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

The partial pressure of nitrogen = 579.54 mmHg.

Step-by-step 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)}* P_(total) \\ \\ Put\text{ }\chi_{_(N_2)}=0.78,\text{ }P_(total)=743\text{ }mmHg \\ \\ P_(N_2)=0.78*743\text{ }mmHg=579.54\text{ }mmHg \end{gathered}`

The partial pressure of nitrogen = 579.54 mmHg.