Question

An automobile tire has a maximum rating of 38.0 psi (gauge pressure). Part A 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 total pressure (gauge pressure + atmospheric pressure) in the tire after warming on a hot day? Express your answer in atmospheres to 3 significant figures.

Answer 1

From gas law P1V1/T1= P2V2/T2

one can calculate the pressrue after expansion to 12.2L

given P1= 36psi = 36+14.7= 50.7 Psi= 50.7/14.7 atm=3.45 atm , V1= 11.8L, T1= 12 deg.c= 12+273= 285K

V2= 12.2L, T2= 65 deg.c= 65+273= 338K

P2* 12.2/338= 3.45*11.8/285

P2= 3.45* 11.8*338/(285*12.2) =3.96 atm,

P2= 3.96*14.7 psi =58psi (g)=

An automobile tire has a maximum rating of 38.0 psi (gauge pressure). Part A 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 total pressure (gauge pressure + atmospheric pressure) in the tire after warming on a hot day? Express your answer in atmospheres to 3 significant figures