How would you calculate the partial pressure of CO2, given an atmospheric pressure of 760 mm Hg and a 0.04% concentration?

Dec 8, 2015

$\text{0.3 mmHg}$

Explanation:

The key to this problem is the fact that each component of a gaseous mixture will contribute to the total pressure exerted by the mixture proportionally to the number of molecules in has in the mixture.

More often than not, you will see the partial pressure of a gas being expresses in terms of its mole fraction.

$\textcolor{b l u e}{{P}_{\text{gas" = chi_"gas" xx P_"mixture}}}$

This is exactly what proportionally to the number of molecules means.

As you know, one mole of any substance is equal to exactly $6.022 \cdot {10}^{23}$ molecules of that substance - this is known as Avogadro's number, ${N}_{A}$.

This means that you an express the number of moles of a gas by using the number of molecules, let's say $x$, and Avogadro's number

$\textcolor{b l u e}{\text{no. of moles" = "no. of molecules} \times {N}_{A}}$

Now, the percent composition of a gaseous mixture tells you how many molecules each gas contributes in $100$ molecules of mixture.

In this case, air is said to be 0.04% carbon dioxide. This means that in every $100$ molecules of air, $0.04$ will be ${\text{CO}}_{2}$ molecules.

For example, the number of moles of carbon dioxide in $100$ molecules of air will be

${n}_{C {O}_{2}} = \text{0.04 molecules} \times {N}_{A} = 0.04 \cdot {N}_{A}$

The total number of moles in this sample of air will be

${n}_{\text{total" = "100 molecules}} \times {N}_{A} = 100 \cdot {N}_{A}$

This means that the mole fraction of carbon dioxide in the mixture will be

${\chi}_{C {O}_{2}} = \frac{0.4 \textcolor{red}{\cancel{\textcolor{b l a c k}{{N}_{A}}}}}{100 \textcolor{red}{\cancel{\textcolor{b l a c k}{{N}_{A}}}}} = 0.00004$

Carbon dioxide's partial pressure in air will thus be

${P}_{C {O}_{2}} = 0.00004 \cdot \text{760 mmHg" = "0.304 mmHg}$

Rounded to one sig fig, the number of sig figs you have for the percent composition of ${\text{CO}}_{2}$, the answer will be

${P}_{C {O}_{2}} = \textcolor{g r e e n}{\text{0.3 mmHg}}$