Combined Gas Law

For the past few days we have been talking about the effects of pressure, volume and temperature on gases. What did Boyle’s Law tell us about the relationship between pressure and volume if temperature is constant? They are inversely related.

And how about Charles’ Law of temperature and volume with constant pressure? They are directly related

So what happens to volume if we increase pressure? It goes down.

And what would that cause the temperature to do? It would go down.

So even though the laws that we’ve talked about assume that one of the variables is held constant, we can see that they are really all related. Pressure and temperature are not independent variables when we are dealing with a sample of gas in a closed container. As the temperature changes, this affects the volume and therefore the pressure. As pressure changes, the volume again will change and this will affect the temperature of the gas sample. To account for this we can actually combine Boyle’s and Charles’ Laws together to get the equation that is on your reference tables, the Combined Gas Law. Like the name says, it’s not a new law, just a combination of other ones:

P1V1/T1 = P2V2/T2

As with Boyle's and Charles's Laws, P1, V1 and T1 are the initial values for the gas sample and P2, V2 and T2 are the final values. Remember that T must be in Kelvin! And actually, you can always use this law, even if one of the variables is constant. What would this equation look like if temperature was constant? Boyle’s Law

Sample Problem

A sample of gas with a volume of 5 m3 at a temperature of 298 K and a pressure of 0.84 atmospheres is left in a closed container where the temperature increases to 323 K and the pressure increases to 0.89 atmospheres. What is the new volume of the gas sample?

P1 = 0.84 atm
V1 = 5 m3
T1 = 298 K
P2 = 0.89 atm
T2 = 323 K
V2 = ?

P1V1/T1 = P2V2/T2

(0.84 atm)(5 m3)/298 K = (0.89 atm)(V2)/323 K

(0.84 atm)(5 m3)(323 K) = (298 K)(0.89 atm)(V2)

(0.84 atm)(5 m3)(323 K)/(298 K)(0.89 atm) = (298 K)(0.89 atm)(V2)/(298 K)(0.89 atm)

(0.84 atm)(5 m3)(323 K)/(298 K)(0.89 atm) = V2

Note that we may cancel both the K and atm units from our equation, leaving only the m3unit.

(0.84)(5 m3)(323)/(298)(0.89) = V2

5.11 m3 = V2

Example 2. A 350 cm3 sample of helium gas is collected at 22.0 oC and 99.3 kPa. What volume would this gas occupy at STP?

Solving:

First, you must change the Celsius temperature to Kelvin.

A 350 cm3 sample of helium gas is collected at 22.0 oC 295 K and 99.3 kPa. What volume would this gas occupy at STP?

Now, list the givens and the unknown

V1 = 350 cm3
P1 = 99.3 kPa
T1 = 295 K
V2 = ?
P2 = 101.3 kPa (standard pressure)
T2 = 273 K (standard temperature)

Now, substitute the values into the equation, which is already set up for volume 2 as the unknown.

V2 = V1P1T2
------
P2T1

V2 = 350 cm3 x 99.3 kPa x 273 K
------
101.3 kPa x 295 K

V2 = 317.5 cm3