absolute temperature and mean molecular kinetic energy

The molecular-kinetic theory describes a gas as a set of molecules moving chaotically, and in this picture temperature stops being just a number on a thermometer: it measures the mean kinetic energy of the translational motion of molecules. The absolute temperature $T$ is measured from absolute zero and expressed in kelvin; its relation to the Celsius scale is $T = t + 273$ (more precisely, $273{,}15$). Experiment shows that the mean kinetic energy of a molecule is directly proportional to the absolute temperature:

$$\bar{E}_k = \frac{3}{2}kT,$$

where $k = 1{,}38 \cdot 10^{-23}$ J/K is the Boltzmann constant. The factor $3/2$ appears because a molecule has three degrees of freedom of translational motion, and each of them carries the energy $\frac{1}{2}kT$. It is important that $\bar{E}_k$ depends only on $T$ and does not depend on the mass of the molecule or on which gas it is: at the same temperature helium and oxygen molecules have the same mean kinetic energy, although their speeds are different. At $T = 0$ K the mean kinetic energy of translational motion vanishes; this is a limiting, unreachable temperature.

A gas molecule is at a temperature $t = 27$ °C. Find the mean kinetic energy of its translational motion. The Boltzmann constant is $k = 1{,}38 \cdot 10^{-23}$ J/K.

The mean kinetic energy of the molecules of a certain gas is $1{,}0 \cdot 10^{-20}$ J. Determine the absolute temperature of the gas and express it in degrees Celsius. $k = 1{,}38 \cdot 10^{-23}$ J/K.

Oxygen ($M = 32$ g/mol) is at $T = 300$ K. Find the root-mean-square speed of its molecules. Compare it with the root-mean-square speed of helium molecules ($M = 4$ g/mol) at the same temperature, and compare the mean kinetic energies of these gases. $k = 1{,}38 \cdot 10^{-23}$ J/K, $N_A = 6{,}02 \cdot 10^{23}$ mol$^{-1}$.

Self-study plan with a simulation (heating a real gas in a closed vessel is hazardous, so we work only in a model).

1. Open the "Gas Properties" simulation (PhET, tab with temperature and the speed distribution). Fix the number of particles and choose one gas, for example neon.
2. Set successively $T = 200$, $400$ and $800$ K. At each temperature record the average particle speed from the simulation scale.
3. Compute the mean kinetic energy with $\bar{E}_k = \frac{3}{2}kT$ for each temperature and fill in a table: $T$, $v_{avg}$, $\bar{E}_k$.
4. Plot $\bar{E}_k$ against $T$. Check that the points lie on a straight line through the origin, and that doubling $T$ increases the average speed only by a factor $\sqrt{2} \approx 1{,}41$.
5. Change the gas (for example, to argon) at the same temperature and verify that the mean kinetic energy is unchanged while the average speed changes. State the conclusion in one sentence.

Show that the mean kinetic energy of molecules is directly proportional to the absolute temperature, is the same for different gases at the same temperature, and vanishes at $T = 0$ K.

Graph: the horizontal axis is the absolute temperature $T$ in kelvin, the vertical axis is the mean kinetic energy $\bar{E}_k$. The line passes through the origin with slope $\frac{3}{2}k$; two coinciding points for helium and oxygen at $T = 300$ K are marked on it ($\bar{E}_k = 6{,}2 \cdot 10^{-21}$ J). Beside it are sketches of the molecular speed distribution (Maxwell curves) at $T_1$ and $T_2 > T_1$: at the higher temperature the curve is lower and wider, while the area under both curves is the same.

The ball was pumped up outdoors at $-10\ ^\circ\text{C}$, that is at $263\ \text{K}$, and carried into a changing room at $+25\ ^\circ\text{C}$ ($298\ \text{K}$). If the volume does not change, the gauge shows the pressure rising by a factor of $\frac{298}{263}\approx1{,}13$, about $13\ \%$. The mean kinetic energy of an air molecule grows from $\bar E_k=\frac{3}{2}k_BT\approx5{,}44\times10^{-21}\ \text{J}$ to $\approx6{,}17\times10^{-21}\ \text{J}$.

The relation $\bar E_k=\frac{3}{2}k_BT$ is at work here: absolute temperature is proportional to the mean kinetic energy of the molecules, so going from $263\ \text{K}$ to $298\ \text{K}$ raises the mean energy of one molecule by exactly the same factor by which the pressure rises at constant volume.

In the engine cylinder the mixture burns, and the gases leave through the exhaust pipe at about $700\ ^\circ\text{C}$, that is $973\ \text{K}$. A nitrogen molecule $N_2$ has a mass of $28\ \text{u}=4{,}65\times10^{-26}\ \text{kg}$, a mean energy $\bar E_k=\frac{3}{2}k_BT\approx2{,}0\times10^{-20}\ \text{J}$ and an rms speed $v_{rms}=\sqrt{3k_BT/m}\approx930\ \text{m/s}$, which is $2{,}7$ times the speed of sound in air.

The scenario uses both the formula $\bar E_k=\frac{3}{2}k_BT$ and its consequence $\bar E_k=\frac{1}{2}m\overline{v^2}$, hence $v_{rms}=\sqrt{3k_BT/m}$: temperature fixes the mean energy, and the molecular mass converts that energy into a speed.

In a saucepan at normal pressure, water boils at $100\ ^\circ\text{C}=373\ \text{K}$. A water molecule has a mass of $18\ \text{u}=2{,}99\times10^{-26}\ \text{kg}$, and in the vapour above the surface its mean kinetic energy is $\bar E_k=\frac{3}{2}k_BT\approx7{,}7\times10^{-21}\ \text{J}$, corresponding to $v_{rms}\approx720\ \text{m/s}$. At the same temperature a light helium molecule ($4\ \text{u}$) would have $v_{rms}\approx1500\ \text{m/s}$ at exactly the same energy.

The formula $\bar E_k=\frac{3}{2}k_BT$ links the boiling temperature to the mean energy of thermal motion: at $373\ \text{K}$ this energy is $\frac{373}{273}\approx1{,}37$ times larger than at $0\ ^\circ\text{C}$. Only molecules with an energy above the average can leave the liquid, which is why evaporation proceeds even at room temperature while boiling occurs at a strictly defined one.