Kinetic Theory • Topic 2 of 3

Kinetic Interpretation of Pressure & Temperature

The great success of the kinetic theory is that it explains pressure and temperature — quantities we measure with gauges and thermometers — entirely in terms of molecular motion. The gas pushes on the container walls because its molecules are constantly striking them. Each collision is perfectly elastic, so a molecule rebounds and reverses its momentum; the steady drum of countless such impacts is what we feel as pressure.

Carefully counting the momentum delivered to a wall per second leads to the central result of the kinetic theory:

  • $P=\frac{1}{3}\frac{Nm\,\overline{v^2}}{V}$, where $N$ is the number of molecules, $m$ the mass of one molecule, $V$ the volume and $\overline{v^2}$ the mean of the squared speeds.
  • Since the density is $\rho=\frac{Nm}{V}$, this can be written compactly as $P=\frac{1}{3}\rho\,\overline{v^2}$.

The factor $\frac{1}{3}$ appears because molecular motion is shared equally among the three perpendicular directions ($x$, $y$, $z$): on average only one-third of the mean-square speed contributes to motion towards any one wall.

Kinetic interpretation of temperature. Rewriting the pressure result as $PV=\frac{1}{3}Nm\,\overline{v^2}=\frac{2}{3}N\left(\frac{1}{2}m\overline{v^2}\right)$ and comparing with the ideal gas law $PV=Nk_BT$, we find that the average translational kinetic energy of a molecule is directly proportional to the absolute temperature:

  • $\frac{1}{2}m\,\overline{v^2}=\frac{3}{2}k_BT$.
  • So temperature is a measure of the average kinetic energy of the molecules: hotter gas means faster molecules. At $T=0\ \text{K}$ all molecular motion would cease.
  • Notice the average KE depends only on $T$, not on the kind of gas — so at the same temperature, molecules of every gas have the same average kinetic energy.

Root-mean-square (rms) speed. The typical speed of the molecules is measured by the square root of $\overline{v^2}$:

  • $v_{rms}=\sqrt{\overline{v^2}}=\sqrt{\frac{3k_BT}{m}}=\sqrt{\frac{3RT}{M}}$, where $M$ is the molar mass (kg/mol).
  • So $v_{rms}\propto\sqrt{T}$ — heating a gas raises its rms speed — and $v_{rms}\propto\frac{1}{\sqrt{M}}$, so at a given temperature lighter molecules move faster. This is why hydrogen molecules outrun oxygen molecules in the same flask.
Gas molecules striking container walls produce pressureP = (1/3) rho v^2elastic wall collisions create pressure
Solved Examples
1
Worked Example
A gas of density $1.25\ \text{kg/m}^3$ exerts a pressure of $1.0\times10^{5}\ \text{Pa}$. Find the rms speed of its molecules.
Solution
  1. Step 1: Use $P=\frac{1}{3}\rho\,\overline{v^2}$, so $\overline{v^2}=\frac{3P}{\rho}$.
  2. Step 2: Substitute: $\overline{v^2}=\frac{3\times1.0\times10^{5}}{1.25}=2.4\times10^{5}$.
  3. Step 3: Then $v_{rms}=\sqrt{2.4\times10^{5}}\approx490\ \text{m/s}$.

Answer: $v_{rms}\approx490\ \text{m/s}$.

2
Worked Example
Calculate the average translational kinetic energy of a gas molecule at 300 K. ($k_B=1.38\times10^{-23}\ \text{J/K}$.)
Solution
  1. Step 1: Average KE per molecule $=\frac{3}{2}k_BT$.
  2. Step 2: Substitute: $\frac{3}{2}\times1.38\times10^{-23}\times300$.
  3. Step 3: Compute: $=6.21\times10^{-21}\ \text{J}$.

Answer: $\approx6.21\times10^{-21}\ \text{J}$.

3
Worked Example
Find the rms speed of oxygen molecules at 300 K. (Molar mass of $\text{O}_2$, $M=0.032\ \text{kg/mol}$; $R=8.314\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$.)
Solution
  1. Step 1: Use $v_{rms}=\sqrt{\frac{3RT}{M}}$.
  2. Step 2: Substitute: $v_{rms}=\sqrt{\frac{3\times8.314\times300}{0.032}}=\sqrt{\frac{7483}{0.032}}$.
  3. Step 3: Compute: $v_{rms}=\sqrt{2.34\times10^{5}}\approx484\ \text{m/s}$.

Answer: $v_{rms}\approx484\ \text{m/s}$.

4
Worked Example
At what temperature will the rms speed of a gas be double its value at 300 K?
Solution
  1. Step 1: Since $v_{rms}\propto\sqrt{T}$, $\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}$.
  2. Step 2: For $v_2=2v_1$: $4=\frac{T_2}{T_1}$, so $T_2=4T_1$.
  3. Step 3: Compute: $T_2=4\times300=1200\ \text{K}$.

Answer: $T_2=1200\ \text{K}$.

5
Worked Example
Compare the rms speeds of hydrogen ($M=2\ \text{g/mol}$) and oxygen ($M=32\ \text{g/mol}$) at the same temperature.
Solution
  1. Step 1: At the same $T$, $v_{rms}\propto\frac{1}{\sqrt{M}}$, so $\frac{v_{H_2}}{v_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}$.
  2. Step 2: Substitute: $\frac{v_{H_2}}{v_{O_2}}=\sqrt{\frac{32}{2}}=\sqrt{16}$.
  3. Step 3: Compute: $\frac{v_{H_2}}{v_{O_2}}=4$.

Answer: Hydrogen molecules move 4 times faster than oxygen molecules.

6
Worked Example
The rms speed of nitrogen molecules at 300 K is about 517 m/s. What will it be at 1200 K?
Solution
  1. Step 1: $v_{rms}\propto\sqrt{T}$, so $\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{1200}{300}}=\sqrt{4}=2$.
  2. Step 2: Therefore $v_2=2v_1=2\times517$.
  3. Step 3: Compute: $v_2\approx1034\ \text{m/s}$.

Answer: $v_{rms}\approx1034\ \text{m/s}$.

Key Points

  • Pressure arises from elastic collisions of molecules with the walls: $P=\frac{1}{3}\frac{Nm\,\overline{v^2}}{V}=\frac{1}{3}\rho\,\overline{v^2}$.
  • The factor $\frac{1}{3}$ comes from sharing motion equally among the three directions $x$, $y$, $z$.
  • Average translational KE of a molecule is $\frac{1}{2}m\,\overline{v^2}=\frac{3}{2}k_BT$ — temperature measures average molecular kinetic energy.
  • At a given temperature all gases have the same average translational KE per molecule.
  • rms speed: $v_{rms}=\sqrt{\frac{3k_BT}{m}}=\sqrt{\frac{3RT}{M}}$; $v_{rms}\propto\sqrt{T}$ and $v_{rms}\propto\frac{1}{\sqrt{M}}$ (lighter molecules are faster).
Quick Check — 4 MCQs
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Q1.The kinetic theory expression for the pressure of an ideal gas is:
Explanation: $P=\frac{1}{3}\frac{Nm\,\overline{v^2}}{V}=\frac{1}{3}\rho\,\overline{v^2}$.
Q2.The average translational kinetic energy of a gas molecule at temperature $T$ is:
Explanation: $\frac{1}{2}m\,\overline{v^2}=\frac{3}{2}k_BT$, depending only on temperature.
Q3.The rms speed of gas molecules is given by:
Explanation: $v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3k_BT}{m}}$.
Q4.At the same temperature, which gas has the highest rms speed?
Explanation: $v_{rms}\propto\frac{1}{\sqrt{M}}$; hydrogen has the smallest molar mass, so it is fastest.
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