Answer:
0.28125
Explanation:
Given:
Loss in the tire pressure = 90 kPa
Gage pressure = 220 kPa
Atmospheric pressure = 100 kPa
Now,
the absolute pressure is given as:
[tex]P_{abs}=P_{gage}+P_{atm}[/tex]
on substituting the values, we get
the initial absolute pressure of the tire as, P₁: 220 + 100 = 320 kPa
also, we know
P×V = nRT
where,
V represents the volume of air in the tire
n is the number of moles of gas in the tire
R is the gas constant
and, T represents temperature
now, the volume and temperature for both cases remains the same
thus, for the initial pressure, we have
P₁V = n₁RT
or
320 × V = n₁RT
now for case 2 we have
P₂ = 320 kPa - 90 kPa = 230 kPa
thus,
P₂V = n₂RT
or
230 × V = n₂RT
now, by dividing 1 by 2, we get
[tex]\frac{320}{230}=\frac{n_1}{n_2}[/tex]
or
[tex]\frac{n_2}{n_1}=0.71875[/tex]
thus,
fraction of air lost is 1 - 0.71875 = 0.28125