Question

If atmospheric pressure on a certain day is 743 mmHg. what is the partial pressure of nitrogen, given that nitrogen makes up about 78% of atmospheric air?

Answer 1

Answer

The partial pressure of nitrogen = 579.54 mmHg.

Explanation

Given that nitrogen makes up about 78 % of atmospheric air.

Mole fraction of nitrogen in the atmosphere = 78/100 = 0.78

Total pressure of the air = 743 mmHg

So according to Raoult's law:

$\begin{gathered} P_{N_2}=\chi_{_{N_2}}\times P_{total} \\ \\ Put\text{ }\chi_{_{N_2}}=0.78,\text{ }P_{total}=743\text{ }mmHg \\ \\ P_{N_2}=0.78\times743\text{ }mmHg=579.54\text{ }mmHg \end{gathered}$

The partial pressure of nitrogen = 579.54 mmHg.