Pressure change from cooling liquid.

[aph]

I have an application where I need to calculate the vacuum (pressure change below atmospheric) generated from a cooling liquid in a semi-rigid container.

If the container is filled near full (e.g. 2 litres) with a liquid at e.g. 70C, sealed and left to cool to ambient (e.g. 20C) there is a pressure differential to atmospheric generated from the volume change of the cooling liquid that equalises when the seal is broken.

How do I calculate what that pressure differential (vacuum effect) is?

How much will it vary depending on the size of the container?

 

[cragar]

(p1V1)/(T1)=(p2V2)/(T2)
p=pressure v=volume T=temperature .
Combined gas law
I don't know if this is quite what you need .

 

[Bob S]

Cragar's answer is correct for liquids with a low vapor pressure. For liquids with a high vapor pressure, like boiling water, the partial pressure of the steam as the liquid cools below 100 degrees centigrade must be taken into account. The vapor pressure of water (steam) drops from 760 mm (Hg) at 100 degrees C to ~355 mm at 80 C.

Bob S

 

[aph]

Thanks very much for your help.

As I never will fill at near boiling point or above I will assume use of the combined gas law without consideration of vapor pressure changes.

So, if P1=101.3kPa, T1=345K, V1=2000ml
and T2=293K, V2=1958ml

(- V2 comes from asssumption that 1ml of water at 20C = 1g and at 72C 1ml = 0.979g so I multiplyed 2000 x 0.979)

I get, P2 = 87.9kPa

Therefore, pressure differential (vacuum effect) is 13.4kPa.

That is, when I break the seal the pressure will equalise and I will hear the air rush in.

The main purpose is to determine a negative pressure to apply to the closed container to see the effect on different container designs.

Am I on the right track or off the mark?

Thanks again.