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In the body, the oxidation of glucose is according to the reaction:

$$\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}$$

Knowing that glucose consumption is $\pu{2.47 mol/d}$, What is the volume of air that must be breathed at $\pu{25 °C}$ to accomplish this oxidation for $24\ \mathrm h$ at the pressure of $1\ \mathrm{atm}$?

We are given:

  • The atmosphere is composed of $21\,\%$ $\ce{O2}$ and $79\,\%$ $\ce{N2}$ (% volume).
  • $R = \pu{8.314 J mol-1 K-1}$; $\pu{1 atm} = \pu{760 mmHg} = \pu{1.014e5 Pa}$.
  • The molar volume of a gas at $\pu{298 K}$ is $\pu{24.45 l mol-1}$.

So, I have been a little stuck on this question. I have been trying to use the formula:

$$pV = nRT,$$

but it seems that I am missing something and I am not getting the right answer. It's possible that it might include molar fraction or partial pressure. I am confused.

The answer is $1\,724\ \mathrm l$. I would like to know how to get to this point.


Following the advice of @matt_black:

Here are my calculations that doesn't give me the right answer of $1\,724$ liters.

$$V_\mathrm m = 24.45\ \mathrm{l/mol}$$

$$n(\ce{C6H12O6}) = 2.47\ \mathrm{mol}$$

$$n(\ce{O2})= 6\times2.47\ \mathrm{mol} = 14.82\ \mathrm{mol}$$

$$V =\ ?$$

By the ideal gas law:

$$V= n\times24.45\ \mathrm{l/mol}$$

$$= 14.82\ \mathrm{mol} \times 24.45\ \mathrm{l/mol}$$

$$= 362.349\ \mathrm l \neq 1\,724\ \mathrm l$$

Tried using $pV=nRT$

$$V= 14.82\ \mathrm{mol} \times 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 293\ \mathrm K / 760\ \mathrm{mmHg}$$

$$= 47.5\ \mathrm l \neq 1\,724\ \mathrm l $$

I don't know how to implicate the fact that the atmosphere is composed of $21\,\%$ of $\ce{O2}$ and $79\,\%$ of $\ce{N2}$. Any ideas?

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    $\begingroup$ Why do you need PV=nRT? The question already gives you the molar volume of a gas. $\endgroup$ – matt_black Feb 6 at 16:44
  • $\begingroup$ @matt_black I have tried calculating without using that formula, by using V = n*24.45 l/mol, however I don't get the right answer, the reason why I thought I should use the PV=nRT... $\endgroup$ – Erika Feb 6 at 16:47
  • $\begingroup$ You might want to tell us your calculations steps in detail as your error may be nothing to do with this. $\endgroup$ – matt_black Feb 6 at 16:49
  • $\begingroup$ In one of your answer, you have used $R = \pu{8.314 J \cdot K^{−1} \cdot mol^{−1}}$. But answer is in L. You may need to convert units appropriately as well. $\endgroup$ – Mathew Mahindaratne Feb 6 at 17:56
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As matt_black commented, there is no need to use ideal gas law as you are given the molar volume already. Volume of the air $V_\mathrm{air}$ can be found from the volume fraction $\varphi(\ce{O2})$:

$$V_\mathrm{air} = \frac{V(\ce{O2})}{\varphi(\ce{O2})}$$

Volume of the oxygen can be found from the molar volume $V_\mathrm{m}(\ce{O2})$ and the amount of oxygen $n(\ce{O2})$ required to oxidize given amount of glucose:

$$V(\ce{O2}) = n(\ce{O2})\cdot V_\mathrm{m}(\ce{O2})$$

$$n(\ce{O2}) = 6n(\ce{C6H12O6})$$

Getting things together:

$$V_\mathrm{air} = \frac{6n(\ce{C6H12O6})\cdot V_\mathrm{m}(\ce{O2})}{\varphi(\ce{O2})} = \frac{6\cdot\pu{2.47 mol}\cdot\pu{24.45 L mol-1}}{0.21} = \pu{1725 L}$$

Note that you have to use appropriate units, e.g. you cannot use all base SI units in the deal gas law and put the pressure in mm Hg! Also, never drop any units — they do participate in calculations as well.

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  • $\begingroup$ thank you! So that's how you implicate the 0.21l of O2. It was so much simpler than I thought. In what cases will the pressure be in Pa or atm? Until now, I never seen the use of the other units besides mmHg. $\endgroup$ – Erika Feb 6 at 17:09
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    $\begingroup$ No, it's not 0.21 L of oxygen, it's volume fraction and is dimensionless! As for pressure, mm Hg is a worthless unit, especially in physical chemistry. If you want to use ideal gas law with SI units, use Pa, as it's very convenient: $$\pu{1 Pa} = \pu{1 J m-3}$$ and goes along with volume in $\pu{m-3}$ and gas constant $R=\pu{8.314 J mol-1 K-1}$. $\endgroup$ – andselisk Feb 6 at 17:14

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