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Chapter 10: Problem 73

At an underwater depth of \(250 \mathrm{ft}\), the pressure is 8.38 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 atm, the same as in air at 1 atm?

Short Answer

Expert verified
The mole percent of oxygen in the diving gas should be approximately 2.51% 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

Identify the given information

We are given: - The total pressure at the underwater depth, \(P_t = 8.38 \, \mathrm{atm}\) - The partial pressure of oxygen in air at 1 atm, \(P_{O_2}^{air} = 0.21 \, \mathrm{atm}\)
02

Define formula for partial pressure

Recall that the partial pressure of a gas in a mixture can be calculated using the formula: \[P_i = X_i \cdot P_t\] Where \(P_i\) is the partial pressure of the gas, \(X_i\) is the mole fraction of the gas, and \(P_t\) is the total pressure of the mixture. In this problem, we need to find the mole fraction of oxygen (\(X_{O_2}\)) in the diving gas, such that the partial pressure of oxygen (\(P_{O_2}\)) is the same as in air at 1 atm.
03

Set up the equation for mole fraction of oxygen

We can write the equation for the partial pressure of oxygen in the diving gas mixture as: \[P_{O_2} = X_{O_2} \cdot P_t\] We know that \(P_{O_2} = P_{O_2}^{air} = 0.21 \, \mathrm{atm}\) and \(P_t = 8.38 \, \mathrm{atm}\). So, we can rewrite the equation as: \[0.21 = X_{O_2} \cdot 8.38\]
04

Solve for the mole fraction of oxygen

Now, we can solve for \(X_{O_2}\) by dividing both sides by 8.38: \[X_{O_2} = \frac{0.21}{8.38} \approx 0.0251\]
05

Calculate the mole percent of oxygen

The mole percent of oxygen is found by multiplying the mole fraction by 100: \[Mole \, Percent \, of \, O_2 = X_{O_2} \cdot 100\] \[Mole \, Percent \, of \, O_2 \approx 0.0251 \cdot 100 = 2.51\%\] Therefore, for the partial pressure of oxygen in the diving gas to be the same as in air at 1 atm, the mole percent of oxygen in the diving gas should be approximately 2.51%.

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Most popular questions from this chapter

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164{ }^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

A fixed quantity of gas at \(21^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of 5.12 L. (a) Calculate the volume the gas will occupy if the pressure is increased to 1.88 atm while the temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(175^{\circ} \mathrm{C}\) while the pressure is held constant.

In the United States, barometric pressures are generally reported in inches of mercury (in. Hg). On a beautiful summer day in Chicago the barometric pressure is 30.45 in. Hg. (a) Convert this pressure to torr. (b) Convert this pressure to atm. (c) A meteorologist explains the nice weather by referring to a "high- pressure area." In light of your answer to parts (a) and (b), explain why this term makes sense.

Consider a mixture of two gases, \(A\) and \(B\), confined in a closed vessel. A quantity of a third gas, \(C,\) is added to the same vessel at the same temperature. How does the addition of gas \(\mathrm{C}\) affect the following: (a) the partial pressure of gas \(A,\) (b) the total pressure in the vessel, \((\mathbf{c})\) the mole fraction of gas \(\mathrm{B} ?\)

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