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Three \(5-\mathrm{L}\) flasks, fixed with pressure gauges and small valves, each contain \(4 \mathrm{~g}\) of gas at \(273 \mathrm{~K}\). Flask A contains \(\mathrm{H}_{2}\), flask B contains He, and flask C contains \(\mathrm{CH}_{4}\). Rank the flask contents in terms of (a) pressure, (b) average kinetic energy of the particles, (c) diffusion rate after the valve is opened, (d) total kinetic energy of the particles, (e) density, and (f) collision frequency.

Short Answer

Expert verified
Gases can be ranked as: (a) pressure: \( \text{H}_2 > \text{He} > \text{CH}_4 \), (b) average kinetic energy: all equal, (c) diffusion rate: \( \text{H}_2 > \text{He} > \text{CH}_4 \), (d) total kinetic energy: \( \text{H}_2 > \text{He} > \text{CH}_4 \), (e) density: all equal, (f) collision frequency: \( \text{H}_2 > \text{He} > \text{CH}_4 \).

Step by step solution

01

Calculate Moles of Each Gas

First, determine the number of moles of each gas using the formula: \( n = \frac{m}{M} \), where \( m \) is the mass of the gas and \( M \) is the molar mass.For \( \text{H}_2 \):\( M(\text{H}_2) = 2 \text{ g/mol} \)\( n(\text{H}_2) = \frac{4 \text{ g}}{2 \text{ g/mol}} = 2 \text{ mol} \)For He:\( M(\text{He}) = 4 \text{ g/mol} \)\( n(\text{He}) = \frac{4 \text{ g}}{4 \text{ g/mol}} = 1 \text{ mol} \)For \( \text{CH}_4 \): \( M(\text{CH}_4) = 16 \text{ g/mol} \)\( n(\text{CH}_4) = \frac{4 \text{ g}}{16 \text{ g/mol}} = 0.25 \text{ mol} \)
02

Calculate Pressure

Use the ideal gas law formula: \( P = \frac{nRT}{V} \)Since \( V = 5 \text{ L} \) and \( T = 273 \text{ K} \) for all flasks, let \( R = 0.0821 \text{ L路atm路K}^{-1}\text{路mol}^{-1} \).For \( \text{H}_2 \):\( P = \frac{2 \text{ mol} \times 0.0821 \times 273}{5} \text{ atm} = 8.95 \text{ atm} \)For He:\( P = \frac{1 \text{ mol} \times 0.0821 \times 273}{5} \text{ atm} = 4.48 \text{ atm} \)For \( \text{CH}_4 \):\( P = \frac{0.25 \text{ mol} \times 0.0821 \times 273}{5} \text{ atm} = 1.12 \text{ atm} \)
03

Average Kinetic Energy

The average kinetic energy of gas particles depends only on the temperature and is given by: \( \text{KE}_{\text{avg}} = \frac{3}{2} k T \), where \( k \) is the Boltzmann constant. Since all gases are at the same temperature, their average kinetic energies are the same.Thus, \( \text{KE}_{\text{avg}}(\text{H}_2) = \text{KE}_{\text{avg}}(\text{He}) = \text{KE}_{\text{avg}}(\text{CH}_4) \)
04

Diffusion Rate

Use Graham's law of diffusion: \( \text{Rate} \propto \frac{1}{\sqrt{M}} \), where M is the molar mass.Rate of \( \text{H}_2 \): \( \frac{1}{\sqrt{2}} \)Rate of He: \( \frac{1}{\sqrt{4}} \)Rate of \( \text{CH}_4 \): \( \frac{1}{\sqrt{16}} \)Thus, \( \text{H}_2 \) diffuses fastest, followed by He, then \( \text{CH}_4 \).
05

Total Kinetic Energy

The total kinetic energy of a gas is given by: \( \text{KE}_{\text{total}} = n \times \text{KE}_{\text{avg}} \).Given the same average kinetic energies and differing moles:For \( \text{H}_2 \): \( \text{KE}_{\text{total}} \text{ (H}_2) = 2 \times \text{K} \)For He: \( \text{KE}_{\text{total}} \text{ (He)} = 1 \times \text{K} \)For \( \text{CH}_4 \): \( \text{KE}_{\text{total}} \text{ (CH}_4) = 0.25 \times \text{K} \)Thus, \( \text{H}_2 > \text{He} > \text{CH}_4 \).
06

Density

Calculate gas density: \( \text{Density} (d) = \frac{\text{mass}}{\text{volume}} \).Given the same volume but differing molar masses:For \( \text{H}_2 \): \( d \text{ (H}_2) = \frac{4 \text{ g}}{5 \text{ L}} = 0.8 \text{ g/L} \)For He: \( d \text{ (He)} = \frac{4 \text{ g}}{5 \text{ L}} = 0.8 \text{ g/L} \)For \( \text{CH}_4 \): \( d \text{ (CH}_4) = \frac{4 \text{ g}}{5 \text{ L}} = 0.8 \text{ g/L} \)Thus, all densities are equal: \( \text{H}_2 = \text{He} = \text{CH}_4 \)
07

Collision Frequency

Collision frequency is proportional to the pressure for a given volume and temperature.From Step 2:\( \text{Pressure(H}_2) > \text{Pressure(He)} > \text{Pressure(CH}_4) \)Thus, \( \text{H}_2 \text{ experiences the highest collision frequency, followed by He, then } \text{CH}_4 \).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is given by \[ PV = nRT \], where \( P \) is the pressure in atmospheres (atm), \( V \) is the volume in liters (L), \( n \) is the number of moles, \( R \) is the ideal gas constant (0.0821 L路atm路K鈦宦孤穖ol鈦宦), and \( T \) is the temperature in Kelvin (K).
This law simplifies the calculation of one variable if the others are known. For example, in our problem, we used this law to find out the pressure in each flask. Since all flasks have the same volume (5 L) and temperature (273 K), the pressures were calculated using their respective moles of gas. The calculated pressures were 8.95 atm for H鈧, 4.48 atm for He, and 1.12 atm for CH鈧.
Kinetic Molecular Theory
The Kinetic Molecular Theory (KMT) helps explain the behavior of gases and their properties based on the motion of their particles. It states that gas particles are in constant random motion, colliding elastically with each other and the walls of their container. According to KMT, the average kinetic energy of gas particles is directly proportional to the temperature in Kelvin.
The average kinetic energy can be expressed as \[ \text{KE}_{\text{avg}} = \frac{3}{2}kT \], where \( k \) is the Boltzmann constant. Since the temperature is the same for all gases in the problem (273 K), their average kinetic energy remains identical, regardless of the gas type.
Graham's Law of Diffusion
Graham's Law of Diffusion describes how the rate at which a gas diffuses is inversely proportional to the square root of its molar mass. This can be mathematically written as \[ \text{Rate} \text{掳}propto \frac{1}{\text{sqrt}(M)} \], where \( M \) is the molar mass of the gas.
In our exercise, the gases have different molar masses: H鈧 (2 g/mol), He (4 g/mol), and CH鈧 (16 g/mol). The fastest diffusing gas is H鈧, followed by He, and the slowest is CH鈧. They diffuse at rates inversely proportional to the square roots of their molar masses. So, H鈧, being the lightest, diffuses the fastest.
Collision Frequency
Collision frequency refers to how often gas molecules collide with each other and the walls of their container. It depends on factors like pressure, volume, and temperature. The higher the pressure, the more collisions occur.
From our calculations under the Ideal Gas Law, H鈧, with the highest pressure of 8.95 atm, experiences the most collisions. He, with 4.48 atm, follows, and CH鈧, with 1.12 atm, has the least collision frequency. This relationship is quite intuitive as more moles of gas in a given volume increases the pressure and collisions.
Gas Density
Gas density is the mass per unit volume of a gas. It can be calculated using the formula \[ d = \frac{\text{mass}}{\text{volume}} \]. In our exercise, each gas sample has the same mass (4 g) and occupies the same volume (5 L). Therefore, the density for each gas is \[ \frac{4 \text{ g}}{5 \text{ L}} = 0.8 \text{ g/L} \].
So, even though the gases are different (H鈧, He, and CH鈧), their densities are the same because of the equal mass and volume provided. However, in real-world scenarios, gas density variations are significant and should be carefully considered for practical applications.

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