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Chapter 10: Problem 67

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

Expert verified
The mole percent of oxygen in the diving gas should be approximately 2.57% for the partial pressure of oxygen in the mixture to be 0.21 atm, the same as in air at 1 atm.

Step by step solution

01

Understand the relationship between total pressure and partial pressure

We can use Dalton's law of partial pressures to understand the relationship between the total pressure of the gas mixture and the partial pressure of individual gases in the mixture. According to Dalton's law: Total pressure = Partial pressure of gas 1 + Partial pressure of gas 2 + ... In this case, we have 2 gases: oxygen and nitrogen (assuming the diving gas is only composed of these 2 gases).
02

Calculate the partial pressure of nitrogen in the air at 1 atm

In air, the mole fraction of nitrogen is approximately 0.78, and the mole fraction of oxygen is approximately 0.21. Therefore, we can calculate the partial pressure of nitrogen in the air at 1 atm. Partial pressure of nitrogen = mole fraction of nitrogen * total pressure in air Partial pressure of nitrogen = 0.78 * 1 atm = 0.78 atm
03

Determine the mole fraction of nitrogen in the diving gas

The mole fraction of nitrogen in the diving gas must be the same as in air because the total pressure of the gas mixture is proportional to the depth. Therefore, the mole fraction of nitrogen in the diving gas is also 0.78.
04

Calculate the partial pressure of nitrogen in the diving gas

Now we can calculate the partial pressure of nitrogen in the diving gas using the mole fraction and total pressure at the depth of 250 ft. Partial pressure of nitrogen in diving gas = mole fraction of nitrogen * total pressure in diving gas = 0.78 * 8.38 atm = 6.5364 atm
05

Determine the partial pressure of oxygen in the diving gas

According to the problem, the partial pressure of oxygen in the diving gas must be 0.21 atm.
06

Calculate the total pressure in the diving gas without oxygen

In this step, we will calculate the total pressure in the diving gas without the pressure contribution from oxygen to determine the total pressure due to the other gases. Total pressure without oxygen = Total pressure in diving gas - Partial pressure of oxygen = 8.38 atm - 0.21 atm = 8.17 atm
07

Determine the mole fraction of oxygen in the diving gas

Now we will calculate the mole fraction of oxygen in the diving gas. Mole fraction of oxygen = Partial pressure of oxygen / Total pressure without oxygen = 0.21 atm / 8.17 atm 鈮� 0.0257
08

Convert the mole fraction to mole percent

Finally, in order to determine the mole percent of oxygen in the diving gas, we can simply multiply the mole fraction by 100: Mole percent of oxygen = 0.0257 * 100 = 2.57 % Thus, the mole percent of oxygen in the diving gas should be approximately 2.57% for the partial pressure of oxygen in the mixture to be 0.21 atm.

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Most popular questions from this chapter

The Goodyear blimps, which frequently fly over sporting events, hold approximately \(175,000 \mathrm{ft}^{3}\) of helium. If the gas is at \(23^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\), what mass of helium is in the blimp?

A fixed quantity of gas at \(21{ }^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of \(4.38 \mathrm{~L}\). (a) Use Boyle's law to calculate the volume the gas will occupy if the pressure is increased to \(1.88 \mathrm{~atm}\) while the temperature is held constant. (b) Use Charles's law to calculate the volume the gas will occupy if the temperature is increased to \(175^{\circ} \mathrm{C}\) while the pressure is held constant.

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A scuba diver's tank contains \(0.29 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of \(2.3\) L. (a) Calculate the gas pressure inside the tank at \(9{ }^{\circ} \mathrm{C}\) (b) What volume would this oxygen occupy at \(26^{\circ} \mathrm{C}\) and \(0.95 \mathrm{~atm} ?\)

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