Work Done by Gas: Pressure, Volume, and Energy

[Ahmed Abdullah]

Homework Statement

There is some gas in a balloon. The volume of the balloon is V and the gas is under a pressure of 2P0 ---where P0 is atmospheric pressure. The gas is allowed to come very slowly out of the balloon and eventually the gas pressure decrease to P0. How much work is done?
Where does the energy required to do work come from?
Thx

Homework Equations

PV=nRT

The Attempt at a Solution

My solution: gas comes out slowly so we can assume the process to be isothermal. So boyles law it applicable. So when gas pressure is halved the volume is doubled (2V).
so work= integration of PdV
= integration of K/V*dV (k=2PoV=boyle's constant)
= K*In2 =2PoV*In2

But the answer is PoV. Can you explain why?

 

[genneth]

Don't forget the balloon.

 

[Ahmed Abdullah]

I can't get it. Someone please help!

 

[genneth]

You are correct that at the end, the volume of the balloon is half of what it used to be. However, the source of the work isn't the gas inside the balloon! The balloon is under tension when it's filled, and it is this tension that does work. Draw a diagram, write some words...

 

[Ahmed Abdullah]

The tension of the balloon decreases as you let some gas out--- I understand that. But how that makes you get the work PoV (V2-V1=V)?
It seems that the gas has expanded under a constant pressure of Po. W=Po*(del V) when net pressure is constant and the process is isothermal. Can you show that these are the case here?

 

[genneth]

Remember that there is already a gas outside of the balloon, and that it's at P0.

 

[Ahmed Abdullah]

Is it the correct soulution to the problem:-
W=delta(PV)
=V delta P (since the volume remains unchanged)
=V (2Po-Po)
=PoV

 

[learningphysics]

No, that doesn't look right to me. work is integral of PdV

For work done on a system... as integral of PdV, P is the external pressure. hence you need to use 2Po as the P here.

 

[Ahmed Abdullah]

Any of you please show me the right way to solve it :-(.

 

[learningphysics]

:( looks like I misunderstood the question... seems the volume stays the same while the gas leaks...

Since work is -Pexternal*change in volume, and since change in volume is 0... wouldn't work be 0?

I don't know how to get PoV.