Question

At an underwater depth of \(250 \mathrm{ft}\), the pressure is \(8.38 \mathrm{~atm}\). What should the mole percent of oxygen be in the diving gas for the partial pressure of oxygen in the mixture to be \(0.21 \mathrm{~atm}\), the same as in air at \(1 \mathrm{~atm}\) ?

Short answer

To maintain a partial pressure of oxygen at 0.21 atm, the same as in air at 1 atm, the mole percent of oxygen in the diving gas at an underwater depth of 250 ft and a pressure of 8.38 atm should be approximately \(2.51 \% \).

Step-by-step solution

1. Dalton's Law of Partial Pressures

Dalton's law states that the total pressure of a gas mixture is equal to the sum of the partial pressures of the individual gases. In mathematical form: \(P_{total} = P_{1} + P_{2} + ... + P_{n}\) In our case, let's consider the partial pressures of oxygen (\(P_{O_2}\)) and the other gases (\(P_{other}\)). Therefore, the total pressure can be expressed as: \(P_{total} = P_{O_2} + P_{other}\)

2. Mole Fractions

Mole fraction of a gas is the ratio of the moles of that gas to the total number of moles of the gas mixture. Let's denote the mole fraction of oxygen as \(X_{O_2}\) and the mole fraction of the other gases as \(X_{other}\). For a binary mixture: \(X_{O_2} + X_{other} = 1\)

3. Mole Fractions and Partial Pressures

A gas' mole fraction is directly related to its partial pressure. In our case, we can relate the mole fractions and partial pressures as follows: \(P_{O_2} = X_{O_2} * P_{total}\) \(P_{other} = X_{other} * P_{total}\)

4. Calculating the Mole Fraction of Oxygen

We are given the total pressure (\(P_{total}\)) and the partial pressure of oxygen (\(P_{O_2}\)). We have to calculate the mole fraction of oxygen (\(X_{O_2}\)): Using the formula we derived in the previous step: \(P_{O_2} = X_{O_2} * P_{total}\) Rewrite the equation to solve for \(X_{O_2}\): \(X_{O_2} = \frac{P_{O_2}}{P_{total}}\) Now, substitute the given values: \(X_{O_2} = \frac{0.21 \mathrm{~atm}}{8.38 \mathrm{~atm}}\)

5. Calculating the Mole Percent of Oxygen

Compute the mole fraction of oxygen (\(X_{O_2}\)): \(X_{O_2} \approx 0.0251\) To express this as a mole percent, simply multiply by 100: Mole Percent of Oxygen = \(0.0251 * 100 \approx 2.51 \% \) So, the mole percent of oxygen in the diving gas should be approximately \(2.51 \% \) to maintain a partial pressure of oxygen at 0.21 atm, the same as in air at 1 atm.