What are the partial pressures in #"bar"# if #"70.6 g"# of #"O"_2# is mixed with #"167.5 g"# of #"Ne"#?

1 Answer
Truong-Son N.
Nov 17, 2017

#P_(Ne) = "19.75 bar"#

#P_(O_2) = "5.25 bar"#

Well, the partial pressure of any ideal gas is based on their mol fraction:

#P_i = chi_iP#

where #P_i# is the partial pressure of gas #i# and #chi_i = (n_i)/(n_1 + n_2 + . . . )# is the mol fraction of gas #i#.

The mols of #"O"_2# are:

#70.6 cancel("g O"_2) xx ("1 mol O"_2)/(31.998 cancel("g O"_2)) = ul("2.21 mols O"_2)#

And for #"Ne"#?

#167.5 cancel("g Ne") xx ("1 mol O"_2)/(20.180 cancel("g Ne")) = ul("8.30 mols Ne")#

So, the mol fraction of #"Ne"# is:

#chi_(Ne) = n_(Ne)/(n_(Ne) + n_(O_2))#

#= "8.30 mols Ne"/("8.30 + 2.21 mols gas") = ul(0.790)#

And thus, the mol fraction of #"O"_2# is #ul(0.210)# (why?). This means their partial pressures are:

#color(blue)(P_(Ne)) = chi_(Ne)P = 0.790 cdot "25 bar" = color(blue)("19.75 bar")#

#color(blue)(P_(O_2)) = 0.210 cdot "25 bar" = color(blue)("5.25 bar")#

or

#color(blue)(P_(O_2)) = "25 bar" - 0.790 cdot "25 bar" = color(blue)("5.25 bar")#

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