Why a gas needs three speeds
Kinetic theory treats a gas as a very large number of molecules in incessant random motion, colliding elastically with one another and the walls. A molecule that is fast now slows on its next collision; another speeds up. At any instant the molecules occupy a broad band of speeds, and that band is steady in time even though individual molecules constantly change speed. To describe such a population with a single figure, we must first decide what we mean by "the speed" of the gas.
Three meaningful averages exist, and they are genuinely different numbers because the average of a squared quantity is not the square of the average. NCERT flags this verbatim: $\langle v^2\rangle$ is not equal to $(\langle v\rangle)^2$. The three characteristic speeds capture three distinct questions — what is the typical energy, what is the arithmetic mean speed, and which speed is most common — and each answers its own kind of problem.
RMS speed from the temperature relation
The root mean square speed is the most fundamental of the three because it is bolted directly to temperature. NCERT's kinetic interpretation of temperature gives the average translational kinetic energy of a molecule as
$$\tfrac{1}{2}m\langle v^2\rangle = \tfrac{3}{2}k_B T.$$
Solving for $\langle v^2\rangle$ and taking the square root defines the rms speed:
$$v_{\text{rms}} = \sqrt{\langle v^2\rangle} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3P}{\rho}}.$$
The three forms are interchangeable. The first uses the Boltzmann constant $k_B$ and the mass $m$ of a single molecule; the second uses the gas constant $R$ and the molar mass $M$, because $R = N_A k_B$ and $M = N_A m$; the third drops temperature entirely, since $P = \tfrac{1}{3}\rho\langle v^2\rangle$ rearranges to $\langle v^2\rangle = 3P/\rho$. NCERT computes the rms speed of nitrogen at $300~\text{K}$ as $516~\text{m s}^{-1}$ — comparable to the speed of sound in air, which is no coincidence, since sound propagates through these same molecular collisions.
The three speed formulas at a glance
The average speed $v_{\text{avg}}$ (the arithmetic mean of all molecular speeds) and the most probable speed $v_{\text{mp}}$ (the peak of the distribution) are obtained from the Maxwell–Boltzmann distribution by integration and by differentiation respectively. Both share the rms structure $\sqrt{(\text{factor})\,RT/M}$; only the numerical factor changes. The three speeds are a perfect parallel and are best held in one table.
| Speed | Formula | Numeric factor | What it measures / when used |
|---|---|---|---|
| Most probable, $v_{\text{mp}}$ | $\sqrt{\dfrac{2k_BT}{m}} = \sqrt{\dfrac{2RT}{M}}$ | $\sqrt{2}\approx 1.414$ | Speed at the peak of the distribution — the most common single speed. Smallest of the three. |
| Average (mean), $v_{\text{avg}}$ | $\sqrt{\dfrac{8k_BT}{\pi m}} = \sqrt{\dfrac{8RT}{\pi M}}$ | $\sqrt{8/\pi}\approx 1.596$ | Arithmetic mean of all speeds. Used in mean free path and collision rate. |
| Root mean square, $v_{\text{rms}}$ | $\sqrt{\dfrac{3k_BT}{m}} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3P}{\rho}}$ | $\sqrt{3}\approx 1.732$ | Tied to kinetic energy: $\tfrac{1}{2}m\,v_{\text{rms}}^2 = \tfrac{3}{2}k_BT$. Largest of the three. |
Compare the factors inside the square root, not after rooting, when you want to remember the ordering quickly: $2 < 8/\pi \,(\approx 2.546) < 3$. The most probable speed has the smallest factor, the rms the largest, so the ordering is locked.
The fixed ratio √2 : √(8/π) : √3
Because all three speeds carry the identical $\sqrt{RT/M}$ block, that block cancels when you take ratios. The result is a pure-number ratio that holds for every gas at every temperature:
$$v_{\text{mp}} : v_{\text{avg}} : v_{\text{rms}} = \sqrt{2} : \sqrt{\tfrac{8}{\pi}} : \sqrt{3} = 1 : 1.128 : 1.224.$$
Dividing through by $\sqrt{2}$ gives the memorable form $1 : 1.128 : 1.224$. The rms speed is the largest, the most probable the smallest, with the average squarely between them. This single inequality, $v_{\text{mp}} < v_{\text{avg}} < v_{\text{rms}}$, is the most frequently tested fact in the whole subtopic.
The kinetic energy relation $\tfrac{1}{2}m\langle v^2\rangle = \tfrac{3}{2}k_BT$ that anchors $v_{\text{rms}}$ is derived in kinetic theory of an ideal gas.
The Maxwell–Boltzmann distribution
The three speeds are summary statistics of one underlying curve: the Maxwell–Boltzmann speed distribution. It plots the number of molecules (per unit speed interval) against speed. The curve starts at zero at $v=0$, rises to a single peak, and then falls away with a long tail toward high speeds — it is distinctly asymmetric, not a symmetric bell. The asymmetry is exactly why the three speeds differ: the tail of fast molecules pulls the average and the rms to the right of the peak.
Two features of the curve carry exam weight. First, the peak locates the most probable speed: more molecules have a speed near $v_{\text{mp}}$ than near any other single value. Second, the total area under the curve equals the total number of molecules in the sample — the curve is a tally of how the fixed population is distributed across speeds, so its area is conserved as long as no molecules are added or removed.
How temperature reshapes the curve
Raising the temperature does not change how many molecules there are; it changes how their speeds are spread. At higher $T$ the entire distribution shifts to the right and flattens: the peak moves to a higher speed, the curve becomes broader and lower, and the high-speed tail fattens. Because the molecule count is unchanged, the area stays fixed — the curve trades height for width.
Dependence on temperature and mass
Every characteristic speed obeys the same proportionality, since the factor in front is a constant for each speed:
$$v \propto \sqrt{\frac{T}{M}}.$$
Two consequences follow. Lighter molecules move faster. At a common temperature, a hydrogen molecule ($M = 2$) outruns an oxygen molecule ($M = 32$) by a factor $\sqrt{32/2} = 4$. NCERT uses exactly this fact to explain isotope separation of uranium hexafluoride and why lighter gases drift to the upper atmosphere. Hotter gas moves faster, but only as the square root of temperature. The mass appears as molar mass $M$ when you use $R$, or as molecular mass $m$ when you use $k_B$ — never mix the two within one calculation.
Worked examples
The rms speed of oxygen molecules ($M = 32~\text{g mol}^{-1}$) at $300~\text{K}$ is $v$. Find the most probable speed and the average speed of the same sample, expressed in terms of $v$.
Use the fixed ratio. All three speeds share the $\sqrt{RT/M}$ block, so $v_{\text{mp}} : v_{\text{avg}} : v_{\text{rms}} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3}$.
Most probable: $v_{\text{mp}} = v_{\text{rms}}\sqrt{2/3} = v\sqrt{2/3} \approx 0.816\,v$.
Average: $v_{\text{avg}} = v_{\text{rms}}\sqrt{\tfrac{8/\pi}{3}} = v\sqrt{8/3\pi} \approx 0.921\,v$. As expected, $v_{\text{mp}} < v_{\text{avg}} < v_{\text{rms}}$.
At what temperature is the rms speed of a gas double its value at $27~^{\circ}\text{C}$?
Convert to kelvin. $T_1 = 27 + 273 = 300~\text{K}$. We require $v_2 = 2v_1$.
Apply $v\propto\sqrt{T}$. $\dfrac{v_2}{v_1} = \sqrt{\dfrac{T_2}{T_1}} \Rightarrow 2 = \sqrt{\dfrac{T_2}{300}} \Rightarrow \dfrac{T_2}{300} = 4 \Rightarrow T_2 = 1200~\text{K}$.
Result. The temperature must rise to $1200~\text{K}$ (i.e. $927~^{\circ}\text{C}$) — four times the absolute temperature, not twice. This is the $\sqrt{T}$ trap in numerical form.
Molecular speeds in one breath
- Three speeds, one structure $\sqrt{(\text{factor})RT/M}$: $v_{\text{mp}}=\sqrt{2RT/M}$, $v_{\text{avg}}=\sqrt{8RT/\pi M}$, $v_{\text{rms}}=\sqrt{3RT/M}$.
- $v_{\text{rms}}$ also equals $\sqrt{3k_BT/m}$ and $\sqrt{3P/\rho}$; it is the speed tied to kinetic energy, $\tfrac{1}{2}m\,v_{\text{rms}}^2=\tfrac{3}{2}k_BT$.
- Fixed ratio $v_{\text{mp}}:v_{\text{avg}}:v_{\text{rms}} = \sqrt{2}:\sqrt{8/\pi}:\sqrt{3} = 1:1.128:1.224$. The rms is always largest.
- Maxwell–Boltzmann curve: asymmetric, peak at $v_{\text{mp}}$, long high-speed tail; area = total number of molecules.
- Higher $T$: curve shifts right and flattens; same area. Lower $T$: taller and narrower.
- $v\propto\sqrt{T/M}$: lighter molecules faster; doubling absolute $T$ multiplies speed by $\sqrt{2}$, not 2. Always work in kelvin.