Problem 86
Question
(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 \(300 \mathrm{~K}\).
Step-by-Step Solution
Verified Answer
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.
1Step 1: 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
2Step 2: 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:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
3Step 3: 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:
\[ v_p = \sqrt{\frac{2RT}{M}} \]
4Step 4: 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.
Key Concepts
Kinetic Molecular TheoryRMS SpeedMolar MassGas Laws
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:
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.
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:
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.
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
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.
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:
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.
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.
Other exercises in this chapter
Problem 84
Suppose you have two 1 -L flasks, one containing \(\mathrm{N}_{2}\) at STP, the other containing \(\mathrm{CH}_{4}\) at STP. How do these systems compare with r
View solution Problem 85
(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}
View solution Problem 87
Explain the difference between effusion and diffusion.
View solution Problem 88
At constant pressure, the mean free path \((\lambda)\) of a gas molecule is directly proportional to temperature. At constant temperature, \(\lambda\) is invers
View solution