We measure the pressure in an automobile tire early in the morning when the temp
ID: 1471114 • Letter: W
Question
We measure the pressure in an automobile tire early in the morning when the temperature is 60F. We read 35.1 psi (lb/in2 ) on the pressure gauge. This is the pressure with respect to atmospheric pressure. To obtain the absolute pressure of the air in the tire we must add to this the pressure of the atmosphere. On that day, the atmospheric pressure is 0.87 × 105 Pa. Later in the day, the temperature is 84.9 F but the atmospheric pressure is unchanged. What will we read (psi) on the pressure gauge if we measure the pressure in the tire then? Assume that the volume of the tire remains the same.
Explanation / Answer
pressure measured = 35.1 psi
= 35.1*6894.75 Pa
=242005.725 Pa
actual Pressure P1 = 242005.725 +0.87x10^5 = 3.29x 10^5 Pa
Temperature T1 = 60F = 15.35C =288.35 K
new temperature T2 = 84.9 F = 29.38 C = 302.38 K
P1/P2 = T1/T2
P2 = 3.29x 10^5*302.38/288.35 = 3.45x10^5
pressure reding in the gauge = 3.45x10^5 -0.87x10^5
= 2.58x10^5 Pa
= 37.4 psi
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