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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$^2$

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  • $\begingroup$ @Qmechanic this wasn't a homework question. $\endgroup$ – naught101 Sep 17 at 0:35
  • $\begingroup$ Hi naught101. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Sep 17 at 4:50
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You need an equation for the density of the gas as a function of temperature and pressure. Assuming the tyre is full of air, this is reasonably close to an ideal gas so the molar volume is given by:

$$V_m = \frac{RT}{P}$$

where R is the ideal gas constant and and the average molecular weight of air (20% O$_2$, 80% N$_2$) is about 28.8. From this you can work out the density at the pressure and temperature of your car tyre, and since you know the volume this immediately gives you the mass.

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    $\begingroup$ Or you could just use the experimental density of air at room TP is around 1.2 kg/m^3. Together with how many Bar you inflate your tire to $\endgroup$ – Martin Beckett Aug 1 '12 at 12:58
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Start with the ideal gas law: $$PV=nRT$$

Let's compare the volume of air in the tyre with a similar volume outside the tyre under normal atm conditions.

We know that $V$, $R$ and $T$ are constant so:

$$\frac{P_{atm}}{n_{atm}}=\frac{P_{tyre}}{n_{tyre}}$$

We know that $$n \propto m \propto \rho$$

So we can say: $$\frac{P_{atm}}{\rho_{atm}}=\frac{P_{tyre}}{\rho_{tyre}}$$

We know that: $$\rho_{tyre} = \frac{m_{tyre}}{V_{tyre}}$$

So: $$m_{tyre}=\frac{\rho_{atm} P_{tyre} V_{tyre}}{P_{atm}}$$

Values:

$\rho_{atm}=1.225 kg/m^3$ ... found on web

$P_{tyre}=73 psi =4.97 atm$ ... your value for pressure , converted to atm on web

$P_{atm} = 1 atm$ ... by definition

$V_{tyre} = 0.0014 m^3$ ... given by you

Sub all values in:

$$m_{tyre}=\frac{(1.225 kg/m^3)(4.97 atm)(0.0014 m^3)}{1 atm}$$

Cancel units $atm$ and $m^3$, and multiply out numbers to get:

$$m_{tyre}=0.0085 kg$$

So you are talking $8.5$ grammes of air.

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  • $\begingroup$ You have a tire that has a volume of only 1.4 liters? $\endgroup$ – David White Sep 15 at 1:18
  • $\begingroup$ @DavidWhite Yes , I came here with a car tyre in mind but the OP specified a bicycle tyre. A quick check online shows that 1.4 L is a reasonable value for a bicycle tyre. $\endgroup$ – Kantura Sep 15 at 9:28

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