any engineers in the house?

[James3shin]

so trying to figure out a physical chem. problem and I think that i'm on the right track. Here's the problem:

Lets say we have a rigid glass globe with a volume of 1L, the globe has 2 openings. we then place a balloon inside the globe via one of the openings (one hole is still open) and try to inflate it. what is the work done on the air in the globe? The initial volume is 15 mL and upon inflation, the volume of the balloon goes to 45 mL.

Here's what I think thus far:

This process is spontaneous (irreversible) because the balloon doesn't stay inflated. Therefore, the equation for work i should be using is:

W = -(Pexternal)Delta V

My questions:

since the glass is still exposed to the the atmosphere, is Pexternal 1atm?

 

[Fenixgoon]

yeah, should be 1 atm IIRC

 

[cirthix]

the work is 0 if the other opening is open. this problem aslo doesnt take into account the fact that the baloon's pressure increases as it is inflated. pretty poor problem :/

 

[imported_Imp]

Ugh... Physical Chemistry, the concept of 'work' brings back painful memories and a crappy textbook.