# Question #48d69

##### 1 Answer

#### 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

#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

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where

*number of moles* of gas

*universal gas constant*, usually given as

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_(He) = n_(He) * (RT)/V -># thepartial pressureof helium

#P_(O_2) = n_(O_2) * (RT)/V -># htepartial pressureof oxygen

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

#n_"total" = n_(He) + n_(O_2)#

You can thus write

#P_"total" = n_"total" * (RT)/V#

Use this equation to write

#(RT)/V = P_"total"/n_"total"#

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

#P_(He) = n_(He) * P_"total"/n_"total" = n_(He)/n_"total" * P_"total"#

and

#P_(O_2) = n_(O_2) * P_"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**,

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

#P_"total" = 6.0 * 10^2"torr" + "1800 torr" = "2400 torr"#

The answers are rounded to two **sig figs**.