Effect of Temperature and Mass on the Distribution
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You should be comfortable with the Maxwell–Boltzmann speed distribution function and the definitions of the most probable, mean, and root-mean-square speeds. Familiarity with the ideal gas law and basic calculus (differentiation and integration) is assumed.
Why this matters
Understanding how temperature and molecular mass reshape the speed distribution is key to predicting reaction rates and diffusion.
In the previous lecture, we derived the Maxwell–Boltzmann distribution, which describes how molecular speeds are spread out in a gas at equilibrium. However, that derivation assumed a fixed temperature and molecular mass. In reality, gases are heated, cooled, or swapped for different chemical species. This lecture explores how changing the temperature \(T\) or the molecular mass \(m\) alters the shape of the distribution curve. Specifically, we will see how the peak shifts, how the curve broadens or narrows, and how the characteristic speeds scale with these parameters. This is crucial for understanding why lighter gases diffuse faster and why heating a gas increases the fraction of molecules with enough energy to react.
Intuitive Picture: Spreading and Shifting
Imagine a crowd of people walking at different speeds. If everyone is the same weight but the "energy" (temperature) increases, they all walk faster. The average speed goes up, and the range of speeds becomes wider. Some walk very slowly, some very fast, but the centre of the crowd moves to higher speeds. Now imagine the energy stays the same, but the people become heavier (higher mass). Heavier people move slower. The average speed drops, and the range of speeds becomes narrower. The crowd is more tightly packed around a lower speed. In a gas, temperature is a measure of the average kinetic energy. Higher temperature means more kinetic energy, which translates directly to higher speeds. Mass is a measure of inertia. Higher mass means more resistance to acceleration, so for the same kinetic energy, the speed must be lower. Therefore, increasing temperature shifts the entire distribution to the right (higher speeds) and broadens it. Increasing mass shifts the distribution to the left (lower speeds) and narrows it. The total area under the curve always remains constant because the total number of molecules does not change. This conservation of area is a critical constraint: as the curve gets taller and narrower (high mass, low temp), or shorter and wider (low mass, high temp), the integral of the function over all speeds must equal the total number density \(n\). This interplay between height, width, and position is the core of this lecture.
Effect of Temperature on the Distribution
Let us examine how the distribution function \(f(v)\) changes when temperature \(T\) increases, keeping mass \(m\) constant. The distribution is given by \(f(v)=4\pi n\left(\tfrac{m}{2\pi kT}\right)^{3/2}v^2e^{-mv^2/2kT}\). There are two temperature-dependent parts: the pre-exponential factor \(\left(\tfrac{m}{2\pi kT}\right)^{3/2}\) and the exponential term \(e^{-mv^2/2kT}\). As \(T\) increases, the pre-exponential factor decreases because \(T\) is in the denominator. This means the peak height of the distribution decreases. Simultaneously, the exponent \(-mv^2/2kT\) becomes less negative (closer to zero) for any given \(v\). This causes the exponential decay to happen more slowly, meaning the tail of the distribution extends further to the right. The result is a curve that is shorter and broader. The most probable speed \(v_p=\sqrt{\frac{2kT}{m}}\) increases with the square root of temperature. If you double the temperature, \(v_p\) increases by a factor of \(\sqrt{2} \approx 1.414\). The mean speed \(\langle v\rangle=\sqrt{\frac{8kT}{\pi m}}\) and the root-mean-square speed \(v_{\text{rms}}=\sqrt{\frac{3kT}{m}}\) also increase by the same factor. This scaling is universal for all characteristic speeds. The broadening of the distribution is particularly important for chemical kinetics. At higher temperatures, a significantly larger fraction of molecules have speeds (and thus kinetic energies) above a certain threshold. This exponential sensitivity explains why reaction rates increase dramatically with temperature, even though the average speed only increases by the square root.
Effect of Molecular Mass on the Distribution
Now consider the effect of changing the molecular mass \(m\) while keeping temperature \(T\) constant. This is relevant when comparing different gases at the same temperature, such as helium versus xenon. The pre-exponential factor \(\left(\tfrac{m}{2\pi kT}\right)^{3/2}\) increases as \(m\) increases. This means the peak height of the distribution increases. The exponential term \(e^{-mv^2/2kT}\) becomes more negative for any given \(v\), causing the distribution to decay much more rapidly. The result is a curve that is taller and narrower. The most probable speed \(v_p=\sqrt{\frac{2kT}{m}}\) decreases with the square root of mass. If you increase the mass by a factor of 4, \(v_p\) decreases by a factor of 2. The mean and rms speeds follow the same inverse square root dependence. For example, helium (\(m \approx 4\ \mathrm{u}\)) has a much higher average speed than xenon (\(m \approx 131\ \mathrm{u}\)) at the same temperature. The ratio of their most probable speeds is \(\sqrt{131/4} \approx 5.7\). This means helium molecules move about 5.7 times faster than xenon molecules on average. This mass dependence is the physical basis for Graham's Law of Effusion, which states that the rate of effusion is inversely proportional to the square root of the molar mass. Lighter gases escape through small pores faster because their molecules move faster. The narrower distribution for heavy gases means that the speeds are more uniform, while the broader distribution for light gases means there is a wider spread of speeds, including a significant tail of very fast molecules.
Comparing Different Gases and Temperatures
To fully appreciate the effects, we must compare distributions for different combinations of \(T\) and \(m\). Consider two gases, A and B, at the same temperature. If \(m_A < m_B\), then gas A has a higher \(v_p\), a higher \(\langle v\rangle\), and a higher \(v_{\text{rms}}\). The distribution for A is shifted to the right and is broader. Now consider the same gas A at two different temperatures, \(T_1\) and \(T_2\), with \(T_2 > T_1\). The distribution at \(T_2\) is shifted to the right and is broader than at \(T_1\). A common point of confusion is comparing a light gas at a low temperature with a heavy gas at a high temperature. For instance, compare helium at \(300\ \mathrm{K}\) with xenon at \(1000\ \mathrm{K}\). The speed scales as \(\sqrt{T/m}\). For helium, \(T/m \propto 300/4 = 75\). For xenon, \(T/m \propto 1000/131 \approx 7.6\). The ratio of speeds is \(\sqrt{75/7.6} \approx 3.1\). So helium is still faster, despite being at a much lower temperature. This highlights that mass is a dominant factor in determining speed. To make the speeds equal, the temperature of the heavier gas must be much higher. Specifically, \(T_B/T_A = m_B/m_A\). This relationship is useful in designing experiments where specific kinetic energies are required. The area under the curve remains \(n\) in all cases, ensuring that the total number of molecules is conserved. The shape changes, but the total population does not. This conservation is a powerful check when sketching distributions: if you shift the peak to the right, you must lower the peak height to keep the area constant.
The result
The distribution of molecular speeds
kinetic-theory-pressure found the mean-square molecular speed but not how speeds are actually distributed among individual molecules. Treating each velocity component as an independent, Boltzmann-weighted Gaussian and converting to spherical (speed) coordinates gives \(f(v)=4\pi n\left(\tfrac{m}{2\pi kT}\right)^{3/2}v^2e^{-mv^2/2kT}\), from which three distinct characteristic speeds emerge in the fixed ratio \(v_p:\langle v\rangle:v_{\text{rms}}=\sqrt2:\sqrt{8/\pi}:\sqrt3\approx1:1.128:1.225\) — the most probable speed \(v_p\), the mean speed \(\langle v\rangle\), and the root-mean-square speed \(v_{\text{rms}}\) (matching kinetic-theory-pressure's result exactly).
A common misconception is that the most probable speed \(v_p\) is the same as the average speed \(\langle v\rangle\). They are not. The distribution is asymmetric (skewed to the right), so the mean is greater than the most probable value. The ratio \(\langle v\rangle/v_p = \sqrt{4/\pi} \approx 1.128\). This difference is significant when calculating fluxes or reaction rates, as the mean speed is often the relevant quantity for transport phenomena, while \(v_p\) is useful for identifying the peak of the distribution. Confusing these two can lead to errors of over 10% in quantitative predictions. C
Worked example
Problem. Helium (4.003 g mol-1) is at 300 K. (a) Find its most probable speed. (b) To what temperature must argon (39.95 g mol-1) be heated to have the same most probable speed? (c) At a common temperature, how do the peak heights of the two speed distributions compare?
- (a) \(v_p(\mathrm{He}) = \sqrt{2RT/M} = \sqrt{2 \times 8.314 \times 300 / 0.004003} = 1116\ \mathrm{m\,s^{-1}}\).
- (b) Equal \(v_p\) means equal \(T/M\): \(T_{\mathrm{Ar}} = 300 \times \dfrac{39.95}{4.003}\).
- So \(T_{\mathrm{Ar}} = 2994\ \mathrm{K}\), nearly ten times hotter.
- (c) The peak value of \(f(v)\) is \(\dfrac{4}{\sqrt{\pi}} \dfrac{1}{e\,v_p}\) (per molecule), so it is inversely proportional to \(v_p\). At the same \(T\), \(v_p \propto 1/\sqrt{M}\), so the argon peak is \(\sqrt{39.95/4.003} = 3.16\) times taller than the helium peak.
- The areas stay equal (both curves describe all of the molecules), so the taller argon curve must also be narrower, by the same factor of 3.16.
Units check. Temperatures scale by a ratio of molar masses (a pure number); \(f(v)\) has units of s m-1, so its peak scales as \(1/v_p\).
Three quick tests
- If the temperature of a gas is quadrupled, by what factor does the most probable speed increase?
Answer
It increases by a factor of \(\sqrt{4} = 2\). - Which gas has a higher mean speed at the same temperature: \(\mathrm{H_2}\) or \(\mathrm{O_2}\)?
Answer
\(\mathrm{H_2}\), because it has a lower molar mass. - Does the area under the Maxwell–Boltzmann distribution curve change when temperature increases?
Answer
No, the area remains constant and equal to the total number density \(n\).
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.