# Question 48d69

Mar 14, 2016

$\left\{\begin{matrix}{P}_{H e} = 6.0 \cdot {10}^{2} \text{torr" \\ P_(O_2) = "1800 torr}\end{matrix}\right.$

#### Explanation:

The idea here is that the two gases will contribute to the total pressure of the mixture proportionally to how many moles each has in the mixture - think Dalton's Law of Partial Pressures.

IF you take ${P}_{H e}$ to be the partial pressure of helium and ${P}_{{O}_{2}}$ to be the partial pressure of oxygen, you can say that

color(blue)(|bar(ul(color(white)(a/a)P_"total" = P_(He) + P_(O_2)color(white)(a/a)|)))" " " "color(orange)("(*)")

Now, to get the partial pressure of each gas you must use the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

The trick here is to realize that you can find the partial pressure of each gas by isolating them in same volume and at the same temperature.

This means that you can write

${P}_{H e} = {n}_{H e} \cdot \frac{R T}{V} \to$ the partial pressure of helium

${P}_{{O}_{2}} = {n}_{{O}_{2}} \cdot \frac{R T}{V} \to$ hte partial pressure of oxygen

Now, the total pressure of the mixture will depend on the total number of moles present.

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

You can thus write

${P}_{\text{total" = n_"total}} \cdot \frac{R T}{V}$

Use this equation to write

$\frac{R T}{V} = {P}_{\text{total"/n_"total}}$

and plug this into the equations for the partial pressures of the two gases. You will find

${P}_{H e} = {n}_{H e} \cdot {P}_{\text{total"/n_"total" = n_(He)/n_"total" * P_"total}}$

and

${P}_{{O}_{2}} = {n}_{{O}_{2}} \cdot {P}_{\text{total"/n_"total" = n_(O_2)/n_"total" * P_"total}}$

The ratio between the number of moles of a gas that's part of a gaseous mixture and the total number of moles of gas present in the mixture will give you that gas' mole fraction, $\chi$.

You will thus have

color(purple)(|bar(ul(color(white)(a/a)color(black)(P_(He) = chi_(He) * P_"total")color(white)(a/a)|)))" " and " "color(purple)(|bar(ul(color(white)(a/a)color(black)(P_(O_2) = chi_(O_2) * P_"total")color(white)(a/a)|)))

Finally, plug in your values to get

P_(He) = (2.0 color(red)(cancel(color(black)("moles"))))/((2.0 + 6.0)color(red)(cancel(color(black)("moles")))) * "2400 torr" = color(green)(|bar(ul(color(white)(a/a)6.0 * 10^2"torr"color(white)(a/a)|)))

P_(O_2) = (6.0 color(red)(cancel(color(black)("moles"))))/((2.0 + 6.0)color(red)(cancel(color(black)("moles")))) * "2400 torr" = color(green)(|bar(ul(color(white)(a/a)"1800 torr"color(white)(a/a)|)))#

Notice that both values satisfy equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$

${P}_{\text{total" = 6.0 * 10^2"torr" + "1800 torr" = "2400 torr}}$

The answers are rounded to two sig figs.