10B: Solving Gas Law Problems
This worksheet will give you practice in solving quantitative problems involving gas laws. The variables
associated with gases are: pressure, volume, number of moles, and temperature. You will begin by solving
simple problems involving only two variables, gradually developing the skills necessary to solve more difficult
problems involving three and four variables.
Example A
A beach ball is inflated to standard atmospheric pressure (1atm). Its volume is measured and found to be 25.8 L.
A diver then takes the ball underwater to a depth of 20m, where the pressure exerted on the ball is known to be
3 atm. Calculate the volume of the ball at this depth.
Solution: This problem can be solved with Boyle’s law, which says that the volume of a gas varies inversely
with pressure, assuming that the amount of gas (moles) and its temperature remain constant. We can predict
qualitatively what should happen, using kinetic molecular theory. As the pressure increases outside the ball, the
gas molecules will be squeezed closer together. Thus, there is a decrease in volume. This example is a twovariable problem. We can solve the problem and obtain a quantitative solution using the following equation.
P1 x V1 = P2 x V2
Where P1 = 1.0 atm, V2 = 25.8 L and P2 = 3.0 atm. V2 is the unknown in the equation. Rearranging terms in
the equationto isolate V2, we have
V2 = (P1 x V1)/ P2= (1.0 atm x 25.8L)/ 3.0 atm = 8.6 L
As you can see, 8.6L is indeed less than 25.8L, and our prediction of a volume decrease has been verified.
You Try It
1) Calculate the final pressure in the cylinder of a car engine if the original volume was 450 mL at a
pressure of 0.80 atm and the final volume is 120 mL. Assume no change in the number of molecules or
temperature.
Example B
The volume of a balloon is 1240 cm cubed when measured at 35 degrees C. Calculate its new volume when it is
cooled to –196 degrees C. Assume no change in the amount of gas or pressure.
Solution: As the gas inside the balloon cools, the average kinetic energy of the molecules decreases. Thus, with
fewer and less forceful collisions, the gas molecules move clower together and occupy a smaller volume than
they previously did. The volume decreases. The equation describing this relationship is called Charles’ law.
V1/T1 = V2/T2 where V2 = 1240 cm cubed
T1 = 35 degrees C +273 = 308K
T2 = -196degrees C +273 = 77K
Remember that temperature measurements must be in Kelvin. Rearranging terms to solve for V2, we have
V2 = (V1 x T2)/ T1 = (1240 x 77K)/ 308 K = 310 cm cubed
You Try It
2. The gas inside a piston =was heated until the volume of gas had increased from 125 mL to 850 mL. If the
temperature inside the piston was originally 15 degrees C, calculate the new temperature in degrees C.
Assuming that amount of the gas and the pressure do not change.
Example C
The beach ball in Example A is once again submerged by a diver, this time to a depth of 25m. The season is
winter and the water in cold! The original volume at 1.0 atm is 25.8 L, and the air temperature is 23 degrees C.
Total external pressure exerted on the ball 25 m underwater is 3.5 atm. Water temperature at 25 m depth is 12
degrees C. Calculate the new volume.
Solution: The problem involves 3 variables; pressure, temperature, and volume. We will assure that the volume
of the beach ball does not change, so the amount of air in it remains constant. The equation for the combined
gas law is as follows.
(P1 x V1)/ T1 = (P2 x V2)/ T2 where P1 = 1 atm
V1 = 25,8L
T1 = 23+273= 296K
P2 = 3.5 atm
T2 = 12+273 = 285 K
V2 = unknown
V2 = (1 x 25.8L x 285 K)/ (3.5x 296K) = 7.1 L