If equal mols of #"N"_2# and #"Ar"# are in the same container with a total pressure of #"10 atm"#, #(A)# what are their partial pressures? #(B)# what is the effusion speed of #"H"_2# if that of #"He"# is #"600 m/s"# at a certain temperature?

#A)# #"5 atm"# each.

#B)# The lighter gas #B# effuses at a speed #v_B# that is #sqrt(M_A/M_B)# times as fast as #v_A#, where #M_A# is the molar mass of #A#.

#A)# If the container is filled with an equal number of each particle, then we know that:

#n_(N_2) = n_(Ar)#

When this is the case, assuming ideal gases, their partial pressures should be identical.

Mathematically, this results in the following mol fractions:

#chi_(N_2) = (n_(N_2))/(n_(Ar) + n_(N_2))#

#chi_(Ar) = (n_(Ar))/(n_(Ar) + n_(N_2))#

But since #n_(N_2) = n_(Ar)#, we expect #chi_(N_2) = chi_(Ar) = 0.5#. Let's just say we had #"1 mol"# of each gas. Then:

#chi_(N_2) = 1/(1 + 1) = 0.5 = chi_(Ar)#

Assuming ideal gases, we can show that both gases have the same partial pressure, the pressure exerted by each gas in the mixture:

#color(blue)(P_(N_2)) = chi_(N_2)P_(t ot)#

#= chi_(Ar)P_(t ot) = color(blue)(P_(Ar))#

#= 0.5("10 atm") = color(blue)("5 atm")#

#B)# I'm not sure what part #B# has to do with part #A#. I will assume the question means to write #"Ar"# and #"N"_2# instead of #"He"# and #"H"_2#.

We are given, then, that argon effuses at #"600 m/s"#. Effusion can be derived from the expression for some sort of speed for each gas. The root-mean-square speed equation is fine:

#v_(rms) = sqrt((3RT)/M)#

where #R# and #T# are known from the ideal gas law, and #M# is the molar mass in #"kg/mol"#.

The rate of effusion, #z_(eff)# is proportional to the speed #v#, so the proportionality constants cancel out in a ratio:

#z_(eff,Ar)/(z_(eff,N_2)) = bb(v_(Ar)/(v_(N_2))) = bb(sqrt(M_(N_2)/(M_(Ar))))#

The relationship of #(z_(eff,i))/(z_(eff,j))# to #sqrt(M_j/M_i)# is known as Graham's law of effusion, but we will instead be using the relationship with #v_i/v_j#, the ratio of the speeds.

We are given that #v_(Ar) = "600 m/s"#, so:

#color(blue)(v_(N_2)) = v_(Ar) sqrt(M_(Ar)/M_(N_2))#

#= "600 m/s" cdot sqrt("0.039948 kg/mol"/"0.028014 kg/mol")#

#=# #color(blue)("716.5 m/s")#

This should make sense, that the lighter gas, #"N"_2#, effused faster.

If it really is supposed to be #"He"# and #"H"_2#, then you should expect:

#color(blue)(v_(H_2)) = v_(He) sqrt(M_(He)/M_(H_2))#

#= "600 m/s" cdot sqrt("0.0040026 kg/mol"/"0.0020158 kg/mol")#

#=# #color(blue)("845.5 m/s")#