You are pumping up a bicycle tire with a compressor. The inner tube is initially

Question

You are pumping up a bicycle tire with a compressor. The inner tube is initially completely deflated, and you inflate it to a final pressure of 100 psig and final volume confined by the tire tread and rim (V_tire = 800 mL). The compressor supplies air steadily at 25 degree C and 180 psig. Assume air is an ideal gas, with C_p = (7/2) R and R = 8.314 J/mol K. Clearly state all assumptions. Is this system open or closed? Explain your reasoning. If the bicycle tire had an initial volume of 200 mL, calculate the work of expansion of the inner tube (against the atmosphere). Assume negligible tensile force from the inner tube. Unfortunately, your bicycle tire is completely deflated. Starting with a general energy balance, calculate the final temperature inside the tire. Assume that the time to inflate the tire is much less than the time for heat dissipation through the tire and rim. For this process, it is possible for the final temperature to be less than the compressor supply temperature? Explain briefly. After some time, the temperature of the tire reaches 25 degree C. What is the final tire pressure in psig

Explanation / Answer

C)

The tire of volume V must end up with n = P V/RT0 moles of air (assumed to be an ideal gas) to be at pressure P at ambient temp T0.

Suppose the tire has pressure P0 to start with, then it already contains n0 = n(P0/P) = P0V/RT0 moles of air.

So, we must add n' = n n0 moles from the tank into the tire, which we can also express as

n' = n(P P0)/P

The volume V' of these n' moles in the tank at pressure Pc and temperature T0 is,

V = n'RT0/Pc = nRT0*[(P P0)/PPc]

As air flows from the tank into the tire through the “one-way” valve, the tank does work Wc at constant (by assumption) pressure Pc given by

Wc = PcV' = nRT0[(P P0)/P],

on the n' moles of air. This work appears as internal energy of the air that is injected into the tire.

Given the assumption that air is an ideal gas, the motion of air through the “one-way” valve into the tire is a free expansion, and no work is done in the latter process. So, the internal energy U of the air in the tire rises by amount WC (assuming that the air transfer is adiabatic).

Therefore, the temperature of the air in the tire rises to Tmax = T0+T , where nCvT = U = Wc. Here, Cv is the molar specific heat of air at constant volume. Thus,

T = Wc/nCv = (RT0/Cv)[(P P0)/P]

In the given case,

P0 = Patm = 14.7 psia (since the tire is completely deflated)

P = 100 psig = 100 + Patm psia = 114.7 psia

Cp = 7R/2 => Cv = Cp - R = 5R/2

T0 = 250C = 298 K

Therefore,

T = (2/5)[1 (P0/P)]T0,

T = 103.92 K

Therefore, the temperature in the tire rises to

Tmax = T0 + T = 401.92 K

After some time the air cools down to T0 and pressure will be the desired value.

D) No, final temperature will always be greater than T0(compressor supply temperature)

E) Final pressure will be 100 psig

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