A scuba diver ascends too quickly and develops the bends.

Question

A scuba diver ascends too quickly and develops the bends. A nitrogen bubble has formed in the patient's elbow. At a depth of 55 ft,55 ft, where the pressure is 2.66 atm,2.66 atm, the bubble had a volume of 0.025 mL.0.025 mL. Assuming a constant temperature and number of moles of nitrogen in the bubble, what volume did the bubble increase to at the surface, where the pressure is 1.00 atm?

Answer 1

Considering you are given only pressure and volume, you must use Boyle's Law (P1V1 = P2V2)

1. Identify your variables
2. P1 = 2.66
3. V1 = 0.025
4. P2 = 1.0
5. V2 = ?
6. plug in variables to law
7. (2.66)(0.025) = 1.0v
8. solve
9. 0.0665 = 1.0v
10. divide both sides by 1.0 to get v by itself

Final answer: 0.0665 mL = V2

Answer 2

This problem can be solved using Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature and number of moles of gas. Mathematically, we can express this as:

P1 x V1 = P2 x V2

where P1 and V1 are the initial pressure and volume of the gas, and P2 and V2 are the final pressure and volume of the gas.

We are given the initial pressure and volume of the nitrogen bubble at a depth of 55 ft:

P1 = 2.66 atm

V1 = 0.025 mL

We are asked to find the final volume of the bubble at the surface, where the pressure is 1.00 atm:

P2 = 1.00 atm

V2 = ?

Using Boyle's Law, we can set up the following equation:

P1 x V1 = P2 x V2

Substituting the given values, we get:

(2.66 atm) x (0.025 mL) = (1.00 atm) x V2

Substituting the given values, we get:

(2.66 atm) x (0.025 mL) = (1.00 atm) x V2

Solving for V2, we get:

V2 = (2.66 atm x 0.025 mL) / 1.00 atm

V2 = 0.0665 mL

Therefore, the nitrogen bubble would increase in volume from 0.025 mL to 0.0665 mL as the scuba diver ascends to the surface.