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:

The molar solubility of nitrogen gas is $1.32\times 10^{-3}mol/L$.

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 gas

Total air pressure = P = 2.71 atm

Percentage of nitrogen in air = 78.09%

Mole fraction of nitrogen ,$\hi_{N_2}= 0.7809$

$p_{N_2}=P\times \chi_{N_2}=2.71 atm\times 0.7809$

Putting values in above equation, we get:

$C_{N_2}=6.26\times 10^{-4}mol/L.atm\times 2.71 atm\times 0.7809\\\\C_{N_2}=1.32\times 10^{-3}mol/L$

Hence, the molar solubility of nitrogen gas is $1.32\times 10^{-3}mol/L$.