Before beginning a long trip on a hot day, a driver inflates an automobile tire

Question

Before beginning a long trip on a hot day, a driver inflates an automobile tire to a gauge pressure of 1.74 atm at 300 K. At the end of the trip, the gauge pressure has increased to 2.04 atm.

(a) Assuming the volume has remained constant, what is the temperature of the air inside the tire?
K

(b) What percentage of the original mass of air in the tire should be released so the pressure returns to its original value? Assume the temperature remains at the value found in part (a) and the volume of the tire remains constant as air is released.
%

Explanation / Answer

a)
P1_gauge = 1.74 atm

Initial absolute pressure,
P1 = Patm + P1_gauge

= 1 + 1.74

= 2.74 atm

T1 = 300 K

P2_gauge = 2.04 atm

final absolute pressure,
P2 = Patm + P2_gauge

= 1 + 2.04

= 3.04 atm

T2 = ?

at constant volume, P/T = constant

so, P1/T1 = P2/T2

==> T2 = T1*(P2/P1)

= 300*(3.04/2.74)

= 333 K <<<<<<<<<-------------------Answer

b)

use

P1*V1 = n1*R*T1 ---(1)

P2*V2 = n2*R*T2 ---(2)

take equation(2)/equation(1)

P2*V2/(P1*V1) = n2*R*T2/(n1*R*T1)

P2/P1 = n2/n1

==> n2/n1 = P2/P1

= 2.74/3.04

= 0.901

n2 = n1*0.901

n2 = n1*(1 - 0.099)

n2 = n1 - 0.099

= n1 - 9.9% of n1

so, 9.9% of mass must be released

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