Work Done by Gas: Pressure, Volume, and Energy

AI Thread Summary
The discussion revolves around calculating the work done by gas escaping from a balloon as its pressure decreases from 2P0 to P0. The initial assumption of an isothermal process leads to confusion regarding the application of Boyle's Law and the work done calculation. It is clarified that the work done is not solely from the gas but also involves the tension of the balloon, which contributes to the work as the gas escapes. The correct formula for work, considering external pressure, is established as W = P0 * ΔV, where the change in volume is effectively V. Ultimately, the resolution highlights the importance of understanding the external pressure and the balloon's tension in calculating work done.
Ahmed Abdullah
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Homework Statement



There is some gas in a baloon. The volume of the baloon 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?
 
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Don't forget the balloon.
 
I can't get it. Someone please help!
 
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...
 
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?
 
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Remember that there is already a gas outside of the balloon, and that it's at P0.
 
Is it the correct soulution to the problem:-
W=delta(PV)
=V delta P (since the volume remains unchanged)
=V (2Po-Po)
=PoV
 
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.
 
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Any of you please show me the right way to solve it :-(.
 
  • #10
:( 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.
 

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