Particle motion and temperature

In the molecular-kinetic model, matter consists of atoms and molecules that move continuously and chaotically, and temperature is the macroscopic measure of how intense that motion is.

For an ideal gas, the average kinetic energy of translational motion of one particle is directly proportional to the absolute temperature:

$$\langle E_k\rangle = \frac{3}{2}k_B T,$$

where $k_B = 1.38 \times 10^{-23}\ \text{J/K}$ and $T$ is expressed in kelvins.

It follows that at the same temperature the average kinetic energies of molecules of any gases are equal, but heavier molecules move more slowly:

$$v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}.$$

The speeds of individual particles are not identical — they follow the Maxwell distribution, so we speak of the average energy rather than of the speed of each molecule. This chaotic motion explains Brownian motion, diffusion and gas pressure on the walls: when heated, particles move faster, their impacts on the walls become stronger and more frequent, and diffusion speeds up.

The temperature of an ideal gas is $T = 300$ K. Find the average kinetic energy of translational motion of one molecule. Use $k_B = 1.38 \times 10^{-23}\ \text{J/K}$.

Calculate the root-mean-square speed of nitrogen molecules $\text{N}_2$ at $T = 300$ K. The molar mass of nitrogen is $M = 28\ \text{g/mol}$, the gas constant is $R = 8.31\ \text{J/(mol·K)}$.

A vessel at $T = 300$ K contains a mixture of hydrogen ($M_1 = 2\ \text{g/mol}$) and oxygen ($M_2 = 32\ \text{g/mol}$). (a) Compare the average kinetic energies of the molecules of these gases. (b) How many times is the root-mean-square speed of hydrogen molecules greater than that of oxygen? (c) How will the $v_{\mathrm{rms}}$ of oxygen change if the temperature is raised to $600$ K?

Self-study plan with a simulation (safe: no heating, no chemicals).

1. Open the PhET "Gas Properties" simulation and choose the constant-volume mode.
2. Add a small number of molecules and record the initial temperature $T$ and the qualitative picture of motion ("slow", "fast").
3. Add heat and note what happens to the speed distribution: the peak shifts to the right and becomes lower, while the average energy grows.
4. Check numerically: at $T = 300$ K compute $\langle E_k\rangle = \frac{3}{2}k_B T$ and compare its order of magnitude with the simulation readings.
5. Switch to a heavier gas (for example, from helium to argon) at the same temperature and note that the average energy stays the same while the speeds become smaller.

Everyday observation (safe). Put a drop of ink into a glass of cold water and into a glass of warm water and compare how quickly the colour spreads: this is a direct demonstration of how diffusion depends on the speed of thermal motion. Do not use boiling water and do not leave the experiment unattended.

Show that temperature sets the average kinetic energy of particles, while the speeds of individual particles are spread over a wide range (the Maxwell distribution).

Sketch, two parts. Above — a vessel with particles: each has a velocity arrow of different length and direction, with the label $\langle E_k\rangle = \frac{3}{2}k_B T$. Below — a Maxwell distribution graph: $v$ on the horizontal axis, fraction of molecules on the vertical axis; two curves for $T_1 < T_2$, where the higher-temperature peak is shifted to the right and lower, while the area under both curves is the same.

In a cool garage at $15\ ^\circ\text{C}$ ($288$ K) a tyre was inflated to $2.2\times10^5$ Pa. After an hour on a summer motorway at $100$ km/h the air inside had warmed to $57\ ^\circ\text{C}$ ($330$ K) and the gauge read $2.5\times10^5$ Pa. The tyre volume and the number of molecules inside stayed practically the same.

Temperature measures the mean kinetic energy of the translational motion of the molecules: $\bar{E}_k=\frac{3}{2}k_BT$. For nitrogen molecules ($m=28$ u$,=4.65\times10^{-26}$ kg) the root-mean-square speed $\sqrt{\overline{v^2}}=\sqrt{3k_BT/m}$ is $506$ m/s at $288$ K and $542$ m/s at $330$ K. Faster molecules strike the wall more often and harder, so the pressure rises: with $V$ and $N$ fixed, $p\propto T$, giving $2.2\times10^5\cdot\frac{330}{288}=2.5\times10^5$ Pa — exactly what the driver reads.

A kettle is filled with $1.5$ L of water at $20\ ^\circ\text{C}$ ($293$ K) and brought to the boil at $100\ ^\circ\text{C}$ ($373$ K). The temperature in kelvin rises by a factor $373/293=1.27$. The mean speed of the water molecules goes up from $637$ to $719$ m/s — only $13\%$.

The mean kinetic energy of a single water molecule grows as $\bar{E}_k=\frac{3}{2}k_BT$, from $6.1\times10^{-21}$ J at $293$ K to $7.7\times10^{-21}$ J at $373$ K, a factor of $1.27$. The speed, by contrast, is proportional to $\sqrt{T}$ and therefore changes more slowly than the energy. This also shows why the count must start from absolute zero: a 'doubled' Celsius temperature ($50\to100$) does not double the particle energy, whereas $293\ \text{K}\to\ 586$ K does exactly that.

A ball is pumped up in a warm changing room at $20\ ^\circ\text{C}$ ($293$ K) to a pressure of $1.6\times10^5$ Pa. After an hour of play outside at $2\ ^\circ\text{C}$ ($275$ K) the ball feels noticeably softer, and at half-time the gauge reads $1.5\times10^5$ Pa, even though no air has escaped.

Cooling lowers the mean kinetic energy of the air molecules: $\bar{E}_k=\frac{3}{2}k_BT$ falls by a factor $275/293=0.94$, and the root-mean-square speed of nitrogen molecules drops from $506$ to $494$ m/s. The molecules hit the casing less often and less hard, and with volume and particle number unchanged the pressure falls in proportion to the absolute temperature: $1.6\times10^5\cdot\frac{275}{293}=1.5\times10^5$ Pa. The number of wall impacts per second, and hence the pressure force, is set by temperature, not by 'how much air' is inside.