Question

I have a question on thermodynamics. It is about a piston cylinder arrangement with gas underneath the piston. My book says if you change the pressure infinitesimally by adding infinitesimally light weights slowly, then the internal pressure will equal the external pressure within an infinitesimal amount. If we add a small weight, then the pressure of the gas would be the weight divided by the area of the piston. But the area the mass takes up is much smaller, so then the external pressure will not equal the internal pressure at all? Since P=F/A. please explain.

Answer 1

I have a question on thermodynamics. It is about a piston cylinder arrangement with gas underneath the piston. My book says if you change the pressure infinitesimally by adding infinitesimally light weights slowly, then the internal pressure will equal the external pressure within an infinitesimal amount.

This is true if you do this you will get the maximum amount of work out of the piston. This is a reversible process

If we add a small weight, then the pressure of the gas would be the weight divided by the area of the piston.

At what point are you adding the weight after you bring the piston up? You are talking about a completely different process adding the weight would be a irreversible process

Answer 2

I should have added quotations sorry

Answer 3

I mean, to get the quasi-static process that is reversible, we would need to add an infinite number of infinitely light weights to the piston. I am talking about the first one you add, or the second one. It doesn't matter which one. I am just saying that once you do add one of those infinitely light weights, then the internal pressure will not equal the external pressure. But the book says that throughout the whole process, the external pressure is equal to the internal pressure. The reason I think they are not equal is because what if you add a weight that has a small area. Then P=F/A. The forces on the piston will be equal when the piston stops moving, but the areas will be different and so the pressures will be different.

Answer 4

What type of system are you taking about?

Answer 5

you are applying work on the system rather then the system applying work on the outside.

http://chemwiki.ucdavis.edu/ @api/deki/files/9998/workdone.JPG?size=bestfit&width=454&height=210&revision=1

I think you have the concept confused.

Answer 6

A reversible process simply is a process that could be going either forward or backward at any point you observe it.

An irreversible process is one that can only go one way.

As far as external pressure and internal pressure always being equal it depends on what kind of system you are working with.

Answer 7

and how you are working with said system.

If say the system had a way of release some of its gas then yes it would maintain the equilibrium

Answer 8

The process you describe would be a reversible process but it would not maintain equilibrium pressure if it was closed

Answer 9

if it was open then it would, if it had the right system set it up to allow infinitesimal gas to release

Answer 10

does that answer your question?

Answer 11

Sorry. No it doesn't. My book just has the expansion or compression of a gas in a piston at constant temperature. There is no gas being released. It says that if the process is carried out slowly and with infinite steps, then the external pressure would equal the gas pressure. I am just asking why that is so?

Answer 12

Assuming the piston can release the gas infinitesimally slowly to compensate then that would be the case

Answer 13

Well, why does the gas have to escape for? I don't understand why the gas has to leave the system.

Answer 14

If the gas doesn't leave the system and it is a piston pressure is going to be applied on the gas,

Assuming nRT is constant

P = nRT/V

you would get a graph like this,



It would be a compression because of gravity and with no way to compensate for this internal pressure would not equal external pressure

Answer 15

Also consider the fact that volume of the outside of the system would be expanding and with expansion of volume reduction of pressure occurs

Answer 16

That sort of makes sense, but then why does my book say that the internal pressure would equal the external pressure?

Answer 17

I'm assuming it is talking about a similar open system

Answer 18

Talk to your teacher or prof about this, I'm sure they will be able to explain it better than I

Answer 19

okay thank you for your help.

Answer 20

No problem

Answer 21

Also a piston works as an open system

Answer 22

but no mass is leaving the system? just energy

Answer 23

I guess that could be a fair reason for it to work, check out this forum it should contain a post that will explain the concepts better than I can http://www.physicsforums.com/showthread.php?t=636487

Answer 24

I literally just read that. But the problem is, it does not use weights. So my problem concerns using weights with small areas.

Answer 25

Does it talk about the type of system this piston is? I think you should talk to your teacher about this. I guess if temperature decreased in the cylinder as volume decreased pressure would remain constant

Answer 26

temperature is constant.

Answer 27

Just refer to my previous explanation

Answer 28

Isothermal open system

Answer 29

Okay. I will. thanks

Answer 30

Without a loss of temperature or moles there is no way internal pressure can remain equal to external pressure.

Answer 31

I would recommend you confirm this with your teacher though

Answer 32

thank you. I will!

Answer 33

look when you put a tiny weight on the piston this weight force the gas to compressed but a little and when gas compressed in constant temperature pressure of gas increased and pressure of weight and gas will be equal.