# 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?

##### 1 Answer

**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

#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")#

We are given, then, that argon effuses at

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

The relationship of **Graham's law of effusion**, but we will instead be using the relationship with

We are given that

#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,

If it really is supposed to be

#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")#