Question

When `"4 g H"_2` and `"32 g O"_2` are mixed, what is the partial pressure of oxygen at a total pressure of `P`?

Answer 1

About `2/3` of `P`.

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

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For ideal gases, the partial pressure is given by

`P_i = chi_iP`,

where:

• `P` is the total pressure.
• `P_i` is the pressure of gas `i` by itself, assuming it is not interacting with anything else.
• `chi_i = (n_i)/(n_1 + n_2 + . . . + n_N)` is the mol fraction of gas `i` in the container.
• `n_i` is the mols of gas `i`.

The fraction of the total pressure `P` is given by

`chi_i = P_i/P`

And so, all we need to do is find the mols of each gas and calculate the mol fraction of `"H"_2`, i.e.:

`chi_(H_2) = (n_(H_2))/(n_(H_2) + n_(O_2))`

The mols of `"H"_2` 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)`.