Answer:
The pressure rise in the tire when the air temperature in the tire rises to 50°C is 337.43 KPa.
The amount of air that must be bled off to restore pressure 0.007 Kg
Explanation:
Knowing
T1 = 25°C = 298 K
T2 = 50°C = 323 K
volume of the tire = V = 0.025 [tex]m^{3}[/tex]
P = 210 kPa (gage)
Pabs = 210 + 101 = 311 KPa
Before the temperature rise
P1 V1 = m1 R1 T1
m1 = [tex]\frac{P1 V1}{R1 T1} = \frac{310 * 10^{3} * 0.25 }{287 - 298} = 0.091 Kg\\ \\[/tex]
After the temperature rise
P2 = [tex]\frac{m2 * R * T2}{V2} = \frac{0.091 *287*323 }{0.025} = 337.43 KPa[/tex]
after bleeding the pressure and the volume returns to its first value
P1 = P2 and V1 = V2
then
[tex]\frac{m2 * R * T2}{V} = \frac{m1 * R * T1}{V}[/tex]
m2 = [tex]\frac{m1*T1}{T2}[/tex]
m2 = [tex]\frac{0.091*298}{332} = 0.084 Kg\\\\[/tex]
mbleed = m1 - m2 --> mbleed = 0.91 - 0.84 = 0.007 Kg
P2 = 337.43 KPa
mbleed = 0.007 Kg