Question

An automobile tire is filled to 32 psi at an air temperature of 40° F. a. After the car is driven for some time, the temperature of the air is 145° F. What is the pressure, if you assume the volume of the tire is constant? b. If a person wants to maintain a pressure of 32 psi, what should the initial air pressure should be put in the tire?

Answer 1

(a) The final pressure will be 38.73 psi

(b) The initial pressure will be 26.44 psi

Explanation :

(a) Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

`P\propto T`

or,

`(P_1)/(T_1)=(P_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 32 psi

`P_2` = final pressure of gas = ?

`T_1` = initial temperature of gas =
`K=(5)/(9)* (^oF-32)+273=(5)/(9)* (40-32)+273=277.44K`

`T_2` = final temperature of gas =
`K=(5)/(9)* (^oF-32)+273=(5)/(9)* (145-32)+273=335.77K`

Now put all the given values in the above equation, we get:

`(32psi)/(277.44K)=(P_2)/(335.77K)`

`P_2=38.73psi`

Thus, the final pressure will be 38.73 psi

(b) Now we have to calculate the initial air pressure should be put in the tire.

`(P_1)/(T_1)=(P_2)/(T_2)`

where,

`P_1` = initial pressure of gas = ?

`P_2` = final pressure of gas = 32 psi

`T_1` = initial temperature of gas =
`K=(5)/(9)* (^oF-32)+273=(5)/(9)* (40-32)+273=277.44K`

`T_2` = final temperature of gas =
`K=(5)/(9)* (^oF-32)+273=(5)/(9)* (145-32)+273=335.77K`

Now put all the given values in the above equation, we get:

`(P_1)/(277.44K)=(32psi)/(335.77K)`

`P_1=26.44psi`

Thus, the initial pressure will be 26.44 psi