# At a total pressure of "1 atm", if the mol percentages of "CO"_2, "O"_2, and "N"_2 are 5%, 15%, and 80% respectively, what are the partial pressures?

Jun 4, 2017

${P}_{C {O}_{2}} \approx \text{0.05 atm}$, ${P}_{{O}_{2}} \approx \text{0.15 atm}$, and ${P}_{{N}_{2}} \approx \text{0.80 atm}$, at the appropriate temperature and at $\text{1 atm}$, where these gases are all ideal enough.

We are given that:

• ${P}_{t o t} = \text{1 atm}$
• %"v/v" ("CO"_2) = 5%
• %"v/v" ("O"_2) = 15%
• %"v/v" ("N"_2) = 80%

By assuming the appropriate conditions (low enough pressure, high enough temperature that intermolecular interactions are minimized), all of these gases are ideal enough that we can get partial pressures from:

$\boldsymbol{{P}_{i} = {\chi}_{i} {P}_{t o t}}$

where:

• ${P}_{i}$ is the partial pressure of gas $i$ in the mixture, the pressure that each gas in the mixture exerts on the walls of the container.
• ${\chi}_{i} = {n}_{i} / \left({n}_{1} + . . . + {n}_{N}\right)$ is the mol fraction of gas $i$ in the mixture of $N$ different gases.

Each mol fraction is just the decimal form of the percentage given, since we are assuming all these gases are ideal, having the same molar volume (whatever it may be at the appropriate $T$ and $P$).

(For example, $V \text{/"n = "22.414 L/mol}$ at ${0}^{\circ} \text{C}$ and $\text{1 atm}$, but this may not be the appropriate temperature for ideality.)

In other words, we are saying:

• (%"v/v" ("CO"_2))/(100%) ~~ chi_(CO_2)
• (%"v/v" ("O"_2))/(100%) ~~ chi_(O_2)
• (%"v/v" ("N"_2))/(100%) ~~ chi_(N_2)

As a result, the partial pressures are:

$\textcolor{b l u e}{{P}_{C {O}_{2}}} = 0.05 {P}_{t o t} = \textcolor{b l u e}{\text{0.05 atm}}$

$\textcolor{b l u e}{{P}_{{O}_{2}}} = 0.15 {P}_{t o t} = \textcolor{b l u e}{\text{0.15 atm}}$

$\textcolor{b l u e}{{P}_{{N}_{2}}} = 0.80 {P}_{t o t} = \textcolor{b l u e}{\text{0.80 atm}}$