Physics

Chapter 2: The Kinetic Theory of Gases

Imron Rosyadi

Chapter 2: The Kinetic Theory of Gases

Topics Covered

• 2.1 Molecular Model of an Ideal Gas
• 2.2 Pressure, Temperature, and RMS Speed
• 2.3 Heat Capacity and Equipartition of Energy
• 2.4 Distribution of Molecular Speeds

Welcome to Chapter 2: The Kinetic Theory of Gases. This presentation will cover the fundamental concepts of gas behavior from a molecular perspective. We’ll start with the ideal gas model, move on to how pressure and temperature relate to molecular motion, discuss heat capacity, and finally, look at the distribution of molecular speeds.

2.1 Molecular Model of an Ideal Gas

Learning Objectives

By the end of this section, you will be able to:

• Apply the ideal gas law to situations involving the pressure, volume, temperature, and the number of molecules of a gas
• Use the unit of moles in relation to numbers of molecules, and molecular and macroscopic masses
• Explain the ideal gas law in terms of moles rather than numbers of molecules
• Apply the van der Waals gas law to situations where the ideal gas law is inadequate

Our learning objectives for this section include applying the ideal gas law, understanding the concept of moles and Avogadro’s number, restating the ideal gas law in terms of moles, and understanding the van der Waals equation for real gases.

Introduction to Gas Behavior

• Gases, liquids, and solids differ primarily in molecular spacing.
• Gas molecules are much farther apart than in liquids or solids.
• Forces between gas molecules are negligible, except during collisions.
• Gas molecules move rapidly and expand to fill available volume.
• This “chaos” simplifies the connection between microscopic and macroscopic properties for gases.

The term “gas” comes from a Flemish word meaning “chaos,” which is quite fitting. Unlike solids and liquids where molecules are close and interact strongly, gas molecules are widely separated. This means intermolecular forces are generally negligible, except when they collide. They move quickly and randomly, filling any container. This randomness actually makes it simpler to link their microscopic behavior to macroscopic properties like pressure and temperature.

Molecular Model of an Ideal Gas

Key Characteristics

• Molecules are widely separated.
• Forces between molecules are weak at these distances.
• Properties depend more on:
  • Number of atoms per unit volume
  • Temperature
• Less dependent on the specific type of atom.

Figure 2.2: Atoms and molecules in a gas are typically widely separated.

As seen in Figure 2.2, gas molecules are typically far apart. Because of these large distances, the attractive or repulsive forces between them are quite weak. This means that the behavior of a gas is largely determined by the number of molecules in a given volume and the temperature, rather than the specific chemical composition of the gas itself. This is a key insight into why all gases behave similarly under certain conditions.

The Gas Laws: Empirical Observations

• Compressibility: Gases are easily compressed due to large intermolecular spacing.
• Volume Expansion: Gases expand and contract rapidly with temperature changes.
  • Most gases expand at nearly the same rate (same coefficient of volume expansion).

Note

Observation: Why do all gases act similarly, unlike liquids and solids?

• Tire Inflation Analogy:
  • Initial inflation: Volume increases proportionally to air added, pressure slightly rises.
  • Full tire: Volume constrained, pressure increases with more air.
  • Driving: Temperature increases, leading to further pressure increase.

In the previous chapter, we observed that gases are easily compressible and have large volume expansion coefficients. What’s striking is that most gases expand and contract at nearly the same rate with temperature changes, a behavior not seen in liquids or solids.

Consider inflating a car tire. Initially, the volume increases with added air without a large pressure jump. Once the tire is nearly full, its walls restrict volume, so adding more air significantly increases pressure. Furthermore, as you drive, the tires heat up, causing the pressure to increase even more. These everyday observations highlight the relationships between pressure, volume, and temperature in gases.

Boyle’s Law: Pressure-Volume Relationship

• Discovered by Robert Boyle (1627–1691).
• Statement: At constant temperature and number of molecules, the absolute pressure of a gas is inversely proportional to its volume.
• Mathematically: \(P \propto \frac{1}{V}\) or \(PV = \text{constant}\)
• When plotted as \(V\) vs. \(1/P\), data shows a linear relationship.

Figure 2.4: Boyle’s Law experimental data

Robert Boyle’s experiments, as shown in Figure 2.4, demonstrated that for a fixed amount of gas at a constant temperature, its absolute pressure is inversely proportional to its volume. This means if you double the pressure, you halve the volume, and vice versa. The graph plotting volume against the inverse of pressure clearly shows this inverse proportionality as a straight line.

Charles’s Law and Gay-Lussac’s Law

Charles’s Law (Jacques Charles, 1746–1823)

• At constant pressure and number of molecules, the volume of a gas is proportional to its absolute temperature.
• Mathematically: \(V \propto T\) or \(\frac{V}{T} = \text{constant}\)

Figure 2.5: Charles’s Law experimental data

Amonton’s (or Gay-Lussac’s) Law

• At constant volume and number of molecules, the pressure of a gas is proportional to its absolute temperature.
• Mathematically: \(P \propto T\) or \(\frac{P}{T} = \text{constant}\)
• This law is the basis of constant-volume gas thermometers.

Building on Boyle’s work, Charles’s Law states that for a gas at constant pressure and number of molecules, its volume is directly proportional to its absolute temperature. Figure 2.5 illustrates this, showing a linear relationship where volume increases with temperature.

Similarly, Amonton’s or Gay-Lussac’s Law states that at constant volume and number of molecules, the pressure of a gas is directly proportional to its absolute temperature. This principle is fundamental to how constant-volume gas thermometers work.

The Ideal Gas Law

• Combines Boyle’s, Charles’s, and Gay-Lussac’s laws.
• Applies to gases at low density and high temperature.
• Constant of proportionality is independent of the gas composition.
• Equation of State of an Ideal Gas:

\[pV = N k_B T\]

• \(p\): absolute pressure (Pa)
• \(V\): volume (m\(^3\))
• \(N\): number of molecules
• \(k_B\): Boltzmann constant (\(1.38 \times 10^{-23}\) J/K)
• \(T\): absolute temperature (K)

Note

The ideal gas law is an excellent approximation for real gases when they are far from liquefaction.

When we combine these empirical gas laws, we arrive at the Ideal Gas Law. This law holds true for any gas under conditions of low density and high temperature – essentially, when molecules are far apart and moving fast, minimizing intermolecular interactions. What’s remarkable is that the constant of proportionality in this combined law is universal, meaning it’s the same for all gases, whether it’s oxygen, helium, or any other gas.

The Ideal Gas Law is expressed as \(pV = N k_B T\). Here, \(p\) is the absolute pressure, \(V\) is the volume, \(N\) is the number of molecules, \(T\) is the absolute temperature in Kelvin, and \(k_B\) is the Boltzmann constant, a fundamental physical constant with a value of \(1.38 \times 10^{-23}\) Joules per Kelvin. This law describes the behavior of any real gas quite well when it’s not close to becoming a liquid.

Ideal Gas Law for Changing States

• For a fixed amount of gas (constant \(N\)), the quantity \(\frac{pV}{T}\) is constant.
• Useful for comparing two states of the same gas:

\[ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \]

• Subscripts 1 and 2 refer to initial and final states.
• Temperature (\(T\)) must be in Kelvin.
• Pressure (\(p\)) must be absolute pressure (\(p_{gauge} + p_{atmospheric}\)).

A powerful application of the ideal gas law is when we’re dealing with a fixed amount of gas, such as in a sealed container. If the number of molecules, \(N\), is constant, then the ratio \(pV/T\) must also be constant. This allows us to easily relate the initial and final states of a gas using the equation: \(p_1V_1/T_1 = p_2V_2/T_2\).

It’s crucial to remember two things here: first, temperature must always be in Kelvin for these calculations, and second, pressure must be absolute pressure, not gauge pressure. Absolute pressure is the sum of the gauge pressure and the ambient atmospheric pressure.

Example 2.1: Calculating Pressure Changes

Problem: A bicycle tire is inflated to an absolute pressure of \(7.00 \times 10^5\) Pa at \(18.0^\circ\text{C}\). What is the pressure after its temperature rises to \(35.0^\circ\text{C}\) on a hot day? Assume no leaks or volume changes.

Strategy:

1. Identify knowns: \(p_0 = 7.00 \times 10^5\) Pa, \(T_0 = 18.0^\circ\text{C}\), \(T_f = 35.0^\circ\text{C}\).
2. Identify unknown: \(p_f\).
3. Since \(N\) and \(V\) are constant, use \(\frac{p_f}{T_f} = \frac{p_0}{T_0}\).
4. Rearrange to solve for \(p_f\): \(p_f = p_0 \frac{T_f}{T_0}\).
5. Convert temperatures to Kelvin.

Let’s work through an example. Imagine your bike tire. Initially, its absolute pressure is \(7.00 \times 10^5\) pascals at \(18^\circ\text{C}\). On a hot day, the temperature inside the tire rises to \(35^\circ\text{C}\). We want to find the new pressure. We assume no air escapes and the tire’s volume doesn’t change significantly.

Our strategy involves using the ideal gas law for changing states. Since the number of molecules and the volume are constant, the equation simplifies to \(p_f/T_f = p_0/T_0\). We’ll rearrange this to solve for the final pressure, \(p_f = p_0 (T_f/T_0)\). The first critical step is to convert all temperatures from Celsius to Kelvin.

Example 2.1: Solution

1. Convert temperatures to Kelvin: \(T_0 = (18.0 + 273) \text{ K} = 291 \text{ K}\) \(T_f = (35.0 + 273) \text{ K} = 308 \text{ K}\)

2. Substitute values into the equation: \(p_f = p_0 \frac{T_f}{T_0} = (7.00 \times 10^5 \text{ Pa}) \left( \frac{308 \text{ K}}{291 \text{ K}} \right) = 7.41 \times 10^5 \text{ Pa}\)

Important

The final pressure is approximately 6% greater than the original pressure, consistent with the 6% increase in absolute temperature. Remember to use absolute pressure and absolute temperature (Kelvin).

First, convert the temperatures: \(18^\circ\text{C}\) becomes \(291 \text{ K}\) and \(35^\circ\text{C}\) becomes \(308 \text{ K}\). Now, plug these values into our equation: \(p_f = (7.00 \times 10^5 \text{ Pa}) \times (308 \text{ K} / 291 \text{ K})\). This calculates to approximately \(7.41 \times 10^5\) Pascals.

The final pressure is about 6% higher, which makes sense because the absolute temperature also increased by about 6%. This example highlights the importance of using absolute units for both temperature and pressure in ideal gas law calculations.

Example 2.2: Calculating Number of Molecules

Problem: How many molecules are in one breath (500 mL) of air at standard temperature and pressure (STP)? - STP: \(0^\circ\text{C}\) and atmospheric pressure (\(1.01 \times 10^5\) Pa).

Strategy:

1. Identify knowns: \(T = 0^\circ\text{C}\), \(p = 1.01 \times 10^5\) Pa, \(V = 500 \text{ mL}\), \(k_B = 1.38 \times 10^{-23}\) J/K.
2. Identify unknown: \(N\).
3. Use the ideal gas law: \(pV = N k_B T\).
4. Rearrange to solve for \(N\): \(N = \frac{pV}{k_B T}\).
5. Convert temperature to Kelvin and volume to cubic meters.

Here’s another example, this time to grasp the sheer number of molecules involved. Let’s calculate how many molecules are in a single breath, which is about 500 mL of air, at standard temperature and pressure (STP). STP is defined as \(0^\circ\text{C}\) and atmospheric pressure, \(1.01 \times 10^5\) Pascals.

Our strategy is to use the ideal gas law \(pV = N k_B T\). We’ll rearrange it to solve for \(N\), the number of molecules: \(N = pV / (k_B T)\). Before we plug in the numbers, we need to convert the temperature to Kelvin and the volume to cubic meters.

Example 2.2: Solution

1. Identify knowns and convert to SI units:

   \(T = 0^\circ\text{C} = 273 \text{ K}\)

   \(p = 1.01 \times 10^5 \text{ Pa}\)

   \(V = 500 \text{ mL} = 5 \times 10^{-4} \text{ m}^3\)

   \(k_B = 1.38 \times 10^{-23} \text{ J/K}\)

2. Substitute and solve for \(N\):

   \(N = \frac{(1.01 \times 10^5 \text{ Pa})(5 \times 10^{-4} \text{ m}^3)}{(1.38 \times 10^{-23} \text{ J/K})(273 \text{ K})} = 1.34 \times 10^{22} \text{ molecules}\)

Important

Even in small volumes like a single breath, the number of molecules (\(N\)) is enormous!

This result is independent of the type of gas, including mixtures.

First, the conversions: \(0^\circ\text{C}\) is \(273 \text{ K}\), and \(500 \text{ mL}\) is \(5 \times 10^{-4}\) cubic meters.

Now, substituting these into the equation for \(N\): \(N = (1.01 \times 10^5 \text{ Pa} \times 5 \times 10^{-4} \text{ m}^3) / (1.38 \times 10^{-23} \text{ J/K} \times 273 \text{ K})\). This calculation gives us approximately \(1.34 \times 10^{22}\) molecules.

This number is incredibly large, highlighting the microscopic nature of gases. For context, a cubic centimeter of gas at STP still contains \(2.68 \times 10^{19}\) molecules. Remember, this result for \(N\) applies to all types of gases, including air, which is a mixture.

Moles and Avogadro’s Number

• Mole (mol): SI unit for amount of substance.
  • Contains Avogadro’s number (\(N_A\)) of molecules.
• Avogadro’s Number (\(N_A\)): \(N_A = 6.02 \times 10^{23} \text{ mol}^{-1}\)
• Relates number of molecules (\(N\)) to number of moles (\(n\)): \(N = n N_A\)
• Unified atomic mass unit (u) or Dalton: \(12 \text{ u} = \text{mass of one carbon-12 atom}\).
• \(N_A\) also converts u to grams: \(6.02 \times 10^{23} \text{ u} = 1 \text{ g}\)
• Molar mass (\(M\)) relation: \(M = N_A m\) (where \(m\) is mass of one molecule).
• Sample mass (\(m_s\)): \(m_s = nM\)

While thinking in terms of individual molecules is useful, chemists and physicists often use a more convenient unit: the mole. One mole is defined as the amount of any substance that contains Avogadro’s number of molecules, which is \(6.02 \times 10^{23}\). So, the total number of molecules \(N\) is simply the number of moles \(n\) multiplied by Avogadro’s number.

Avogadro’s number also connects the microscopic atomic mass unit (u) to macroscopic grams. For example, \(6.02 \times 10^{23}\) atomic mass units equals one gram. The molar mass \(M\) of a substance (in grams per mole) is numerically equal to the mass of one molecule in atomic mass units, and it’s also Avogadro’s number times the mass of a single molecule (\(m\)). Finally, the total mass of a sample, \(m_s\), is the number of moles times the molar mass.

The Ideal Gas Law Restated using Moles

• Substitute \(N = n N_A\) into \(pV = N k_B T\):

  \(pV = (n N_A) k_B T\)

• Define the universal gas constant \(R = N_A k_B\).

Ideal Gas Law (in terms of moles)

\[ pV = nRT \]

• \(p\): absolute pressure (Pa)
• \(V\): volume (m\(^3\))
• \(n\): number of moles (mol)
• \(R\): universal gas constant
• \(T\): absolute temperature (K)

Values of \(R\): - SI units: \(R = 8.31 \text{ J/(mol}\cdot\text{K)}\) - Other common units: - \(R = 1.99 \text{ cal/(mol}\cdot\text{K)}\) - \(R = 0.0821 \text{ L}\cdot\text{atm/(mol}\cdot\text{K)}\)

We can express the ideal gas law in terms of moles, which is often more convenient in calculations. By substituting \(N = n N_A\) into our original equation, \(pV = N k_B T\), we get \(pV = (n N_A) k_B T\).

We then define a new constant, the universal gas constant, \(R\), as the product of Avogadro’s number and the Boltzmann constant: \(R = N_A k_B\). This gives us the ideal gas law in its most common form: \(pV = nRT\).

The value of \(R\) in SI units is \(8.31 \text{ J/(mol}\cdot\text{K)}\). You’ll also encounter it in other units, like \(1.99 \text{ cal/(mol}\cdot\text{K)}\) or \(0.0821 \text{ L}\cdot\text{atm/(mol}\cdot\text{K)}\). The choice of \(R\) depends on the units given in the problem.

Problem-Solving Strategy for Ideal Gas Law

1. Examine the situation: Confirm an ideal gas model is appropriate (low density, high temperature, far from boiling point).
2. List knowns: Identify all given or inferable quantities.
3. Identify unknowns: Clearly state what needs to be determined.
4. Choose equation: Decide between \(pV = Nk_BT\) (for number of molecules) or \(pV = nRT\) (for moles).
5. Convert units: Use proper SI units (K, Pa, m\(^3\), molecules, moles) or consistent non-SI units if using a corresponding \(R\) value. Use absolute temperature and pressure.
6. Solve: Rearrange the equation for the unknown. Use ratios for changing states if quantities are fixed.
7. Substitute and calculate: Plug in values with units and get the numerical solution.
8. Check reasonableness: Does the answer make physical sense?

To effectively solve problems using the ideal gas law, follow these steps. First, ensure that the ideal gas model is a valid approximation for your scenario. Next, list all known and unknown quantities. Then, choose the appropriate form of the ideal gas law based on whether you’re dealing with the number of molecules (\(N\)) or moles (\(n\)). Unit conversion to SI units (Kelvin, Pascals, cubic meters) is critical, especially using absolute temperature and pressure. Finally, solve for your unknown, perform the calculations carefully, and always check if your answer is reasonable in a physical context.

The Van der Waals Equation of State

• An improved model for real gases.
• Accounts for:
  1. Attractive forces between molecules: Adds \(a(n/V)^2\) to pressure term.
     • Stronger at higher density, reduces pressure.
  2. Volume of molecules: Subtracts \(nb\) from total volume.
     • \(b\) is the volume occupied by one mole of molecules.
• Equation:

\[ \left[p + a\left(\frac{n}{V}\right)^2\right](V - nb) = nRT \]

• Constants \(a\) and \(b\) are experimentally determined for each gas.
• At low density (small \(n\)), \(a\) and \(b\) terms become negligible, reducing to ideal gas law.

The ideal gas law is an approximation. For real gases, especially at higher densities or lower temperatures, we need a more accurate model. This is where the van der Waals equation of state comes in. It improves upon the ideal gas law by accounting for two key factors that the ideal model ignores.

First, it includes a term that accounts for the attractive forces between molecules. These forces reduce the effective pressure a gas exerts, especially at higher densities. This is represented by adding \(a(n/V)^2\) to the pressure term, where ‘a’ is a constant specific to the gas. Second, the van der Waals equation considers that gas molecules themselves occupy a finite volume. This molecular volume is subtracted from the total container volume, meaning the actual free space for molecules to move in is slightly smaller than V. This is represented by subtracting \(nb\) from \(V\), where ‘b’ is another gas-specific constant.

The constants ‘a’ and ‘b’ are experimentally determined for each gas. Importantly, if the gas density is low (meaning \(n\) is small), both the ‘a’ and ‘b’ terms become negligible, and the van der Waals equation simplifies back to the ideal gas law, as it should. This equation provides a much better description of real gas behavior and can even predict gas-liquid phase transitions.

pV Diagrams and Isotherms

• A pV diagram plots pressure versus volume.
• Isotherm: A curve showing \(p\) as a function of \(V\) at a fixed temperature and number of molecules.

Ideal Gas Isotherms

• \(pV = \text{constant}\) (for constant \(T, N\)).
• Graphs are hyperbolas.
• Volume decreases as pressure increases.

Figure 2.7: pV diagram for a Van der Waals gas at various temperatures.

A pV diagram is a powerful tool to visualize gas behavior, plotting pressure against volume. An isotherm is a curve on this diagram representing states where the temperature and number of molecules are held constant. For an ideal gas, where \(pV = \text{constant}\) (at constant T and N), isotherms are perfect hyperbolas, showing that as pressure increases, volume decreases.

However, when we use the van der Waals equation, the isotherms become more complex, as seen in Figure 2.7. At high temperatures, they still look like hyperbolas, indicating approximately ideal behavior. But at lower temperatures, they deviate significantly. Below a certain critical temperature (\(T_c\)), the curves show a “hump,” where increasing volume would seemingly increase pressure. This unphysical behavior is actually interpreted as a liquid-gas phase transition.

pV Diagrams: Phase Transitions

• The “hump” in van der Waals isotherms (below \(T_c\)) represents a liquid-gas phase transition.
• Unphysical oscillating parts are replaced by horizontal lines:
  • Pressure remains constant during boiling/condensation.
  • This region represents liquid-gas coexistence.
• Critical Temperature (\(T_c\)): Above this temperature, liquid cannot exist, regardless of pressure.
• Critical Pressure: Maximum pressure at which liquid can exist.
• Critical Point: Point on pV diagram at \(T_c\) and critical pressure.

Figure 2.8: More realistic pV diagrams showing liquid-gas transition.

In more realistic pV diagrams like Figure 2.8, the oscillating part of the van der Waals isotherm below the critical temperature is replaced by a horizontal line. This flat segment represents the liquid-gas phase transition, where a substance can change phase (boil or condense) at constant temperature and pressure. The regions defined by these flat lines signify where liquid and gas phases can coexist.

Above a critical temperature, \(T_c\), there is no liquid-gas transition; the substance exists as a supercritical fluid, having properties of both gas and liquid. The critical pressure is the maximum pressure at which a liquid can exist, and the critical point is where the critical isotherm meets this pressure. For example, carbon dioxide has no liquid phase above \(31^\circ\text{C}\). This understanding of pV diagrams is crucial for studying phase changes and real gas behavior.

2.2 Pressure, Temperature, and RMS Speed

Learning Objectives

By the end of this section, you will be able to:

• Explain the relations between microscopic and macroscopic quantities in a gas
• Solve problems involving mixtures of gases
• Solve problems involving the distance and time between a gas molecule’s collisions

In this section, we’ll delve into the microscopic origins of macroscopic properties like pressure and temperature. We’ll explore how these relate to the motion of individual gas molecules, including their root-mean-square (RMS) speed. We’ll also cover the behavior of gas mixtures using Dalton’s law and discuss molecular collisions through concepts like mean free path and mean free time.

Kinetic Theory of Gases: Assumptions

To connect microscopic molecular motion to macroscopic gas properties, we make some assumptions for an ideal gas:

1. Large Number of Identical Molecules: Many molecules (\(N\)), all with mass \(m\).
2. Newtonian Motion: Molecules obey Newton’s laws and move continuously, randomly, and isotropically (same in all directions).

For a gas in a rigid container, we add:

3. Negligible Molecular Volume: Molecules are much smaller than average distance between them; their total volume is negligible compared to container volume (\(V\)).
4. Perfectly Elastic Collisions: Collisions with container walls and other molecules are perfectly elastic. Other forces (gravity, intermolecular attractions) are negligible.

The Kinetic Theory of Gases provides a powerful framework to understand gas behavior. To build this model, we start with several key assumptions for an ideal gas. First, we assume there’s a very large number of identical molecules, each with mass ‘m’, moving in a continuous, random, and isotropic manner according to Newton’s laws.

For a gas confined in a rigid container, we add two more assumptions: the total volume of the molecules themselves is negligible compared to the container’s volume. And critically, all collisions—both with the container walls and between molecules—are perfectly elastic, meaning kinetic energy is conserved. We also assume other forces, like gravity and attractions between molecules, are negligible. These simplifications allow us to derive fundamental relationships between microscopic molecular motion and macroscopic gas properties.

Pressure from Molecular Collisions

• Pressure arises from gas molecules colliding with container walls.
• Each collision exerts a force on the wall (Newton’s third law).
• More molecules \(\implies\) more collisions \(\implies\) higher pressure.
• Higher average molecular velocity \(\implies\) stronger collisions \(\implies\) higher pressure.

Figure 2.9: Collision of a gas molecule with a wall

Figure 2.10: Gas in a box exerts an outward pressure

Pressure in a gas is fundamentally a consequence of countless molecular collisions with the container walls. As shown in Figure 2.9, each molecule striking a wall exerts an impulse, and by Newton’s third law, the wall exerts an equal and opposite force on the molecule. When summed over a vast number of molecules, this translates to a steady force, and thus pressure, on the container.

Figure 2.10 further illustrates this: if there are more molecules, there are more collisions, leading to higher pressure. Similarly, if molecules move faster on average, they hit the walls with greater force and more frequently, also increasing the pressure. While individual collisions are random, the sheer number of them results in a stable, observable macroscopic pressure.

Derivation: Pressure, Temperature, and RMS Speed

• Consider a molecule with velocity \(v_x\) colliding elastically with a wall.
  • Momentum change: \(\Delta p = 2mv_x\).
  • Average force on molecule: \(F_i = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{2l/v_x} = \frac{mv_x^2}{l}\).
• Total force on wall from \(N\) molecules: \(F = \sum F_i = \frac{m}{l} \sum v_{ix}^2 = \frac{N m \bar{v_x^2}}{l}\).
• Due to isotropy, \(\bar{v^2} = \bar{v_x^2} + \bar{v_y^2} + \bar{v_z^2} = 3\bar{v_x^2}\). So, \(\bar{v_x^2} = \frac{1}{3}\bar{v^2}\).
• Total force: \(F = \frac{N m \bar{v^2}}{3l}\).
• Pressure \(p = F/A = \frac{N m \bar{v^2}}{3Al} = \frac{N m \bar{v^2}}{3V}\).

Key Result: Pressure and Average Molecular Kinetic Energy

\[ pV = \frac{1}{3} N m \bar{v^2} \]

Let’s trace the derivation that links microscopic molecular motion to macroscopic pressure and temperature. Imagine a single molecule with velocity \(v_x\) colliding elastically with a container wall. Its momentum in the x-direction reverses, so the change in momentum is \(2mv_x\). The average force exerted by this molecule on the wall is its change in momentum divided by the average time between collisions with that wall. For a box of length \(l\), this time is \(2l/v_x\), leading to a force of \(mv_x^2/l\).

Summing this force over all \(N\) molecules gives the total force on the wall. Because molecular motion is isotropic (random in all directions), the average squared velocity in the x-direction (\(\bar{v_x^2}\)) is one-third of the total average squared velocity (\(\bar{v^2}\)). Substituting this into the total force equation, and then dividing by the wall area (\(A\)) to get pressure (\(p=F/A\)), we arrive at the crucial result: \(pV = \frac{1}{3} N m \bar{v^2}\). This equation connects the macroscopic variables of pressure and volume to the microscopic properties of the number of molecules, their mass, and their average squared speed.

Average Kinetic Energy per Molecule

• From the ideal gas law (\(pV = N k_B T\)) and \(pV = \frac{1}{3} N m \bar{v^2}\), we get:

  \(\frac{1}{3} N m \bar{v^2} = N k_B T\)

• Dividing by \(N\) and rearranging for average kinetic energy:

\[ \bar{K} = \frac{1}{2} m \bar{v^2} = \frac{3}{2} k_B T \]

Note

The average kinetic energy of a molecule depends only on its absolute temperature, not on its mass, pressure, or any other property.

• Internal Energy of a Monatomic Ideal Gas:
  • For monatomic gases, internal energy (\(E_{int}\)) is purely translational kinetic energy.
  • \(E_{int} = N \bar{K} = \frac{3}{2} N k_B T = \frac{3}{2} nRT\)

By combining our derived expression for pressure (\(pV = \frac{1}{3} N m \bar{v^2}\)) with the ideal gas law (\(pV = N k_B T\)), we can establish a profound relationship between temperature and molecular kinetic energy. Equating the two \(pV\) expressions, we find \(\frac{1}{3} N m \bar{v^2} = N k_B T\).

A simple rearrangement shows that the average kinetic energy of a molecule, \(\bar{K} = \frac{1}{2} m \bar{v^2}\), is directly proportional to its absolute temperature: \(\bar{K} = \frac{3}{2} k_B T\). This is a fundamental result: the average kinetic energy of a molecule depends only on the absolute temperature, not on its mass, the gas type, or pressure.

For a monatomic ideal gas, where molecules only have translational kinetic energy, the total internal energy (\(E_{int}\)) of the gas is simply the sum of the average kinetic energies of all its molecules. Therefore, \(E_{int} = N \bar{K} = \frac{3}{2} N k_B T\), or in terms of moles, \(E_{int} = \frac{3}{2} nRT\).

RMS Speed of a Molecule

• The root-mean-square (rms) speed is the square root of the average of the square of the speed:

\[ v_{rms} = \sqrt{\bar{v^2}} = \sqrt{\frac{3 k_B T}{m}} \]

-   Where $m$ is the mass of a single molecule.

• In terms of molar mass (\(M\)) and universal gas constant (\(R\)):

\[ v_{rms} = \sqrt{\frac{3 R T}{M}} \]

-   Where $M$ is the molar mass in kg/mol.

Note

The rms speed is a good estimate of typical molecular speeds and is directly related to kinetic energy. It is not the average speed or most likely speed, but often very close.

From the average kinetic energy, we can define a characteristic speed for gas molecules: the root-mean-square, or rms speed. This is calculated as the square root of the average of the squared velocities. The formula for the rms speed is \(v_{rms} = \sqrt{\frac{3 k_B T}{m}}\), where \(m\) is the mass of a single molecule.

We can also express this using the molar mass (\(M\)) and the universal gas constant (\(R\)), which is often more practical: \(v_{rms} = \sqrt{\frac{3 R T}{M}}\). Note that \(M\) must be in kilograms per mole for this equation.

The rms speed provides a useful estimate of the typical speed of molecules in a gas and is directly linked to their kinetic energy. While not precisely the average or most probable speed, it’s a very good approximation and a key metric derived from kinetic theory.

Example 2.4: Kinetic Energy and Speed

Problem:

1. What is the average kinetic energy of a gas molecule at \(20.0^\circ\text{C}\) (room temperature)?
2. Find the rms speed of a nitrogen molecule (\(\text{N}_2\)) at this temperature.

Strategy:

1. Use \(\bar{K} = \frac{3}{2} k_B T\). Convert \(20.0^\circ\text{C}\) to Kelvin.
2. Use \(v_{rms} = \sqrt{\frac{3 k_B T}{m}}\).
   • Need to find the mass (\(m\)) of one \(\text{N}_2\) molecule: \(m = \frac{M}{N_A}\).
   • Molar mass of \(\text{N}_2 = 2 \times 14.0067 \text{ g/mol} = 28.0134 \text{ g/mol}\).

Let’s apply these concepts to a practical problem. First, we’ll calculate the average kinetic energy of any gas molecule at room temperature, \(20^\circ\text{C}\). Then, we’ll find the rms speed specifically for a nitrogen molecule (\(\text{N}_2\)) at that same temperature.

For part (a), the average kinetic energy formula \(\bar{K} = \frac{3}{2} k_B T\) is all we need, after converting the temperature to Kelvin. For part (b), we’ll use the rms speed formula \(v_{rms} = \sqrt{\frac{3 k_B T}{m}}\). However, we’ll need the mass of a single nitrogen molecule, \(m\). We can find this by dividing nitrogen’s molar mass (which we get from the periodic table) by Avogadro’s number.

Example 2.4: Solution

1. Convert temperature: \(T = (20.0 + 273) \text{ K} = 293 \text{ K}\)

2. Calculate mass of \(\text{N}_2\) molecule:

   \(M_{\text{N}_2} = 28.0134 \times 10^{-3} \text{ kg/mol}\)

   \(m = \frac{M_{\text{N}_2}}{N_A} = \frac{28.0134 \times 10^{-3} \text{ kg/mol}}{6.02 \times 10^{23} \text{ mol}^{-1}} = 4.65 \times 10^{-26} \text{ kg}\)

1. Average Kinetic Energy:

\(\bar{K} = \frac{3}{2} k_B T = \frac{3}{2} (1.38 \times 10^{-23} \text{ J/K})(293 \text{ K}) = 6.07 \times 10^{-21} \text{ J}\)

2. RMS Speed of Nitrogen Molecule:

\(v_{rms} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3(1.38 \times 10^{-23} \text{ J/K})(293 \text{ K})}{4.65 \times 10^{-26} \text{ kg}}} = 511 \text{ m/s}\)

Important

• Average kinetic energy depends only on absolute temperature.
• The rms speed of nitrogen at room temperature is surprisingly high (511 m/s), close to the speed of sound!

First, convert \(20^\circ\text{C}\) to \(293 \text{ K}\). Then, calculate the mass of one nitrogen molecule: its molar mass of \(28.0134 \text{ g/mol}\) becomes \(28.0134 \times 10^{-3} \text{ kg/mol}\), and dividing by Avogadro’s number gives \(4.65 \times 10^{-26} \text{ kg}\) for one molecule.

Now for the calculations:

1. The average kinetic energy is \(\frac{3}{2} \times (1.38 \times 10^{-23} \text{ J/K}) \times (293 \text{ K})\), which equals \(6.07 \times 10^{-21}\) Joules. This energy is tiny, which is why we don’t feel air molecules hitting our skin.
2. The rms speed is \(\sqrt{3 \times (1.38 \times 10^{-23} \text{ J/K}) \times (293 \text{ K}) / (4.65 \times 10^{-26} \text{ kg})}\), resulting in \(511 \text{ m/s}\). This is quite fast—faster than the speed of sound! However, molecules don’t travel far due to frequent collisions (short mean free path). This high speed does explain why sound travels quickly through air.

Vapor Pressure, Partial Pressure, and Dalton’s Law

• Partial Pressure: The pressure a gas would create if it alone occupied the total volume.
• Dalton’s Law of Partial Pressures: In a mixture of ideal gases, the total pressure (\(p_{total}\)) is the sum of the partial pressures (\(p_i\)) of the component gases. \(p_{total} = p_1 + p_2 + p_3 + \dots\)
• Each gas in a mixture obeys the ideal gas law separately (\(p_i V = N_i k_B T\)).
• For any two gases in equilibrium: \(\frac{p_1}{n_1} = \frac{p_2}{n_2}\) (at constant \(T, V\)).
• Vapor Pressure: Partial pressure of a vapor in equilibrium with its liquid (or solid) phase.
  • When partial pressure reaches vapor pressure, condensation occurs (e.g., dew).
  • Dew Point: Temperature at which condensation occurs.
• Relative Humidity (R.H.): \(\text{R.H.} = \frac{\text{Partial pressure of water vapor at } T}{\text{Vapor pressure of water at } T} \times 100\%\)

When dealing with mixtures of gases, we introduce the concept of partial pressure. The partial pressure of a gas is the pressure it would exert if it were the only gas present in the container. Dalton’s Law of Partial Pressures states that the total pressure of a gas mixture is simply the sum of the partial pressures of all its component gases, assuming ideal gas behavior and no chemical reactions. This is consistent with Pascal’s principle, where pressures add up.

A significant application is vapor pressure, which is the partial pressure of a substance’s vapor when it’s in equilibrium with its liquid (or solid) phase. The air’s partial pressure of water vapor cannot exceed the vapor pressure of water at that temperature; once it reaches this point, condensation occurs, like dew formation. The temperature at which this happens is called the dew point.

Relative humidity quantifies how saturated the air is with water vapor. It’s the ratio of the actual partial pressure of water vapor to the maximum possible vapor pressure at a given temperature, expressed as a percentage. This concept is vital for meteorology and even human comfort.

Mean Free Path and Mean Free Time

• Mean Free Path (\(\lambda\)): Average distance a molecule travels between collisions with other molecules.

  • Assuming molecules are spheres with radius \(r\):

  \[ \lambda = \frac{V}{4\pi r^2 N \sqrt{2}} = \frac{k_B T}{4\pi r^2 p \sqrt{2}} \]

• Mean Free Time (\(\tau\)): Average time between collisions of a molecule.

  • Calculated by dividing mean free path by rms speed (\(v_{rms}\)):

  \[ \tau = \frac{\lambda}{v_{rms}} = \frac{k_B T}{4\pi r^2 p v_{rms} \sqrt{2}} \]

Now, let’s explicitly consider molecular collisions. The mean free path, denoted by \(\lambda\), is the average distance a molecule travels before colliding with another molecule. If we model molecules as spheres of radius ‘r’, the formula for mean free path is derived by considering the effective collision cross-section. The final simplified expression, after accounting for the motion of all molecules, is \(\lambda = \frac{k_B T}{4\pi r^2 p \sqrt{2}}\).

Closely related is the mean free time, \(\tau\), which is the average time between these collisions. It’s simply the mean free path divided by a typical molecular speed, usually the rms speed. So, \(\tau = \frac{\lambda}{v_{rms}}\). These concepts help us understand why gas molecules, despite their high speeds, don’t travel very far before encountering another molecule.

2.3 Heat Capacity and Equipartition of Energy

Learning Objectives

By the end of this section, you will be able to:

• Solve problems involving heat transfer to and from ideal monatomic gases whose volumes are held constant
• Solve similar problems for non-monatomic ideal gases based on the number of degrees of freedom of a molecule
• Estimate the heat capacities of metals using a model based on degrees of freedom

In this section, we’ll explore heat capacity, particularly for ideal gases at constant volume. We’ll learn how to calculate it for monatomic gases and then extend this understanding to non-monatomic gases by introducing the concept of “degrees of freedom.” Finally, we’ll see how this framework can even be used to estimate the heat capacities of solid metals.

Molar Heat Capacity at Constant Volume (\(C_V\))

• For gases, we define heat capacity in terms of moles.

• Molar Heat Capacity at Constant Volume (\(C_V\)):

  \[ C_V = \frac{1}{n} \frac{Q}{\Delta T} \quad (\text{with } V \text{ held constant}) \]

• At constant volume, no work is done (\(\Delta V = 0\)).

  • Change in internal energy equals heat flow: \(\Delta E_{int} = Q\).

• For a monatomic ideal gas, \(E_{int} = \frac{3}{2} nRT\).

  • Therefore, \(\Delta E_{int} = \frac{3}{2} nR\Delta T\).

• Molar Heat Capacity for Monatomic Ideal Gas:

  \[ C_V = \frac{3}{2} R \]

Note

This value is independent of temperature and agrees well with experimental results for monatomic gases.

When discussing heat capacity for gases, it’s more convenient to use moles rather than mass. We define the molar heat capacity at constant volume, \(C_V\), as the heat \(Q\) required to change the temperature of one mole of gas by \(\Delta T\), while keeping the volume constant.

When the volume is constant, the gas does no work, so any heat added directly changes its internal energy (\(Q = \Delta E_{int}\)). For a monatomic ideal gas, we know \(E_{int} = \frac{3}{2} nRT\). Taking the change in internal energy, we get \(\Delta E_{int} = \frac{3}{2} nR\Delta T\). Substituting this back into the definition of \(C_V\), we find that for a monatomic ideal gas, \(C_V = \frac{3}{2} R\). This value is independent of temperature and is well-supported by experimental observations for gases like helium or argon.

Degrees of Freedom and Equipartition Theorem

• Degree of Freedom: An independent way a molecule can store energy (e.g., translational motion in x, y, z directions).
• For a monatomic ideal gas, there are 3 translational degrees of freedom.
  • Each contributes \(\frac{1}{2} k_B T\) to the average energy per molecule.
  • Total average kinetic energy: \(\bar{K} = 3 \times \frac{1}{2} k_B T = \frac{3}{2} k_B T\).

Equipartition Theorem

• In a thermodynamic system in equilibrium, energy is partitioned equally among its degrees of freedom.
• Each degree of freedom contributes \(\frac{1}{2} k_B T\) to the average energy per molecule.
• Molar Heat Capacity at Constant Volume:

\[ C_V = \frac{d}{2} R \]

-   Where $d$ is the number of degrees of freedom.

To generalize heat capacity beyond monatomic gases, we introduce the concept of degrees of freedom. A degree of freedom represents an independent way a molecule can store energy, such as translation along an axis. For a monatomic gas, there are three translational degrees of freedom (x, y, and z directions), each contributing \(\frac{1}{2} k_B T\) to the molecule’s average energy.

The Equipartition Theorem, a cornerstone of classical statistical mechanics, states that in thermal equilibrium, the total energy of a system is equally distributed among all active degrees of freedom. Each active degree of freedom contributes \(\frac{1}{2} k_B T\) to the average energy per molecule. This powerful theorem implies that the molar heat capacity at constant volume, \(C_V\), for any ideal gas is simply \(\frac{d}{2} R\), where \(d\) is the total number of active degrees of freedom.

Degrees of Freedom for Diatomic and Polyatomic Gases

• Diatomic Gas (e.g., \(\text{N}_2, \text{O}_2\)):
  • 3 translational degrees of freedom.
  • 2 rotational degrees of freedom (perpendicular to molecular axis).
  • 2 vibrational degrees of freedom (kinetic + potential energy from vibration).
  • Classically: \(d=7\).
• Quantum Mechanical Effects:
  • At low temperatures (\(\lesssim 60 \text{ K}\)), only translational degrees are active (\(d=3\)).
  • At room temperature (\(\approx 300 \text{ K}\) to \(600 \text{ K}\)), translational + rotational are active (\(d=5\)).
  • At high temperatures (\(\gtrsim 3000 \text{ K}\)), vibrational degrees become active (\(d=7\)).

Figure 2.13: Molar heat capacity of hydrogen vs. temperature.

• Polyatomic Gas (e.g., \(\text{CO}_2, \text{H}_2\text{S}\)):
  • Typically 3 translational + 3 rotational degrees of freedom (\(d=6\)) at room temperature.
  • Additional vibrational modes at higher temperatures.

The number of active degrees of freedom changes with temperature, especially for molecules with more than one atom. For a diatomic gas, like \(\text{N}_2\) or \(\text{O}_2\), besides the three translational degrees, it also has two rotational degrees of freedom (rotations perpendicular to its axis) and two vibrational degrees of freedom (one for kinetic and one for potential energy of vibration). Classically, this would mean \(d=7\).

However, quantum mechanics dictates that these degrees of freedom only become “active” and contribute to heat capacity when the thermal energy (\(k_B T\)) is sufficient to excite them. As shown in Figure 2.13 for hydrogen: At very low temperatures (below about \(60 \text{ K}\)), only translational degrees are active, so \(d=3\). From room temperature (around \(300 \text{ K}\)) up to about \(600 \text{ K}\), both translational and rotational degrees are active, making \(d=5\). Only at very high temperatures (above \(3000 \text{ K}\)) do the vibrational degrees of freedom become active, reaching the classical prediction of \(d=7\).

Polyatomic molecules, due to their more complex structures, typically have three translational and three rotational degrees of freedom at room temperature (\(d=6\)), with additional vibrational modes activated at higher temperatures. It’s important to consider these temperature-dependent effects when calculating heat capacity for polyatomic gases.

Molar Heat Capacity of Solid Elements

• Model: Atoms in a solid are attached to neighbors by springs.
• Each atom has 6 degrees of freedom:
  • 3 for kinetic energy (one for each x, y, z direction).
  • 3 for potential energy (one for each x, y, z direction, like a spring).
• Therefore, \(d=6\) for atoms in a solid.
• Law of Dulong and Petit: Predicts molar specific heat of a metal should be \(C_V = 3R\).
• This works reasonably well at room temperature for many elements.

Caution

This law fails at low temperatures due to quantum-mechanical reasons, and for some light or heavy elements even at room temperature.

We can extend the concept of degrees of freedom to solids. Imagine atoms in a solid lattice connected by “springs” to their neighbors. Each atom can vibrate in three dimensions. For each dimension of vibration, it has both kinetic and potential energy associated with that motion. This gives each atom six degrees of freedom: three for kinetic energy along x, y, z, and three for potential energy along x, y, z.

According to the equipartition theorem, if each degree of freedom contributes \(\frac{1}{2} k_B T\) of energy, then each atom in a solid has an average energy of \(6 \times \frac{1}{2} k_B T = 3 k_B T\). This leads to a predicted molar heat capacity at constant volume of \(C_V = 3R\) for solid elements. This is known as the Law of Dulong and Petit, and it holds fairly well for many metals at room temperature. However, it fails at very low temperatures due to quantum mechanical effects, and for some light or unusually heavy elements, even at room temperature.

2.4 Distribution of Molecular Speeds

Learning Objectives

By the end of this section, you will be able to:

• Describe the distribution of molecular speeds in an ideal gas
• Find the average and most probable molecular speeds in an ideal gas

In the final section of this chapter, we’ll dive into the concept that not all molecules in a gas move at the same speed. We’ll learn about the Maxwell-Boltzmann distribution, which describes the range of molecular speeds, and how to identify the most probable and average speeds within this distribution.

The Maxwell-Boltzmann Distribution

• Molecules in an ideal gas move at a range of speeds, not just one average speed.
• The Maxwell-Boltzmann distribution describes this predictable distribution of molecular speeds.

Figure 2.15: Maxwell-Boltzmann distribution of molecular speeds.

Distribution Function \(f(v)\): - The number of particles with speeds between \(v_1\) and \(v_2\) is \(N(v_1, v_2) = N \int_{v_1}^{v_2} f(v)dv\). - \(f(v)dv\) is the probability that a molecule’s speed is between \(v\) and \(v+dv\).

Maxwell-Boltzmann Distribution of Speeds

\[ f(v) = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{(-mv^2/(2k_B T))} \]

• \(v^2\) term: Causes \(f(0)=0\) and an initial parabolic rise.
• \(e^{(-mv^2/(2k_B T))}\) term: Causes an exponential decay at high speeds.
• Interaction of terms creates the single-peaked shape.

While we’ve discussed average speeds, individual gas molecules actually travel at a wide range of speeds. The Maxwell-Boltzmann distribution, illustrated in Figure 2.15, mathematically describes this distribution.

We define a distribution function \(f(v)\) such that the expected number of particles with speeds between \(v_1\) and \(v_2\) is found by integrating \(f(v)\) over that range, multiplied by the total number of molecules \(N\). Essentially, \(f(v)dv\) is the probability that a molecule has a speed between \(v\) and \(v+dv\).

The formula for \(f(v)\) is complex, but its key features are due to two main terms: the \(v^2\) term causes the distribution to start at zero and rise quadratically, while the exponential term \(e^{(-mv^2/(2k_B T))}\) causes it to decay rapidly at very high speeds. The combination of these terms gives the characteristic single-peaked shape, indicating that most molecules are clustered around a particular speed, but some move much slower and some much faster.

Characteristic Speeds from the Distribution

The Maxwell-Boltzmann distribution allows us to calculate specific characteristic speeds:

1. Average speed (\(\bar{v}\)):

   \[ \bar{v} = \int_0^\infty v f(v) dv = \sqrt{\frac{8 k_B T}{\pi m}} = \sqrt{\frac{8 R T}{\pi M}} \]

2. RMS speed (\(v_{rms}\)): (Already derived from kinetic energy)

   \[ v_{rms} = \sqrt{\bar{v^2}} = \sqrt{\int_0^\infty v^2 f(v) dv} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}} \]

3. Most probable speed (\(v_p\)): Speed at the peak of the distribution (\(f'(v)=0\)).

   \[ v_p = \sqrt{\frac{2 k_B T}{m}} = \sqrt{\frac{2 R T}{M}} \]

Note

Generally, \(v_p < \bar{v} < v_{rms}\).

The distribution shifts to higher speeds and broadens at higher temperatures.

From the Maxwell-Boltzmann distribution function, we can calculate various characteristic speeds.

First, the average speed (\(\bar{v}\)) is found by integrating \(v \times f(v)\) over all possible speeds. This yields \(\sqrt{\frac{8 k_B T}{\pi m}}\).

Second, the rms speed (\(v_{rms}\)), which we encountered earlier, is \(\sqrt{\frac{3 k_B T}{m}}\). This is derived from the average kinetic energy and is also calculated by integrating \(v^2 \times f(v)\).

Third, the most probable speed (\(v_p\)) is the speed at which the distribution function peaks. This is found by taking the derivative of \(f(v)\) with respect to \(v\) and setting it to zero. The result is \(\sqrt{\frac{2 k_B T}{m}}\).

It’s important to note the hierarchy: the most probable speed is always less than the average speed, which in turn is less than the rms speed (\(v_p < \bar{v} < v_{rms}\)). As temperature increases, the entire distribution shifts towards higher speeds and becomes broader, reflecting the increased kinetic energy of the molecules.

Key Takeaways

• Ideal Gas Law: \(pV = Nk_BT\) or \(pV = nRT\). Describes behavior of gases at low density and high temperature.
• Boltzmann Constant (\(k_B\)): \(1.38 \times 10^{-23}\) J/K.
• Universal Gas Constant (\(R\)): \(8.31\) J/(mol\(\cdot\)K). \(R = N_A k_B\).
• Average Kinetic Energy: \(\bar{K} = \frac{3}{2} k_B T\). Depends only on absolute temperature.
• RMS Speed: \(v_{rms} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}}\).
• Degrees of Freedom (\(d\)): Independent ways a molecule can store energy.
  • Monatomic: \(d=3\).
  • Diatomic: \(d=5\) (at room temperature).
  • Polyatomic: \(d=6\) (at room temperature).
• Equipartition Theorem: Each degree of freedom contributes \(\frac{1}{2} k_B T\) to average energy.
• Molar Heat Capacity at Constant Volume: \(C_V = \frac{d}{2} R\).
• Dalton’s Law: \(p_{total} = \sum p_i\).
• Maxwell-Boltzmann Distribution: Describes the range of molecular speeds.
  • Most probable speed (\(v_p\)), average speed (\(\bar{v}\)), and rms speed (\(v_{rms}\)) are distinct: \(v_p < \bar{v} < v_{rms}\).

Let’s summarize the key points from this chapter.

The Ideal Gas Law, \(pV = Nk_BT\) or \(pV = nRT\), is fundamental for describing gas behavior under specific conditions. Remember the Boltzmann constant (\(k_B\)) and the universal gas constant (\(R\)), which links \(k_B\) to Avogadro’s number (\(N_A\)).

A crucial result is that the average kinetic energy of a molecule, \(\bar{K} = \frac{3}{2} k_B T\), depends only on the absolute temperature. This directly leads to the formula for the RMS speed, \(v_{rms} = \sqrt{\frac{3 k_B T}{m}}\).

The concept of degrees of freedom (\(d\)) helps explain heat capacity for different types of gases. We saw that monatomic gases have \(d=3\), diatomic gases typically have \(d=5\) at room temperature, and polyatomic gases often have \(d=6\). The Equipartition Theorem states that each degree of freedom contributes \(\frac{1}{2} k_B T\) to the average energy, leading to the molar heat capacity at constant volume formula, \(C_V = \frac{d}{2} R\).

We also discussed Dalton’s Law of Partial Pressures, which states that the total pressure in a gas mixture is the sum of the individual partial pressures.

Finally, the Maxwell-Boltzmann Distribution provides a detailed picture of molecular speeds, showing that molecules move at a range of velocities, with distinct values for the most probable speed (\(v_p\)), average speed (\(\bar{v}\)), and rms speed (\(v_{rms}\)), where \(v_p < \bar{v} < v_{rms}\).

Key Equations

Equation	Description
\(pV = Nk_BT\)	Ideal Gas Law (number of molecules)
\(pV = nRT\)	Ideal Gas Law (number of moles)
\(\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}\)	Ideal Gas Law (changing states)
\([p + a(n/V)^2](V - nb) = nRT\)	Van der Waals Equation of State
\(\bar{K} = \frac{3}{2} k_B T\)	Average Kinetic Energy per Molecule
\(E_{int} = \frac{3}{2} nRT\)	Internal Energy of a Monatomic Ideal Gas
\(v_{rms} = \sqrt{\frac{3 k_B T}{m}}\)	RMS Speed of a Molecule
\(p_{total} = \sum p_i\)	Dalton’s Law of Partial Pressures
\(\lambda = \frac{k_B T}{4\pi r^2 p \sqrt{2}}\)	Mean Free Path
\(\tau = \frac{\lambda}{v_{rms}}\)	Mean Free Time
\(C_V = \frac{1}{n} \frac{Q}{\Delta T}\)	Molar Heat Capacity at Constant Volume
\(C_V = \frac{d}{2} R\)	Molar Heat Capacity (Equipartition Theorem)
\(f(v) = 4\pi (\frac{m}{2\pi k_B T})^{3/2} v^2 e^{(-mv^2/(2k_B T))}\)	Maxwell-Boltzmann Distribution of Speeds
\(v_p = \sqrt{\frac{2 k_B T}{m}}\)	Most Probable Speed
\(\bar{v} = \sqrt{\frac{8 k_B T}{\pi m}}\)	Average Speed

This slide provides a concise table of the key equations discussed in Chapter 2. It serves as a quick reference for the mathematical relationships governing the kinetic theory of gases. Each equation is paired with a brief description to remind you of its context and application. Be sure to familiarize yourself with these formulas for problem-solving.

Key Terms

Term	Definition
Ideal Gas Law	Equation relating pressure, volume, temperature, and number of molecules (or moles) for an ideal gas.
Boltzmann Constant (\(k_B\))	Fundamental constant relating energy to temperature, \(k_B = 1.38 \times 10^{-23}\) J/K.
Mole (mol)	SI unit for the amount of substance, containing Avogadro’s number of molecules.
Avogadro’s Number (\(N_A\))	The number of molecules in one mole of a substance, \(N_A = 6.02 \times 10^{23} \text{ mol}^{-1}\).
Universal Gas Constant (\(R\))	Constant relating pressure, volume, temperature, and moles in the ideal gas law, \(R = N_A k_B = 8.31 \text{ J/(mol}\cdot\text{K)}\).
Van der Waals Equation	An improved equation of state for real gases, accounting for intermolecular forces and molecular volume.
Critical Temperature (\(T_c\))	The temperature above which a substance cannot exist as a liquid, regardless of pressure.
Isotherm	A curve on a pV diagram representing states at a fixed temperature.
Kinetic Theory of Gases	Theory relating macroscopic properties of gases to the motion of their molecules.
Average Kinetic Energy	The average translational kinetic energy of a molecule, \(\frac{3}{2} k_B T\).
RMS Speed (\(v_{rms}\))	The root-mean-square speed of molecules, \(\sqrt{\overline{v^2}}\).
Partial Pressure	The pressure a gas would exert if it alone occupied the total volume.
Dalton’s Law of Partial Pressures	The total pressure of a gas mixture is the sum of the partial pressures of its components.
Vapor Pressure	The partial pressure of a vapor in equilibrium with its liquid (or solid) phase.
Dew Point	The temperature at which water vapor in the air condenses.
Relative Humidity (R.H.)	Ratio of actual water vapor partial pressure to saturation vapor pressure, expressed as a percentage.
Mean Free Path (\(\lambda\))	The average distance a molecule travels between collisions.
Mean Free Time (\(\tau\))	The average time between molecular collisions.
Molar Heat Capacity at Constant Volume (\(C_V\))	Heat required per mole to raise temperature by \(1 \text{ K}\) at constant volume.
Degree of Freedom (\(d\))	An independent possible motion or way a molecule can store energy.
Equipartition Theorem	States that energy is equally partitioned among degrees of freedom, each contributing \(\frac{1}{2} k_B T\).
Maxwell-Boltzmann Distribution	Describes the distribution of molecular speeds in an ideal gas at a given temperature.
Most Probable Speed (\(v_p\))	The speed at which the Maxwell-Boltzmann distribution function peaks.

This slide lists the essential terms and their definitions encountered in Chapter 2. Understanding these terms is crucial for grasping the concepts of the kinetic theory of gases. Review this table to solidify your vocabulary and conceptual understanding.