Conservation of mass question ??

Question

THE QUESTION:
Initially, a rigid 6-m^3 tank contains helium (assumed to be an ideal gas) at 300 K and 0.10 MPa. The tank is then heated until the temperature reaches 400 K.

A pressure-relief valve at the top of the tank opens when the pressure reaches 0.15 MPa, and the helium flows out, preventing the pressure from exceeding 0.15 MPa. Determine the final mass (kg) of helium in the tank.


MY ANSWER:
So, in hoping that people don't think I just posted this up here without trying, this is what I've done so far--

I couldn't really come up with a way to go use PV=nRT a couple of times to keep transforming. What I noticed though is that I **THINK** I have everything I need for the final state.

I just looked at the state after the process. It MUST have the same V. And it tells us the P&T. I used V=6 P=.15 T=400. As the question described, it sounds like this is the final state. I calculated n from this using PV=nRT, and converted it to kg using its amu.

thoughts?

Answer 1

I don't think the pressure ever gets to .15 MPa. Since the temperature only increases 33%, I think the pressure increases proportionally to .133 MPa.

I could be wrong, of course.

Answer 2

The question is what is the final mass in the tank. You have Temperature, Pressure and Volume at the final conditions.

Solve PV=nRT for n

That is your answer. Note you need to make sure that the units of R you are using are consistent with the pressure, temperature, volume and kg-mols. The inital conditionsl don't matter because you aren't concerned with mass lost just the mass in the tank at final conditions.

Answer 3

Mass will be the same. The pressure wont reach 15. You have to heat to 450 to get 15 and then more to release helium in to atmosphere.
v,n,R are constant
P1/p2 = T1/T2
Go from here

Answer 4

Compare n's of each state to figure how much was released.

Answer 5

huh??? that the first thing i said when i read dat !!!! lol