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 1.80 atm at 300 K. At the end of the trip the gauge pressure has increased to 2.27 atm. (a) Assuming the volume has remained constant, what is the temperature of the air inside the tire? (b) What percentage of the original mass of air in the tire should be released so the pressure returns to the original value? Assume the temperature remains at the value found in (a), and the volume of the tire remains constant as air is released. Also assume that the atmospheric pressure is 1.00 atm and remains constant. Caution: Gauge pressure is measured relative to the atmospheric pressure. The absolute pressure in the tire at the beginning of the trip is 2.80 atm.

A is between330 and 380 K

B is between 10 and 20%.

Explanation / Answer

Straightforward application of pV = nRT (a) Rearrange as p/T = nR/V, since only p and T change. Initially, p/T = 1.8atm / 300ºK. Finally, p/T = 2.2atm / T T = (2.2/1.8)·300ºK = 367ºK (b) now rearrange as p/n = RT/V, since only p and n change. Initially, p/n = 2.2atm/n1 Finally, p/n = 1.8atm/n2 Since these initial states both = RT/V, n2/ n1= 1.8/2.2 = 0.82, so 18% of the mass should be released.

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