States of Matter • Topic 2 of 3

Kinetic Theory & Molecular Speeds

The gas laws are experimental facts. The kinetic molecular theory of gases explains why they hold by modelling a gas as a vast crowd of tiny molecules in ceaseless, random motion. The model rests on a few postulates:

A gas consists of a large number of identical molecules whose total volume is negligible compared with the container.
Molecules are in constant random motion, travelling in straight lines until they collide.
There are no attractive or repulsive forces between molecules except during collisions.
Collisions (with each other and the walls) are perfectly elastic — no kinetic energy is lost.
The average kinetic energy of the molecules is directly proportional to the absolute temperature.

Working out the force of wall collisions gives the kinetic-theory pressure relation $PV=\frac{1}{3}mN\,\overline{u^2}$, where $m$ is the molecular mass, $N$ the number of molecules and $\overline{u^2}$ the mean-square speed. Comparing with $PV=nRT$ shows that the average translational kinetic energy per molecule is $\overline{KE}=\frac{3}{2}k_BT$, where $k_B=\frac{R}{N_A}=1.38\times10^{-23}\ \text{J/K}$. So temperature is a direct measure of molecular kinetic energy — raise $T$ and the molecules move faster.

Because molecules collide constantly, they do not all share one speed; instead there is a spread. Three average speeds are useful:

Most probable speed $u_{mp}=\sqrt{\frac{2RT}{M}}$ — the peak of the distribution, the speed possessed by the largest fraction of molecules.
Average speed $u_{avg}=\sqrt{\frac{8RT}{\pi M}}$ — the simple arithmetic mean.
Root-mean-square speed $u_{rms}=\sqrt{\frac{3RT}{M}}$ — the square root of the mean-square speed; it links directly to kinetic energy.

Their fixed ratio is $u_{mp}:u_{avg}:u_{rms}=1:1.128:1.224$, so $u_{mp}

The full spread of speeds is described by the Maxwell–Boltzmann distribution. Its curve rises from zero, peaks at $u_{mp}$, then tails off at high speed. As temperature rises the peak shifts to higher speed and broadens and flattens — the molecules speed up and the spread widens — while the total area (total number of molecules) stays constant. Lighter gases (smaller $M$) move faster at the same temperature, which is why hydrogen escapes a balloon faster than nitrogen.

Solved Examples

Worked Example

Calculate the rms speed of $\text{N}_2$ molecules ($M=0.028\ \text{kg/mol}$) at 300 K. ($R=8.314$)

Solution

Step 1: $u_{rms}=\sqrt{\frac{3RT}{M}}$.
Step 2: $u_{rms}=\sqrt{\frac{3\times8.314\times300}{0.028}}=\sqrt{\frac{7482.6}{0.028}}$.
Step 3: $u_{rms}=\sqrt{267236}\approx517\ \text{m/s}$.

Answer: $u_{rms}\approx517\ \text{m/s}$.

Worked Example

Find the most probable speed of $\text{O}_2$ ($M=0.032\ \text{kg/mol}$) at 300 K.

Solution

Step 1: $u_{mp}=\sqrt{\frac{2RT}{M}}$.
Step 2: $u_{mp}=\sqrt{\frac{2\times8.314\times300}{0.032}}=\sqrt{\frac{4988.4}{0.032}}$.
Step 3: $u_{mp}=\sqrt{155888}\approx395\ \text{m/s}$.

Answer: $u_{mp}\approx395\ \text{m/s}$.

Worked Example

For a gas the rms speed is $500\ \text{m/s}$. Find its average speed.

Solution

Step 1: Use the fixed ratio $\frac{u_{avg}}{u_{rms}}=\sqrt{\frac{8}{3\pi}}\approx0.921$.
Step 2: $u_{avg}=0.921\times500$.
Step 3: $u_{avg}\approx460\ \text{m/s}$.

Answer: $u_{avg}\approx460\ \text{m/s}$.

Worked Example

By what factor does the rms speed change when the temperature of a gas is raised from 300 K to 1200 K?

Solution

Step 1: $u_{rms}\propto\sqrt{T}$, so $\frac{u_2}{u_1}=\sqrt{\frac{T_2}{T_1}}$.
Step 2: $\frac{u_2}{u_1}=\sqrt{\frac{1200}{300}}=\sqrt{4}$.
Step 3: $\frac{u_2}{u_1}=2$.

Answer: The rms speed doubles.

Worked Example

Find the average translational kinetic energy of one gas molecule at 300 K. ($k_B=1.38\times10^{-23}\ \text{J/K}$)

Solution

Step 1: $\overline{KE}=\frac{3}{2}k_BT$.
Step 2: $\overline{KE}=\frac{3}{2}\times1.38\times10^{-23}\times300$.
Step 3: $\overline{KE}=6.21\times10^{-21}\ \text{J}$.

Answer: $\overline{KE}\approx6.21\times10^{-21}\ \text{J}$ per molecule.

Worked Example

At the same temperature, compare the rms speeds of $\text{He}$ ($M=4$) and $\text{O}_2$ ($M=32$).

Solution

Step 1: At fixed $T$, $u_{rms}\propto\frac{1}{\sqrt{M}}$, so $\frac{u_{He}}{u_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{He}}}$.
Step 2: $\frac{u_{He}}{u_{O_2}}=\sqrt{\frac{32}{4}}=\sqrt{8}$.
Step 3: $\frac{u_{He}}{u_{O_2}}\approx2.83$.

Answer: Helium moves about $2.83$ times faster than oxygen.

Key Points

Kinetic theory models a gas as many tiny molecules in constant random motion, with negligible volume, no inter-molecular forces except at collisions, and perfectly elastic collisions.
Kinetic pressure relation: $PV=\frac{1}{3}mN\,\overline{u^2}$; average translational KE per molecule $=\frac{3}{2}k_BT$, so temperature measures molecular kinetic energy.
Molecular speeds: $u_{mp}=\sqrt{\frac{2RT}{M}}$, $u_{avg}=\sqrt{\frac{8RT}{\pi M}}$, $u_{rms}=\sqrt{\frac{3RT}{M}}$.
Their ratio is fixed: $u_{mp}:u_{avg}:u_{rms}=1:1.128:1.224$, so $u_{mp}
The Maxwell–Boltzmann curve peaks at $u_{mp}$; raising $T$ shifts the peak right and flattens the curve, while lighter gases move faster at the same temperature.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.According to kinetic theory, the average kinetic energy of gas molecules depends only on:

Explanation: $\overline{KE}=\frac{3}{2}k_BT$ depends only on $T$, not on the gas’s identity.

Q2.The correct order of molecular speeds is:

Explanation: The fixed ratio $1:1.128:1.224$ gives $u_{mp}

Q3.If the absolute temperature of a gas is quadrupled, its rms speed becomes:

Explanation: $u_{rms}\propto\sqrt{T}$, so a 4× increase in $T$ doubles the speed.

Q4.On the Maxwell–Boltzmann curve, raising temperature causes the peak to:

Explanation: Higher $T$ moves the most-probable speed up and broadens (flattens) the distribution.

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