Gaseous State: A Detailed M.Sc. Level Tutorial

Hello! Welcome to this detailed breakdown of the Gaseous State. We're going to move from the microscopic world of individual molecules up to the macroscopic properties you observe, like pressure and viscosity. Let's start with the foundation of it all: how molecules are moving.

1. The Maxwell-Boltzmann Distribution of Velocities

Instead of tracking every single molecule (which is impossible!), we use statistics to describe the distribution of velocities. James Clerk Maxwell and Ludwig Boltzmann developed a model based on these key assumptions (postulates) of the Kinetic Theory of Gases:

• Gases are composed of a large number of identical molecules in continuous, random motion.
• The volume of the molecules themselves is negligible compared to the total volume of the container.
• Intermolecular forces (attraction/repulsion) are negligible, except during brief collisions.
• Collisions (molecule-molecule and molecule-wall) are perfectly elastic.
• The average kinetic energy of the molecules is directly proportional to the absolute temperature ($T$).

Derivation of the 3D Velocity Distribution ($f(v)$)

The full derivation is quite involved, but we can build it from a key insight: a molecule's motion in one dimension (say, $x$) is independent of its motion in the other dimensions ($y, z$). This allows us to find the probability distribution for a single velocity component, $f(v_x)$, and then combine them.

The probability that a molecule has an $x$-component of velocity between $v_x$ and $v_x + dv_x$ is given by a Gaussian density function (a "bell curve"):

$$f(v_x) dv_x = \left(\frac{m}{2\pi k_B T}\right)^{1/2} e^{-mv_x^2 / 2k_B T} dv_x$$

Here, $m$ is the mass of one molecule, $k_B$ is the Boltzmann constant, and $T$ is the absolute temperature. The term $e^{-mv_x^2 / 2k_B T}$ is a form of the famous Boltzmann factor, $e^{-E/k_B T}$, where $E$ is the kinetic energy in the $x$-direction ($E_x = \frac{1}{2}mv_x^2$).

Since the $x, y,$ and $z$ motions are independent, the probability of having a specific velocity vector $(v_x, v_y, v_z)$ is the product of the individual probabilities:

$$f(v_x, v_y, v_z) dv_x dv_y dv_z = f(v_x)f(v_y)f(v_z) dv_x dv_y dv_z = \left(\frac{m}{2\pi k_B T}\right)^{3/2} e^{-m(v_x^2 + v_y^2 + v_z^2) / 2k_B T} dv_x dv_y dv_z$$

This is great, but we usually don't care about the direction (the vector), just the speed, $v$. The speed is the magnitude of the velocity vector: $v^2 = v_x^2 + v_y^2 + v_z^2$. To find the probability $f(v)dv$ that a molecule has a speed between $v$ and $v+dv$, we switch from Cartesian coordinates ($v_x, v_y, v_z$) to spherical polar coordinates. We are asking for the probability of the velocity vector ending up in a thin spherical shell of radius $v$ and thickness $dv$. The "volume" of this shell in "velocity space" is $4\pi v^2 dv$.

Substituting $v^2$ for $(v_x^2 + v_y^2 + v_z^2)$ and $4\pi v^2 dv$ for $dv_x dv_y dv_z$, we get the Maxwell's Law of Distribution of Speeds:

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

The term $f(v)$ is the probability distribution function. $f(v)dv$ gives the fraction of molecules with speeds between $v$ and $v+dv$.

Features of the Distribution Curve

If you plot $f(v)$ (y-axis) against $v$ (x-axis), you get a characteristic curve:

• It starts at zero (no molecules have zero speed).
• It rises to a peak (the most probable speed).
• It falls off and approaches zero at high speeds, but never truly reaches it (there's always a tiny probability of a very fast molecule).
• The total area under the curve is 1 (the probabilities of all possible speeds must sum to 1).

The shape of this curve changes with Temperature ($T$) and Molar Mass ($M$):

• Effect of Temperature: As $T$ increases, the molecules have more kinetic energy. The curve flattens and broadens, and the peak shifts to a higher speed. The gas has a wider range of speeds.
• Effect of Molar Mass: At a given $T$, heavier molecules (larger $M$ or $m$) move slower. Their distribution is narrower and the peak is shifted to a lower speed compared to lighter molecules.

Tutor Tip: Think of it like this: The $v^2$ term in $f(v)$ tries to push the curve up (more "space" at higher $v$), while the exponential term $e^{-mv^2 / 2k_B T}$ (the Boltzmann factor) tries to slam it down (it's very improbable to have energy much greater than $k_B T$). The peak is where these two competing effects balance out!

2. Characteristic Molecular Velocities

The distribution curve gives us three important "landmark" speeds that characterize the gas.

a) Most Probable Velocity ($v_{mp}$)

This is the speed possessed by the largest fraction of molecules. It's the peak of the $f(v)$ curve. To find it, we find the maximum by taking the derivative of $f(v)$ with respect to $v$ and setting it to zero: $\frac{df(v)}{dv} = 0$.

Solving this gives:

$$v_{mp} = \sqrt{\frac{2k_B T}{m}} = \sqrt{\frac{2RT}{M}}$$

(Remember $R = N_A k_B$ and $M = N_A m$, where $N_A$ is Avogadro's number).

b) Average Velocity ($\bar{v}$ or $v_{avg}$)

This is the simple arithmetic mean of the speeds of all molecules. To find it, we integrate $v$ times its probability $f(v)$ over all possible speeds:

$$\bar{v} = \int_{0}^{\infty} v f(v) dv$$

This requires solving a standard integral. The result is:

$$\bar{v} = \sqrt{\frac{8k_B T}{\pi m}} = \sqrt{\frac{8RT}{\pi M}}$$

c) Root-Mean-Square Velocity ($v_{rms}$)

This one is the most important for kinetic energy, as $E_k = \frac{1}{2}mv^2$. The $v_{rms}$ is the square root of the average of the squares of the velocities. It's found by calculating the mean-square velocity, $\overline{v^2}$, and then taking the square root.

$$\overline{v^2} = \int_{0}^{\infty} v^2 f(v) dv$$

The result is $\overline{v^2} = \frac{3k_B T}{m}$. Taking the square root gives:

$$v_{rms} = \sqrt{\overline{v^2}} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3RT}{M}}$$

The Ratio

If you look at the formulas, you'll see $v_{rms} > \bar{v} > v_{mp}$. The $v_{rms}$ is the fastest because the $v^2$ term gives more "weight" to the faster-moving molecules in the long tail of the distribution.

Their ratio is a constant:

$$v_{mp} : \bar{v} : v_{rms} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3}$$

$$v_{mp} : \bar{v} : v_{rms} \approx 1 : 1.128 : 1.224$$

3. Distribution of Energy and the Barometric Formula

a) Maxwell's Distribution of Kinetic Energy ($f(\epsilon)$)

We can convert the speed distribution $f(v)$ into an energy distribution $f(\epsilon)$. The kinetic energy is $\epsilon = \frac{1}{2}mv^2$.

To do this, we substitute $v = \sqrt{2\epsilon/m}$ and $dv = \frac{1}{\sqrt{2m\epsilon}}d\epsilon$ into the $f(v)dv$ expression.

After simplifying, we get the Maxwell-Boltzmann Distribution of Energy:

$$f(\epsilon)d\epsilon = 2\pi \left(\frac{1}{\pi k_B T}\right)^{3/2} \epsilon^{1/2} e^{-\epsilon/k_B T} d\epsilon$$

Notice this equation is dominated by the $\epsilon^{1/2}$ term (which starts at zero) and the Boltzmann factor $e^{-\epsilon/k_B T}$. This exponential decay shows that the probability of a molecule having an energy much larger than $k_B T$ is extremely small.

b) The Barometric Formula (Law of Atmospheres)

This formula describes how the pressure of a gas (like our atmosphere) decreases with height $h$ in a uniform gravitational field $g$. It's a direct consequence of the Boltzmann distribution!

Derivation:

1. Consider a thin slice of air $dh$ at height $h$. The pressure difference $dP$ across this slice is due to the weight of the air in it.
2. $dP = -(\text{Force}/\text{Area}) = -(\text{Weight}/\text{Area}) = -(\frac{mg}{A}) = -(\frac{(\rho A dh)g}{A}) = -\rho g dh$. The negative sign means pressure decreases as $h$ increases.
3. Now, we assume the gas is ideal. From the ideal gas law $PV = nRT$, we get $P = \frac{n}{V}RT = \frac{m_{total}}{M V}RT = \frac{\rho RT}{M}$, where $\rho$ is the density and $M$ is the molar mass.
4. Rearranging for density: $\rho = \frac{PM}{RT}$.
5. Substitute this $\rho$ into our $dP$ equation: $dP = - \left(\frac{PM}{RT}\right) g dh$.
6. Separate the variables: $\frac{dP}{P} = - \left(\frac{Mg}{RT}\right) dh$.
7. Now, integrate this from height $h=0$ (where pressure is $P_0$) to a height $h$ (where pressure is $P$):

$$\int_{P_0}^{P} \frac{dP}{P} = - \frac{Mg}{RT} \int_{0}^{h} dh$$

$$\ln(P) - \ln(P_0) = - \frac{Mgh}{RT}$$

$$\ln\left(\frac{P}{P_0}\right) = - \frac{Mgh}{RT}$$

8. Taking the exponential of both sides gives the Barometric Formula:

$$P = P_0 e^{-Mgh/RT}$$

Connection to Boltzmann: The potential energy of a molecule at height $h$ is $E_p = mgh$. For one mole, this is $E_p = Mgh$. The formula $P = P_0 e^{-E_p/RT}$ shows that the pressure (which is proportional to the number of molecules) follows a Boltzmann distribution based on potential energy. It's much less probable to find molecules at high potential energy (high $h$)!

4. The Law of Equipartition of Energy

This is a powerful (but ultimately flawed) classical law. It provides a simple way to calculate the total internal energy, and thus the heat capacity, of a gas.

The Law: For a classical system in thermal equilibrium at temperature $T$, the average energy associated with each quadratic degree of freedom is $\frac{1}{2}k_B T$.

A "quadratic degree of freedom" is any term in the total energy expression that is proportional to a coordinate or a momentum squared. Examples:

• $\frac{1}{2}mv_x^2$ (kinetic energy of translation in $x$)
• $\frac{1}{2}I_x\omega_x^2$ (kinetic energy of rotation around the $x$-axis)
• $\frac{1}{2}kx^2$ (potential energy of a harmonic oscillator/vibration)

Application to Molar Heat Capacity ($C_v$)

The molar internal energy $U_m$ is $N_A \times (\text{total average energy per molecule})$. The molar heat capacity at constant volume is $C_{v,m} = \left(\frac{\partial U_m}{\partial T}\right)_V$. Since $k_B T \times N_A = RT$, each degree of freedom contributes $\frac{1}{2}RT$ to the molar internal energy $U_m$, and thus $\frac{1}{2}R$ to the molar heat capacity $C_{v,m}$.

• Monoatomic Gas (e.g., He, Ar):
  • Degrees of freedom: 3 (translation in $x, y, z$).
  • $U_m = 3 \times (\frac{1}{2}RT) = \frac{3}{2}RT$.
  • $C_{v,m} = \frac{3}{2}R$. (This works perfectly!)
• Diatomic Gas (e.g., N$_2$, O$_2$ at moderate T):
  • Degrees of freedom: 3 (translation) + 2 (rotation). Rotation about the bond axis is negligible.
  • $U_m = (3 + 2) \times (\frac{1}{2}RT) = \frac{5}{2}RT$.
  • $C_{v,m} = \frac{5}{2}R$. (This works well at room temperature).
• Diatomic Gas (at very high T):
  • Degrees of freedom: 3 (trans) + 2 (rot) + 2 (vibration: 1 for kinetic, 1 for potential).
  • $U_m = (3 + 2 + 2) \times (\frac{1}{2}RT) = \frac{7}{2}RT$.
  • $C_{v,m} = \frac{7}{2}R$.
• Non-linear Polyatomic Gas (e.g., H$_2$O):
  • Degrees of freedom: 3 (trans) + 3 (rotation).
  • $U_m = (3 + 3) \times (\frac{1}{2}RT) = 3RT$.
  • $C_{v,m} = 3R$ (plus vibrational contributions at high T).

Limitations and Failure of Equipartition

The classical equipartition law predicts that $C_v$ should be constant, independent of temperature (e.g., $\frac{5}{2}R$ for N$_2$ at all $T$). Experiments show this is wrong.

For $H_2$, $C_{v,m}$ is $\frac{3}{2}R$ at low T, rises to $\frac{5}{2}R$ at moderate T, and then rises again towards $\frac{7}{2}R$ at very high T. This is called the "freezing out" of degrees of freedom.

The Resolution: Quantum Mechanics! Rotational and vibrational energies are quantized. A molecule cannot have *any* rotational energy; it can only have discrete values. To "activate" a rotational or vibrational mode, the molecule must gain a minimum quantum of energy ($\Delta E$).

Tutor Tip: The average thermal energy "kick" available is about $k_B T$.
- If $k_B T \ll \Delta E_{rot/vib}$, the molecule can't "pay the toll" to get to the first excited state. That degree of freedom is frozen out and contributes 0 to the heat capacity.
- If $k_B T \gg \Delta E_{rot/vib}$, the energy levels are so close together relative to the thermal energy that the mode behaves classically, and Equipartition works (it contributes $\frac{1}{2}R$ or $R$).
- This is why translation (tiny $\Delta E$) is always active, rotation (moderate $\Delta E$) activates at low T, and vibration (large $\Delta E$) only activates at high T.

5. Collision Parameters

Now let's zoom in on the collisions that are fundamental to the kinetic theory. We model molecules as rigid spheres with a diameter $d$.

a) Mean Free Path ($\lambda$)

This is the average distance a molecule travels between collisions. Let's derive it.

1. First, a simple (but slightly wrong) model: Imagine one molecule (Molecule 1) moving with speed $\bar{v}$ through a gas of stationary molecules.
2. A collision occurs if the center of another molecule comes within a distance $d$ of Molecule 1's center. This means Molecule 1 effectively "sweeps out" a cylinder of cross-sectional area $\sigma = \pi d^2$ (the collision cross-section).
3. In time $t$, the cylinder's volume is $V_{cyl} = (\text{Area}) \times (\text{Length}) = \sigma \bar{v} t$.
4. The number of stationary molecules in this cylinder is $\mathcal{N} V_{cyl}$, where $\mathcal{N} = N/V$ is the number density (molecules per $m^3$).
5. This is the number of collisions in time $t$. So, the collision frequency ($Z_1$, collisions per second) is $\frac{\text{Collisions}}{t} = \mathcal{N} \sigma \bar{v}$.
6. The mean free path $\lambda = \frac{\text{Total Distance}}{\text{Collisions}} = \frac{\bar{v} t}{\mathcal{N} \sigma \bar{v} t} = \frac{1}{\mathcal{N} \sigma}$.
7. The Correction: This model is flawed because all molecules are moving. When we account for the relative motion, we must use the average relative speed, $\bar{v}_{rel}$, which for like molecules is $\sqrt{2}\bar{v}$.
8. The corrected collision frequency for one molecule is $Z_1 = \mathcal{N} \sigma \bar{v}_{rel} = \sqrt{2} \mathcal{N} \sigma \bar{v}$.
9. The corrected Mean Free Path is $\lambda = \frac{\text{Total Distance}}{\text{Collisions}} = \frac{\bar{v} t}{Z_1 t} = \frac{\bar{v}}{Z_1} = \frac{\bar{v}}{\sqrt{2} \mathcal{N} \sigma \bar{v}}$.

$$\lambda = \frac{1}{\sqrt{2} \mathcal{N} \pi d^2}$$

We can express this in terms of $P$ and $T$. Since $P = (N/V)k_B T = \mathcal{N}k_B T$, we have $\mathcal{N} = P/k_B T$.

$$\lambda = \frac{k_B T}{\sqrt{2} P \pi d^2}$$

This shows that:

• $\lambda$ is inversely proportional to Pressure. (More pressure $\rightarrow$ higher density $\mathcal{N}$ $\rightarrow$ more collisions $\rightarrow$ shorter $\lambda$).
• $\lambda$ is directly proportional to Temperature. (At constant $P$, higher $T \rightarrow$ lower density $\mathcal{N}$ $\rightarrow$ fewer collisions $\rightarrow$ longer $\lambda$).

b) Collision Frequency ($Z_1, Z_{11}, Z_{12}$)

$Z_1$ (Frequency for one molecule): We just found this! It's the number of collisions one molecule experiences per second.

$$Z_1 = \sqrt{2} \mathcal{N} \sigma \bar{v} = \frac{\bar{v}}{\lambda}$$

$Z_{11}$ (Total frequency for like molecules): This is the total number of collisions per unit volume, per second between identical molecules.

• We can take the collision frequency for one molecule, $Z_1$, and multiply it by the number of molecules per unit volume, $\mathcal{N}$.
• This gives $Z_1 \times \mathcal{N} = (\sqrt{2} \mathcal{N} \sigma \bar{v}) \mathcal{N} = \sqrt{2} \mathcal{N}^2 \sigma \bar{v}$.
• Wait! This double-counts every collision (A hitting B is counted, and B hitting A is counted separately). We must divide by 2.

$$Z_{11} = \frac{1}{2} (\sqrt{2} \mathcal{N}^2 \sigma \bar{v}) = \frac{1}{\sqrt{2}} \mathcal{N}^2 \sigma \bar{v} = \frac{1}{\sqrt{2}} \mathcal{N}^2 \pi d^2 \bar{v}$$

$Z_{12}$ (Total frequency for unlike molecules): This is the total collisions per unit volume, per second between molecules of type 1 ($\mathcal{N}_1$) and type 2 ($\mathcal{N}_2$).

• Here, we don't divide by 2, as a 1-2 collision is distinct.
• Collision cross-section: $\sigma_{12} = \pi d_{12}^2$, where $d_{12} = \frac{d_1 + d_2}{2}$ is the average collision diameter.
• Average relative speed: $\bar{v}_{rel} = \sqrt{\frac{8k_B T}{\pi \mu}}$, where $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass.

$$Z_{12} = \mathcal{N}_1 \mathcal{N}_2 \sigma_{12} \bar{v}_{rel} = \mathcal{N}_1 \mathcal{N}_2 \pi \left(\frac{d_1 + d_2}{2}\right)^2 \sqrt{\frac{8k_B T}{\pi \mu}}$$

6. Transport Phenomena

This is the payoff! We use all the concepts above (collisions, $\lambda$, $\bar{v}$) to explain macroscopic properties: viscosity, thermal conductivity, and diffusion. These are all transport processes, where the random motion of molecules causes a net "transport" of some property down a concentration gradient.

The "One-Third" Model:

A simple (but powerful) model for all three properties is to consider a plane at $z=0$. There is a gradient in some property $G$ (momentum, energy, or number) in the $z$-direction. Molecules move randomly. A molecule crossing the plane from below last collided (on average) at $z = -\frac{2}{3}\lambda$ and carries the property $G(z=-\frac{2}{3}\lambda)$. A molecule crossing from above last collided at $z = +\frac{2}{3}\lambda$ and carries $G(z=+\frac{2}{3}\lambda)$. The net flux (net transport) across the plane is the difference.

The flux of molecules crossing the plane from one side (per area, per time) is $\frac{1}{4}\mathcal{N}\bar{v}$. The net flux of the property $G$ becomes:

$$J(G) = \frac{1}{4}\mathcal{N}\bar{v} \left[ G(z=-\frac{2}{3}\lambda) - G(z=+\frac{2}{3}\lambda) \right]$$

Using a Taylor expansion $G(z) \approx G(0) + z \frac{dG}{dz}$, we get:

$$J(G) = \frac{1}{4}\mathcal{N}\bar{v} \left[ (G(0) - \frac{2}{3}\lambda \frac{dG}{dz}) - (G(0) + \frac{2}{3}\lambda \frac{dG}{dz}) \right] = \frac{1}{4}\mathcal{N}\bar{v} \left[ -\frac{4}{3}\lambda \frac{dG}{dz} \right]$$

$$J(G) = -\frac{1}{3}\mathcal{N}\bar{v}\lambda \frac{dG}{dz}$$

This is our master equation! The coefficient of the transport property (e.g., $\eta, \kappa, D$) is just the part in front of the gradient.

a) Viscosity ($\eta$) - Transport of Momentum

• Property transported ($G$): $x$-momentum, $p_x = mv_x$.
• Gradient: A velocity gradient, $\frac{dv_x}{dz}$. (Imagine gas flowing faster at the top, like in a pipe).
• Defining Law (Newton's): $J_{mom} = F/A = -\eta \frac{dv_x}{dz}$. This flux of momentum is just Force/Area.
• Derivation:
  • $G = mv_x$, so $\frac{dG}{dz} = m \frac{dv_x}{dz}$.
  • $J_{mom} = -\frac{1}{3}\mathcal{N}\bar{v}\lambda \left( m \frac{dv_x}{dz} \right) = - \left( \frac{1}{3} m \mathcal{N} \bar{v} \lambda \right) \frac{dv_x}{dz}$.
  • Compare this to $F/A = -\eta \frac{dv_x}{dz}$. We find the coefficient of viscosity, $\eta$:

  $$\eta = \frac{1}{3} m \mathcal{N} \bar{v} \lambda$$

  • Since $m\mathcal{N} = \rho$ (the mass density), we get the common form:

  $$\eta = \frac{1}{3} \rho \bar{v} \lambda$$

• Effect of T and P:
  • Pressure: Let's substitute $\lambda = \frac{1}{\sqrt{2} \mathcal{N} \pi d^2}$. $\eta = \frac{1}{3} (m\mathcal{N}) \bar{v} \left( \frac{1}{\sqrt{2} \mathcal{N} \pi d^2} \right) = \frac{m \bar{v}}{3\sqrt{2} \pi d^2}$. The number density $\mathcal{N}$ cancels out! This means, surprisingly, that gas viscosity is independent of pressure (at moderate P).

    Why? At higher pressure, more molecules ($\rho \uparrow$) carry momentum, but they carry it over a shorter distance ($\lambda \downarrow$). The two effects exactly cancel.

  • Temperature: Since $\bar{v} \propto \sqrt{T}$, the equation $\eta \propto m \bar{v} / d^2$ shows that $\eta \propto \sqrt{T}$. The viscosity of a gas INCREASES with temperature. (This is the opposite of liquids!)

b) Thermal Conductivity ($\kappa$) - Transport of Energy

• Property transported ($G$): Thermal energy, $\epsilon = c_v T$, where $c_v$ is the heat capacity per molecule.
• Gradient: A temperature gradient, $\frac{dT}{dz}$.
• Defining Law (Fourier's): $J_{energy} = -\kappa \frac{dT}{dz}$. This is the flux of heat energy.
• Derivation:
  • $G = \epsilon$, so $\frac{dG}{dz} = \frac{d\epsilon}{dT}\frac{dT}{dz} = c_v \frac{dT}{dz}$.
  • $J_{energy} = -\frac{1}{3}\mathcal{N}\bar{v}\lambda \left( c_v \frac{dT}{dz} \right) = - \left( \frac{1}{3} \mathcal{N} c_v \bar{v} \lambda \right) \frac{dT}{dz}$.
  • Compare to $J_{energy} = -\kappa \frac{dT}{dz}$. We find the coefficient of thermal conductivity, $\kappa$:

  $$\kappa = \frac{1}{3} \mathcal{N} c_v \bar{v} \lambda$$

  • We can write this using molar heat capacity $C_{v,m} = N_A c_v$ and density $\rho = m \mathcal{N}$:

  $$\kappa = \frac{1}{3} \frac{C_{v,m}}{N_A} \frac{\rho}{m} \bar{v} \lambda = \frac{1}{3} \frac{C_{v,m}}{M} \rho \bar{v} \lambda$$

• Relation to Viscosity:
  • We know $\eta = \frac{1}{3} \rho \bar{v} \lambda$. Let's just substitute that right in!

  $$\kappa = \frac{C_{v,m}}{M} \left( \frac{1}{3} \rho \bar{v} \lambda \right) \implies \kappa = \frac{\eta C_{v,m}}{M}$$

  • This is a beautiful, simple relation connecting two different transport properties.

c) Diffusion ($D$) - Transport of Mass

• Property transported ($G$): The molecules themselves (their "identity" or number). We'll look at self-diffusion, where (hypothetical) labeled molecules diffuse through unlabeled ones.
• Gradient: A number density (concentration) gradient, $\frac{d\mathcal{N}}{dz}$.
• Defining Law (Fick's First): $J_{molecules} = -D \frac{d\mathcal{N}}{dz}$. This is the flux of molecules.
• Derivation:
  • Here, the flux is $J(G) = -\frac{1}{3}\bar{v}\lambda \frac{dG}{dz}$. (We don't multiply by $\mathcal{N}$ because we are calculating the flux of $\mathcal{N}$).
  • $G = \mathcal{N}$, so $\frac{dG}{dz} = \frac{d\mathcal{N}}{dz}$.
  • $J_{molecules} = - \left( \frac{1}{3} \bar{v} \lambda \right) \frac{d\mathcal{N}}{dz}$.
  • Compare to $J_{molecules} = -D \frac{d\mathcal{N}}{dz}$. We find the coefficient of self-diffusion, $D$:

  $$D = \frac{1}{3} \bar{v} \lambda$$

• Relation to Viscosity:
  • We have $D = \frac{1}{3} \bar{v} \lambda$ and $\eta = \frac{1}{3} \rho \bar{v} \lambda$.
  • We can write $D = \frac{\eta}{\rho}$.
  • This means $D\rho = \eta$, another key relationship.

And that completes our tour! We've gone from the random motion of a single molecule all the way to the macroscopic flow of the entire gas. I hope this helps you build a strong foundation!