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Why do you need to add air to your tires when the temperature drops? The temperature inside and outside of the tire is the same (let's assume that you aren't currently driving, and your car has been sitting for a few hours). I understand that the absolute pressure of the tire drops, because P=nRT/V but why doesn't that apply to the outside atmosphere as well?

What I don't understand is why relative pressure of the tire (to the atmospheric temperature) doesn't remain the same. I'm thinking that it might have something to do with the weight of the car, or that the atmosphere doesn't exactly have a fixed volume. Any insight would be much appreciated!

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  • $\begingroup$ Analyze the problem as constant volume inside the tire and constant pressure outside the tire. $\endgroup$ – casey Oct 17 '14 at 0:32
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I believe the confusion is that you believe pressure will always increase as temperature increases. This is only the case in a closed environment such as inside the tire. In an environment such as the atmosphere which is, essentially, in an unconfined environment, the density will decrease with temperature as well. This does not happen inside of a closed environment since $\Large density = \frac{mass}{volume}$ and neither the mass nor the volume change.

However, you are correct in that if you were to drive your car in a closed environment, the tires wouldn't need to be changed (assuming an equal temperature throughout the containers.

$$P=\frac{nrt}{v}$$ so, logically: $$v\to 0, P\to \infty\\ v\to \infty, P\to 0$$ However, as I said, this requires an unchanging volume. Instead, the gass will expand and the density will decrease leaving the pressure, $P$, more or less the same.

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