A gas mixture consists of 0.4mole of N2, 0.6 mole of O2 and 0.2 mole of argon at a total pressure of 300kPa. Calculate the partial pressure of each gas in the mixture?
1 Answer
Use the mole fraction of each gas to find its partial pressure.
Explanation:
The idea here is that when you have a mixture of two or more gases, the partial pressure of one component of that mixture will depend on how many moles said gas contributes to the total number of moles present.
In other words, the bigger the number of moles a gas has in a mixture, the higher its partial pressure will be.
Moreover, the total pressure of the mixture will be equal to the sum of the partial pressures of each individual gas that's part of that mixture.
So, the first thing you need to do is figure out how many moles of gas you have in the mixture.
#n_"total" = n_(N_2) + n_(O_2) + n_(Ar)#
#n_"total" = 0.4 + 0.6 + 0.2 = "1.2 moles"#
Now, the partial pressure of each component will depend on its mole fraction and on the total pressure of the mixture - this is known as Raoult's Law.
#color(blue)(P_"partial" = chi * P_"total"" "# , where
The mole fraction is simply the ratio between the number of moles of one gas and the total number of moles in the mixture.
So, for nitrogen, you have
#chi_(N_2) = (0.4color(red)(cancel(color(black)("moles"))))/(1.2color(red)(cancel(color(black)("moles")))) = 1/3#
The partial pressure of nitrogen will be
#P_(N_2) = chi_(N_2) * P_"total"#
#P_(N_2) = 1/3 * "300 kPa" = color(green)("100 kPa")#
Do the same for oxygen and argon to get
#P_(O_2) = chi_(O_2) * P_"total"#
#P_(O_2) = (0.6color(red)(cancel(color(black)("moles"))))/(1.2color(red)(cancel(color(black)("moles")))) * "300 kPa" = color(green)("150 kPa")#
and
#P_(Ar) = chi_(Ar) * P_"total"#
#P_(Ar) = (0.2color(red)(cancel(color(black)("moles"))))/(1.2color(red)(cancel(color(black)("moles")))) * "300 kPa" = color(green)("50 kPa")#
The partial pressures must add up to give the total pressure
#P_"total" = 100 + 150 + 50 = "300 kPa"#
