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 2.0 atm at 274 K. At the end of the trip, the gauge pressure in the tire has increased to 2.4 atm. Assuming the volume of the air inside the tire has remained constant, what is its tem­perature at the end of the trip? Answer in units of K. Air is released from the tire during a short time interval, so that the temperature remains at the value found in part 1. Assume that the amount of air released is small enough for the tire's volume to be treated as constant. What quantity of air (as a fraction of N_i, the initial number of particles) must be re­leased from the tire so that the pressure re­turns to its initial value? Answer in units of N_i.

Explanation / Answer

007 (part 1 of 2) -

Assuming the volume of the air inside the tire has remained constant, its temperature at the end of the trip which will be given as :

using a gay-lussac's law, we have

P1 / T1 = P2 / T2

(2 atm) / (274 K) = (2.4 atm) / T2

T2 = [(2.4 atm) / (2 atm)] (274 K)

T2 = 328.8 K

008 (part 2 of 2) -

Assume that the amount of air released is small enough for the tire's volume to be treated as constant.

using an ideal gas law, we have

for an initial conditions ; P1 V = ni R T

(2.4 atm) V = ni R (328.8 K)                                                           { eq.1 }

for final conditions ; P2 V = nf R T

(2 atm) V = nf R (328.8 K)                                                               { eq.2 }

dividing eq.1 by eq.2 & we get -

(2.4 atm) V / (2 atm) V = ni R (328.8 K) / nf R (328.8 K)     

(2.4) / (2) = ni / nf

nf / ni = (2 / 2.4)

nf = (0.833) ni

where, n = number of moles

and which is directly proportional to the number of particles.

So that, we have

Nf = (0.833) Ni

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