Consider a steel container filled with 40.0 g of helium gas and 40.0 g of argon gas. What is the ratio of pressures that each gas exerts (helium:argon)? a) 1:1 b) 10:1 c) 1:10 d) 9:1

1 Answer
Mar 10, 2015

The answer is b) 10:1.

When dealing with gas mixtures, it's important to remember that the partial pressures of the gases that make up the mixture depend on the number of moles each gas contributes to the total.

In your case, the total pressure in the container can be used to express the partial pressures of the two gases

#P_("He") = chi_("He") * P_("total")# (1)

#P_("Ar") = chi_("Ar") * P_("total")# (2)

where #chi_("He")# is the mole fraction of the helium gas and #chi_("Ar")# is the mole fraction of the argon gas. The mole fraction of a gas is the ratio between the number of moles of that gas and the total number of moles present in the mixture.

So, use helium and argon's molar masses to determine how many moles you have in the mixture

#"40.0 g" * "1 mole He"/"4.00 g" = "10.0 moles He"#

#"40.0 g" * "1 mole Ar"/"40.0 g" = "1.00 mole Ar"#

This means you'll have a total number of moles equal to

#n_("total") = n_("He") + n_("Ar") = 10.0 + 1.00 = "11.0 moles gas"#

The mole fractions for each of the two gases will be

#chi_("He") = n_("He")/n_("total") = "10.0"/"11.0"#

#chi_("Ar") = n_("Ar")/n_("total") = "1.00"/"11.0"#

Finally, divide equation (1) by equation (2) to get the pressure ratio you want

#P_("He")/P_("Ar") = chi_("He")/chi_("Ar") = ("10.0"/"11.0")/("1.00"/"11.0") = 10.0/11.0 * 11.0/1.00 = 10/1#