Question

At sea level, P = 1 atm, therefore gas solubility is x = 1atm * molar tion of gas. For O₂ = 0.2 atm/773 atmkgH₂O/mol and N₂ = 0.8 atm/1610 atmkgH₂O/mol, 40m below sea level, P = 6 atm. What is the concentration of O₂ and N₂?

a) O₂ = 0.8 atm, N₂ = 0.4 atm
b) O₂ = 0.4 atm, N₂ = 0.8 atm
c) O₂ = 0.2 atm, N₂ = 0.6 atm
d) O₂ = 0.6 atm, N₂ = 0.2 atm

Answer 1

Using Henry's Law, the concentration of O₂ and N₂ at 40 meters below sea level with a pressure of 6 atm are calculated. The correct concentrations are O₂ = 0.4 atm and N₂ = 0.8 atm.

Step-by-step explanation:

To calculate the concentration of O₂ and N₂ at 40 meters below sea level where the pressure is 6 atm, we need to use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The solubility (x) of a gas is given by the product of its partial pressure (P) and its solubility coefficient (k), which is the solubility of the gas at a standard pressure (1 atm in this case).

For O₂, the solubility at 1 atm is 0.2 atm/773 atmkgH₂O/mol. At 6 atm pressure, the solubility of O₂ will be:

6 atm * (0.2 atm/773 atmkgH₂O/mol) = 1.2 atm/773 atmkgH₂O/mol

For N₂, the solubility at 1 atm is 0.8 atm/1610 atmkgH₂O/mol. At 6 atm pressure, the solubility of N₂ will be:

6 atm * (0.8 atm/1610 atmkgH₂O/mol) = 4.8 atm/1610 atmkgH₂O/mol

Therefore, the correct concentrations of O₂ and N₂ are option b) O₂ = 0.4 atm, N₂ = 0.8 atm.