Molecular Speeds: RMS, Average, and Most Probable
Real-Life Hook
Ever wondered why helium balloons rise while air-filled ones don’t? It’s all about molecular speeds! Helium atoms, being lighter, move at average speeds of about 1,350 m/s at room temperature - that’s faster than the speed of sound (343 m/s)! This is why helium changes your voice when you inhale it - sound waves travel faster through the lighter, faster-moving helium molecules. Similarly, this is why hydrogen is the fastest-escaping gas from Earth’s atmosphere, while heavier gases like CO₂ stay trapped. Understanding molecular speeds also explains why refrigerators work - slower-moving molecules mean lower temperature!
Core Concepts
Why Three Different Speeds?
In a gas, molecules don’t all move at the same speed. The Maxwell-Boltzmann distribution shows that molecules have a wide range of speeds, from nearly zero to very high values. We use three statistical measures:
- Most Probable Speed ($v_p$): The speed at which maximum molecules are found
- Average Speed ($\overline{v}$): The arithmetic mean of all molecular speeds
- Root Mean Square Speed ($v_{rms}$): The square root of the mean of squared speeds
Maxwell-Boltzmann Distribution
The distribution of molecular speeds follows:
Maxwell-Boltzmann Distribution Function
$$f(v) = 4\pi n \left(\frac{m}{2\pi kT}\right)^{3/2} v^2 e^{-\frac{mv^2}{2kT}}$$Where:
- $f(v)$ = number of molecules with speed $v$
- $n$ = total number of molecules
- $m$ = mass of one molecule
- $k$ = Boltzmann constant
- $T$ = absolute temperature
Key Features:
- Asymmetric curve (not Gaussian)
- Peak shifts right with increasing temperature
- Area under curve = total number of molecules
- No molecules at v = 0 (v² term) or v → ∞ (exponential decay)
The Three Speeds: Formulas and Derivations
1. Most Probable Speed ($v_p$)
The speed at which the distribution function is maximum (peak of the curve).
Most Probable Speed
$$v_p = \sqrt{\frac{2kT}{m}} = \sqrt{\frac{2RT}{M}}$$Where:
- $k$ = Boltzmann constant = 1.38 × 10⁻²³ J/K
- $R$ = Gas constant = 8.314 J/(mol·K)
- $m$ = mass of one molecule (kg)
- $M$ = molar mass (kg/mol)
Derivation: Found by setting $\frac{df(v)}{dv} = 0$
Physical Meaning: The speed at which you’ll find the maximum number of molecules.
2. Average Speed ($\overline{v}$)
The arithmetic mean of all molecular speeds.
Average (Mean) Speed
$$\overline{v} = \sqrt{\frac{8kT}{\pi m}} = \sqrt{\frac{8RT}{\pi M}}$$Or:
$$\overline{v} = \sqrt{\frac{8}{\pi}} \sqrt{\frac{RT}{M}} \approx 1.596\sqrt{\frac{RT}{M}}$$
Derivation: $\overline{v} = \frac{\int_0^\infty v \cdot f(v) \, dv}{\int_0^\infty f(v) \, dv}$
Physical Meaning: The average speed if you measured every single molecule.
3. Root Mean Square Speed ($v_{rms}$)
The square root of the average of squared speeds.
RMS Speed
$$v_{rms} = \sqrt{\overline{v^2}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$$
Derivation: From kinetic theory, $\frac{1}{2}m\overline{v^2} = \frac{3}{2}kT$
Physical Meaning: Related to average kinetic energy; used in kinetic energy calculations.
Comparing the Three Speeds
Speed Ratio
$$v_p : \overline{v} : v_{rms} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$$ $$v_p : \overline{v} : v_{rms} = 1 : 1.128 : 1.224$$Or approximately: 1 : 1.13 : 1.22
Important Relation:
$$v_p < \overline{v} < v_{rms}$$Always in this order!
Memory Tricks
“2-8-3 for Pi-Pie-Pie” Mnemonic
- $v_p$ has 2 (and no π): $v_p = \sqrt{\frac{2RT}{M}}$
- $\overline{v}$ has 8/π: $\overline{v} = \sqrt{\frac{8RT}{\pi M}}$
- $v_{rms}$ has 3 (and no π): $v_{rms} = \sqrt{\frac{3RT}{M}}$
“PAR” Order
- Probable is smallest
- Average is middle
- RMS is largest
Speed Ratio Trick
Remember: 1 : 1.13 : 1.22
- Or think: 1, then add ~0.13, then add ~0.09 more
- For quick estimates: 10 : 11.3 : 12.2 (multiply by 10)
Temperature Dependence
All three speeds are proportional to $\sqrt{T}$
- “Speed ↑ when Temperature ↑, but as square root!”
- Double temperature → speeds increase by $\sqrt{2} \approx 1.414$ times
Molar Mass Dependence
All three speeds are proportional to $\frac{1}{\sqrt{M}}$
- “Lighter gases move faster!”
- $\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{32}{2}} = 4$ (H₂ is 4× faster than O₂)
Common Unit Conversion Mistakes
Molar Mass Units
Critical Error: Using g/mol instead of kg/mol
- Correct: For O₂, $M = 32 \times 10^{-3}$ kg/mol
- Wrong: $M = 32$ kg/mol (gives answer 1000× too small!)
Quick Check: If your speed comes out ~1000 m/s for room temperature, you’re correct. If it’s ~1 m/s, you forgot to convert!
Temperature Units
Always use Kelvin, never Celsius!
- Room temperature: 300 K (not 25 K!)
- STP: 273 K (not 0 K!)
Gas Constant Values
Choose the right R:
- $R = 8.314$ J/(mol·K) when using SI units
- $R = 0.0821$ L·atm/(mol·K) for gas law problems (not for speed calculations!)
Common Calculation Errors
Forgetting the square root: $v_{rms} = \sqrt{\frac{3RT}{M}}$ not $\frac{3RT}{M}$
Using molecular mass instead of molar mass:
- With k, use m (mass of one molecule)
- With R, use M (molar mass)
Speed ratio confusion: $v_p$ is smallest, not largest!
Important Formulas Summary
Speed Formulas
| Speed Type | Formula (with k) | Formula (with R) | Numerical Factor |
|---|---|---|---|
| Most Probable | $\sqrt{\frac{2kT}{m}}$ | $\sqrt{\frac{2RT}{M}}$ | $\sqrt{2} \approx 1.414$ |
| Average | $\sqrt{\frac{8kT}{\pi m}}$ | $\sqrt{\frac{8RT}{\pi M}}$ | $\sqrt{\frac{8}{\pi}} \approx 1.596$ |
| RMS | $\sqrt{\frac{3kT}{m}}$ | $\sqrt{\frac{3RT}{M}}$ | $\sqrt{3} \approx 1.732$ |
Temperature Dependence
At constant molar mass:
$$v \propto \sqrt{T}$$ $$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$$Molar Mass Dependence
At constant temperature:
$$v \propto \frac{1}{\sqrt{M}}$$ $$\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}}$$Energy Relations
Average kinetic energy per molecule:
$$\overline{KE} = \frac{1}{2}m\overline{v^2} = \frac{1}{2}mv_{rms}^2 = \frac{3}{2}kT$$Note: Use $v_{rms}$ for energy calculations, not $v_p$ or $\overline{v}$!
Interactive Demo: Visualize Molecular Speeds
Explore how molecular speeds vary with temperature and molar mass in this interactive simulation.
3-Level Practice Problems
Level 1: JEE Main Basics
Problem 1: Calculate the rms speed of oxygen molecules at 27°C. (Molar mass of O₂ = 32 g/mol, R = 8.314 J/(mol·K))
Solution
Given:
- $T = 27 + 273 = 300$ K
- $M = 32 \times 10^{-3}$ kg/mol (convert to kg!)
Answer: 484 m/s (approximately)
Quick Check: ~500 m/s for O₂ at room temperature is reasonable!
Problem 2: If the rms speed of H₂ molecules at a certain temperature is 1920 m/s, find the rms speed of O₂ molecules at the same temperature. (M(H₂) = 2 g/mol, M(O₂) = 32 g/mol)
Solution
At the same temperature:
$$\frac{v_{rms}(O_2)}{v_{rms}(H_2)} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$$ $$v_{rms}(O_2) = \frac{1}{4} \times 1920 = 480 \text{ m/s}$$Answer: 480 m/s
Level 2: JEE Main Advanced
Problem 3: Calculate all three speeds (most probable, average, and rms) for nitrogen molecules at 127°C. (M(N₂) = 28 g/mol, R = 8.314 J/(mol·K))
Solution
Given:
- $T = 127 + 273 = 400$ K
- $M = 28 \times 10^{-3}$ kg/mol
Most Probable Speed:
$$v_p = \sqrt{\frac{2RT}{M}} = \sqrt{\frac{2 \times 8.314 \times 400}{28 \times 10^{-3}}}$$ $$v_p = \sqrt{\frac{6651.2}{0.028}} = \sqrt{237542.86} = 487.5 \text{ m/s}$$Average Speed:
$$\overline{v} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8}{\pi}} \times v_p \times \sqrt{\frac{M_{v_p}}{M_{v_p}}}$$Or directly:
$$\overline{v} = \sqrt{\frac{8 \times 8.314 \times 400}{\pi \times 0.028}} = \sqrt{\frac{26604.8}{0.0880}} = 550.1 \text{ m/s}$$RMS Speed:
$$v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 400}{0.028}}$$ $$v_{rms} = \sqrt{\frac{9976.8}{0.028}} = 596.3 \text{ m/s}$$Verification of ratio: $487.5 : 550.1 : 596.3 = 1 : 1.128 : 1.223$ ✓
Answer:
- $v_p = 488$ m/s
- $\overline{v} = 550$ m/s
- $v_{rms} = 596$ m/s
Problem 4: At what temperature will the average speed of O₂ molecules be equal to the rms speed of H₂ molecules at 300 K? (M(H₂) = 2 g/mol, M(O₂) = 32 g/mol)
Solution
For H₂ at 300 K:
$$v_{rms}(H_2) = \sqrt{\frac{3RT_1}{M_{H_2}}}$$For O₂ at T₂:
$$\overline{v}(O_2) = \sqrt{\frac{8RT_2}{\pi M_{O_2}}}$$Given: $\overline{v}(O_2) = v_{rms}(H_2)$
$$\sqrt{\frac{8RT_2}{\pi M_{O_2}}} = \sqrt{\frac{3RT_1}{M_{H_2}}}$$Squaring both sides:
$$\frac{8RT_2}{\pi M_{O_2}} = \frac{3RT_1}{M_{H_2}}$$ $$T_2 = \frac{3\pi M_{O_2} T_1}{8 M_{H_2}} = \frac{3 \times \pi \times 32 \times 300}{8 \times 2}$$ $$T_2 = \frac{28800\pi}{16} = 1800\pi = 5654.9 \text{ K}$$Answer: 5655 K (or about 5382°C)
Level 3: JEE Advanced
Problem 5: A vessel contains a mixture of 7 g of N₂ and 11 g of CO₂ at 300 K. Calculate: (a) The average kinetic energy per molecule (b) The rms speed of the mixture (M(N₂) = 28 g/mol, M(CO₂) = 44 g/mol, k = 1.38 × 10⁻²³ J/K, R = 8.314 J/(mol·K))
Solution
(a) Average KE per molecule:
The average KE per molecule depends only on temperature:
$$\overline{KE} = \frac{3}{2}kT = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300$$ $$\overline{KE} = 6.21 \times 10^{-21} \text{ J}$$This is the same for all molecules regardless of their mass!
(b) RMS speed of mixture:
Number of moles:
- $n_{N_2} = \frac{7}{28} = 0.25$ mol
- $n_{CO_2} = \frac{11}{44} = 0.25$ mol
- Total: $n = 0.5$ mol
For a mixture, we need to find the average molar mass:
$$M_{avg} = \frac{m_{total}}{n_{total}} = \frac{7 + 11}{0.5} = \frac{18}{0.5} = 36 \text{ g/mol} = 0.036 \text{ kg/mol}$$ $$v_{rms} = \sqrt{\frac{3RT}{M_{avg}}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.036}}$$ $$v_{rms} = \sqrt{\frac{7482.6}{0.036}} = \sqrt{207850} = 455.9 \text{ m/s}$$Answer:
- (a) $6.21 \times 10^{-21}$ J
- (b) 456 m/s
Problem 6: Show that the ratio of average speed to rms speed is independent of temperature and molar mass. Calculate this ratio.
Solution
Average speed:
$$\overline{v} = \sqrt{\frac{8RT}{\pi M}}$$RMS speed:
$$v_{rms} = \sqrt{\frac{3RT}{M}}$$Taking the ratio:
$$\frac{\overline{v}}{v_{rms}} = \frac{\sqrt{\frac{8RT}{\pi M}}}{\sqrt{\frac{3RT}{M}}}$$ $$= \sqrt{\frac{8RT}{\pi M} \times \frac{M}{3RT}}$$ $$= \sqrt{\frac{8}{\pi} \times \frac{1}{3}} = \sqrt{\frac{8}{3\pi}}$$Note that T and M cancel out! The ratio is a constant.
$$\frac{\overline{v}}{v_{rms}} = \sqrt{\frac{8}{3\pi}} = \sqrt{\frac{8}{9.42}} = \sqrt{0.849} = 0.922$$Or: $\frac{\overline{v}}{v_{rms}} = \frac{\sqrt{8/\pi}}{\sqrt{3}} = \frac{1.596}{1.732} = 0.922$
Answer: $\frac{\overline{v}}{v_{rms}} = \sqrt{\frac{8}{3\pi}} \approx 0.922$ (independent of T and M)
Proved
Effect of Temperature on Distribution
Temperature Increase Effects:
- Peak shifts right: Most probable speed increases
- Peak height decreases: Same number of molecules spread over wider range
- Curve broadens: Greater variation in speeds
- Area remains constant: Total number of molecules unchanged
Mathematical Expression:
- At $T_2 = 4T_1$: All speeds double ($v \propto \sqrt{T}$)
- Peak shifts from $v_p$ to $2v_p$
- Peak height reduces by factor of 2 (curve stretches horizontally)
Cross-Links to Other Topics
Related to Kinetic Theory
- Kinetic Theory Assumptions - Foundation of molecular speeds
- Ideal Gas Equation - Macroscopic relations
- Mean Free Path - Uses average speed
Related to Thermodynamics
- Internal Energy - Related to rms speed
- Temperature Scale - Why we use Kelvin
- Specific Heat - Energy storage modes
Applications
- Graham’s law of effusion: Rate ∝ 1/√M
- Sound speed in gases: Related to molecular speeds
- Atmospheric escape velocity: Light gases escape Earth’s gravity
Key Takeaways
- Three speeds exist: Most probable < Average < RMS (always in this order)
- Speed ratio is constant: 1 : 1.128 : 1.224 (independent of T and M)
- Temperature dependence: All speeds ∝ √T
- Molar mass dependence: All speeds ∝ 1/√M (lighter gases move faster)
- Use $v_{rms}$ for energy: $\overline{KE} = \frac{1}{2}mv_{rms}^2$
- Use $\overline{v}$ for collisions: Mean free path and collision frequency
Exam Tips
- JEE Main: Focus on calculating individual speeds, speed ratios, and temperature/mass comparisons
- JEE Advanced: Expect mixture problems, Maxwell distribution concepts, and derivation-based questions
- Common trap: Using $v_p$ or $\overline{v}$ for KE calculations (must use $v_{rms}$!)
- Unit check: Speeds should be ~500 m/s for common gases at room temperature
- Quick ratio: Remember 1:1.13:1.22 for instant comparisons
- Conceptual clarity: Understand why $v_{rms} > \overline{v} > v_p$ from distribution shape
Last updated: February 22, 2025