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When air is inhaled, it enters the alveoli of the lungs, and varying amounts of the component gases exchange with dissolved gases in the blood. The resulting alveolar gas mixture is quite different from the atmospheric mixture. The following table presents selected data on the composition and partial pressure of four gases in the atmosphere and in the alveoli:

 Gas  Mole %  Partial  Pressure (torr)  Mole %  Partial  Pressure (torr) N278.6569O220.9104CO200.0440H2O00.4647\begin{array}{|c|c|c|c|c|} \hline \text { Gas } & \text { Mole \% } & \begin{array}{l} \text { Partial } \\ \text { Pressure (torr) } \end{array} & \text { Mole \% } & \begin{array}{l} \text { Partial } \\ \text { Pressure (torr) } \end{array} \\ \hline \mathrm{N}_2 & 78.6 & - & - & 569 \\ \hline \mathrm{O}_2 & 20.9 & - & - & 104 \\ \hline \mathrm{CO}_2 & 00.04 & - & - & 40 \\ \hline \mathrm{H}_2 \mathrm{O} & 00.46 & - & - & 47 \\ \hline \end{array}

If the total pressure of each gas mixture is 1.00 atm1.00 \mathrm{~atm}, calculate:

(a) The partial pressure (in torr) of each gas in the atmosphere

(b) The mole % of each gas in the alveoli

(c) The number of O2\mathrm{O}_2 molecules in 0.50 L0.50 \mathrm{~L} of alveolar air (volume of an average breath of a person at rest) at 37C37^{\circ} \mathrm{C}

Question

A gas mixture contains 1.5×1051.5 \times 10^{-5} mole CO\mathrm{CO} and has a total of 1 mole. What is the CO\mathrm{CO} concentration in ppmv_{\mathrm{v}}?

Solution

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In the given example, we have:

  • 1.5 ×\times 105^{-5} mole of CO contained in a gas mixture.

  • The total of the gas mixture is 1 mole.

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