An Introduction to the Gas Laws Using the Jeep TPMS CW gas laws using Jeep TPMS 050112.

My Jeep has an “on-board computer”, the EVIC (Electronic Vehicle Information Center), that includes the TPMS (Tire Pressure Monitoring System), which displays the pressure in each tire.

One Friday afternoon, with the air temperature at about 60F, I made sure that all four tires had pressures of 33 PSI. But early the next morning after the temperature had plunged to 28 F, the pressure on all four tires was 30 PSI. Why the drop?

The Ideal Gas Equation

The ideal gas equation, PV = nRT, describes the behavior of an ideal gas. An ideal gas is one that follows the assumptions of the kinetic molecular theory. The nitrogen and oxygen in air at STP will behave as ideal gases.

The ideal gas equation is PV=nRT …. where P is pressure, V is volume, n is the number of moles, R is a constant and T is the pressure in absolute pressure units.

When P is given in atmospheres, V in liters, n in moles and T in Kelvins, then R = 0.0821 Latm/molK. This is the most common value of R. R can also commonly be given in two other pressure units. The value of R depends upon the choice of pressure units. R = 8.314 LkPa/molK when pressure is in kilopascals. R = 62.4 Ltorr/mol K when pressure is given in torr or its equivalent, millimeters of mercury, or mm Hg.

The Combined Gas Law

The kissing cousin of the ideal gas equation is the This says that the pressure in the tire is directly Combined Gas Law. The combined gas law can be proportional to the temperature. As the temperature derived from the ideal gas equation. The combined drops, so does the pressure, just as was indicated on gas law is actually the combination of two of the my TPMS. So, does Amonton’s law allow us to original mathematical relationships for gases: model the pressure in my tires? A temperature of Boyle’s law and Charles’s law. Boyle’s law says that 60F = 16C = 289K. And 28F = -2C = 271K. A for any confined gas at a constant pressure, that pressure of 33 PSI is what is called the gauge changes in pressure are inversely proportional to pressure and does not include the atmospheric changes in volume. Charles’s law says that for any pressure. The tire gauge only reads the pressure in confined gas at a constant pressure that the volume is excess of atmospheric pressure. directly proportional to the absolute temperature. A confined gas means that the number of moles remains So we really need to know the atmospheric pressure constant even though the other variables can be was on the day I topped off the tire pressure. The changing. average air pressure is about 13.1 lb/in2 (uncorrected to sea level). So, to the pressure of 33 psi we will add The combined gas law is usually given as: 13 psi to get an actual pressure of the tire of 46 psi.

P1V1/T1 = P2V2/T2 This makes use of another named relationship, Dalton’s law of partial pressures. This law says that The air in a car tire is a confined gas, and the volume the pressure of a mixture of gases is the sum of the of the air in a tire is essentially constant. Therefore, partial pressures of the gases in the mixture. We can the combined gas law boils down to a third law, use Dalton’s law to determine the actual pressure of called Amonton’s law: P/T = k, or P1/T1 = P2/T2. the gases in the tire. The total pressure is the atmospheric pressure plus the gauge pressure. This the pressure in the tire was 30 psi. Therefore the gives us a pressure of 33 + 13 or 46 psi. When the actual pressure is 43 psi. Is this predicted by temperature dropped, the gauge pressure indicating Amonton’s law?

P1/T1 = P2/T2 …. Compute the new pressure, P2, when the temperature falls from 60F to 28F. P2 = P1T2/T1 P2 = 46 psi x 271K / 289K P2 = 43 psi …. Yes! Amonton’s law allows us to predict the lower pressure at the lower temperatures.

Questions

1. After driving the car for 30 minutes, the pressure rose to about 32 PSI (from 30 PSI). Why?

2. Pressure can be defined in two ways (a) force per unit area, (b) pressure is proportional to the number of collisions between the gas molecules and the walls of the container. Explain how these two descriptions of pressure are consistent with each other.

3. Explain how the pressure of a confined gas will be affected by changes in (a) volume, (b) temperature and (c) the number of molecules (in moles).

4. Use the proportionalities from the preceding question to justify the ideal gas equation, PV = nRT.

5. Look up the “kinetic molecular theory” and list all of its premises.

6. We speak of “ideal” gases and “real” gases. Which two premises of the KMT are not true for a “real” gas.

7. Determine the volume of 1.00 mole of SO3 gas at STP.

8. Find the pressure of 88.0 grams of CO2 gas in a 3.50L container at 30.0C.

9. What is the volume of a 0.350 moles of ammonia gas at a pressure of 1.20 atm and a temperature of -15.0C.

10. Find the mass of 500.0 mL of oxygen gas at a temperature of 25C and a pressure of 715 mm Hg.

11. A confined gas is in a cylinder with a movable piston. Initially the gas has a volume of 100.0 cm3 at a pressure of 0.95 atm and 20.0C. What will be the pressure in the cylinder if the volume is decreased to 65.0 cm3 while keeping the temperature constant?

12. Consider 2.00 L of argon gas at STP. What will be the volume of the gas at 2.50 atm and 400.K

13. A confined gas is in a cylinder with a movable piston. When the piston is withdrawn from 40.0 mL to 100.0 mL the temperature falls from 20.0 C to 18.0C. If the pressure at 100.0 mL is 600. torr, what was the pressure at 40.0 mL?

14. Determine the volume of O2 gas at STP that can be produced from the decomposition of 20.0 grams of potassium chlorate.

2KClO3(s)  2KCl(s) + 3O2(g)

15. Assuming the reaction goes to completion, determine the volumes of N2 and H2 gases needed to make 10.0 L of ammonia gas when all gases are at the same temperature and pressure.

N2(g) + 3H2(g)  2NH3(g)