If you mixed three gases: 8 g of helium, 2 g of nitrogen, and 16 g of oxygen and the total pressure was 10.0 atm, what is the partial pressure of helium?

Jan 10, 2016

${P}_{H e} = \text{6 atm}$

Explanation:

The idea here is that according to Dalton's Law of Partial Pressures, the partial pressure of a gas that's part of a gaseous mixture depends on two things

• the mole fraction of said gas in the mixture
• the total pressure of the mixture

Mathematically, this is written as

color(blue)(P_A = chi_A xx P_"total")" ", where

${P}_{A}$ - the partial pressure of gas $A$
${\chi}_{A}$ - the mole fraction of gas $A$ in the mixture
${P}_{\text{total}}$ - the total pressure of the mixture

Simply put, the partial pressures of the gases that are part of a gaseous mixture are proportional to their number of moles. The more moles a specific gas will have, the higher its partial pressure will be.

So, in essence, all you have to do here is figure out how many moles of each gas you have present in the mixture, since the total pressure is said to be equal to $\text{10.0 atm}$.

So, use the molar masses of the three gases to determine how many moles of each are present. Keep in mind that nitrogen gas and oxygen gas are diatomic molecules, ${\text{N}}_{2}$ and ${\text{O}}_{2}$, respectively.

You will thus have

8 color(red)(cancel(color(black)("g"))) * "1 mole He"/(4.003color(red)(cancel(color(black)("g")))) = "2.00 moles He"

2 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(14.007color(red)(cancel(color(black)("g")))) = "0.0714 moles N"_2

16 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.500 moles O"_2

The total number of moles present in the mixture will be

${n}_{\text{total}} = {n}_{H e} + {n}_{{N}_{2}} + {n}_{{O}_{2}}$

${n}_{\text{total" = 2.00 + 0.714 + 0.500 = "3.214 moles}}$

The mole fraction of helium, which is defined as the number of moles of helium divided by the total number of moles present in the mixture, will be equal to

${\chi}_{H e} = \left(2.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(3.214color(red)(cancel(color(black)("moles}}}}\right) = 0.6223$

Therefore, the partial pressure of helium will be

${P}_{H e} = 0.6223 \times \text{10.0 atm}$

${P}_{H e} = \text{6.223 atm}$

Rounded to one sig fig, the number of sig figs you have for the masses of helium and nitrogen gas, the answer will be

${P}_{H e} = \textcolor{g r e e n}{\text{6 atm}}$