# Creating a temperature differential on SF6 to evacuate a cylinder

1. Aug 12, 2014

### co0ldood

I have a 50KG cylinder with 1-2LB of SF6 at 15PSI. I want to transfer it into another cylinder by temperature differential.

Keeping it simple. I try to model it with two cylinders that has a valve on each cylinder. Connecting the cylinder with a 1/4" hose. Both valves are closed. The cylinder with the SF6 is at ambient temperature (25C). I cool the other cylinder to -40C (~-50C is solidifying point).. Open both valves.

How would I go about calculating the flow rate?

I try to find the pressure differential using ideal gas law

PV=nRT; P1/T1 = P2/T2, 15 PSI / 298.15K = P2/233.15K; P2 = 11.73PSI.

This is the wrong way since I know it is a compressible gas and the pressure should be much lower. Any help or direction would greatly be appreciated.

Thanks!

2. Aug 12, 2014

### Staff: Mentor

Weird unit system :(.

PV=nRT takes compression into account, but you won't get a pressure difference if your valves are open. You just get a density difference, while the total amount of SF6 is constant (and known).

Your calculation would require the same amount of SF6 in both cylinders and closed valves.