# When "4 g H"_2 and "32 g O"_2 are mixed, what is the partial pressure of oxygen at a total pressure of P?

Jul 6, 2017

About $\frac{2}{3}$ of $P$.

Without your calculator, what is the fraction of $P$ that oxygen gas exerts?

For ideal gases, the partial pressure is given by

${P}_{i} = {\chi}_{i} P$,

where:

• $P$ is the total pressure.
• ${P}_{i}$ is the pressure of gas $i$ by itself, assuming it is not interacting with anything else.
• ${\chi}_{i} = \frac{{n}_{i}}{{n}_{1} + {n}_{2} + . . . + {n}_{N}}$ is the mol fraction of gas $i$ in the container.
• ${n}_{i}$ is the mols of gas $i$.

The fraction of the total pressure $P$ is given by

${\chi}_{i} = {P}_{i} / P$

And so, all we need to do is find the mols of each gas and calculate the mol fraction of ${\text{H}}_{2}$, i.e.:

${\chi}_{{H}_{2}} = \frac{{n}_{{H}_{2}}}{{n}_{{H}_{2}} + {n}_{{O}_{2}}}$

The mols of ${\text{H}}_{2}$ are given by:

4 cancel("g H"_2) xx "1 mol H"_2/(2.0158 cancel("g H"_2))

$=$ $\text{1.984 mols}$

The mols of ${\text{O}}_{2}$ are given by:

32 cancel("g O"_2) xx "1 mol O"_2/(31.998 cancel("g O"_2))

$=$ $\text{1.000 mols}$

Therefore, the fraction of $P$ that ${\text{H}}_{2}$ exerts is given by

$\textcolor{b l u e}{{\chi}_{{H}_{2}}} = \left({\text{1.984 mols")/("1.984 mols H"_2 + "1.000 mols O}}_{2}\right)$

$= \textcolor{b l u e}{0.665}$

This is about $\textcolor{b l u e}{\frac{2}{3}}$ $\textcolor{b l u e}{\text{of}}$ $\textcolor{b l u e}{P}$.