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Chapter 10: Problem 78

(a) Place the following gases in order of increasing average molecular speed at \(300 \mathrm{~K}: \mathrm{CO}, \mathrm{SF}_{6}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{Cl}_{2}\), HBr. (b) Calculate and compare the \(\mathrm{rms}\) speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at \(300 \mathrm{~K}\).

Short Answer

Expert verified
The gases can be arranged in increasing molecular speed order at 300 K as: CO, H2S, Cl2, HBr, SF6. The rms speeds of CO and Cl2 molecules at 300 K are approximately 404.42 m/s and 238.54 m/s, respectively, with CO having a higher rms speed.

Step by step solution

01

Arrange gases in increasing molecular speed order

Since mass directly affects molecular speed (lighter molecules have higher molecular speeds), we need to arrange the gases in order of increasing molar mass. First, let's find the molar mass of each gas: 1. CO: 12 (C) + 16 (O) = 28 g/mol 2. SF6: 32 (S) + 6 * 19 (F) = 146 g/mol 3. H2S: 2 * 1 (H) + 32 (S) = 34 g/mol 4. Cl2: 2 * 35.5 (Cl) = 71 g/mol 5. HBr: 1 (H) + 80 (Br) = 81 g/mol Now, arrange these gases in order of increasing molar mass: CO, H2S, Cl2, HBr, SF6.
02

Calculate rms speeds of CO and Cl2

To calculate the rms speeds of CO and Cl2 molecules, we will use the following formula: \[v_{rms} = \sqrt{\frac{3RT}{M}}\] Where: \(v_{rms}\) is the root-mean-square speed, \(R\) is the gas constant (8.314 J/(mol路K)), \(T\) is the temperature (300 K), \(M\) is the molar mass of the molecule (converted to kg/mol). For CO: M = 28 g/mol = 0.028 kg/mol \(v_{rms} = \sqrt{\frac{3 * 8.314 * 300}{0.028}}\) \(v_{rms} \approx 404.42 m/s\) For Cl2: M = 71 g/mol = 0.071 kg/mol \(v_{rms} = \sqrt{\frac{3 * 8.314 * 300}{0.071}}\) \(v_{rms} \approx 238.54 m/s\) Comparing the rms speeds, CO has a higher rms speed than Cl2 at 300 K.

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Key Concepts

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

Molecular Speed
Gases are composed of molecules in constant motion, exhibiting various speeds. The average molecular speed is influenced by the mass of the gas molecules. At a given temperature, lighter gas molecules move faster than heavier ones. This behavior stems from the kinetic energy equation, which relates energy to mass and speed:
  • The kinetic energy ( ext{KE}) of a gas molecule is given by ext{KE} = \( \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the speed of the molecule.
  • For gases at the same temperature, their molecules have the same average kinetic energy.
  • Thus, molecules with less mass must move faster to maintain the same kinetic energy as heavier molecules.
This principle explains why, in the exercise, lighter gases like CO and H\(_2\)S, typically have higher molecular speeds when compared to heavier gases such as SF\(_6\), Cl\(_2\), and HBr.
RMS Speed
The root-mean-square (rms) speed provides a useful measure of the speed of molecules in a gas. It considers the variability in speeds due to continuous molecular collisions.
The rms speed formula is: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
  • \( R \) is the gas constant, which helps relate various thermodynamic quantities.
  • \( T \) is the absolute temperature of the gas in Kelvin.
  • \( M \) is the molar mass in kg/mol, converted from grams per mole.
Applying this in calculations provides a single value that reflects the typical speed of an average gas molecule at a given temperature. As shown in the solution:
  • CO molecules have an \( v_{\text{rms}} \) of 404.42 m/s, faster than Cl\(_2\) molecules at 238.54 m/s, due to CO's lower molar mass.
Molar Mass
The molar mass of a gas is a crucial factor in determining its molecular speed. Molar mass is the mass of one mole of a substance, generally expressed in grams per mole (g/mol).
  • Lighter molar masses lead to higher speeds, as indicated by the rms speed formula where \( M \) appears in the denominator.
Determining the molar mass is quite straightforward. It's the sum of the atomic masses of all atoms in the molecule:
  • For example, carbon monoxide (CO) has a molar mass of \(28 \text{ g/mol} \), combined from carbon's \(12 \text{ g/mol} \) and oxygen's \(16 \text{ g/mol} \).
  • Molar mass influences the order of gas speeds, such that lighter gases like CO exhibit higher speeds compared to heavier gases such as SF\(_6\) at \(146 \text{ g/mol} \).
Kinetic Theory of Gases
The kinetic theory of gases provides insights into the behavior of gases, connecting macroscopic properties, like pressure and temperature, to the motion of molecules.
  • The theory postulates that gas molecules are in constant, random motion.
  • The speed of the molecules is related to the temperature: increasing temperature increases the speed of the molecules.
  • The pressure exerted by a gas arises from molecules colliding with the walls of its container.
Overall, the kinetic theory aids in understanding how changes in temperature and molar mass influence molecular speeds and pressure, assisting in problem-solving within gas laws. It builds the foundation for calculations involving the molecular speeds, as discussed in the exercise, by relating temperature and molar mass to kinetic energy and rms speeds.

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Most popular questions from this chapter

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{Ne}, \mathrm{HBr}, \mathrm{SO}_{2}, \mathrm{NF}_{3}, \mathrm{CO}\). (b) Calculate the rms speed of \(\mathrm{NF}_{3}\) molecules at \(25^{\circ} \mathrm{C}\).

(a) Calculate the density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr.

The planet Jupiter has a surface temperature of \(140 \mathrm{~K}\) and a mass 318 times that of Earth. Mercury has a surface temperature between \(600 \mathrm{~K}\) and \(700 \mathrm{~K}\) and a mass \(0.05\) times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.

Nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4}\), is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8 -hr workday is 1 part in \(10^{9}\) parts by volume, which means that there is one mole of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for every \(10^{9}\) moles of gas. Assume \(24{ }^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory that is \(54 \mathrm{~m}^{2}\) in area, with a ceiling height of \(3.1 \mathrm{~m}\) ?

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N}\), and \(\mathrm{Cl}\). The complete combustion of a \(100.0\) -mg sample of the herbicide in excess oxygen produces \(83.16 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(73.30 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor at STP. A separate analysis shows that the sample also contains \(16.44 \mathrm{mg}\) of \(\mathrm{Cl}\). (a) Determine the percent composition of the substance. (b) Calculate its empirical formula.
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