The pressure in an automobile tire depends on thetemperature of the air in the tire. When the air temperature is25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire whenthe air temperature in the tire rises to 50°C. Also, determinethe amount of air that must be bled off to restore pressure toits original value at this temperature. Assume the atmosphericpressure to be 100 kPa.

Respuesta :

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

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