Question

Consider separate \(1.0-\mathrm{L}\) gaseous samples of \(\mathrm{H}_{2}, \mathrm{Xe}, \mathrm{Cl}_{2}\), and \(\mathrm{O}_{2}\) all at STP. a. Rank the gases in order of increasing average kinetic energy. b. Rank the gases in order of increasing average velocity. c. How can separate \(1.0-\mathrm{L}\) samples of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) each have the same average velocity?

Short answer

In summary, at STP: a. All four gases have the same average kinetic energy, as the average kinetic energy is directly proportional to temperature: H₂ ≈ Xe ≈ Cl₂ ≈ O₂. b. The ranking of the gases in terms of increasing average velocity is: Xe < O₂ < Cl₂ < H₂, as the average velocity is inversely proportional to the square root of the molar mass of the gas. c. Separate 1.0-L samples of O₂ and H₂ can have the same average velocity if the product of the moles (n) and molar mass (M) in the expression \( \large υ = \sqrt{\frac {2 \cdot \frac{3}{2}nRT}{M}} "\) is equal for both gases. This implies that they have different numbers of moles, but the product of moles and molar mass is the same.

Step-by-step solution

Recall the Kinetic Theory of Gases.

We need to remember a few important relationships from the Kinetic Theory of Gases: - The average kinetic energy of a gas is given by \(\large \frac 1 2 mυ^2 \) where m is the mass of a molecule and υ is the velocity of the gas. - The average kinetic energy of a molecule in a gas is directly proportional to the temperature: \(\large \frac 3 2 nRT = \frac 1 2 mυ^2 \), where n is the number of moles, R is the gas constant, and T is the temperature.

Rank the gases in order of increasing average kinetic energy.

As per the Kinetic Theory of Gases, the average kinetic energy of a gas is directly proportional to its temperature. All four gases are at STP (273K), so their temperatures are the same. Therefore, their average kinetic energies will also be the same. The ranking in terms of increasing average kinetic energy is: H₂ ≈ Xe ≈ Cl₂ ≈ O₂

Compute the molecular mass for each gas.

To find the average velocity of each gas, we need to determine their molecular masses (mass per mole) in atomic mass units (amu). Here are the molecular masses of each gas: - H₂ : 2 amu - Xe : 131 amu - Cl₂ : 2 * 35.5 = 71 amu - O₂ : 2 * 16 = 32 amu

Rank the gases in order of increasing average velocity.

We know that the average kinetic energy \(\large \frac 1 2 mυ^2 \) of gases is proportional to temperature and is the same for all gases at STP. Rewrite the expression for average velocity, which gives us: \( \large υ = \sqrt{\frac {2 \cdot \frac{3}{2}nRT}{M}} ") where M is the molar mass of the gas. Since the temperature, R, and n are constant, the average velocity is inversely proportional to the square root of the molar mass of the gas. Now we can rank the gases in terms of increasing average velocity: - H₂ has the smallest molar mass and will have the highest average velocity among the gases. - Xe has the highest molar mass and will have the lowest average velocity. Therefore, the ranking in terms of increasing average velocity is: Xe < O₂ < Cl₂ < H₂

Understand how separate samples of O₂ and H₂ can have the same average velocity.

For two different gases to have the same average velocity, their average kinetic energies must be the same and they must have the same temperature. Since all gases at STP have the same temperature, the average kinetic energies will be the same at STP. However, to have the same average velocity, the two gases should have equal expression in this equation: \( \large υ = \sqrt{\frac {2 \cdot \frac{3}{2}nRT}{M}} ") Thus, for the same average velocity, the product of the moles (n) and molar mass (M) should be equal for both O₂ and H₂. This means that they must have different numbers of moles such that the product of moles and molar mass is equal.

Key concepts

Average Kinetic Energy

When discussing the kinetic theory of gases, one of the fundamental ideas is the average kinetic energy of gas particles. According to the kinetic theory, all gas particles are in constant, random motion and possess kinetic energy, which is energy due to their motion. In a sample of gas, each particle may have different kinetic energy, but we can calculate an average value for all particles.

The average kinetic energy (\( KE_{avg} \) of gas particles is directly related to the absolute temperature of the gas. The mathematical expression of this relationship is given by \( KE_{avg} = \frac{3}{2}kT \) where \( k \) is Boltzmann's constant and \( T \) is the temperature in kelvins (K).

• At standard temperature and pressure (STP), which is at 0 degrees Celsius or 273.15 K, all ideal gases have the same average kinetic energy because the temperature is the same.
• The mass of the particles or the type of gas is not a factor in determining average kinetic energy at a given temperature—this is why different gases at STP were ranked with equal average kinetic energies in the provided example.

This concept is crucial because it helps us understand that at the same temperature, all gases, regardless of their molecular composition, have the same energy on average due to random motion of their particles.

Molecular Mass

Molecular mass plays a significant role in understanding gas behavior, particularly when discussing gas velocity and diffusion. Molecular mass, often expressed in atomic mass units (amu), is the sum of the individual masses of all the atoms in a molecule.

For instance, in our exercise, different gases have different molecular masses:

• Hydrogen (\( H_2 \) has a molecular mass of 2 amu since it consists of two hydrogen atoms, each of roughly 1 amu.
• Xenon (\( Xe \) has a molecular mass of 131 amu, being a monoatomic gas.
• Chlorine (\( Cl_2 \) has a molecular mass of 71 amu, derived from two atoms of chlorine, each about 35.5 amu.
• Oxygen (\( O_2 \) has a molecular mass of 32 amu, with two oxygen atoms each at 16 amu.

Understanding the molecular mass is vital because it helps us predict the behavior of gases in terms of velocity and how they will respond under different conditions.

Average Gas Velocity

When examining the properties of gases, average gas velocity is a measure of how fast the gas particles are moving on average. Due to the kinetic theory of gases, we understand that particles are in perpetual motion, and this motion is related to the temperature and mass of the gas particles.

To quantify the average velocity, the root-mean-square (rms) velocity equation is commonly used: \( v_{rms} = \sqrt{\frac{3RT}{M}} \) where \( R \) is the ideal gas constant, \( T \) is the temperature in kelvin, and \( M \) is the molar mass of the gas. Because all gases at a given temperature have the same kinetic energy, their temperatures and the number of particles (n) are constant, and the formula simplifies to an inverse relationship between the rms velocity and the square root of the molecular mass.

In our textbook problem, Hydrogen, with the lowest molecular mass, has the highest average velocity, and Xenon, with the highest mass, has the lowest. This concept is vital when working with gas mixtures and when analyzing molecular diffusion through mediums.

STP Conditions

Standard Temperature and Pressure (STP) are standard sets of conditions used for scientific calculations involving gases to ensure consistency and comparability of data. At STP, the temperature is 273.15 K (0 degrees Celsius), and the pressure is 1 atmosphere (atm). Under these conditions, one mole of an ideal gas occupies 22.414 liters.

One key point from our exercise is the fact that at STP, the average kinetic energy across different gases is the same because the temperature is consistent. STP allows chemists and physicists to have a basis for predicting gas behavior and for performing calculations that can be universally understood.

Important STP Notes:

• All calculations assuming STP conditions must use the temperature of 273.15K, not room temperature or any other value.
• Many gas laws provide more accurate results with real gases when these standard conditions are assumed.
• STP conditions are part of the international agreement for scientific temperature and pressure definitions.

Understanding STP is crucial for students not only in solving problems but also in grasping the predictability and behavior of gases under controlled conditions.