What are the partial pressures in "bar" if "70.6 g" of "O"_2 is mixed with "167.5 g" of "Ne"?
1 Answer
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 gasi andchi_i = (n_i)/(n_1 + n_2 + . . . ) is the mol fraction of gasi .
The mols of
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
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
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
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")