Question

As a scuba diver descends under water, the pressure increases. At a total air pressure of 2.71 atm and a temperature of 25.0 C, what is the solubility of N2 in a diver's blood?[Use the value of the Henry's law constant k calculated , 6.26 x 10^{-4} (mol/(L*atm).]Assume that the composition of the air in the tank is the same as on land and that all of the dissolved nitrogen remains in the blood.Express your answer with the appropriate units.

Answer 1

Answer:

$1.32\times 10^{-3} mol/L$is the solubility of nitrogen gas in a diver's blood.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{N_2}=K_H\times p_{liquid}$

where,

$K_H$ = Henry's constant = $6.26 \times 10^{-4}mol/L.atm$

$p_{N_2}$ = partial pressure of nitrogen

$p_{N_2}= P\times \chi_{N_2}$ (Raoult's law)

$=2.71 atm\times 0.78=2.1138 atm$

$C_{N_2}=6.26 \times 10^{-4}mol/L.atm\times 2.1138 atm$

$C_{N_2}=1.32\times 10^{-3} mol/L$

$1.32\times 10^{-3} mol/L$is the solubility of nitrogen gas in a diver's blood.