RMS Speed, Degrees of Freedom, and Equipartition

RMS Speed of Gas Molecules

The kinetic theory of gases provides a microscopic explanation for the macroscopic properties of gases, such as pressure, temperature, and volume. A key concept within this theory is the motion of gas molecules. Gas molecules are in constant, random motion, colliding with each other and with the walls of their container. These collisions are responsible for the pressure exerted by the gas.

Because the molecules are moving randomly, their individual speeds vary greatly. Some move very fast, while others move slowly. To describe this distribution of speeds, we often use an average speed. However, a more physically significant average is the root-mean-square (RMS) speed. The RMS speed is defined as the square root of the average of the squares of the speeds of all the molecules in a gas.

The formula for the RMS speed ($v_{rms}$) of gas molecules is derived from the kinetic theory and is given by:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

Where:

• $R$ is the ideal gas constant (8.314 J/mol·K).
• $T$ is the absolute temperature of the gas in Kelvin.
• $M$ is the molar mass of the gas in kg/mol.

Alternatively, the RMS speed can be expressed in terms of the Boltzmann constant ($k_B$) and the number of molecules ($N$) and volume ($V$):

$v_{rms} = \sqrt{\frac{3k_B T}{m}}$

Where:

• $k_B$ is the Boltzmann constant ($1.38 \times 10^{-23}$ J/K).
• $m$ is the mass of a single molecule in kg.

This formula shows that the RMS speed of gas molecules increases with temperature and decreases with increasing molar mass. At a given temperature, lighter gases like hydrogen move faster than heavier gases like oxygen.

Let's consider an example. For oxygen gas ($O_2$) at room temperature (approximately 300 K). The molar mass of oxygen is about 0.032 kg/mol.

$v_{rms} = \sqrt{\frac{3 \times 8.314 \text{ J/mol·K} \times 300 \text{ K}}{0.032 \text{ kg/mol}}}$ $v_{rms} = \sqrt{\frac{7482.6}{0.032}} \text{ m/s}$ $v_{rms} = \sqrt{233831.25} \text{ m/s}$ $v_{rms} \approx 483.56 \text{ m/s}$

This means that oxygen molecules at 300 K are moving with an RMS speed of about 483.56 meters per second.

Degrees of Freedom (DoF)

Degrees of freedom refer to the number of independent ways in which a system, such as a molecule, can store energy. For a molecule, these ways correspond to its possible types of motion. The number of degrees of freedom depends on the structure of the molecule (monatomic, diatomic, polyatomic) and the temperature.

At moderate temperatures, the main types of motion considered are:

• Translational motion: Movement of the molecule's center of mass through space. A molecule can move along the x, y, and z axes. Thus, there are 3 translational degrees of freedom.
• Rotational motion: Rotation of the molecule about its center of mass.
• Vibrational motion: Oscillation of atoms within a molecule about their equilibrium positions.

Let's look at the degrees of freedom for different types of molecules:

Monatomic Gases:

Monatomic molecules, like Helium (He), Neon (Ne), and Argon (Ar), consist of single atoms. These atoms are treated as point masses.

• Translational DoF: 3 (movement along x, y, z axes).
• Rotational DoF: 0 (rotation of a point mass has negligible moment of inertia and energy).
• Vibrational DoF: 0 (no internal bonds to vibrate).

Total DoF for monatomic gas = 3.

Diatomic Gases:

Diatomic molecules, like Hydrogen ($H_2$), Nitrogen ($N_2$), and Oxygen ($O_2$), consist of two atoms bonded together. At moderate temperatures, they exhibit both translational and rotational motion.

• Translational DoF: 3 (movement along x, y, z axes).
• Rotational DoF: 2 (rotation about two axes perpendicular to the bond. Rotation about the bond axis is negligible due to small moment of inertia).
• Vibrational DoF: 0 (at moderate temperatures, vibrational modes are usually 'frozen out' and do not contribute significantly to energy storage).

Total DoF for diatomic gas (moderate temperature) = 3 + 2 = 5.

At very high temperatures, vibrational modes become active.

• Vibrational DoF: 2 (one mode for stretching/compressing the bond). Each vibrational mode contributes 2 degrees of freedom (one kinetic and one potential).

Total DoF for diatomic gas (high temperature) = 3 (trans) + 2 (rot) + 2 (vib) = 7.

Polyatomic Gases:

Polyatomic molecules, like water ($H_2O$) or methane ($CH_4$), consist of three or more atoms. Their structure can be linear or non-linear.

For non-linear polyatomic molecules (e.g., $H_2O$):

• Translational DoF: 3.
• Rotational DoF: 3 (can rotate about three perpendicular axes).
• Vibrational DoF: Contribute at higher temperatures. The number of vibrational modes depends on the number of atoms. For a molecule with $n$ atoms, there are $3n-6$ vibrational modes for non-linear molecules. Each mode contributes 2 DoF.

Total DoF for non-linear polyatomic gas (moderate temperature) = 3 + 3 = 6.

For linear polyatomic molecules (e.g., $CO_2$):

• Translational DoF: 3.
• Rotational DoF: 2 (rotation about the bond axis is negligible).
• Vibrational DoF: Contribute at higher temperatures. For linear molecules, there are $3n-5$ vibrational modes. Each mode contributes 2 DoF.

Total DoF for linear polyatomic gas (moderate temperature) = 3 + 2 = 5.

Law of Equipartition of Energy

The Law of Equipartition of Energy, a fundamental principle of statistical mechanics, states that in thermal equilibrium, the total energy of a system is shared equally among all its accessible degrees of freedom, and each degree of freedom contributes an average energy of $\frac{1}{2}k_B T$.

Here, $k_B$ is the Boltzmann constant and $T$ is the absolute temperature. This law applies to systems where the energy can be distributed among various modes of motion (translational, rotational, vibrational).

Let's break down the energy contribution:

• Translational Energy: A molecule has 3 translational degrees of freedom. The average translational kinetic energy per molecule is $3 \times (\frac{1}{2}k_B T) = \frac{3}{2}k_B T$.
• Rotational Energy: For a diatomic molecule with 2 rotational degrees of freedom, the average rotational kinetic energy per molecule is $2 \times (\frac{1}{2}k_B T) = k_B T$. For a non-linear polyatomic molecule with 3 rotational degrees of freedom, it's $\frac{3}{2}k_B T$.
• Vibrational Energy: Each vibrational mode contributes $\frac{1}{2}k_B T$ for its kinetic energy and $\frac{1}{2}k_B T$ for its potential energy, totaling $k_B T$ per vibrational mode. So, if a molecule has $v$ vibrational modes, the total vibrational energy is $v \times k_B T$.

The total average energy of a molecule is the sum of the average energies from all its active degrees of freedom.

Internal Energy of an Ideal Gas

The internal energy ($U$) of an ideal gas is the sum of the kinetic energies of all its molecules. Since an ideal gas has no intermolecular forces, its potential energy is zero. Therefore, the internal energy is solely due to the random motion of its molecules.

For $N$ molecules of an ideal gas, the total internal energy is $U = N \times (\text{average energy per molecule})$.

Using the equipartition theorem:

• Monatomic Gas: Each molecule has 3 DoF. Average energy per molecule = $\frac{3}{2}k_B T$. Total internal energy for $N$ molecules: $U = N \times \frac{3}{2}k_B T$. Since $N k_B = n N_A k_B = nR$, where $n$ is the number of moles and $N_A$ is Avogadro's number, $U = \frac{3}{2}nRT$.
• Diatomic Gas (Moderate Temperature): Each molecule has 5 DoF (3 trans + 2 rot). Average energy per molecule = $\frac{3}{2}k_B T + 2 \times \frac{1}{2}k_B T = \frac{5}{2}k_B T$. Total internal energy: $U = N \times \frac{5}{2}k_B T = \frac{5}{2}nRT$.
• Diatomic Gas (High Temperature, including vibration): Each molecule has 7 DoF (3 trans + 2 rot + 2 vib). Average energy per molecule = $\frac{3}{2}k_B T + k_B T + 2 \times k_B T = \frac{7}{2}k_B T$. Total internal energy: $U = N \times \frac{7}{2}k_B T = \frac{7}{2}nRT$.
• Polyatomic Gas (Non-linear, Moderate Temperature): Each molecule has 6 DoF (3 trans + 3 rot). Average energy per molecule = $\frac{3}{2}k_B T + \frac{3}{2}k_B T = \frac{6}{2}k_B T = 3 k_B T$. Total internal energy: $U = N \times 3 k_B T = 3nRT$.

In general, if a molecule has $f$ degrees of freedom, its average energy is $\frac{f}{2}k_B T$. The total internal energy of $N$ molecules is $U = N \times \frac{f}{2}k_B T = \frac{f}{2}nRT$.

Specific Heats of Gases

The specific heat capacity of a gas is related to how much heat energy is required to raise its temperature. For gases, we often talk about molar specific heats at constant volume ($C_v$) and constant pressure ($C_p$).

The First Law of Thermodynamics states $\Delta Q = \Delta U + W$.

At constant volume ($V$), $W = P\Delta V = 0$. So, $\Delta Q_V = \Delta U$. The molar specific heat at constant volume is defined as $C_v = \frac{1}{n} \frac{\Delta Q_V}{\Delta T} = \frac{1}{n} \frac{\Delta U}{\Delta T}$. For an ideal gas, $\Delta U = n C_v \Delta T$.

Using the internal energy formulas derived from the equipartition theorem:

• Monatomic Gas: $U = \frac{3}{2}nRT$. So, $\Delta U = \frac{3}{2}nR\Delta T$. $C_v = \frac{1}{n} \frac{\Delta U}{\Delta T} = \frac{1}{n} \frac{\frac{3}{2}nR\Delta T}{\Delta T} = \frac{3}{2}R$.
• Diatomic Gas (Moderate Temp): $U = \frac{5}{2}nRT$. $C_v = \frac{5}{2}R$.
• Diatomic Gas (High Temp): $U = \frac{7}{2}nRT$. $C_v = \frac{7}{2}R$.
• Polyatomic Gas (Non-linear, Moderate Temp): $U = 3nRT$. $C_v = 3R$.

The relationship between $C_p$ and $C_v$ for an ideal gas is given by Mayer's relation: $C_p - C_v = R$.

Therefore, we can find $C_p$ for each type of gas:

• Monatomic Gas: $C_p = C_v + R = \frac{3}{2}R + R = \frac{5}{2}R$.
• Diatomic Gas (Moderate Temp): $C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$.
• Diatomic Gas (High Temp): $C_p = C_v + R = \frac{7}{2}R + R = \frac{9}{2}R$.
• Polyatomic Gas (Non-linear, Moderate Temp): $C_p = C_v + R = 3R + R = 4R$.

The ratio of specific heats, $\gamma = \frac{C_p}{C_v}$, is also an important parameter.

• Monatomic Gas: $\gamma = \frac{5/2 R}{3/2 R} = \frac{5}{3} \approx 1.67$.
• Diatomic Gas (Moderate Temp): $\gamma = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4$.
• Diatomic Gas (High Temp): $\gamma = \frac{9/2 R}{7/2 R} = \frac{9}{7} \approx 1.29$.
• Polyatomic Gas (Non-linear, Moderate Temp): $\gamma = \frac{4R}{3R} = \frac{4}{3} \approx 1.33$.

These values of $\gamma$ are crucial for understanding processes like adiabatic expansion and for calculating the speed of sound in gases.

It is important to remember that the equipartition theorem and the derived specific heats are based on the assumption of classical mechanics. At very low temperatures, quantum effects become significant, and some degrees of freedom (especially rotational and vibrational) may become 'frozen out', meaning they do not contribute to energy storage. This explains why experimental values of specific heats can deviate from theoretical predictions, particularly at low temperatures.

The RMS speed is a measure of the average kinetic energy of the molecules, which is directly related to temperature. Degrees of freedom describe how molecules can store this energy, and the equipartition theorem tells us how this energy is distributed among the available modes. These concepts are fundamental to understanding the thermal behavior of gases.

Example Problem:

Calculate the RMS speed of Helium atoms at 127°C. The molar mass of Helium is approximately 4 g/mol.

First, convert the temperature to Kelvin: $T = 127^\circ C + 273.15 = 400.15$ K. Convert molar mass to kg/mol: $M = 4 \text{ g/mol} = 0.004 \text{ kg/mol}$. Use the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.

$v_{rms} = \sqrt{\frac{3 \times 8.314 \text{ J/mol·K} \times 400.15 \text{ K}}{0.004 \text{ kg/mol}}}$ $v_{rms} = \sqrt{\frac{9980.9}{0.004}} \text{ m/s}$ $v_{rms} = \sqrt{2495225} \text{ m/s}$ $v_{rms} \approx 1579.6 \text{ m/s}$

The RMS speed of Helium atoms at 127°C is approximately 1580 m/s.

Another example: What is the ratio of the specific heat at constant pressure to that at constant volume for nitrogen gas at room temperature?

Nitrogen ($N_2$) is a diatomic molecule. At room temperature, we consider its degrees of freedom to be 5 (3 translational + 2 rotational). Therefore, $C_v = \frac{5}{2}R$. Using Mayer's relation, $C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$. The ratio $\gamma = \frac{C_p}{C_v} = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4$.