Question

The solubility of N2 in blood at 37°C and a partial pressure of 0.80 atm is 5.6 ✕ 10−4 mol·L−1. A deep-sea diver breathes compressed air with a partial pressure of N2 equal to 3.8 atm. Assume that the total volume of blood in this diver's body is 6.0 L. Calculate the amount of N2 gas released (in liters) when the diver returns to the surface of water, where the partial pressure of N2 is 0.80 atm.

Answer 1

Answer:

0.0126 moles are released

Explanation:

Using Henry's law, where the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid:

S = k×P

Where S is solubility (5.6x10⁻⁴molL⁻¹), k is Henry's constant and P is partial pressure (0.80atm)

Replacing:

5.6x10⁻⁴molL⁻¹ / 0.80atm = 7x10⁻⁴molL⁻¹atm⁻¹

Thus, with Henry's constant, solubility of N₂ when partial pressure is 3.8atm is:

S = 7x10⁻⁴molL⁻¹atm⁻¹ × 3.8atm

S = 2.66x10⁻³molL⁻¹

Thus, when the deepd-sea diver has a pressure of 3.8amt, moles dissolved are:

6.0L × 2.66x10⁻³molL⁻¹ = 0.01596 moles of N₂

At the surface, pressure is 0.80atm and solubility is 5.6x10⁻⁴molL⁻¹, moles dissolved are:

6.0L × 5.6x10⁻⁴molL⁻¹ = 3.36x10⁻³mol

Thus, released moles are:

0.01596mol - 3.36x10⁻³mol = 0.0126 moles are released