# Question #e7a85

##### 1 Answer

#### Answer:

#### Explanation:

**SIDE NOTE** *Yes, your answer is correct, the partial pressure of helium is indeed 0.8 atm. However, to make sure that other students benefit from this answer as well, I'll solve it completely.*

The idea here is that you need to use the information the problem provides about the ratio between the number of moles of each gas to find the **mole fractions** of the two gases.

The partial pressure of each gas will depend on its *mole fraction* and on the total pressure of the mixture.

#color(blue)(P_"i" = chi_"i" xx P_"total")#

The mole fraction of a gas that's part of a gaseous mixture is defined as the number of moles of that gas divided by the **total number of moles** of gas found in the mixture.

Let's say that the mixture contains **twice as many** moles of helium than it does of neon, so you can say that

#x = 2 * y implies y = x/2#

The *total number* of moles your sample contains will be

#n_"total" = x + y#

This means that the mole fraction of helium will be

#chi_"He" = "no. of moles of He"/"total number of moles"#

#chi_"He" = (xcolor(red)(cancel(color(black)("moles"))))/( (x + y)color(red)(cancel(color(black)("moles")))) = x/(x+y)#

The mole fraction of neon will be

#chi_"Ne" = (x/2color(red)(cancel(color(black)("moles"))))/((x+y)color(red)(cancel(color(black)("moles")))) = 1/2 * x/(x+y) = 1/2 * chi_"He"#

This means that the partial pressures of the two gases will be

#P_"He" = x/(x+y) * P_"total"#

#P_"Ne" = 1/2 overbrace(x/(x+y) * P_"total")^(color(blue)(=P_"He")) = 1/2 * P_"He"#

So, you've found that the partial pressure of helium is **twice as high** as the partial pressure of neon. This means that you have

#P_"He" + P_"Ne" = P_"total" -># Dalton's Law of Partial Pressures

and

#P_"He" = 2 xx P_"Ne"#

This comes down to

#2 xx P_"Ne" + P_"Ne" = "1.2 atm"#

#3 xx P_"Ne" = "1.2 atm" implies P_"Ne" = "1.2 atm"/3 = color(green)("0.4 atm")#

This means that you have

#P_"He" 2 * "0.4 atm" = color(green)("0.8 atm")#