While tire pressure is critical for getting optimal fuel economy and reduced tir
ID: 1717516 • Letter: W
Question
While tire pressure is critical for getting optimal fuel economy and reduced tire wear, the pressure in your car tires depends on the temperature. If you fill your tires at 25°C to 250 kPa (gauge), calculate the gauge pressure in them when the temperature drops to -40°C. If the air volume of one of your tires is 0.5 m³, what mass of air is required to get the pressure back up to 250 kPa (gauge) at -40°C? Assume the atmospheric pressure is 90 kPa. Neglect the change in volume in your tires due to the elasticity of the rubber.
Explanation / Answer
>> Now, let P1 and T1 are the initial pressure and Temperature
and, P1g = 250kPa ( Gauge)
>> As, Pa = Atmospheric Pressure = 90 kPa
=> P1 = Pa + P1g
=> P1 = 340 kPa
also, T1 = 25 degree C = 298 K
>> Now, let P2 and T2 are the final pressure and Temperature
and, P2 = P2g + Pa = ?
also, T2 = -40 degree C = 233 K
>> As, Volume remains Constant
So, applying Charle's Law, i.e. P is directly proportional to T
=> P1/T1 = P2/T2
=> P2 = P1*(T2/T1)
=> P2 = 340*(233/298)
=> P2 = 265.54 kPa
=> P2 = Pa + P2g = 265.54
=> P2g = Final gauge Pressure = 175.84 kPa .....ANSWER......
>> Now, let mass of air required = m
As, PV = m*RT
Now, as P and m is changing
=> P3/m1 = P3/m2
As, mass = Volume*density
and density of air will remin same
=> m3/m2 = V3/V2
=> Above equation becomes:
=> P3/P2 = V3/V2
=> V3 = 0.5*(340/265.54)
=> V3 = New Volume of air = 0.64 m3
[ NOTE:- these V2, V3 represents the amount of air inside tyre, not the volume of tyre, as air in stage 3 has been compressed ]
So, V3 - V2 = 0.64 - 0.5 = 0.14 m3
So, mass of air is required to get the pressure back up to 250 kPa (gauge) at -40°C = (V3 - V2)*density
= 0.14*1.225 = 0.172 Kg
=> Mass of air is required to get the pressure back up to 250 kPa (gauge) at -40°C = 0.172 Kg ...ANSWER...
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