Temperature Of the Air in a Tire

[sona1177]

Homework Statement

Before starting out on a long drive, you check the air in your tires to make sure they are properly inflated. The pressure gauge reads 214 kPa and the temperature is 15 degrees celsius. After a few hours of highway driving, you stop and check the pressure again. Now the gauge reads 241 kPa. a) What is the temperature of the air in the tires now? b) Suppose you now (unwisely) decide to bleed air from the tires. If you let out enough air so that the pressure returns to 214 kPa, what percentage of the air molecules did you let out of the tires? c) What is the gauge pressure after the tires cool back to 15 degrees celsius?

Homework Equations

I figured parts a and b out. I am little confused what they mean in part c because isn't the pressure gauge at 15 degrees celsius 214 kPa? The answer is 189 kPa but I don't know why. Thanks.

The Attempt at a Solution

 

[gneill]

I figured parts a and b out. I am little confused what they mean in part c because isn't the pressure gauge at 15 degrees celsius 214 kPa? The answer is 189 kPa but I don't know why. Thanks.

The idea is that, since you've let some air out of the tire in the previous step when the temperature is higher, when it cools back to the starting temperature (15 C) the pressure will drop below the original starting pressure.

 

[sona1177]

Thanks but if I use the pressure when the temperature is higher and the the temperature when the pressure is lower, I end up with the samw gauge pressure given in the problem. I'm still confused :(

 

[Borek]

Thanks but if I use the pressure when the temperature is higher and the the temperature when the pressure is lower, I end up with the samw gauge pressure given in the problem. I'm still confused :(

You start with 15 deg C and 214 kPa.

When you stop you have T and 241 kPa. You have already stated you calculated T.

You let some air out and you have T and 214 kPa.

Now the temperature goes down to 15 deg C. Calculate new pressure knowing T, 214 kPa, and 15 deg C.

 

[rwisz]

a) To find the temperature of the air in the tires after a few hours of highway driving, we can use the ideal gas law: P1V1/T1 = P2V2/T2. We know that the pressure has increased from 214 kPa to 241 kPa, and the volume of air in the tires has not changed. We can also assume that the number of moles of air in the tires remains constant. Therefore, we can rearrange the equation to solve for T2:

T2 = (P2/P1) x T1 = (241 kPa/214 kPa) x (15 + 273.15 K) = 288.24 K

b) To find the percentage of air molecules that were let out of the tires when the pressure was reduced from 241 kPa to 214 kPa, we can use the ideal gas law again. We know that the pressure, volume, and number of moles of air in the tires have changed, but the temperature remains constant at 288.24 K. Therefore, we can rearrange the equation to solve for the number of moles:

n2 = (P2V2)/(RT) = (214 kPa)(V)/(8.314 J/mol*K)(288.24 K) = 0.0221 V

Where V is the initial volume of air in the tires. Therefore, the percentage of air molecules let out is:

(0.0221 V - V)/V x 100% = -97.79%

c) In part c, we are asked to find the gauge pressure after the tires cool back to 15 degrees celsius. This means that the temperature is returning to its original value, but the volume and number of moles of air in the tires remain the same. Therefore, we can use the ideal gas law again to solve for the pressure:

P = (nRT)/V = [(0.0221 V)(8.314 J/mol*K)(15 + 273.15 K)]/V = 189 kPa

The reason why the gauge pressure is now 189 kPa instead of the original 214 kPa is because when we let out air from the tires, we decreased the number of moles of air in the tires. This means that there are fewer air molecules in the same volume, resulting in a decrease in pressure. When the tires cool back to 15 degrees c