I may be totally off base here, but it occurs to me that this formula applies
to a solid container that has no give the way tires do. If you put an
equal amount of gas in a tire and a balloon and heat them up the same amount,
the balloon will allow for more expansion than the tire and therefore
have less pressure change from the same temperature increase. By the same
token, it seems to me that the tire would give a bit more than a solid
cylinder and therefore have a smaller pressure increase than said solid
container.
Am I all wet here, or does this seem likely to others?
Joe (C)
Larry Young wrote:
>
> Ok, I tried to be quiet, but my chemical engineering background won't let me.
> Air in your tires (or nitrogen, argon, hydrogen, hellium, whatever) obeys the
> ideal gas law, PV=nRT. The tire volume is fixed, so the absolute pressure
> increases in direct proportion to the absolute temperature. Let's say you
> have 20 psi in your tires at 70 degrees F. The absolute temperature is 70 +
> 460 = 530 degrees R and at sea level the absolute pressure is 20 + 14.7 =
> 34.7. If the temperature drops to 50 deg F (510 deg R), the pressure will
> drop to 34.7(510/530) - 14.7 = 18.7 psi. If the temperature increases to 130
> deg F (590 deg R) during a race, the pressure goes to 34.7(590/530) - 14.7 =
> 23.9 psi.
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