Work Done by Gas: Pressure, Volume, and Energy

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Homework Help Overview

The problem involves a gas in a balloon, initially at a pressure of 2P0, which is allowed to escape slowly until the pressure drops to P0. The main questions raised are about the work done during this process and the source of energy required for the work.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the applicability of isothermal processes and Boyle's law, with one participant attempting to integrate pressure over volume to find work done. Others question the assumptions about pressure and volume changes during the gas's escape.

Discussion Status

There is an ongoing exploration of different interpretations regarding the work done by the gas and the role of external pressure. Some participants have offered insights about the tension in the balloon and the external pressure, while others express confusion about the calculations and assumptions being made.

Contextual Notes

Participants note that the volume of the balloon may remain constant while gas escapes, leading to questions about how this affects the work done. There is also mention of the external pressure being at P0, which influences the calculations of work.

Ahmed Abdullah
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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?
 
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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?
 
Last edited:
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.
 
Last edited:
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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