of the change in volume.

I think you are confused about the direction of the forces...bear with me...

Consider a car jack/lift and define up as positive...

When you raise the car the force applied by the jack is in the same direction as the motion eg both upwards. No problem with that. The work done by the jack s positive.

When you lower the car slowly at constant velocity the jack still applies an upward force on the car but the direction of motion is downwards. The work done by the jack is negative.

Its similar with a piston in a syringe...

If you let the gas push the piston out slowly at constant velocity you will still be applying a push force on the plunger eg opposite to the direction of motion. So the work you do will be negative.

You could pull the piston really fast. But there will be a net force on the piston that causes it to accelerate. You would have to take into account the work done accelerating the piston. Both you and the gas would do positive work.
 
  • Like
Likes BvU
  • #7
CWatters said:
I think you are confused about the direction of the forces...bear with me...

Consider a car jack/lift and define up as positive...

When you raise the car the force applied by the jack is in the same direction as the motion eg both upwards. No problem with that. The work done by the jack s positive.

When you lower the car slowly at constant velocity the jack still applies an upward force on the car but the direction of motion is downwards. The work done by the jack is negative.

Its similar with a piston in a syringe...

If you let the gas push the piston out slowly at constant velocity you will still be applying a push force on the plunger eg opposite to the direction of motion. So the work you do will be negative.

You could pull the piston really fast. But there will be a net force on the piston that causes it to accelerate. You would have to take into account the work done accelerating the piston. Both you and the gas would do positive work.

This solved the mystery for me... Thank you so much.
 

Related to Work done on the system through a piston?

What is work done on a system through a piston?

Work done on a system through a piston refers to the energy transferred to the system by the movement of a piston. This can occur in various systems, such as a gas trapped in a cylinder with a piston, where the work done on the gas causes it to expand or compress.

How is the amount of work done through a piston calculated?

The work done on a system through a piston can be calculated by multiplying the force exerted by the piston by the distance it moves. This is known as the work-energy theorem and is represented by the equation W = F*d, where W is work, F is force, and d is distance.

What factors can affect the amount of work done through a piston?

The amount of work done on a system through a piston can be affected by several factors, including the force applied to the piston, the distance it moves, and the speed at which it moves. The properties of the system, such as temperature and pressure, can also impact the work done.

How does work done on a system through a piston relate to its internal energy?

When work is done on a system through a piston, the system's internal energy can change. This is because the work done causes a change in the system's physical state, such as an increase in temperature or a change in volume. The internal energy of a system is the sum of its kinetic and potential energies.

Can work be done by a system through a piston?

Yes, a system can also do work on its surroundings through a piston. This occurs when the system's internal energy decreases, causing the piston to move and exert a force on its surroundings. This is known as work done by the system and is represented by a negative value in the work-energy equation.

Similar threads

  • Thermodynamics
Replies
8
Views
658
Replies
5
Views
1K
Replies
5
Views
2K
Replies
15
Views
2K
Replies
5
Views
618
Replies
5
Views
1K
  • Mechanics
Replies
9
Views
2K
Replies
5
Views
729
Replies
4
Views
1K
Replies
16
Views
1K
Back
Top