Unit 1: Gaseous State I

Table of Contents

Kinetic Theory of Gases (KTG)

Postulates of Kinetic Theory of Gases

This theory explains the macroscopic properties of gases (like pressure, temperature) by considering the motion of their microscopic components (molecules). The key postulates for an ideal gas are:

  1. Molecular Nature: Gases consist of a large number of tiny particles (molecules) that are in continuous, random, and rapid motion.
  2. Negligible Volume: The actual volume of the gas molecules is negligible compared to the total volume of the container. (Molecules are treated as point masses).
  3. No Intermolecular Forces: There are no forces of attraction or repulsion between the gas molecules.
  4. Elastic Collisions: Collisions between molecules, and with the walls of the container, are perfectly elastic. (Total kinetic energy is conserved).
  5. Pressure: The pressure exerted by a gas is due to the continuous bombardment of its molecules on the walls of the container.
  6. Kinetic Energy and Temperature: The average kinetic energy of the gas molecules is directly proportional to the absolute temperature (in Kelvin).

Derivation of the Kinetic Gas Equation

This equation relates the macroscopic pressure (P) to the microscopic properties of the molecules (m, N, c2).

Step-by-step derivation:

  1. Consider a cubic container of side length l containing N molecules, each of mass m.
  2. Consider one molecule moving with velocity c1 and components u1, v1, w1 in the x, y, and z directions.
  3. When the molecule collides with the wall perpendicular to the x-axis, its x-velocity becomes -u1.
    Change in momentum = (Final) - (Initial) = (-mu1) - (mu1) = -2mu1.
    Momentum transferred to the wall = +2mu1.
  4. The time between two successive collisions on the same wall is Δ t = (2l) / (u1).
  5. Force exerted by this one molecule on the wall is (Change in Momentum) / (Time):
    F1 = (2mu1) / (2l/u1) = (mu12) / (l).
  6. Pressure from this one molecule is (Force) / (Area):
    P1 = (F1) / (l2) = (mu12/l) / (l2) = (mu12) / (l3) = (mu12) / (V) (since l3 = V, volume of cube).
  7. The total pressure P from all N molecules in the x-direction is the sum:
    Px = (m) / (V)(u12 + u22 + ... + uN2).
  8. We define the mean square velocity in the x-direction as u2 = (u12 + u22 + ... + uN2) / (N).
    So, Px = (mNu2) / (V).
  9. The total mean square velocity c2 is given by c2 = u2 + v2 + w2.
  10. Because motion is random, the average velocity in all three directions is equal: u2 = v2 = w2.
    Therefore, u2 = (1) / (3)c2.
  11. Substituting this into the pressure equation (and knowing Px = Py = Pz = P):
    P = (mN) / (V) ( (1) / (3)c2 ).
Kinetic Gas Equation: PV = (1) / (3)mNc2
Where: P = Pressure, V = Volume, m = mass of one molecule, N = total number of molecules, c2 = mean square velocity.
(c is often used for root-mean-square velocity, so c2 is c2 or vrms2).

This can also be written in terms of total Kinetic Energy (KE = (1) / (2)mNc2):
PV = (2) / (3) ( (1) / (2)mNc2 ) ⇒ PV = (2) / (3)KEtotal.

Collision Properties and Viscosity

Collision Properties

  • Collision Diameter (σ): The distance of closest approach between the centers of two colliding molecules. It's the effective diameter for collisions.
  • Collision Frequency (Z1): The average number of collisions experienced by a single molecule per unit time (per second).
  • Mean Free Path (λ): The average distance a molecule travels between two successive collisions.
    Formula: λ = (1) / (√(2)πσ2 n*)
    Where n* is the number density (N/V).

Viscosity of Gases

Definition: The internal resistance to flow in a fluid. For gases, it arises from the transfer of momentum between adjacent layers of gas flowing at different speeds.

Relation between Mean Free Path and Coefficient of Viscosity (η)

The coefficient of viscosity (η) is given by the formula:

Formula: η = (1) / (3)ρvλ
Where: ρ = density of the gas (mn*), v = average molecular speed, λ = mean free path.

By substituting the formula for λ, we can get a relation for η in terms of molecular parameters:
η = (1) / (3)(mn*) v ( (1) / (√(2)πσ2 n*) ) = (mv) / (3√(2)πσ2)

Temperature and Pressure Dependence of Viscosity

  • Temperature Dependence:
    • We know that average speed v is proportional to √(T).
    • From the formula η = (mv) / (3√(2)πσ2), we see that η is directly proportional to v.
    • Conclusion: η \propto √(T). The viscosity of a gas increases with an increase in temperature.
    • Reason: Hotter molecules move faster and transfer more momentum between layers, increasing the internal "friction".
  • Pressure Dependence:
    • The formula η = (1) / (3)ρvλ contains density (ρ) and mean free path (λ).
    • As pressure increases (at constant T), density (ρ) increases.
    • As pressure increases, n* increases, so the mean free path (λ) decreases.
    • These two effects (increase in ρ, decrease in λ) cancel each other out.
    • Conclusion: The viscosity of a gas is largely independent of pressure (at moderate pressures).

Maxwell Distribution of Velocities

The molecules in a gas do not all travel at the same speed. The Maxwell-Boltzmann distribution describes the probability distribution of molecular speeds in a gas at a given temperature.

Evaluating Molecular Velocities

From the distribution curve, we can evaluate three important types of molecular speeds:

  1. Most Probable Velocity (cmp or vmp): The speed possessed by the maximum fraction of molecules. This is the peak of the curve.
    Formula: cmp = √((2kT) / (m)) = √((2RT) / (M))
  2. Average Velocity (cavg or v): The arithmetic mean of the speeds of all molecules.
    Formula: cavg = √((8kT) / (π m)) = √((8RT) / (π M))
  3. Root Mean Square (RMS) Velocity (crms or c): The square root of the mean of the squares of the speeds of all molecules. (This is the one from the KTG derivation).
    Formula: crms = √((3kT) / (m)) = √((3RT) / (M))

Key Relationship: crms > cavg > cmp (Ratio ≈ 1.224 : 1.128 : 1)

Average Kinetic Energy

Using the KTG equation (PV = (1) / (3)mN crms2) and the Ideal Gas Law (PV = nRT = (N/NA)RT), we can find the average kinetic energy.

(1) / (3)mN crms2 = (N) / (NA)RT
(1) / (3)m crms2 = (RT) / (NA) = kT (where k = R/NA is the Boltzmann constant)
(1) / (2)m crms2 = (3) / (2)kT

Average Kinetic Energy (per molecule): KE = (3) / (2)kT
Crucial Point: The average kinetic energy of a gas depends only on the absolute temperature, not on the mass or nature of the gas.

Law of Equipartition of Energy

The Law: For a system in thermal equilibrium, the total energy is divided equally among all its degrees of freedom. Each quadratic degree of freedom (like (1) / (2)mvx2 or (1) / (2)Iωy2) has an average energy of (1) / (2)kT per molecule.

Degrees of Freedom (DOF)

The number of independent ways a molecule can store energy.

Degrees of Freedom and Energy Contribution (at moderate temperatures)
Molecule Type Translational DOF Rotational DOF Vibrational DOF Total DOF (KE) Avg. Energy (per molecule)
Monatomic
(e.g., He, Ar)		 3 (vx, vy, vz) 0 0 3 3 × ((1) / (2)kT) = \mathbf{(3) / (2)kT}
Diatomic
(e.g., N2, O2, HCl)		 3 (vx, vy, vz) 2 (rotation ⊥ to bond) 0 (Frozen out) 5 5 × ((1) / (2)kT) = \mathbf{(5) / (2)kT}
Polyatomic (Linear)
(e.g., CO2, C2H2)		 3 (vx, vy, vz) 2 (rotation ⊥ to bond) 0 (Frozen out) 5 5 × ((1) / (2)kT) = \mathbf{(5) / (2)kT}
Polyatomic (Non-linear)
(e.g., H2O, CH4)		 3 (vx, vy, vz) 3 (rotation about x, y, z) 0 (Frozen out) 6 6 × ((1) / (2)kT) = \mathbf{3kT}
Note on Vibrations: Vibrational degrees of freedom exist, but they are "frozen out" (do not contribute) at normal temperatures. They only become active at very high temperatures. Each vibrational mode, when active, contributes kT ((1) / (2)kT for kinetic and (1) / (2)kT for potential energy).