Answer: Partial pressure of [tex]N_{2}[/tex] at a depth of 132 ft below sea level is 2964 mm Hg.
Explanation:
It is known that 1 atm = 760 mm Hg.
Also, [tex]P_{N_{2}} = x_{N_{2}}P[/tex]
where, [tex]P_{N_{2}}[/tex] = partial pressure of [tex]N_{2}[/tex]
P = atmospheric pressure
[tex]x_{N_{2}}[/tex] = mole fraction of [tex]N_{2}[/tex]
Putting the given values into the above formula as follows.
[tex]P_{N_{2}} = x_{N_{2}}P[/tex]
[tex]593 mm Hg = x_{N_{2}} \times 760 mm Hg[/tex]
[tex]x_{N_{2}}[/tex] = 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 [tex]N_{2}[/tex] is as follows.
[tex]P_{N_{2}} = x_{N_{2}}P[/tex]
= [tex]0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}[/tex]
= 2964 mm Hg
Therefore, we can conclude that partial pressure of [tex]N_{2}[/tex] at a depth of 132 ft below sea level is 2964 mm Hg.
