# If a mixture of gases with a total pressure of 1.0 atm contains 2 mol He, 4 mol Ne, and 2 mol O_2, what is the partial pressure of the He in the mixture?

Jan 19, 2016

$\text{0.25 atm}$

#### Explanation:

The partial pressure of a gas that's part of a gaseous mixture will depend on two things

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

As you know, Dalton's Law of Partial Pressures allows you to calculate the partial pressure of gas that's part of a gaseous mixture by using the number of moles of that gas and the total number of moles present in the mixture - this is known as the mole fraction

color(blue)(P_i - chi_i xx P_"total")" ", where

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

In your case, you know that the mixture contains

• $\text{2 moles}$ of helium
• $\text{4 moles}$ of neon
• $\text{2 moles}$ of oxygen gas

The *total number of moles8 present in the mixture will thus be

${n}_{\text{total" = 2 + 4 + 2 = "8 moles}}$

The mole fraction of helium will be

${\chi}_{H e} = \left(2 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles "))))/(8color(red)(cancel(color(black)("moles}}}}\right) = \frac{1}{4} = 0.25$

Therefore, the partial pressure of helium will be

${P}_{H e} = {\chi}_{H e} \times {P}_{\text{total}}$

P_(He) = 0.25 xx "1.0 atm" = color(green)("0.25 atm")

I'll leave the answer rounded to two sig figs.