When "4 g H"_24 g H2 and "32 g O"_232 g O2 are mixed, what is the partial pressure of oxygen at a total pressure of PP?

1 Answer
Truong-Son N.
Jul 6, 2017

About 2/323 of PP.

Without your calculator, what is the fraction of PP that oxygen gas exerts?

For ideal gases, the partial pressure is given by

P_i = chi_iPPi=χiP,

where:

  • PP is the total pressure.
  • P_iPi is the pressure of gas ii by itself, assuming it is not interacting with anything else.
  • chi_i = (n_i)/(n_1 + n_2 + . . . + n_N)χi=nin1+n2+...+nN is the mol fraction of gas ii in the container.
  • n_ini is the mols of gas ii.

The fraction of the total pressure PP is given by

chi_i = P_i/Pχi=PiP

And so, all we need to do is find the mols of each gas and calculate the mol fraction of "H"_2H2, i.e.:

chi_(H_2) = (n_(H_2))/(n_(H_2) + n_(O_2))χH2=nH2nH2+nO2

The mols of "H"_2H2 are given by:

4 cancel("g H"_2) xx "1 mol H"_2/(2.0158 cancel("g H"_2))

= "1.984 mols"

The mols of "O"_2 are given by:

32 cancel("g O"_2) xx "1 mol O"_2/(31.998 cancel("g O"_2))

= "1.000 mols"

Therefore, the fraction of P that "H"_2 exerts is given by

color(blue)(chi_(H_2)) = ("1.984 mols")/("1.984 mols H"_2 + "1.000 mols O"_2)

= color(blue)(0.665)

This is about color(blue)(2/3) color(blue)("of") color(blue)(P).

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