Question #e7a85

1 Answer
Nov 8, 2015


#"0.8 atm"#


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 #x# moles of helium and #y# moles of neon. You know from the information given 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


#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")#