Question

When the temperature of an automobile tire is 20°C, the pressure in the tire reads 32 PSI on a tire gauge. (The gauge measures the difference between the pressures inside and outside the tire.) What is the pressure when the tire heats up to 40°C while driving? You may assume that the volume of the tire remains the same and that atmospheric pressure is a steady 14 PSI.

Answer 1

Final pressure = = 49.13 PSI,

Pressure shown by gauge = 35.13 PSI

Step-by-step explanation:

Given data:

Initial temperature = T1 = 20°C

Final temperature = T2 = 40°C

Initial gauge pressure = P1 = 32 PSI

Final pressure = P2 =?

Volume of gas remains constant

Solution:

As the volume of the gas in constant so the ration of pressure and temperature of gas remains same.

It means:

P1/T1 = P2/T2

Because the gauge measures the pressure difference, the initial actual pressure inside the tire is:

P₁ = 32 + 14 = 46 PSI

Temperature:

T1 = 20 + 273 = 293 K

T2 = 40 +273 = 313 K

P1 = 46 PSI =

From the formula:

P1/T1 = P2/T2

By putting values:

= 46/293 = P2/313

= 0.156 = P2/313

= P2 = 49.13 PSI

Now the pressure shown by the gauge:

= 49.13 - 14 = 35.13 PSI.

Final pressure = 35.13 PSI

Answer 2

The initial pressure in the tire is given by the difference between the gauge pressure and the atmospheric pressure:

`p_1 = 32 PSI - 14 PSI = 18 PSI = 1.24 \cdot 10^5 Pa`

Let's also convert the initial and final temperature of the tire into Kelvin:

`T_1 = 20^(\circ)C=293 K`

`T_2 = 40^(\circ)C = 313 K`

We know that the volume of the tire remains the same, so we can use the fact that the ratio between pressure and temperature of the tire remains constant:

`(P)/(T)=constant`

which can be rewritten as

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

from which we can find the value of the final pressure, p2:

`P_2 = T_2 (P_1)/(T_1)=(313 K) (1.24 \cdot 10^5 Pa)/(293 K)=1.32 \cdot 10^5 Pa`

And converted back into PSI,

`p_2 = 19.1 PSI`