How much does tire pressure affect the weight of the wheels Kind of inspired by this question, but something I've been thinking about for a while. 

How much does the air in a bike tire weigh?  Is it an appreciable amount? Is there a point where using a wider tire, like at 28c at 80 psi would be lighter than a 25c tire at 100psi? Obviously, this depends on the specific tires used.  I don't have a scale precise enough to measure, and I don't have the math/physics knowledge to figure this out.
The ideal gas law (which is a good approximation in this case) says PV=nRT where P is pressure, V is volume, n is mols of gas, R is the ideal gas law constant, and T is temperature in Kelvin. 

Thus, solving for n, we see n = (PV)/(RT). Then, assuming air is made up of {gas1, gas2,...} with fractions {p1,p2,...} (so p1+p2+...=1) and corresponding molar masses {m1,m2,...}, the mass of air in a tire is (PV/(RT))(p1*m1+p2*m2+...). So, what we see is that the mass of air in a tire is directly proportional to the volume of the tire and directly proportional to the pressure in the tire, and inversely proportional to the temperature of the air in the tire. 

We will make the following (reasonable) assumptions: Assume the temperature is around room temperature (293 Kelvin) and the volume of the tire regardless of pressure is the same (primarily determined by the shape of the rubber, assuming not severely under/over inflated). For convenience, air is about {nitrogen, oxygen} with {p1,p2}= {0.8,0.2} and molar masses {28 g/mol,32 g/mol}. Thus, under these assumptions (V is fixed, and T is fixed), the mass of the air in tire grows linearly with pressure. 

So, the mass of air in a tire of volume V and pressure P and temperature T is about (PV/RT)(0.8*28+0.2*32) grams. It may be better to write it as "P ((V/(RT)) (0.8*28+0.2*32)) grams" noting that V/(RT) is a constant for us. 

Since I don't want to put the units into wolfram alpha carefully, you can put in the entry "(7 bar* 10 gallons)/(ideal gas constant*293 Kelvin)*(0.8*28+0.2*32)" and read the result off in grams (ignoring the unit it says there) to get an estimate for the weight of air in a 7 bar (~100 psi), 10 gallon volume tire as around 313 grams. Is 10 gallons reasonable? No. 

Lets be crude about estimating the volume of a tube using a torus. The volume of a torus is V=(pi*r^2)(2*pi*R) where R is the major radius and r is the minor radius. Google will calculate it for you (and has a picture of what major and minor radius is).

I can't be bothered to actually go outside and measure these things, but lets be crude and use a massive tire. Say the minor radius is 2 inches, and the major radius is 15 inches (this is probably larger the size of the tire on something like a Surly Moonlander). This has a volume of about 5 gallons. If you were a nutcase and running this at 7 bar, it would be around 150 grams of air. At a more reasonable 1 bar or 2 bar, youd be at 45 or 90 grams. 

What about a thin road bike tire? Lets also assume the major radius is around 15 inches, and the minor radius is around a half inch. Thats around 0.3 gallons of volume. Plugging into our formula, at 7 bar, we see that this is about 9 grams. At 10 bar, a whopping 13.5 grams.