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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