# Pulling a piston

1. Aug 22, 2011

### Tabeia

Hi, I'm studying thermodynamics and I got a question, it's not a homework or anything, it simply isn't covered in my book and it got me thinking
I know that if I have cylinder filled with gas, with a piston on top and there are weights on it maintaining an equilibrium if I remove the weights slowly the gas will expand doing work on the surroundings.
After all the gas molecules will be hitting the piston wall, and as the gas pressure is higher than the external pressure it will expand.

But what if I have a gas cylinder at let's say 1atm, with an external pressure of 1 atm, at equilibrium. No weights on it.
Let's say I pull the piston, I grab it with my hand and pull it.
What will happen?
At first I thought, ok, it will just do expansion work, reversible if I do it slowly, irreversible if I do it quickly...
But then I thought, no that isn't possible, the gas is not doing any work, it's not like with the weights where the only upward force is caused by the gas molecules, now I'm actually pulling the piston.
But my book says that if a gas expands against any moveable surface it's doing work...

Could somebody help me?

2. Aug 22, 2011

### xts

Grab a plastic syringe and try!
Then compare your first thoughts with an experiment.
And ask again who (you or gas?) is doing a work (or more work...).

3. Aug 22, 2011

### Tabeia

Well I tried that lol
Yes, it seems like I'm doing a good amount of work, but isn't the gas doing anything? Maybe something I can't detect with such crude experiment

4. Aug 22, 2011

### K^2

Start with plunger all the way down. Plug hole and pull. Now start with some air in the cylinder. Plug hole and pull. Which is easier? Does that explain why gas is still doing work?

5. Aug 22, 2011

### Tabeia

Hmm, I get it.
Now let's see with my math is right
If the pressure outside is 1atm, and I'm doing part of the work I can't calculate the work done by the gas simply
integrating(or multiplying since it's constant) the dV times the P.
So could I calculate it by removing the pressure caused by my force/area of the piston, and integrating?
Like if I'm doing a pulling force equal to 1N, the piston area is $1m^2$ the gas will be doing a work equals to
$dW = (1atm-1Pa) * dV$
Yes, I realize I would need to pull harder with time since the pressure of the gas would decrease, but in the first infinitesimal moment I think it would be that.

6. Aug 22, 2011

### xts

You may take it that way: you are making a hard work against 1 atm pressure (times syringe piston area). And gas makes his work - not that hard - just its pressure is working.

7. Aug 22, 2011

### cjl

Your book is correct. The gas is doing work against the piston. The reason you also have to do work against the piston is because you have to consider the entire system. When you pull on the piston, it has the gas inside the piston on one side, and the atmosphere on the other. If the gas is at a lower pressure than the atmosphere, then the gas is not doing enough work on the piston to offset the work the piston has to do on the atmosphere while it is moving. The extra work the piston has to do against the atmosphere comes from you.

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