$\overline{){\mathbf{PV}}{\mathbf{=}}{\mathbf{nRT}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{P}\overline{)\mathrm{V}}}{\overline{)\mathrm{V}}}=\frac{{\displaystyle \frac{\mathrm{m}}{\mathrm{M}}}\mathrm{RT}}{\mathrm{V}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{P}}{\mathbf{=}}\frac{\mathbf{mRT}}{\mathbf{VM}}}$

Cyclists sometimes use pressurized carbon dioxide inflators to inflate a bicycle tire in the event of a flat. These inflators use metal cartridges that contain 16.0 g of carbon dioxide.

At 297 K , to what pressure (in psi) can the carbon dioxide in the cartridge inflate a 3.51 L mountain bike tire? (Note: The *gauge pressure * is the *difference * between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi.)

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