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A sample of fluorine has a volume of 6.46 L when the temperature is -20.1 degC and the pressure is 855 torr. What will be the volume when the pressure is 99.15 kPa and the temperature is 120.9 degC?

Feb 6th, 2015

I'm reasonably certain that the fluorine remains a gas under all these conditions, and the Ideal Gas Law holds.

(P1*V1)/T1 = (P2*V2)/T2

Where P = absolute pressure, V = volume, T = absolute temperature

Rewriting this for the value we wish to solve for, V2:

V2 = (P1 * V1 * T2)/(P2*T1)

First let's put the temperatures in an absolute form, using Kelvin (= ÂșC + 273.15)

T1 = -20.1 + 273.15 = 253.05 K

T2 = 120.9 + 273.15 = 394.05 K

Then we put our known values into the formula with their units.  (Always include the units.)

V2 = ( 855 torr * 6.46 L * 394.05 K) / (99.15 kPa * 253.05 K)

For any units that didn't cancel, we need to find the proper conversion factor.  I prefer to write the units so that they will cancel, and then look up the number to put it on the correct side of the fraction.

V2 = [( 855 torr * 6.46 L * 394.05 K) / (99.15 kPa * 253.05 K)]  * ____ kPa/  ___ torr

V2 = [( 855 torr * 6.46 L * 394.05 K) / (99.15 kPa * 253.05 K)] * 0.1333 kPa/ 1 torr

Solving this gives us V2 = 11.56 L.   Intuitively, this makes sense: the pressure didn't change much but the temperature went up a fair amount, so an increase in volume is expected. [Please check the values yourself; I don't intentionally make any mistakes but it can always happen.]

The formula used is a simplification of the Ideal Gas Law used for comparing the same amount (mass) of gas under two sets of conditions.  This is also known as the 'Combined Gas Law'.  It can be derived from the Ideal Gas Law by solving for n, which is the same in both cases (the constants drop out as well).

Feb 6th, 2015

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Feb 6th, 2015
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Feb 6th, 2015
Oct 22nd, 2017
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