Temperature and Mean Kinetic Energy
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You should be familiar with the ideal gas law and the concept of pressure arising from molecular collisions with container walls, as covered in the previous lecture.
Why this matters
Temperature is not just a number on a thermometer; it is a direct measure of the average kinetic energy of particles.
Understanding the link between temperature and molecular motion provides the microscopic foundation for thermodynamics. It explains why heating a gas increases its pressure if volume is constant, why diffusion rates change with temperature, and why different gases at the same temperature have the same average kinetic energy despite having different molecular masses. This kinetic picture transforms temperature from an empirical scale into a physical quantity describing the intensity of random molecular motion.
The Kinetic Picture of Temperature
Imagine a sealed container filled with gas molecules. These molecules are in constant, random motion, bouncing off each other and the walls of the container. In the previous lecture, we established that pressure arises from the momentum transfer when these molecules collide with the walls. Now, consider what happens when we heat this gas. Heating adds energy to the system. In the kinetic model, this added energy does not sit still; it is distributed among the molecules as translational kinetic energy. Consequently, the molecules move faster on average. Faster molecules hit the walls harder and more frequently, increasing the pressure if the volume is fixed. Conversely, if we cool the gas, the molecules slow down, the momentum transfer per collision decreases, and the pressure drops. Thus, temperature is a macroscopic manifestation of the microscopic average speed of the particles. It is not the total energy of the gas, which depends on the number of molecules, but the average energy per molecule. This distinction is crucial: a cup of boiling water has a higher temperature than a swimming pool of warm water, but the pool contains vastly more total thermal energy. The temperature tells us about the intensity of the motion, not the extent of it. This perspective allows us to connect the visible behaviour of gases—expansion, contraction, pressure changes—to the invisible, chaotic dance of their constituent atoms and molecules.
Deriving the Energy-Temperature Relationship
To quantify the relationship between temperature and molecular speed, we start with the result from kinetic theory: \(PV=\tfrac13Nm\langle v^2\rangle\). Here, \(N\) is the number of molecules, \(m\) is the mass of a single molecule, and \(\langle v^2\rangle\) is the mean-square speed. We also know from the ideal gas law that \(PV=NkT\), where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature in kelvins. By equating these two expressions for \(PV\), we can eliminate the macroscopic variables \(P\) and \(V\) to find a direct link between molecular properties and temperature. Setting \(\tfrac13Nm\langle v^2\rangle = NkT\) and simplifying by dividing both sides by \(N\), we obtain \(\tfrac13m\langle v^2\rangle = kT\). Rearranging this gives \(\tfrac12m\langle v^2\rangle = \tfrac32kT\). The left-hand side, \(\tfrac12m\langle v^2\rangle\), is precisely the average translational kinetic energy of a single molecule. This equation reveals that the average kinetic energy is directly proportional to the absolute temperature. The proportionality constant is \(\tfrac32k\), which is independent of the type of gas. This means that at a given temperature, a hydrogen molecule and a xenon atom have the same average kinetic energy, even though their masses and speeds differ significantly. The lighter hydrogen molecule must move much faster to possess the same energy as the heavier xenon atom.
Implications for Molecular Speeds
Since the average kinetic energy is fixed by the temperature, the speed of the molecules is inversely related to the square root of their mass. We define the root-mean-square speed, \(v_{\text{rms}}\), as \(\sqrt{\langle v^2\rangle}\). From the energy equation, we can express this speed as \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\). To make this more practical for chemists, we often work with moles rather than individual molecules. Using the molar mass \(M\) (in \(\mathrm{kg\,mol^{-1}}\)) and the gas constant \(R\), the relationship becomes \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\). This formula highlights two key dependencies. First, \(v_{\text{rms}}\) is proportional to \(\sqrt{T}\). Doubling the absolute temperature does not double the speed; it increases the speed by a factor of \(\sqrt{2}\). Second, \(v_{\text{rms}}\) is inversely proportional to \(\sqrt{M}\). Lighter gases have higher average speeds. For example, at room temperature (\(298\ \mathrm{K}\)), hydrogen gas (\(M \approx 2\ \mathrm{g\,mol^{-1}}\)) has a much higher \(v_{\text{rms}}\) than carbon dioxide (\(M \approx 44\ \mathrm{g\,mol^{-1}}\)). This difference in speed explains why lighter gases diffuse and effuse more rapidly than heavier ones, a phenomenon described by Graham's law. The distribution of speeds is not uniform; while the average energy is fixed, individual molecules have a range of speeds described by the Maxwell-Boltzmann distribution. However, the mean kinetic energy remains a robust predictor of the overall energetic state of the gas.
The result
Kinetic theory of gases
Modelling a gas as a large number of point-like molecules undergoing elastic collisions with a container's walls, the rate of momentum transfer per unit wall area gives \(PV=\tfrac13Nm\langle v^2\rangle\), where \(N\) is the number of molecules, \(m\) their mass, and \(\langle v^2\rangle\) the mean-square speed. Comparing this to the empirically established ideal-gas-law (\(PV=NkT\), with Boltzmann constant \(k=R/N_A\)) identifies temperature directly with average molecular translational kinetic energy: \(\tfrac12m\langle v^2\rangle=\tfrac32kT\) per molecule.
A common misconception is that temperature measures the total kinetic energy of the gas. In reality, temperature measures the average kinetic energy per molecule. A small sample of gas at a high temperature has less total kinetic energy than a large sample at a lower temperature. When comparing two gases at the same temperature, their average kinetic energies are identical, but their root-mean-square speeds differ based on molecular mass. Confusing total energy with average energy leads to incorrect predictions about heat capacity and thermal equilibrium. C
Worked example
Problem. At 300 K, find (a) the average translational kinetic energy of one gas molecule, (b) the total translational kinetic energy of one mole, and (c) how many times faster an \(\mathrm{H_2}\) molecule (2.016 g mol-1) moves than a xenon atom (131.29 g mol-1), in root-mean-square terms.
- (a) \(\langle E_k \rangle = \tfrac32 k_B T = 1.5 \times 1.381 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\ \mathrm{J}\), the same for every gas.
- (b) Per mole: \(\tfrac32 RT = 1.5 \times 8.314 \times 300 = 3741\ \mathrm{J\,mol^{-1}} \approx 3.74\ \mathrm{kJ\,mol^{-1}}\) (equal to \(N_A \times\) the answer to (a)).
- (c) Equal average kinetic energies mean \(M_{\mathrm{H_2}} v_{\mathrm{H_2}}^2 = M_{\mathrm{Xe}} v_{\mathrm{Xe}}^2\), so \(v_{\mathrm{H_2}}/v_{\mathrm{Xe}} = \sqrt{M_{\mathrm{Xe}}/M_{\mathrm{H_2}}}\).
- Substitute: \(\sqrt{131.29/2.016} = \sqrt{65.13} = 8.07\).
- Check with the speeds themselves: \(v_{\mathrm{rms}}(\mathrm{H_2}) = \sqrt{3 \times 8.314 \times 300 / 0.002016} = 1927\ \mathrm{m\,s^{-1}}\) and \(v_{\mathrm{rms}}(\mathrm{Xe}) = 239\ \mathrm{m\,s^{-1}}\); \(1927/239 = 8.06\), the same within rounding.
Units check. \(\mathrm{J\,K^{-1}} \times \mathrm{K} = \mathrm{J}\) per molecule; \(\mathrm{J\,mol^{-1}\,K^{-1}} \times \mathrm{K} = \mathrm{J\,mol^{-1}}\); the speed ratio is a ratio of molar masses under a square root, so it has no units.
Three quick tests
- What is the relationship between the average kinetic energy of a gas molecule and the absolute temperature?
Answer
The average kinetic energy is directly proportional to the absolute temperature: \(\langle E_k \rangle = \tfrac32 kT\). - If the temperature of a gas is doubled, by what factor does the root-mean-square speed increase?
Answer
The speed increases by a factor of \(\sqrt{2}\), because \(v_{\text{rms}} \propto \sqrt{T}\). - Two gases, helium and oxygen, are at the same temperature. Which has the higher average kinetic energy per molecule?
Answer
Both have the same average kinetic energy per molecule, as this depends only on temperature, not molecular mass.
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Lecture notes checked line by line by the editor; every number in the worked example was recomputed.