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Consider three identical flasks filled with different gases. Flask \(\mathrm{A} : \mathrm{CO}\) at 760 torr and \(0^{\circ} \mathrm{C}\) Flask \(\mathrm{B} : \mathrm{N}_{2}\) at 250 torr and \(0^{\circ} \mathrm{C}\) Flask \(\mathrm{C} : \mathrm{H}_{2}\) at 100 torr and \(0^{\circ} \mathrm{C}\) a. In which flask will the molecules have the greatest average kinetic energy? b. In which flask will the molecules have the greatest average velocity?

Short Answer

Expert verified
a. The molecules will have the greatest average kinetic energy in all the flasks since they are at the same temperature (\(0^{\circ}\mathrm{C}\)). b. The molecules will have the greatest average velocity in Flask C (hydrogen gas) with an average velocity of approximately \(1930.23 \,\text{m/s}\).

Step by step solution

01

a. Calculating average kinetic energy

According to the Kinetic Theory of Gases, the average kinetic energy of an ideal gas depends only on its temperature. All the flasks are at \(0^{\circ}\mathrm{C}\), which means they all have the same temperature. Therefore, the average kinetic energy of the gas molecules in all three flasks will be the same. So, the answer to question (a) is that the molecules will have the greatest average kinetic energy in all the flasks.
02

b. Calculating average velocity

The average velocity of the gas molecules can be determined using the following formula: \(v_{rms} = \sqrt{\frac{3RT}{M}}\) Where \(v_{rms}\) is the root mean square velocity of the gas molecules, \(R\) is the ideal gas constant (8.314 J K⁻¹ mol⁻¹), \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas in kg/mol. Since the temperature in all three flasks is \(0^{\circ}\mathrm{C}\) or 273.15 K, we can calculate the average velocity of the molecules in each flask by plugging in the values for each gas's molar mass (in kg/mol) and solving for \(v_{rms}\). For Flask A (carbon monoxide, CO): Molar mass of CO is 28.01 g/mol, which is 0.02801 kg/mol. \(v_{rms,CO} = \sqrt{\frac{3(8.314)(273.15)}{0.02801}}\) For Flask B (nitrogen, N₂): Molar mass of N₂ is 28.02 g/mol, which is 0.02802 kg/mol. \(v_{rms,N_2} = \sqrt{\frac{3(8.314)(273.15)}{0.02802}}\) For Flask C (hydrogen, H₂): Molar mass of H₂ is 2.02 g/mol, which is 0.00202 kg/mol. \(v_{rms,H_2} = \sqrt{\frac{3(8.314)(273.15)}{0.00202}}\) Solving the equations, we get: \(v_{rms,CO} ≈ 510.41 \,\text{m/s}\) \(v_{rms,N_2} ≈ 510.55 \,\text{m/s}\) \(v_{rms,H_2} ≈ 1930.23 \,\text{m/s}\) As we can see, the molecules in Flask C (hydrogen gas) have the greatest average velocity. So, the answer to question (b) is that the molecules will have the greatest average velocity in Flask C.

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Key Concepts

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

Average Kinetic Energy
The average kinetic energy of gas molecules is a key concept in understanding gas behavior, particularly under the framework of the Kinetic Theory of Gases. At the heart of this theory is the idea that the average kinetic energy of a gas depends only on its temperature, not the type of gas or the pressure it is under. This is because temperature is a measure of the average energetic movement of particles.

- **Dependence on Temperature**: Regardless of the type of gas, if the temperature remains constant, the average kinetic energy among gas molecules will also be constant. For instance, in the original exercise, all gases are at the same temperature of 0°C or 273.15 K. Therefore, the gases in flasks A, B, and C have the same average kinetic energy.
- **Implications in Real World**: This concept is crucial for applications in chemistry and physics, especially in processes involving heat transfer. It implies that all gases at the same temperature have molecules moving with the same average energy, though they may differ in speed due to their mass.
Average Velocity
Understanding average velocity helps in analyzing how gas molecules move within a container. It reflects the speed at which gas molecules travel, which varies according to their mass and the temperature of the environment. The average velocity can be calculated using the root mean square velocity formula, which takes into account these factors.

- **Formula for Calculation**: The mathematical expression used is: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] Here, \(v_{rms}\) represents the root mean square velocity, \(R\) is the ideal gas constant, \(T\) is the temperature, and \(M\) is the molar mass of the gas. This formula helps determine the speed of molecules based on their mass.
- **Effect of Molar Mass**: Lighter molecules, such as hydrogen in flask C in the original exercise, move faster because they have lower mass. Despite having the same temperature as heavier gases like carbon monoxide (flask A) and nitrogen (flask B), hydrogen molecules exhibit a higher average velocity.
Root Mean Square Velocity
The root mean square velocity (RMS) is an important statistical measure used to determine the speed distribution of gas particles within a container. This measure provides insight into how fast molecules are moving on average in relation to their temperature and mass.

- **Computation Details**: The RMS velocity is derived from the average kinetic energy and involves the molar mass of the gas and the temperature. For example, in solving for RMS in the original solution: - Carbon monoxide \( (v_{rms,CO}) \) results in approximately 510.41 m/s, - Nitrogen \( (v_{rms,N_2}) \) gives about 510.55 m/s, - Hydrogen \( (v_{rms,H_2}) \) shoots up to 1930.23 m/s.
- **Importance in Gas Dynamics**: The RMS velocity is crucial in predicting how gases will behave under different conditions. It also helps in understanding diffusion and effusion rates, where lighter gases typically move faster than heavier gases.

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Most popular questions from this chapter

Ethene is converted to ethane by the reaction $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)\stackrel{\text { Catalyst }}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g)$$ \(\mathrm{C}_{2} \mathrm{H}_{4}\) flows into a catalytic reactor at 25.0 \(\mathrm{atm}\) and \(300.^{\circ} \mathrm{C}\) with a flow rate of \(1000 . \mathrm{L} / \mathrm{min}\) . Hydrogen at 25.0 \(\mathrm{atm}\) and \(300 .^{\circ} \mathrm{C}\) flows into the reactor at a flow rate of \(1500 . \mathrm{L} / \mathrm{min}\) . If 15.0 \(\mathrm{kg}\) \(\mathrm{C}_{2} \mathrm{H}_{6}\) is collected per minute, what is the percent yield of the reaction?

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