The initial pressure in the tire is given by the difference between the gauge pressure and the atmospheric pressure:
[tex]p_1 = 32 PSI - 14 PSI = 18 PSI = 1.24 \cdot 10^5 Pa[/tex]
Let's also convert the initial and final temperature of the tire into Kelvin:
[tex]T_1 = 20^{\circ}C=293 K[/tex]
[tex]T_2 = 40^{\circ}C = 313 K[/tex]
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:
[tex]\frac{P}{T}=constant[/tex]
which can be rewritten as
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
from which we can find the value of the final pressure, p2:
[tex]P_2 = T_2 \frac{P_1}{T_1}=(313 K) \frac{1.24 \cdot 10^5 Pa}{293 K}=1.32 \cdot 10^5 Pa[/tex]
And converted back into PSI,
[tex]p_2 = 19.1 PSI[/tex]
Final pressure = = 49.13 PSI,
Pressure shown by gauge = 35.13 PSI
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
