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 1.70 atm at 300 K. At the end of the trip, the gauge pressure has increased to 2.60 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

part a:

using ideal gas law:

P1*V1/T1=P2*V2/T2

where P1=initial pressure=atmospheric pressure+gauge pressure=1+1.7=2.7 atm

P2=final pressure=1+2.6=3.6 atm

T1=initial temperature=300 K

T2=final temperature

V1=V2 (volume remains constant)

then P1/T1=P2/T2

==>2.7/300=3.6/T2

==>T2=3.6*300/2.7=400 K

hence temperature of air inside the tire is 400 K

part b:

as P*V=n*R*T

where n is number of moles and R=gas constant

==>n=(P*V)/(R*T)

initially number of moles=P1*V1/(R*T1)=P1*V1/(R*300)

final number of moles with P2=P1 (final pressure=original pressure)=P1*V2/(R*T2)=P1*V1/(R*400)

then percentage of moles to be released=(initial number of moles-final number of moles)/initial number of moles

=((1/300)-(1/400))/(1/300)=0.25=25%

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