Question

Dry air is 78.08% nitrogen, 20.95% oxygen and 0.93% argon with the 0.04% being other gases. if the atmospheric pressure is 760.0 torr, what is the partial pressure of nitrogen in dry air?

Answer 1

Explanation:

Let us assume that volume of air present is 100.

So, when 78.08% nitrogen is present then it means that volume of nitrogen is 78.08.

Similarly, volume of oxygen is 20.95, volume of argon is 0.93 and volume of other gases is 0.04.

Hence, calculate the mole fraction of given gases as follows.

         Mole fraction of $N_{2}$ = $\frac{78.08}{100}$

                                               = 0.7808

         Mole fraction of $O_{2}$ = $\frac{20.95}{100}$

                                               = 0.2095

        Mole fraction of Ar = $\frac{0.93}{100}$

                                               = 0.0093

        Mole fraction of other gases = $\frac{0.04}{100}$

                                               = 0.0004

As the mole fraction of nitrogen is 0.7808 and total pressure is 760.0 torr. Therefore, calculate the partial pressure of nitrogen as follows.

                         $p_{i} = x_{i} \times P_{total}$

                                      = $0.7808 \times 760 torr$

                                       = 593.408 torr

Therefore, we can conclude that the partial pressure of nitrogen in dry air is 593.408 torr.

Answer 2

Partial pressure (N2) = mole fraction * total pressure
 
{    1 mole of any ideal gas occupy same volume of 1 mole of any other ideal gas under same condition of temperature and pressure so mole fraction in the sample is simply 78.08%   =   0.7808   this is because equal volume of each gas has equal moles

partial pressure N2 = 0.7808 * 760 .0
partial pressure =  593.4 mmhg    (   1 torr = 1mmhg  )