Maxwell–Boltzmann DistributionJEE Main

Concept

In a gas at temperature $T$, molecules share a distribution of speeds, not one speed. The Maxwell–Boltzmann curve $f(v)$ gives the fraction of molecules near each speed. It rises as $v^2$, peaks at the most probable speed, then falls off exponentially.

Key formula

Derivation

The three characteristic speeds come from the distribution:

- Most probable (peak of $f$): $v_p = \sqrt{\dfrac{2RT}{M}}$ - Mean: $\bar v = \sqrt{\dfrac{8RT}{\pi M}}$ - Root-mean-square: $v_{rms} = \sqrt{\dfrac{3RT}{M}}$

Their fixed ratio is $v_p : \bar v : v_{rms} = \sqrt2 : \sqrt{8/\pi} : \sqrt3 \approx 1 : 1.13 : 1.22$, so $v_p < \bar v < v_{rms}$ always.

Scenarios to explore

• Maxwell–Boltzmann — Molecular speed distribution — vₚ, mean & rms.

Real-world applications

• Explaining evaporation — only the fast tail escapes a liquid.
• Reaction rates: only molecules above an activation speed react.
• Isotope separation by diffusion (lighter molecules are faster).

JEE exam tips

• All three speeds scale as $\sqrt{T/M}$ — double $T$ multiplies every speed by $\sqrt2$.
• Remember the ordering $v_p < \bar v < v_{rms}$ and the ratio $1 : 1.13 : 1.22$.

Common mistakes

• Thinking all molecules move at $v_{rms}$ — it is only one average of a spread.
• Using $M$ in g/mol instead of kg/mol in SI formulas.
• Believing the peak height stays fixed as $T$ rises — area is conserved, so a wider curve is also shorter.

Exam traps to avoid

• The distribution is asymmetric (a long high-speed tail), so the mean lies to the right of the peak.
• $v_{rms} \neq \bar v$ — squaring then averaging weights fast molecules more.