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Chapter 11: Problem 99

A \(1.0-\) - flask contains 10.0 g each of \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) (a) Which gas has the greater partial pressure, \(\mathbf{O}_{2}\) or \(\mathrm{CO}_{2},\) or are they the same? (b) Which molecules have the greater rms speed, or are they the same? (c) Which molecules have the greater average kinetic energy, or are they the same?

Short Answer

Expert verified
(a) \(\mathrm{O}_{2}\) has greater partial pressure. (b) \(\mathrm{O}_{2}\) has greater rms speed. (c) Their average kinetic energy is the same.

Step by step solution

01

Calculate Molar Mass

First, calculate the molar mass of \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\). The molar mass of \(\mathrm{O}_{2}\) (oxygen) is approximately 32.0 g/mol (16.0 g/mol for each atom) and for \(\mathrm{CO}_{2}\) (carbon dioxide), it is approximately 44.0 g/mol (12.0 g/mol for carbon and 16.0 g/mol for each oxygen atom).
02

Determine the Moles of Each Gas

Use the formula to find the number of moles: \(n = \frac{\text{mass}}{\text{molar mass}}\). Calculate for \(\mathrm{O}_{2}\): \(n = \frac{10.0 \text{ g}}{32.0 \text{ g/mol}} = 0.3125\text{ mol}\). For \(\mathrm{CO}_{2}\): \(n = \frac{10.0 \text{ g}}{44.0 \text{ g/mol}} = 0.2273\text{ mol}\).
03

Apply Ideal Gas Law to Find Partial Pressure

Use the ideal gas law \(PV = nRT\) with \(P = \frac{nRT}{V}\). Assume \(R = 0.0821\text{ L atm/mol K}\) and temperature is \(298\text{ K}\). The volume of the flask is \(1.0 \text{ L}\). Calculate partial pressure for \(\mathrm{O}_{2}\): \(P = \frac{0.3125 \times 0.0821 \times 298}{1.0} = 7.66 \text{ atm}\). For \(\mathrm{CO}_{2}\): \(P = \frac{0.2273 \times 0.0821 \times 298}{1.0} = 5.55 \text{ atm}\).
04

Analyze Root Mean Square Speed (rms)

The rms speed is given by \(\sqrt{\frac{3RT}{M}}\), where \(M\) is the molar mass in kg/mol. Since \(\mathrm{O}_{2}\) (molar mass 32.0 g/mol = 0.032 kg/mol) is lighter than \(\mathrm{CO}_{2}\) (44.0 g/mol = 0.044 kg/mol), \(\mathrm{O}_{2}\) has a greater rms speed. This is because speed increases as molar mass decreases.
05

Determine Average Kinetic Energy

The average kinetic energy of a molecule is given by \(\frac{3}{2}kT\) where \(k\) is the Boltzmann constant, and is the same for all gases at the same temperature. Therefore, both \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) have the same average kinetic energy at \(25^{\circ} \mathrm{C}\).

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Key Concepts

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

Partial Pressure
In a gas mixture, each gas exerts a pressure as if it were alone in the container. This is known as partial pressure. The ideal gas law, \(PV = nRT\), tells us that the pressure \(P\) is proportional to the number of moles \(n\) of the gas. Thus, partial pressure depends on the amount of gas and its volume and temperature.

For \(\mathrm{O}_{2}\), with 0.3125 moles, the partial pressure is 7.66 atm. For \(\mathrm{CO}_{2}\), with 0.2273 moles, it is 5.55 atm. This shows \(\mathrm{O}_{2}\) has a greater partial pressure since it has more moles at the same temperature and volume.
Root Mean Square Speed
The root mean square speed (rms speed) considers the velocities of molecules in a gas. It indicates the average speed of gas molecules based on their mass and temperature. The formula is \(\sqrt{\frac{3RT}{M}}\), where \(M\) stands for molar mass.

Since \(\mathrm{O}_{2}\) has a molar mass of 32.0 g/mol compared to \(\mathrm{CO}_{2}\)'s 44.0 g/mol, \(\mathrm{O}_{2}\) molecules are lighter, allowing them to travel faster. Therefore, \(\mathrm{O}_{2}\) has a higher rms speed due to its lower molar mass.
Kinetic Energy
Kinetic energy in gases arises from their motion. The formula \(\frac{3}{2}kT\) suggests that temperature \(T\) solely determines average kinetic energy in gases. \(k\) is Boltzmann鈥檚 constant, a fundamental constant.

All gases at the same temperature, like our \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\), share the same average kinetic energy. Even if their speeds differ, their energy, dictated by temperature, remains alike.
Ideal Gas Law
The ideal gas law \(PV = nRT\) serves as the cornerstone of gas behavior in ideal conditions. \(P\) stands for pressure, \(V\) for volume, \(n\) for moles of gas, \(R\) for the gas constant (0.0821 L atm/mol K), and \(T\) for temperature in Kelvin.

This equation links macroscopic properties of gases to molecular behavior. By knowing any three properties, you can find the fourth. In our problem, knowing the moles, temperature, and volume allows us to determine each gas's partial pressure, demonstrating the law's utility in practical scenarios.

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Most popular questions from this chapter

The average barometric pressure at an altitude of \(10 \mathrm{km}\) is 210 mm Hg. Express this pressure in atmospheres, bars, and kilopascals.

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A Chlorine gas \(\left(\mathrm{Cl}_{2}\right)\) is used as a disinfectant in municipal water supplies, although chlorine dioxide \(\left(\mathrm{ClO}_{2}\right)\) and ozone are becoming more widely used. \(\mathrm{ClO}_{2}\) is a better choice than \(\mathrm{Cl}_{2}\) in this application because it leads to fewer chlorinated by-products, which are themselves pollutants. (a) How many valence electrons are in \(\mathrm{ClO}_{2} ?\) (b) The chlorite ion, \(\mathrm{ClO}_{2}^{-}\), is obtained by reducing ClO \(_{2} .\) Draw a possible electron dot structure for \(\mathrm{ClO}_{2}^{-} .\) (Cl is the central atom.) (c) What is the hybridization of the central Cl atom in \(\mathrm{ClO}_{2}^{-} ?\) What is the shape of the ion? (d) Which species has the larger bond angle, \(\mathbf{O}_{3}\) or \(\mathrm{ClO}_{2}^{-} ?\) Explain briefly. (e) Chlorine dioxide, \(\mathrm{ClO}_{2},\) a yellow-green gas, can be made by the reaction of chlorine with sodium chlorite: \(2 \mathrm{NaClO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})+2 \mathrm{ClO}_{2}(\mathrm{g})\) Assume you react \(15.6 \mathrm{g}\) of \(\mathrm{NaClO}_{2}\) with chlorine gas, which has a pressure of \(1050 \mathrm{mm}\) Hg in a \(1.45-\mathrm{L}\). flask at \(22^{\circ} \mathrm{C}\). What mass of \(\mathrm{ClO}_{2}\) can be produced?

A 1.50 L constant volume calorimeter (Figure 5.12 ) contains \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) The partial pressure of \(\mathrm{C}_{3} \mathrm{H}_{8}\) is 0.10 atm and the partial pressure of \(\mathrm{O}_{2}\) is 5.0 atm. The temperature is \(20.0^{\circ} \mathrm{C}\). A reaction occurs between the two compounds, forming \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\ell) .\) The heat from the reaction causes the temperature to rise to \(23.2^{\circ} \mathrm{C}\) (a) Write a balanced chemical equation for the reaction. (b) How many moles of \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) are present in the flask initially? (c) What is the mole fraction of \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) in the flask before reaction? (d) After the reaction, the flask contains excess oxygen and the products of the reaction, \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\ell) .\) What amount of unreacted \(\mathrm{O}_{2}(\mathrm{g})\) remains? (e) After the reaction, what is the partial pressure exerted by the \(\mathrm{CO}_{2}(\mathrm{g})\) in this system? (f) What is the partial pressure exerted by the excess oxygen remaining after the reaction?

A \(1.25-\mathrm{g}\) sample of \(\mathrm{CO}_{2}\) is contained in a \(750 .\) mI. flask at \(22.5^{\circ} \mathrm{C} .\) What is the pressure of the gas?
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