Question

The tire is inflated (while cold) to a volume of 11.8 L and a gauge pressure of 36.0 psi (Note: The gauge pressure is the difference between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi.) at a temperature of 12.0 ∘C. While driving on a hot day, the tire warms to 65.0 ∘C and its volume expands to 12.2 L. What is the pressure in the tire after warming on a hot day?

Express your answer in pound-force per square inch to 3 significant figures.

(My answer was 41.4, but its not right. I am missing something here......)

Answer 1

Pressure in tire = 24.4 psi

Explanation

initial pressure in tire = gauge pressure - atmospheric pressure

initial pressure in tire = 36.0 psi - 14.7 psi

initial pressure in tire = 21.3 psi

initial volume of tire = 11.8 L

initial temperature of tire = 12.0 ^oC = (12.0 + 273) K = 285 K

final volume of tire = 12.2 L

final temperature of tire = 65.0 ^oC = (65.0 + 273) K = 338 K

According to combined gas law,

(P1 * V1) / T1 = (P2 * V2) / T2

where

P1 = initial pressure = 21.3 psi

V1 = initial volume = 11.8 L

T1 = initial temperature = 285 K

P2 = final pressure

V2 = final volume = 12.2 L

T2 = final temperature = 338 K

P2 = P1 * (V1 / V2) * (T2 / T1)

P2 = 21.3 psi * (11.8 L / 12.2 L) * (338 K / 285 K)

P2 = 24.4 psi

Pressure inside the tire = 24.4 psi