Answer
0.074, or 7.4 percent of the original air.
Work Step by Step
Use the ideal gas model for air. The volume of the tire remains constant. Let “1” denote the situation before, and “2” denote the situation after.
$$PV=n_1RT_1$$
$$PV=n_2RT_2$$
$$\frac{n_2}{n_1}=\frac{T_1}{T_2}=\frac{273.15+15}{273.15+38}=0.926$$
The fraction that must be removed is 1-0.926 = 0.074.