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How much pressure will a 16 gram cartridge of $\ce{CO2}$ (carbon dioxide) provide in a bicycle tire? Remembering my high school chemistry (forty years ago) and googling, I came up with this answer:

Figure the volume of the tube (a torus): area of a cross section times the circumference, for a $30\:\mathrm{mm}$ tube on a $70\:\mathrm{cm}$ wheel, I got $1,554\:\mathrm{cm^3}$, or $1.554\:\mathrm{L}$. One mole is $24.47\:\mathrm{L}$. $1.554 / 24.47 = .06353$. One mole of $\ce{CO2}$ is $44\:\mathrm{g}$ (using atomic weights of carbon, 12, and oxygen, 16). $.06353 \cdot 44 = 2.795\:\mathrm{g}$. So the volume of the tube at one atmosphere is $2.795\:\mathrm{g}$ of $\ce{CO2}$. $16 / 2.795 = 5.7\:\mathrm{atm}$. One atmosphere is $14.7\:\mathrm{psi}$. $5.7 \cdot 14.7\:\mathrm{psi} = 84\:\mathrm{psi}$.

Is this correct?

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    $\begingroup$ I strongly suggest to search data for density of $\ce{CO2}$ at different pressures. Most real gases do not behave as ideal gases at any moderate or highter pressure. $\endgroup$ – permeakra Aug 15 '14 at 6:43
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    $\begingroup$ Are you working at 25 degrees Celsius? If not, the 24.47 is incorrect...' $\endgroup$ – DJohnM Aug 15 '14 at 7:20
  • $\begingroup$ Don't forget to subtract 14.7 to convert your absolute pressure answer to the gauge pressure you'll measure after filling your tire... $\endgroup$ – DJohnM Aug 15 '14 at 7:22
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    $\begingroup$ I did not understand why you said the "volume of tube at one atmosphere is 2.795g..." you divide g/g and result atm ? $\endgroup$ – ordinary chemistry student Aug 15 '14 at 10:10
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Assumptions. The $\ce{CO2}$ will cool down a bit as it expands out of the cartridge (due to the Joule-Thomson effect). But let's assume that you're interested in the pressure in the tire when it warms back up to room temperature, $25~^\circ\mathrm{C}$, so there is essentially a zero temperature change before and after filling. Let's also assume that the full $16~\mathrm{g}$ of $\ce{CO2}$ is contained by the cartridge and tire when they are connected; none of it is vented during filling.

Estimating the volume of the tire and cartridge. A $30 \times 700$ bicycle tire has a $30~\mathrm{mm}$ outside tube diameter and a $70~\mathrm{cm}$ wheel diameter. Let's say the rubber in the tire is $3~\mathrm{mm}$ thick. You then have an inside diameter of about $24~\mathrm{mm}$. If we assume the tube has a circular cross-section, the volume inside the tire when fully inflated is.

$$V = \pi \left(\frac{2.4~\mathrm{cm}}{2}\right)^2\cdot (\pi \cdot 70~\mathrm{cm}) \cdot \frac{1~\mathrm{L}}{1000~\mathrm{cm^3}} = 0.99_5~\mathrm{L}$$

Let's assume the $\ce{CO2}$ cartridge has a volume of $21~\mathrm{cm^3}$. The total volume of the $\ce{CO2}$ in the attached cartridge and tire will then be about $1.0_2~\mathrm{L}$.

If you have $16~\mathrm{g}$ $\ce{CO2}$, and the tire is fully deflated to begin with, the moles of gas is.

$$n = 16~\mathrm{g} \cdot \frac{1~\mathrm{mol}}{44.01~\mathrm{g}} = 0.36_4~\mathrm{mol}$$

Estimating the pressure. Since we're dealing with a gas with significant intermolecular interactions, which is also at pressures above $1~\mathrm{atm}$, let's compare an ideal gas calculation of the pressure with a van der Waals calculation to see if nonideality makes any difference.

$$\begin{align} P_{ideal} &= \frac{n R T}{V}\\ &= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}}) (298~\mathrm{K})}{1.02~\mathrm{L}}\cdot \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\ &\approx 129~\mathrm{psi}\\ P_{vdw} &= \frac{n R T}{V - n b} - \frac{n^2 a}{V^2}\\ &= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}}) (298~\mathrm{K})}{1.02~\mathrm{L} - (0.364~\mathrm{mol}) (0.04267~\mathrm{L~mol^{-1}})}\\ &~~~~~~ - \frac{(0.364~\mathrm{mol})^2(3.592~\mathrm{L^2~atm~mol^{-2}})} {(1.02~\mathrm{L})^2}\\ &= 8.43~\mathrm{atm} \cdot \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\ &\approx 124~\mathrm{psi}\\ \end{align}$$

A tire gauge reads (gas pressure - atmospheric pressure), so you'd read about about $109~\mathrm{psi}$ if you started with a fully deflated tire. If you started with a tire filled with air at $1~\mathrm{atm}$, though, you'd get about $124~\mathrm{psi}$.

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