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At an underwater depth of \(100 \mathrm{~m}\), the pressure is \(1.106 \mathrm{MPa}\). What should the partial pressure of oxygen be in the diving gas for the mole fraction of oxygen in the mixture to be 0.21 , the same as in air?

Short Answer

Expert verified
The partial pressure of oxygen should be 232.26 kPa.

Step by step solution

01

Calculate the Total Pressure Underwater

First, let's establish that the underwater pressure at a depth of 100 meters is given as 1.106 Megapascals (MPa). This is the absolute pressure acting on the diving gas mixture.
02

Convert the Pressure to Kilopascals

The pressure is provided in MPa. To convert that to kilopascals (kPa), we use the conversion: 1 MPa = 1000 kPa. Therefore, \(1.106 \text{ MPa} = 1106 \text{ kPa}\).
03

Apply Raoult's Law for Partial Pressure

Raoult’s Law states that the partial pressure of a gas in a mixture is the product of the total pressure and the mole fraction of the gas. Here, the mole fraction \(x\) of oxygen is given as 0.21. Therefore, the partial pressure of oxygen is: \(P_{O_2} = 1106 \text{ kPa} \times 0.21\).
04

Calculate the Partial Pressure of Oxygen

Perform the multiplication from Step 3: \(P_{O_2} = 1106 \text{ kPa} \times 0.21 = 232.26 \text{ kPa}\).
05

Conclusion

The partial pressure of oxygen at an underwater depth of 100 meters, for a mole fraction of 0.21, should be 232.26 kPa.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Raoult's Law
Raoult's Law is a fundamental principle used to determine the partial pressure of a component in a mixture. This law is particularly helpful for understanding solutions like diving gases. The law states that the partial pressure of a component gas is directly proportional to its mole fraction in the mixture. In simpler terms, this means:
  • Partial pressure = Total pressure × Mole fraction of the gas
This relationship allows us to find out how much pressure a specific gas in a mixture exerts on its own. For divers, understanding Raoult's Law ensures safety by controlling the amounts of gases like oxygen, avoiding potential complications like decompression sickness. When diving, the partial pressure needs to be calculated carefully to match conditions similar to those on the surface, thus keeping divers safe and breathing normally underwater.
Mole Fraction
The mole fraction is a way to express the concentration of a component in a mixture. It's a simple ratio, showing how many moles of a particular component are present compared to the total number of moles of all components combined. In mathematical terms:
  • Mole fraction = Moles of the component / Total moles of the mixture
For divers, maintaining the same mole fraction of oxygen as in air (which is 0.21) is crucial. This mole fraction assures that the amount of oxygen remains adequate for breathing efficiently, whether underwater or at the surface. By keeping this ratio consistent, divers can rely on the breathing apparatus to provide predictable and safe levels of oxygen in high-pressure environments.
Underwater Pressure
When we delve underwater, the pressure increases significantly compared to the surface. At a depth of 100 meters, the pressure is approximately 1.106 MPa, far greater than the atmospheric pressure of 0.1013 MPa at sea level. This increased pressure affects the gases divers breathe in their tanks. It compresses the gas, thus increasing its density and requiring divers to manage these conditions. Underwater pressure directly affects the partial pressures of the gases in a diving mixture. Divers must adjust their gear to ensure that the partial pressures remain within safe levels to prevent conditions like nitrogen narcosis or oxygen toxicity. Specifically, using Raoult’s Law, divers calculate necessary partial pressures to ensure safe breathing mixtures, avoiding excessive or insufficient oxygen supply while underwater.

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Most popular questions from this chapter

Does the effect of intermolecular attraction on the properties of a gas become more significant or less significant if (a) the gas is compressed to a smaller volume at constant temperature; (b) the temperature of the gas is increased at constant volume?

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