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

Short Answer

Expert verified
The gases in order of increasing average molecular speeds at 300 K are: CO, H₂S, HBr, Cl₂, SF₆. The root-mean-square speed of CO is approximately 13.83 m/s, while for Cl₂, it is approximately 8.051 m/s. The most probable speed of CO is approximately 12.12 m/s, whereas for Cl₂, it is approximately 7.096 m/s.

Step by step solution

01

Calculate the molar masses of the gases

First, we need to calculate the molar mass of each gas molecule: - CO: 12 g/mol (C) + 16 g/mol (O) = 28 g/mol - SF₆: 32 g/mol (S) + 6 * 19 g/mol (F) = 146 g/mol - H₂S: 2 g/mol (H) + 32 g/mol (S) = 34 g/mol - Cl₂: 2 * 35.5 g/mol = 71 g/mol - HBr: 1 g/mol (H) + 80 g/mol (Br) = 81 g/mol
02

Order gases based on their average molecular speeds

Since the average molecular speed is inversely proportional to the square root of the molar mass, we can order the gases: SF₆ > Cl₂ > HBr > H₂S > CO (in decreasing molecular speeds) Therefore, the gases ordered in increasing molecular speeds are: CO, H₂S, HBr, Cl₂, SF₆. For part (b), we have to calculate and compare the root-mean-square (rms) speeds of CO and Cl₂ at a temperature of 300 K. The formula for the root-mean-square speed is as follows: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
03

Calculate the root-mean-square speeds of CO and Cl₂

First, convert the molar masses of CO and Cl₂ from g/mol to kg/mol: - CO: 28 g/mol = 0.028 kg/mol - Cl₂: 71 g/mol = 0.071 kg/mol Next, use the given temperature, T = 300 K, and the universal gas constant, R = 8.314 J/(mol∙K). Calculate the root-mean-square speeds for both gases: \( v_{rms,CO} = \sqrt{\frac{3(8.314)(300)}{0.028}} \) \( v_{rms,CO} = 13.83\,\mathrm{m/s} \) \( v_{rms,Cl_2} = \sqrt{\frac{3(8.314)(300)}{0.071}} \) \( v_{rms,Cl_2} = 8.051\,\mathrm{m/s} \) So, the root-mean-square speed of CO is approximately 13.83 m/s, and the root-mean-square speed of Cl₂ is approximately 8.051 m/s. For part (c), we need to calculate and compare the most probable speeds of CO and Cl₂ molecules at 300 K. The formula for the most probable speed is as follows: \[ v_p = \sqrt{\frac{2RT}{M}} \]
04

Calculate the most probable speeds of CO and Cl₂

Using the given temperature, T = 300 K, and the universal gas constant, R = 8.314 J/(mol∙K). Calculate the most probable speeds for both gases: \( v_{p,CO} = \sqrt{\frac{2(8.314)(300)}{0.028}} \) \( v_{p,CO} = 12.12\,\mathrm{m/s} \) \( v_{p,Cl_2} = \sqrt{\frac{2(8.314)(300)}{0.071}} \) \( v_{p,Cl_2} = 7.096\,\mathrm{m/s} \) So, the most probable speed of CO is approximately 12.12 m/s, and the most probable speed of Cl₂ is approximately 7.096 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 is a fundamental concept that explains the behavior of gases in terms of motion. It states that gas molecules are in constant random motion, colliding with each other and the walls of their container. These collisions result in pressure, which we can measure. The theory assumes that gases consist of tiny particles moving in straight lines until they collide, and these collisions are perfectly elastic. This means no energy is lost, just transferred.

Key points of the Kinetic Molecular Theory include:
  • Gas molecules move freely and occupy any available space.
  • The average kinetic energy of gas molecules is directly proportional to the temperature of the gas.
  • Heavier molecules move slower than lighter molecules at the same temperature.
These principles help us understand gas behavior and facilitate calculations involving gas speeds, like root-mean-square (RMS) speed and most probable speed.
RMS Speed
RMS Speed, or root-mean-square speed, is a useful concept in gases that represents the average speed of gas molecules. It provides insight into the molecular motion in a gas. The formula for calculating RMS speed is:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

This involves:
  • \(R\), the universal gas constant, which is 8.314 J/(mol·K).
  • \(T\), the temperature in Kelvin.
  • \(M\), the molar mass of the gas in kg/mol.
Since RMS speed depends on both the mass and temperature of the gas, heavier gases have lower RMS speeds at the same temperature.

For example, calculating RMS speeds for CO and Cl₂ at 300 K shows CO has a higher speed due to its lower molar mass. This illustrates how different gases behave under identical conditions.
Molar Mass
Molar Mass is an essential concept in chemistry, referring to the mass of one mole of a substance, usually expressed in grams/mol. It significantly influences the properties of gases, particularly their speed and behavior.

In gas calculations:
  • Molar mass helps determine how gas particles move. Lighter gases like hydrogen move faster than heavier gases like chlorine.
  • It is crucial in obtaining related speeds, such as RMS and most probable speeds. The formula for these speeds includes molar mass in a way that is inversely related to these speeds.
To calculate molar mass, sum up the atomic masses of each element in the molecule. For example, CO has a molar mass of 28 g/mol, which affects how we rank various gases by speed under the gas laws.
Gas Laws
Gas Laws are mathematical relationships that describe the behavior of gases. Understanding these laws helps to predict how gases will respond to changes in pressure, volume, and temperature. Some fundamental gas laws include:

  • Boyle’s Law: At constant temperature, the pressure of a gas is inversely proportional to its volume.
  • Charles's Law: At constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin.
  • Ideal Gas Law: Combines the laws into one equation \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature.
These laws guide how we derive the speed equations used in molecular calculations. By knowing these principles, we can understand the behavior of gases under various conditions and determine different speed metrics, highlighting the versatile nature of gas particles in response to environmental changes.

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Most popular questions from this chapter

Newton had an incorrect theory of gases in which he assumed that all gas molecules repel one another and the walls of their container. Thus, the molecules of a gas are statically and uniformly distributed, trying to get as far apart as possible from one another and the vessel walls. This repulsion gives rise to pressure. Explain why Charles's law argues for the kineticmolecular theory and against Newton's model.

To minimize the rate of evaporation of the tungsten filament, \(1.4 \times 10^{-5} \mathrm{~mol}\) of argon is placed in a \(600-\mathrm{cm}^{3}\) lightbulb. What is the pressure of argon in the lightbulb at \(23^{\circ} \mathrm{C} ?\)

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2},\) which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and 0.970 atm when \(24.5 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at 1.00 atm and \(298 \mathrm{~K}\), to completely oxidize \(50.0 \mathrm{~g}\) of glucose.

It turns out that the van der Waals constant \(b\) equals four times the total volume actually occupied by the molecules of a mole of gas. Using this figure, calculate the fraction of the volume in a container actually occupied by Ar atoms (a) at STP, (b) at 200 atm pressure and \(0^{\circ} \mathrm{C}\). (Assume for simplicity that the ideal-gas equation still holds.)

Complete the following table for an ideal gas: $$ \begin{array}{llll} \hline \boldsymbol{P} & \boldsymbol{v} & \boldsymbol{n} & \boldsymbol{T} \\ \hline 2.00 \mathrm{~atm} & 1.00 \mathrm{~L} & 0.500 \mathrm{~mol} & ? \mathrm{~K} \\ 0.300 \mathrm{~atm} & 0.250 \mathrm{~L} & ? \mathrm{~mol} & 27^{\circ} \mathrm{C} \\ 650 \text { torr } & ? \mathrm{~L} & 0.333 \mathrm{~mol} & 350 \mathrm{~K} \\ ? \mathrm{~atm} & 585 \mathrm{~mL} & 0.250 \mathrm{~mol} & 295 \mathrm{~K} \\ \hline \end{array} $$

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