Question

The concentration of N2 in blood at 37 °C (body temperature) and atmospheric pressure (partial pressure of N2 = 0.80 atm) is 0.00056 mol/L. A deep-sea diver breathes compressed air with the partial pressure of N2 equal to 4.0 atm. Assume that the total volume of blood in the body is 5.0 L. Calculate the mass and volume (at 37°C and atmospheric pressure (p(N2) = 0.80 atm) of N2 gas released as a diver surfaces.

Answer 1

Answer:

volume  = 0.285 L

Explanation:

Henry constant is given as

$Henry constant = \frac{C}{P}$

Where C  is concentration

P is atmospheric pressure

$Henry\ constant = \frac{0.00056}{0.80} = 0.0007 M/atm$

when atmospheric pressure 4 atm

$solubility = K_H\times P$

              $= 0.0007 \times 4 = 0.0028 M$

In 5 litere blood , moles of N_2 $= Molarity \times volume$

                                                     $= 0.0028 \times 5 = 0.014$

At surface moles  $= 0.0056 \times 5 = 0.0028$

Moles of N_2 release = 0.014 - 0.0028 = 0.0112

Mass of $N_2 = 0.0112 \times 28 = 0.314 g$

T = 37 + 273 = 310 K

$VOLUME = \frac{nRT}{P}$

$= \frac{0.0112\times 0.0821\times 310}{1}$

volume  = 0.285 L