A tire is filled with air at 15°C to a gauge pressure of 210 kPa.?
A tire is filled with air at 15°C to a gauge pressure of 210 kPa. If the tire reaches a temperature of 38°C, what fraction of the original air must be removed if the original pressure of 210 kPa is to be maintained?
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Hello
use the gas law:
n1 = p*V/(R*T1)
n2 = p*V/(R*T2)
where p, V, R = constants
now find the ratio n2/n1
n2/n1 = (pV/(R*288))/((pV/(R*311))
n2/n1 = 288/311 = 0.926 = 92.6/100 %
so 7.4 % or 7.4/100 or 1/13.5 must be removed
Regards