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?

(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.

[tex]P\propto T[/tex]

or,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 32 psi

[tex]P_2[/tex] = final pressure of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]K=\frac{5}{9}\times (^oF-32)+273=\frac{5}{9}\times (40-32)+273=277.44K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]K=\frac{5}{9}\times (^oF-32)+273=\frac{5}{9}\times (145-32)+273=335.77K[/tex]

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

[tex]\frac{32psi}{277.44K}=\frac{P_2}{335.77K}[/tex]

[tex]P_2=38.73psi[/tex]

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.

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = ?

[tex]P_2[/tex] = final pressure of gas = 32 psi

[tex]T_1[/tex] = initial temperature of gas = [tex]K=\frac{5}{9}\times (^oF-32)+273=\frac{5}{9}\times (40-32)+273=277.44K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]K=\frac{5}{9}\times (^oF-32)+273=\frac{5}{9}\times (145-32)+273=335.77K[/tex]

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

[tex]\frac{P_1}{277.44K}=\frac{32psi}{335.77K}[/tex]

[tex]P_1=26.44psi[/tex]

Thus, the initial pressure will be 26.44 psi