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(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{O}_{2}, \mathrm{Ar}, \mathrm{CO}, \mathrm{HCl}, \mathrm{CH}_{4} \cdot(\mathbf{b})\) Calculate the rms speed of CO molecules at \(25^{\circ} \mathrm{C}\). (c) Calculate the most probable speed of an argon atom in the stratosphere, where the temperature is \(0^{\circ} \mathrm{C}\).

Short Answer

Expert verified
(a) Ar, HCl, O2, CO, CH4. (b) 515.4 m/s. (c) 337.0 m/s.

Step by step solution

01

Understanding Molecular Speeds

The average molecular speed of gases is inversely related to their molar masses, as lighter molecules move faster. You'll use the formula for root mean square (rms) speed, \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the gas constant 8.314 J/mol·K, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass in kg/mol.
02

Convert Temperatures to Kelvin

Convert the given temperatures to Kelvin. \( 25^{\circ}C = 298 K \) and \( 0^{\circ}C = 273 K \).
03

Molar Mass Calculation

Calculate the molar masses of each gas: \( \mathrm{O}_2 = 32\, g/mol \), \( \mathrm{Ar} = 40\, g/mol \), \( \mathrm{CO} = 28\, g/mol \), \( \mathrm{HCl} = 36.5\, g/mol \), and \( \mathrm{CH}_4 = 16\, g/mol \). Convert these to kg/mol by dividing by 1000.
04

Order of Increasing Molecular Speed

Order the gases from highest to lowest molar mass, which gives the order of increasing average molecular speed at \( 25^{\circ}C \): \( \mathrm{Ar}, \mathrm{HCl}, \mathrm{O}_2, \mathrm{CO}, \mathrm{CH}_4 \).
05

Calculate RMS Speed of CO

Use the formula \( v_{rms} = \sqrt{\frac{3RT}{M}} \) to calculate the root mean square speed of CO at 298 K. Substitute \( R = 8.314 \), \( T = 298 \), \( M = 0.028 \). \( v_{rms} = \sqrt{\frac{3 \times 8.314 \times 298}{0.028}} = 515.4 \text{ m/s} \).
06

Calculate Most Probable Speed of Argon

Use the formula for the most probable speed, \( v_p = \sqrt{\frac{2RT}{M}} \), at 273 K. Substitute \( R = 8.314 \), \( T = 273 \), \( M = 0.040 \). \( v_p = \sqrt{\frac{2 \times 8.314 \times 273}{0.040}} = 337.0 \text{ m/s} \).

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Key Concepts

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

Kinetic Molecular Theory
The Kinetic Molecular Theory provides a fundamental explanation of the behaviors of gases. This theory posits that gases consist of small particles in constant, random motion. These particles move freely and collide elastically with each other and the walls of their container.

There are several assumptions underpinning this theory:
  • The gas particles are in constant, random motion.
  • The volume of individual gas particles is negligible compared to the total volume of the gas.
  • There are no intermolecular forces acting between particles; collisions are perfectly elastic.
  • The average kinetic energy of the gas particles is directly proportional to the temperature of the gas in Kelvin.
Understanding these principles allows us to predict and explain the behavior of gases under different conditions, such as changes in temperature and pressure. It also forms the basis for further exploration into related concepts such as root mean square speed and gas laws.
Root Mean Square Speed
Root Mean Square (RMS) speed is a meaningful measure of the average speed of gas particles, which considers both their mass and temperature. It can be calculated using the formula:\[v_{rms} = \sqrt{\frac{3RT}{M}}\]where:
  • \( R \) is the universal gas constant, 8.314 J/mol·K.
  • \( T \) is the absolute temperature in Kelvin.
  • \( M \) is the molar mass of the gas in kg/mol.

The RMS speed is particularly useful in comparing the speed of different types of gas molecules at a given temperature. It gives insight into how fast the molecules are moving on average, which influences factors like diffusion, effusion, and reaction rates.

For example, lighter gas molecules, such as methane (\( CH_4 \)), will have higher RMS speeds compared to heavier molecules like argon (\( Ar \)) when both are at the same temperature.
Molar Mass Calculation
To accurately calculate gas properties and behaviors, determining the correct molar mass is crucial. Molar mass is the mass of one mole of a given substance, expressed in grams per mole (g/mol). For calculations involving physical gas dynamics, it is often necessary to convert it into kilograms per mole (kg/mol) by dividing by 1000.

To find a gas's molar mass, sum up the atomic masses of all the atoms in a molecule. For instance, oxygen (\( O_2 \)) consists of two oxygen atoms, each with an atomic mass of approximately 16 g/mol, totaling 32 g/mol. Carbon monoxide (\( CO \)) combines one carbon atom (~12 g/mol) and one oxygen atom (~16 g/mol) for a total molar mass of 28 g/mol.

Calculating molar masses is a critical step when dealing with gas laws and calculating molecular speeds, as it directly influences the outcomes of these calculations.
Gas Laws
Gas laws are a series of principles that describe the behavior of gases based on the relationships between pressure, volume, temperature, and quantity. These laws help us predict how gases will react under different conditions.
  • **Boyle’s Law** states that the pressure of a gas is inversely proportional to its volume, assuming temperature and number of molecules are held constant: \( PV = constant \).
  • **Charles’s Law** explains that the volume of a gas is directly proportional to its temperature in Kelvin, assuming pressure and quantity remain constant: \( V/T = constant \).
  • **Avogadro’s Law** indicates that the volume of a gas is directly proportional to the number of moles, assuming temperature and pressure are held constant: \( V/n = constant \).
  • **Ideal Gas Law** combines these three laws into a single equation: \( PV = nRT \), where \( n \) is the number of moles and \( R \) is the gas constant.
Understanding these laws allows scientists and engineers to manipulate and predict gas behavior in various applications, from industrial processes to natural phenomena.

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Most popular questions from this chapter

In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In 30 minutes the average cockroach (running at \(0.08 \mathrm{~km} / \mathrm{h})\) consumed \(1.0 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at \(101.33 \mathrm{kPa}\) pressure and \(20^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in 1 day by a 6.3 -g cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 2.0-L fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, how much of the available \(\mathrm{O}_{2}\) will the cockroach consume in 1 day? (Air is \(21 \mathrm{~mol} \% \mathrm{O}_{2}\).)

Suppose you are given two flasks at the same temperature, one of volume \(2 \mathrm{~L}\) and the other of volume \(3 \mathrm{~L}\). The 2 -L flask contains \(4.8 \mathrm{~g}\) of gas, and the gas pressure is \(x \mathrm{kPa}\). The 3 -L flask contains \(0.36 \mathrm{~g}\) of gas, and the gas pressure is \(0.1 x\). Do the two gases have the same molar mass? If not, which contains the gas of higher molar mass?

A \(6.53-g\) sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces \(1.72 \mathrm{~L}\) of carbon dioxide gas at \(28^{\circ} \mathrm{C}\) and \(99.06 \mathrm{kPa}\) pressure. (a) Write balanced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture. (b) Calculate the total number of moles of carbon dioxide that forms from these reactions. (c) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ; \mathrm{in}\) other words, rate is the amount that diffuses over the time it takes to diffuse.)

A rigid vessel containing a \(3: 1 \mathrm{~mol}\) ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of \(202.7 \mathrm{kPa}\). If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

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