Chapter 6: Gases – Kinetic Molecular Theory, Molecular Speeds, Graham’s Law, and Real Gases

Chapter 6: Gases

Kinetic Molecular Theory of Gases

The kinetic molecular theory provides a molecular-level explanation for the behavior of gases, describing their motion and interactions.

• Particles as Point Masses: Gas molecules are treated as point masses in constant, random, straight-line motion.

• Large Separation: Molecules are separated by great distances compared to their size.

• Collisions: Molecules collide only fleetingly; most of the time, they are not colliding.

• No Intermolecular Forces: It is assumed that there are no forces between molecules (ideal gas assumption).

• Energy Conservation: Individual molecules may gain or lose energy during collisions, but the total energy of the system remains constant.

Example: The random motion of gas molecules explains properties such as diffusion and pressure exerted on container walls.

Distribution of Molecular Speeds

The speeds of molecules in a gas are not all the same; they follow a statistical distribution described by the Maxwell-Boltzmann distribution.

• Maxwell-Boltzmann Distribution: The probability of finding a molecule with a particular speed u is given by:

$F(u) = 4\pi \left( \frac{M}{2\pi RT} \right)^{3/2} u^2 e^{-(Mu^2/2RT)}$

• Most Probable Speed: $u_m = \sqrt{\frac{2RT}{M}}$

• Average Speed: $u_{av} = \sqrt{\frac{8RT}{\pi M}}$

• Root-Mean-Square Speed: $u_{rms} = \sqrt{\frac{3RT}{M}}$

• Variables: R = 8.3145 J·K-1·mol-1; M = molar mass in kg/mol; T = temperature in K.

Example: Hydrogen gas at 0°C has a distribution of molecular speeds, with most molecules near the most probable speed, but some much faster or slower.

Effect of Mass and Temperature on Speed Distribution

The distribution of molecular speeds depends on both the molar mass of the gas and the temperature.

• Higher Temperature: Increases the average speed and broadens the distribution.

• Lighter Molecules: Have higher average speeds at the same temperature compared to heavier molecules.

Example: O2 at 273 K has a lower average speed than H2 at the same temperature; increasing temperature shifts the distribution to higher speeds.

Experimental Determination of Speed Distribution

The distribution of molecular speeds can be measured experimentally using techniques such as the rotating disk method, which separates molecules by speed and counts their number.

• Method: Molecules pass through collimators and rotating disks; their speed is determined by the time taken to reach a detector.

Example: The resulting data matches the Maxwell-Boltzmann distribution curve.

Meaning of Temperature

Temperature is a measure of the average translational kinetic energy of gas molecules.

• Kinetic Energy per Molecule: $E_k = \frac{1}{2}mu^2 = \frac{1}{2} \times \frac{M}{N_A} \times (u_{rms})^2$

• Root-Mean-Square Speed: $u_{rms} = \sqrt{\frac{3RT}{M}}$

• Average Kinetic Energy per Mole: $E_k = \frac{3}{2} \left( \frac{RT}{N_A} \right)$

• Pressure Relation: $P = \frac{1}{3} \frac{Nm\overline{u^2}}{V}$

• Temperature Proportionality: The Kelvin temperature (T) is directly proportional to the average translational kinetic energy ($E_k$) of molecules: $T \propto E_k$.

Example: Molecules in a hotter object have, on average, higher kinetic energies than those in a colder object.

Gas Properties Relating to the Kinetic-Molecular Theory

Two important phenomena explained by kinetic theory are diffusion and effusion.

• Diffusion: The mixing of gases due to random molecular motion. Lighter gases diffuse faster than heavier ones.

• Effusion: The passage of gas molecules through a tiny orifice into a vacuum. Lighter gases effuse faster than heavier ones.

Example: H2 diffuses and effuses faster than N2 due to its lower molar mass.

Graham’s Law of Effusion

Graham’s Law quantifies the relationship between the rate of effusion and the molar mass of a gas.

• Law Statement: The rate of effusion of a gas is inversely proportional to the square root of its molar mass.

$\text{Rate}_{\text{effusion}} \propto \frac{1}{\sqrt{M_w}}$

• Conditions: Applies only to gases at low pressure and through a tiny orifice (no collisions).

• Not for Diffusion: Graham’s Law does not apply to diffusion.

Ratio Form:

$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}$

• Other Quantities: The same ratio applies to molecular speeds, effusion times, distances traveled, and amounts of gas effused.

Example: If Kr effuses in 87.3 s and an unknown gas effuses in 42.9 s, Graham’s Law can be used to find the molar mass of the unknown gas.

Real Gases: Intermolecular Forces

Real gases deviate from ideal behavior due to intermolecular forces and finite molecular volume, especially at high pressures and low temperatures.

• Ideal Gas Law: $PV = nRT$ is an equation of state relating pressure, volume, temperature, and moles.

• Deviations: At high pressure, Boyle’s Law ($V = C$) is not satisfied; at low temperature, Charles’s Law ($V \propto T$) breaks down.

• Compressibility Factor: $z = \frac{PV}{nRT}$; for ideal gases, $z = 1$. Deviations indicate non-ideal behavior.

Example: Polar molecules like water vapor and ammonia show deviations from ideality even at atmospheric pressure.

The van der Waals Equation of State

The van der Waals equation introduces corrections for molecular volume and intermolecular forces, providing a more accurate description of real gases.

• Equation:

$(P + \frac{n^2a}{V^2})(V - nb) = nRT$

$P = \frac{nRT}{V - nb} - \frac{n^2a}{V^2}$

• Repulsive Forces (b): Molecules exclude volume; $b$ is the excluded volume per mole. This increases the effective pressure.

• Attractive Forces (a): Molecules attract each other, reducing the number of wall collisions and thus the pressure. $a$ quantifies the strength of these attractions.

• Units: $a$: atm·L2·mol-2; $b$: L·mol-1.

Example: For CO2, using the van der Waals equation gives a more accurate pressure than the ideal gas law, especially at high pressure or low temperature.

Compressibility Factor and van der Waals Parameters

The compressibility factor $z$ for the van der Waals equation is:

$z = \frac{PV}{nRT} = \frac{V}{V - nb} - \frac{a n}{RT V}$

• Repulsive Forces (b): Increase $z$ above 1.

• Attractive Forces (a): Decrease $z$ below 1.

Table: van der Waals Constants and Compressibility Factors

The following table summarizes van der Waals constants and compressibility factors for various gases at 10 bar and 300 K:

Gas	a (bar·L2·mol-2)	b (L·mol-1)	Compressibility Factor
He	0.0346	0.0238	1.005
H2	0.244	0.0266	1.016
N2	1.370	0.0387	0.991
O2	1.360	0.0318	0.995
CO2	3.590	0.0427	0.857
CH4	2.253	0.0428	0.982
Ideal gas	0	0	1.000

Example Problem: van der Waals vs. Ideal Gas Law

Calculate the pressure exerted by 50.0 g of CO2 in a 1.00-L vessel at 25°C using (a) the ideal gas law and (b) the van der Waals equation. Determine whether attractive or repulsive forces dominate.

• Step 1: Calculate moles of CO2: $n = \frac{50.0\,g}{44.01\,g/mol} \approx 1.136\,mol$

• Step 2: Use $P = \frac{nRT}{V}$ for ideal gas law.

• Step 3: Use $P = \frac{nRT}{V - nb} - \frac{n^2a}{V^2}$ for van der Waals equation, with $a$ and $b$ from the table.

• Step 4: Compare results; if $z > 1$, repulsive forces dominate; if $z < 1$, attractive forces dominate.

Additional info: This example illustrates the importance of using the correct equation of state for real gases under non-ideal conditions.