AP Chemistry 3.5-3.6: Kinetic Molecular Theory and Gas Deviations

Kinetic-molecular theory (KMT)

A microscopic model that explains the macroscopic behavior of gases, including the gas laws, in terms of particle motion, collisions, kinetic energy, and temperature.

What are the five postulates of the kinetic-molecular theory for an ideal gas?

Gas particles are in continuous, straight-line motion; their individual volumes are negligible; gas pressure results from particle-wall collisions; particles exert no intermolecular attractions or repulsions and collide elastically; and average kinetic energy is proportional to kelvin temperature.

Why does a gas exert pressure according to the kinetic-molecular theory?

Moving gas particles collide with the container walls and transfer momentum to them. The combined effect of these collisions produces the measured pressure.

Elastic collision

A collision in which the total kinetic energy is conserved. In the ideal-gas model, gas particles collide elastically with one another and with container walls.

How does the KMT explain Amontons's law?

At constant volume and amount of gas, increasing temperature increases particles' average kinetic energy and speed. Collisions with the walls become more frequent and more forceful, so pressure increases: $P \propto T$.

How does the KMT explain Charles's law?

At constant pressure and amount of gas, increasing temperature makes collisions more forceful. The gas must expand, increasing the distance between wall collisions and the wall area so that pressure remains constant: $V \propto T$.

How does the KMT explain Boyle's law?

At constant temperature and amount of gas, decreasing volume reduces the wall area available to the particles. Wall collisions occur more frequently per unit area, so pressure increases: $P \propto \frac{1}{V}$.

How does the KMT explain Avogadro's law?

At constant temperature and pressure, adding gas particles increases the frequency of wall collisions. The volume must increase proportionally to keep the collision rate per unit area constant: $V \propto n$.

How does the KMT explain Dalton's law of partial pressures?

Gas particles are far apart and, in the ideal model, do not interact significantly. Each gas contributes pressure independently, so $P_{\text{total}}=P_1+P_2+\cdots$.

What does the ideal-gas assumption imply about particle size and intermolecular forces?

Ideal-gas particles have negligible volume compared with the container and exert no attractive or repulsive forces on one another. Real gases approximate these assumptions best at low pressure and high temperature.

Kinetic energy of a single gas particle

The kinetic energy of a particle of mass $m$ moving at speed $u$ is $KE=\frac{1}{2}mu^2$. Using kilograms and meters per second gives energy in joules.

Maxwell-Boltzmann speed distribution

A distribution showing the relative number of gas particles moving at each speed. Most particles have intermediate speeds, while relatively few have extremely low or extremely high speeds.

How does increasing temperature affect a Maxwell-Boltzmann speed distribution?

The average kinetic energy increases, and the distribution shifts toward higher speeds and becomes broader and lower in its peak. At lower temperature, it shifts toward lower speeds and becomes narrower and taller.

Root-mean-square speed, $u_{\mathrm{rms}}$

A measure of molecular speed defined by $u_{\mathrm{rms}}=\sqrt{\overline{u^2}}$. It is the square root of the average of the squared speeds, not the simple arithmetic average speed.

Average kinetic energy of one mole of ideal-gas particles

The average kinetic energy is $KE_{\mathrm{avg}}=\frac{3}{2}RT$, where $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ and $T$ is in kelvins.

How are molar mass, root-mean-square speed, and temperature related?

Combining kinetic-energy relationships gives $u_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$, where $M$ is molar mass in kilograms per mole and $T$ is in kelvins.

How do gases with different molar masses compare at the same temperature?

All ideal gases have the same average kinetic energy at a given temperature, but lighter particles move faster on average. Since $u_{\mathrm{rms}}\propto\frac{1}{\sqrt{M}}$, heavier gases have lower characteristic speeds.

Why must temperature be expressed in kelvins in molecular-speed and gas-law equations?

Kelvin temperature is proportional to average molecular kinetic energy and begins at absolute zero. Celsius values cannot be used directly because they do not have a proportional relationship to kinetic energy.

How do the most probable speed and root-mean-square speed compare?

For a Maxwell-Boltzmann distribution, the most probable speed is the speed at the peak of the distribution. It is lower than the root-mean-square speed because faster particles contribute disproportionately to the squared-speed average.

Diffusion

The spontaneous dispersal and mixing of gas particles through space due to random molecular motion. Net movement occurs from higher concentration to lower concentration until concentrations become uniform.

Effusion

The escape of gas particles through a very small opening into a vacuum or region of much lower pressure. It differs from diffusion, which involves dispersal and mixing throughout available space.

Mean free path

The average distance a gas particle travels between collisions. It generally increases as pressure decreases because particles are farther apart.

What factors affect the rate of gas diffusion?

Diffusion rate depends on temperature, particle molar mass, the concentration gradient, the available surface area, and the distance particles must travel. Higher temperature and a steeper concentration gradient generally increase the rate, while greater distance decreases it.

How are diffusion and effusion related but different?

Diffusion is the unrestricted dispersal of gas particles through space, whereas effusion is passage through a very small opening. Their rates are not generally equal, although their dependence on molar mass follows the same inverse-square-root relationship.

Graham's law of effusion

At the same temperature and pressure, effusion rate is inversely proportional to the square root of molar mass: $\text{rate}\propto\frac{1}{\sqrt{M}}$.

How can the relative effusion rates of two gases be calculated?

For gases A and B at the same temperature and pressure, $\frac{\text{rate}_A}{\text{rate}_B}=\sqrt{\frac{M_B}{M_A}}$. The gas with lower molar mass effuses faster.

How are effusion time and molar mass related for equal amounts of gas?

Because rate equals amount divided by time, the time required for equal amounts to effuse is proportional to the square root of molar mass: $\frac{t_A}{t_B}=\sqrt{\frac{M_A}{M_B}}$. A heavier gas takes longer.

Why does a lighter gas effuse faster than a heavier gas at the same temperature?

Both gases have the same average kinetic energy at the same temperature. Therefore, the lighter gas must have a greater characteristic speed, and its particles pass through a small opening more rapidly.

Non-ideal gas behavior

Behavior in which a real gas deviates measurably from the relationships predicted by $PV=nRT$. Deviations become more significant when particles are crowded or moving slowly enough for intermolecular forces to matter.

Under what conditions do real gases most closely approximate ideal gases?

Real gases behave most ideally at low pressure and high temperature. Low pressure separates particles, while high temperature makes intermolecular attractions relatively less important.

Why does high pressure cause gases to deviate from ideal behavior?

High pressure brings particles close together, so their own volumes become significant compared with the container volume. Intermolecular attractions may also become important.

How do intermolecular attractions affect a real gas's pressure and compressibility?

Attractions reduce the force and frequency of particle-wall collisions, lowering the pressure relative to an ideal gas at the same $T$ and $V$. They can also make a gas more compressible, especially at relatively low temperatures and moderate pressures.

Why does a real gas become less compressible at very high pressure?

The particles themselves occupy an appreciable portion of the total volume and cannot be compressed significantly. Further pressure increases therefore produce less volume decrease than the ideal-gas model predicts.

Compressibility factor, $Z$

A measure of deviation from ideal-gas behavior defined by $Z=\frac{PV_m}{RT}$, where $V_m$ is the measured molar volume. An ideal gas has $Z=1$; values above or below 1 indicate non-ideal behavior.

How does a plot of $Z$ versus pressure reveal the effects of intermolecular forces?

If attractions dominate, $Z$ falls below 1 because the gas is more compressible than ideal. At very high pressure, particle volume dominates and $Z$ rises above 1 because the gas is less compressible than ideal.

van der Waals equation

A real-gas equation that accounts for particle volume and intermolecular attractions: $\left(P+\frac{n^2a}{V^2}\right)(V-nb)=nRT$. The constant $a$ measures attractive-force strength, and $b$ represents molar particle volume.

What do the two correction terms in the van der Waals equation represent?

The pressure correction $\frac{n^2a}{V^2}$ compensates for pressure lowered by intermolecular attractions. The volume correction $nb$ subtracts the space occupied by gas particles, leaving less free volume for motion.

When does the van der Waals equation approach the ideal gas law?

When $V$ is large relative to $n$—typically at low pressure—the corrections $\frac{n^2a}{V^2}$ and $nb$ become small. The equation then approaches $PV=nRT$.

How do the van der Waals constants $a$ and $b$ affect a gas's non-ideal behavior?

A larger $a$ indicates stronger intermolecular attractions and a greater tendency for $Z<1$. A larger $b$ indicates larger particles and a greater tendency for excluded-volume effects and $Z>1$ at high pressure.

Why are attractions especially important for non-ideal behavior at low temperature?

At lower temperature, particles have less kinetic energy and are less able to overcome attractive forces. Consequently, attractions produce larger reductions in pressure or volume than they would at higher temperature.

A gas sample has $Z<1$. What does this indicate about its behavior?

The gas has a smaller molar volume or lower pressure than predicted by the ideal-gas law under the same conditions. This usually indicates that intermolecular attractions dominate.

A gas sample has $Z>1$. What does this indicate about its behavior?

The gas is less compressible than an ideal gas, generally because the finite volume of its particles is significant. This effect is especially important at high pressure.