Comparing pressure of air in a tire installed in a car & a free tire

[scoutfai]

Imagine an ordinary car wheel with rubber tire, installed in a car axle. So the weight of the car (some of the weight of course, because there are 4 wheels) is pressing on it. Let's denote this pressure of the air in the tire as $P_{c}$.

Now the same wheel (with its rim and tire) remove from the car axle, so it is now freely lying on the floor. Hence no car weight is pressing on it. Let's denote the pressure of the air in the tire as $P_{f}$.

Question: Is $P_{c}$ different than $P_{f}$ ? How to show it by using ideal gas law equation? You may assume the weight of the rim as negligible if needed.

My personal guess: $P_{c}$ is more than $P_{f}$ because with the weight of the car pressing on the tire, volume of tire (hence volume of gas) reduces, thus pressure increases. In other words, if a tire pressure is 30psi while installed on car, once take it out from the car axle, its pressure will drop.

 

[mrspeedybob]

You are correct, but the difference is so small that you will not notice it on a tire gauge.

Suppose your car is 4000 pounds with 1000 pounds on each wheel. Also suppose your tires are 10 inches wide. At 30 psi your contact patch for each wheel will be about 33 square inches. Since the tires are 10 inches wide that means your contact patch is only 3.3 inches long. Deforming the tire with a typical 6.5+ foot circumfrence so that there is a 3.3 inch flat spot will change it's volume (and pressure) only very slightly.

 

[scoutfai]

You are correct, but the difference is so small that you will not notice it on a tire gauge.

Suppose your car is 4000 pounds with 1000 pounds on each wheel. Also suppose your tires are 10 inches wide. At 30 psi your contact patch for each wheel will be about 33 square inches. Since the tires are 10 inches wide that means your contact patch is only 3.3 inches long. Deforming the tire with a typical 6.5+ foot circumfrence so that there is a 3.3 inch flat spot will change it's volume (and pressure) only very slightly.

Hence, if I was to pump air to a tire but I can only carry a free wheel instead of driving the car to the air pump, says I wish to have 30psi in my tire when it is installed on axle, then I should just directly pump in 30psi instead of a slightly lower pressure (for instance, 29psi) because the increase in pressure after the tire is installed is just so insignificant?

 

[mrspeedybob]

Correct.

 

[PJC01]

I would like to clarify that the pressure of air in a tire is not affected by the weight of the car or any external force. The pressure of air in a tire is solely determined by the temperature and volume of the gas inside the tire.

According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. This equation shows that the pressure is directly proportional to the temperature and inversely proportional to the volume of the gas. This means that if the temperature remains constant, the pressure of the gas will increase as the volume decreases.

In the case of a tire installed in a car, the weight of the car does not affect the volume of the gas inside the tire. The tire is designed to withstand the weight of the car and maintain its volume, hence the pressure remains constant. Therefore, Pc and Pf will be the same.

If we remove the tire from the car axle, the volume of the gas inside the tire will not change as there is no external force acting on it. Therefore, the pressure will remain the same. Pc and Pf will still be the same.

In conclusion, the pressure of air in a tire does not change with the weight of the car or any external force. It is solely determined by the temperature and volume of the gas inside the tire. Therefore, Pc and Pf will be the same.