### Re: Calculating water (or is it air) pressure versus temperature

Date: Tue Mar 1 11:59:03 2005
Posted By: Leslie Allen, Staff, Laboratory Chemist, Valero Refining Company
Area of science: Physics
ID: 1109577036.Ph
Message:

Here are my questions:
1) Can you show me a formula and an example of how to calculate the amount
of water pushed out of the bottle when the temp. of the air inside of the
bottle is raised by 10 and 20 deg. C?

The amount of water pushed from the bottle would be the equivalent of the
increased volume of gas. According to the "Ideal Gas Law", the volume of
the gas will be proportionate to the temperature.

Ideal Gas Law              pV = nRT

p= pressure in atm: V = the volume:  n = moles of gas: T= temperature in
kelvin (C +273): R is the molar gas constant 0.082058 L*atm*mol-1*K-1.

In this case, the pressures will be 1 atm and constant.  The number of
moles of gas (n) will be constant. Since you have a 2 Liter bottle, fill ½
with water. That leaves us with 0.5 L of air.
Now I'll call on Charles's Law : V1/V2 = T1/T2:
V1 = 0.5 L air
T 1= (20C +273)=293K
T2 = (30C+273)=303K
V2 = X L
V1/T1 = V2/T2
0.5L/293K = X/303K
Rearrange our formula and 303K*0.5L/293K=0.517L
Since we started out with 0.5L of air and our new volume at +10C measured
0.517 L ,  we see that the increase in temperature added 0.017L of volume
to the gas. That's 17mL of water in our measuring device. I'll leave you

2) Does the amount of air in the bottle matter? ie 3/4 air vs. 1/4 water
or 1/4 Air vs. 3/4 water?
In "ideal" terms, it doesn't matter the amount of air and water. However,
the more air that's in the bottle, the easier it will be to measure the
amount of water leaving the bottle.
While your calculating, try 0.75L of air and then use 0.25L of air and see
the amount of water that you'll be measuring.
The gas laws are very handy. They are found in any chemistry or physics
text books.
Have fun! That's what it's all about.

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