What are the partial pressures of methane and argon in a mixture containing #"2.20 g"# methane and #"7.48 g"# argon if the total pressure is #"988 mm Hg"#?

#color(blue)(P_(CH_4)) = 0.423("988 mm Hg") = color(blue)("418 mm Hg")#

#color(blue)(P_(Ar)) = 0.577("988 mm Hg") = color(blue)("570 mm Hg")#

Dalton stated his law of partial pressures for ideal gases as:

#P_(t ot) = P_1 + P_2 + . . . P_N#

where #P_i# is the partial pressure of gas #i# in a set of #N# gases.

From the ideal gas law:

#P_(t ot) = (n_(t ot)RT)/V = (n_1RT)/V + . . . + (n_NRT)/V#

#= (n_1 + . . + n_N)(RT)/V#

From here, the partial pressures are then given by:

#P_A = n_A/n_(t ot) P_(t ot) = chi_A P_(t ot)#

where #chi_A# is the mol fraction of #A#, with the property that the sum of all the mol fractions for a given system is #1# (the amount of everything in one system adds up to #100%#).

So, what we'll need to do first is find the mols of each gas and the total mols of gas.

For methane:

#n_(CH_4) = "2.20 g CH"_4 xx ("1 mol CH"_4)/(12.011 + 4 xx 1.007"9 g CH"_4)#

#=# #ul("0.137 mols CH"_4)#

For argon:

#n_(Ar) = "7.48 g Ar" xx ("1 mol Ar")/("39.948 g Ar")#

#=# #ul("0.187 mols Ar")#

From here, the total mols of gas is #ul"0.324 mols"#, so the mol fraction of each gas is:

#chi_(CH_4) + chi_(Ar) = 1#

#chi_(CH_4) = "0.137 mols CH"_4/"0.324 mols gas" = ul0.423#

#chi_(Ar) = 1 - chi_(CH_4) = ul0.577#

This therefore means that the partial pressures of each gas are:

#color(blue)(P_(CH_4)) = 0.423("988 mm Hg") = color(blue)("418 mm Hg")#

#color(blue)(P_(Ar)) = 0.577("988 mm Hg") = color(blue)("570 mm Hg")#