How can one calculate the mass of air inside a tyre, given a particular tyre size; a pressure, in $kPa = \frac{1000kg}{m\cdot s^2}$; and assuming room temperature, and normal air composition? I can't quite work out what part of the equations I'm missing to remove the $s^2$ component.
I realise that surface/volume ratio is important, but for the purposes and approximation to a torus would be fine. For example, assuming a 700cx25 bicycle tyre, we might assume a torus where the diameter between centre of the two cross-section circles is about 630mm, and the diameter of the circles themselves is about 30mm. Let's assume a tyre pressure of 500kPa (~73psi).
Rough volume would be $630$mm$\times\pi \times \pi (15$mm$)^2 = 1.4\times10^6$mm$^3 = 0.0014$m$^3$
Rough surface area: $630$mm$\times\pi \times 30$mm$\times\pi = 1.87\times10^5$mm$^2 = 0.187$m$^2$