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Ideal Gas Law and Tire Pressure

In an Ideal Gas, collisions between atoms or molecules are elastic; there are no intermolecular attractive forces. You can also see it as hard spheres colliding and hitting each other but which don’t interact with each other. One mole of an Ideal Gas at STP (standard temperature pressure) occupies 22.4 L

The Ideal Gas Law is PV= nRT or


P(M/m)= RT/V

PM= RT(m/V)


n = number of moles

R = universal gas constant = 8.3145 J/mol K

T= Absolute Temperature (Kelvin)

In such kinds of gases, where the molecules and atoms have no intermolecular attractions, all the internal energy is in the form of kinetic energy. This means that the very microscopic energy of the molecules exists in the form of kinetic energy, moving freely with great speed. If there is a change in the internal energy of the molecules, it results in a change of temperature.

In the winter time our tires loose air and look kind of flat because of temperature fluctuations. Well, we know that tires are usually filled from about 35 psi to 50 psi. It has been estimated that tires loose 1 psi per month. This cannot be proven with a physics equation, but this is kind of consistent with the saying that tires loose air over time. Let's suppose the change in temperature is 10 deg. Celsius, and the volume of the tire is almost constant for small changes in temperature and pressure. When we look at the change in pressure, the number of moles of the gas inside will be assumed to be constant. So this can now be solved with the Ideal Gas Law. Change in pressure is directly proportional to the change in the absolute temperature when the number of moles and volume doesn’t change significantly. A drop in temperature of 10 deg. F corresponds to a relative change of about = 1.9%. This means that the total tire pressure will decrease by about 2% as well so decreased Pressure = 0.02(50 psi) = 1 psi. The actual pressure drop will be less than this since the volume of the tire will also decrease slightly maybe 10 %, so the drop in pressure might be more like 0.9 psi.


An air sample containing only nitrogen and oxygen gases has a density of 1.3393 g / L at STP. Find the weight and mole percentages of nitrogen and oxygen in the sample.

Answer: From the density d, we can find the Molar Mass.

P M = d R T
M = 22.4 * d
= 22.4 L/mol * 1.3393 g/L
= 30.0 g / mol

If we have 1.0 mol of gas, and x mol of nitrogen, then (1 - x) is the amount of oxygen.

28.0 x + 32.0 (1 - x) = 30.0
- 4 x = - 2
x = 0.50 mol of N2, and 1.0 - 0.50 = 0.50 mol O2

Now, to find the weight percentage, find the amounts of nitrogen and oxygen in 1.0 mol. (30.0 g) of the mixture.

Mass of 0.5 mol nitrogen = 0.5 * 28.0 = 14.0 g
Mass of 0.5 mol oxygen = 0.5 * 32.0 = 16.0 g

So, Percentage of nitrogen = 100 (14.0 / 30.0) = 46.7 %

Percentage of oxygen = 100 (16.0 / 30.0) = 100 - 46.7 = 53.3 %

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Mohit Ghai

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