Interpretation of Temperature in Kinetic Theory

From pressure to energy

The kinetic theory expression for pressure can be rewritten so that the average kinetic energy of a molecule appears explicitly. Multiplying $P = \dfrac{1}{3}\dfrac{N}{V}m\overline{v^{2}}$ by $V$,

$$PV = \dfrac{1}{3}Nm\overline{v^{2}} = \dfrac{2}{3}N\left(\dfrac{1}{2}m\overline{v^{2}}\right) \qquad ....(1)$$

The quantity $\dfrac{1}{2}m\overline{v^{2}}$ is the average translational kinetic energy of a molecule. In an ideal gas the molecules are non-interacting, so there is no potential energy term, and the internal energy of an ideal gas is purely kinetic. The average total energy is therefore

$$E = N\left(\dfrac{1}{2}m\overline{v^{2}}\right)$$

and equation (1) becomes

$$PV = \dfrac{2}{3}E \qquad ....(2)$$

Temperature as a measure of molecular kinetic energy

Using the ideal gas equation $PV = Nk_{B}T$ in equation (2),

$$Nk_{B}T = \dfrac{2}{3}E \qquad \Rightarrow \qquad E = \dfrac{3}{2}Nk_{B}T \qquad \text{or} \qquad \dfrac{E}{N} = \dfrac{3}{2}k_{B}T$$

This means that the average energy per molecule is proportional to the absolute temperature $T$ of the gas, and is the same for every gas at a given temperature. It is this equation that relates the macroscopic parameter $T$ to the kinetic energy of a molecule, and it is the kinetic-theory interpretation of temperature.

Equivalently, $\dfrac{1}{2}m\overline{v^{2}} = \dfrac{3}{2}k_{B}T$.

Why a gas heats up when heat is supplied

When heat is supplied to a gas, the energy goes into increasing the average translational kinetic energy of its molecules, since an ideal gas has no potential energy. Because $\dfrac{1}{2}m\overline{v^{2}} = \dfrac{3}{2}k_{B}T$, an increase in the average kinetic energy is precisely an increase in the absolute temperature. Hence the temperature of the gas rises.

Deduction of Boyle's law from kinetic theory

Statement of Boyle's law: at constant temperature, the pressure exerted by a fixed mass of a gas is inversely proportional to the volume occupied by the gas.

Let $P$ be the pressure, $V$ the volume, $N$ the number of molecules and $m$ the mass of each molecule, so the total mass of the gas is $M = mN$.

i. From kinetic theory, $P = \dfrac{1}{3}\dfrac{mN}{V}\overline{v^{2}}$.

ii. This can be written as $P = \dfrac{2}{3}\dfrac{N}{V}\left(\dfrac{1}{2}m\overline{v^{2}}\right)$.

iii. But $\dfrac{1}{2}m\overline{v^{2}}$ is the average kinetic energy of a molecule, which is constant for all gases at a given temperature; and $N$ is constant for a given mass of the gas.

$$\therefore P = \dfrac{\text{constant}}{V} \qquad \Rightarrow \qquad P \propto \dfrac{1}{V}$$

Hence, at constant temperature, the pressure of the gas increases if its volume is reduced. This is Boyle's law, and it is also the kinetic-theory explanation of how the pressure of a gas changes when its volume is reduced at constant temperature: the molecules have to travel a shorter distance between collisions with the walls, so they strike the walls more often and the rate of momentum transfer per unit area increases.

Kinetic energy per unit volume

From equation (1), $PV = \dfrac{2}{3}N\left(\dfrac{1}{2}m\overline{v^{2}}\right)$, and the total energy of the gas is $E = N\left(\dfrac{1}{2}m\overline{v^{2}}\right)$. Substituting,

$$PV = \dfrac{2}{3}E$$

Hence the kinetic energy per unit volume of the gas is

$$\dfrac{E}{V} = \dfrac{3}{2}P$$

which the board writes simply as $E = \dfrac{3}{2}P$, $E$ here meaning the energy per unit volume.

The forms of the energy used in numericals

• Per molecule: $\dfrac{3}{2}k_{B}T$.

• Per mole (or per kilomole, with $R$ in J kmol$^{-1}$ K$^{-1}$): $\dfrac{3}{2}RT$.

• Per unit mass: $\dfrac{3}{2}\dfrac{RT}{M_{0}}$.

• For a mass $M$ of gas: $\dfrac{3}{2}\dfrac{M}{M_{0}}RT$.

• Per unit volume: $\dfrac{3}{2}P$.

Since the kinetic energy is proportional to $T$ in every one of these forms, ratio problems reduce to $\dfrac{(\text{K.E.})_{2}}{(\text{K.E.})_{1}} = \dfrac{T_{2}}{T_{1}}$ for a fixed mass of a given gas.

Standard results

$$PV = \dfrac{2}{3}E, \qquad E = \dfrac{3}{2}Nk_{B}T, \qquad \dfrac{E}{N} = \dfrac{3}{2}k_{B}T, \qquad \dfrac{1}{2}m\overline{v^{2}} = \dfrac{3}{2}k_{B}T$$

$$\text{K.E. per unit volume} = \dfrac{3}{2}P, \qquad P \propto \dfrac{1}{V} \ \text{at constant } T$$