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 1.65 atm at 300 K. At the end of the trip, the gauge pressure has increased to 2.05 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

P1 = 1.65 + 1 atm
P2 = 2.05 + 1 atm
T1 = 300 K
P1/T1 = P2/T2
T2 = 3.05*300/2.65 = 345.283 K

now, P2V = nRT2
3.05V = nR*345.283
nR/V = 0.00883

P1V = mRT2
2.65V = mR*345.283
mR/V = 0.00767

m/n = 0.86918
(n - m)/n = 0.13081 *100 = 13.0819 percent

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