Question

The partial pressure of N2 in the air is 593 mm Hg at 1 atm. What is the partial pressure of N2 in a bubble of air a scuba diver breathes when he is 132 ft below the surface of the water where the pressure is 5.00 atm?

Answer 1

Answer: Partial pressure of $N_{2}$ at a depth of 132 ft below sea level is 2964 mm Hg.

Explanation:

It is known that 1 atm = 760 mm Hg.

Also,   $P_{N_{2}} = x_{N_{2}}P$

where,    $P_{N_{2}}$ = partial pressure of $N_{2}$

                 P = atmospheric pressure

            $x_{N_{2}}$ = mole fraction of $N_{2}$

Putting the given values into the above formula as follows.

      $P_{N_{2}} = x_{N_{2}}P$

    $593 mm Hg = x_{N_{2}} \times 760 mm Hg$

       $x_{N_{2}}$ = 0.780

Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of $N_{2}$ is as follows.

         $P_{N_{2}} = x_{N_{2}}P$

                  = $0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}$

                  = 2964 mm Hg

Therefore, we can conclude that partial pressure of $N_{2}$ at a depth of 132 ft below sea level is 2964 mm Hg.