N_A = 6.022 × 10²³ molecules/mol. Avogadro's law: equal volumes of gases at same T and P contain equal numbers of molecules. Universal gas constant R = N_A k.
-- NCERT, p. 3Avogadro Number
Lesson
Avogadro's number, Nₐ = 6.022 × 10²³ mol⁻¹, is the bridge between microscopic molecular quantities and macroscopic molar quantities. It tells you how many molecules (or atoms, ions, or particles) sit in exactly one mole of any substance.
Where it comes from in NCERT: Class 11 Physics Chapter 13 (Kinetic Theory), page 3 states this as a fundamental physical constant linking the gas constant R and the Boltzmann constant k through R = Nₐk. This relation is how the ideal gas law PV = nRT converts to PV = NkT, where N is the actual number of molecules.
The key relation to internalise:
Nₐ = R / k = 8.314 J mol⁻¹ K⁻¹ / 1.381 × 10⁻²³ J K⁻¹ ≈ 6.022 × 10²³ mol⁻¹
This means:
- n (moles) × Nₐ = N (number of molecules)
- Any per-mole quantity divided by Nₐ gives the per-molecule quantity (and vice versa)
Common confusions in NEET context:
-
R vs k mix-up. R is per mole; k is per molecule. When a problem gives molecular-level data, use k. When it gives molar data, use R. Swapping them changes the answer by a factor of ~10²³.
-
Forgetting the mole–molecule conversion. Problems sometimes give mass and ask for the number of molecules. You must go mass → moles (using molar mass M) → molecules (multiplying by Nₐ). Skipping the mole step is a direct route to a wrong option.
-
Confusing Nₐ with N. Nₐ is a fixed constant. N is the actual number of molecules in a given sample, which depends on how many moles you have: N = nNₐ.
Watch-out: When a NEET stem says "number of molecules in 2 moles," the answer is 2 × 6.022 × 10²³ = 1.204 × 10²⁴ — not 6.022 × 10²³. Read whether the problem asks per mole or for the given sample.
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Avogadro's number (Nₐ) represents:
The Boltzmann constant k is related to Avogadro's number Nₐ and the gas constant R by:
The numerical value of Avogadro's number is approximately:
How many molecules are present in 3.0 moles of an ideal gas?
The gas constant R = 8.314 J mol⁻¹ K⁻¹ and Avogadro's number Nₐ = 6.022 × 10²³ mol⁻¹. The Boltzmann constant k is:
An ideal gas sample contains N = 1.204 × 10²⁴ molecules. The number of moles in the sample is:
A container holds n moles of an ideal gas at pressure P and temperature T. If the gas constant is R and Avogadro's number is Nₐ, which expression correctly gives the number of molecules per unit volume?
Two ideal gas samples — one containing helium (M = 4 g/mol) and the other nitrogen (M = 28 g/mol) — each have exactly one mole at the same temperature. Which statement about the number of molecules in each sample is correct?
Quick recall before you leave
Worked Example
- 1
Given
- R = 8.314 J mol⁻¹ K⁻¹ - k = 1.381 × 10⁻²³ J K⁻¹ - n = 5.0 mol
- 2
Required
(a) Avogadro's number Nₐ (b) Total molecules N in 5.0 moles
- 3
Concept
Avogadro's number bridges the molar scale (R) and the molecular scale (k). From R = Nₐk, we extract Nₐ. Then N = nNₐ converts moles to molecules.
- 4
Formula
(a) Nₐ = R / k (b) N = nNₐ
- 5
Substitution
(a) Nₐ = 8.314 / 1.381 × 10⁻²³ (b) N = 5.0 × Nₐ
- 6
Calculation
(a) Nₐ = (8.314 / 1.381) × 10²³ = 6.021 × 10²³ mol⁻¹ (Rounding: 8.314 / 1.381 = 6.0203… ≈ 6.021 to 4 significant figures, matching the precision of the given data.) (b) N = 5.0 × 6.022 × 10²³ = 3.011 × 10²⁴ Note on exact values: the factor 5.0 is a given counting quantity in this problem context and does not limit significant figures. The answer precision is governed by Nₐ (4 sig figs).
- 7
Final answer
(a) **Nₐ ≈ 6.022 × 10²³ mol⁻¹** (b) **N = 3.011 × 10²⁴ molecules**
- 8
Common trap
Confusing R and k: using k in place of R (or vice versa) changes the answer by ~10²³. If you get a Nₐ on the order of 10⁰ or 10⁴⁶, you have swapped R and k.
- 9
Similar NEET-style question
"The Boltzmann constant is 1.38 × 10⁻²³ J K⁻¹ and R = 8.31 J mol⁻¹ K⁻¹. How many molecules are present in 0.50 moles of oxygen at STP?" (Answer: N = 0.50 × (8.31/1.38 × 10⁻²³) ≈ 0.50 × 6.02 × 10²³ = 3.01 × 10²³.) ---
Before solving, remember these
Formulas
5 formulas — click to collapse
Average translational KE per molecule
Microscopic interpretation of temperature: T is direct measure of average translational kinetic energy.
| Symbol | Quantity | SI Unit |
|---|---|---|
| k | Boltzmann constant | J/K |
| T | absolute temperature | K |
Valid when
- Translational degrees of freedom only
- Ideal gas
Cv from degrees of freedom
Each quadratic DoF contributes (1/2)R to molar Cv. Mono: f=3, Cv=3R/2; di-rigid: f=5, Cv=5R/2; poly-rigid: f=6, Cv=3R.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Cv | molar specific heat | J/mol/K |
| f | degrees of freedom | - |
| R | gas constant | J/mol/K |
Valid when
- Equipartition holds (temperature high enough)
- Quadratic energy modes
Ideal gas equation
Fundamental equation of state of ideal gas relating pressure, volume, temperature.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | pressure | Pa |
| V | volume | m^3 |
| n | moles | mol |
| R | 8.314 | J/mol/K |
| N | molecule count | - |
| k | Boltzmann 1.38e-23 | J/K |
| T | temp | K |
Valid when
- Gas obeys ideal gas approximation (low pressure, high temperature relative to phase transitions)
Mean free path of gas molecule
Average distance between successive molecular collisions.
| Symbol | Quantity | SI Unit |
|---|---|---|
| lambda | mean free path | m |
| n | number density | 1/m^3 |
| d | molecular diameter | m |
Valid when
- Hard-sphere model
- Equilibrium gas
RMS speed of gas molecules
Root-mean-square molecular speed; depends on T and molar mass M (or molecular mass m).
| Symbol | Quantity | SI Unit |
|---|---|---|
| R | gas constant | J/mol/K |
| T | temp | K |
| M | molar mass | kg/mol |
| k | Boltzmann | J/K |
| m | molecular mass | kg |
Valid when
- Ideal gas
- Maxwell-Boltzmann distribution
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student treats v_rms ∝ T instead of √T. Doubling T does NOT double v_rms; it multiplies by √2.
When it triggers
Question asks for new v_rms after T change.
How to avoid
v_rms = √(3RT/M). v_rms ∝ √T. To double v_rms, T must quadruple.
Root cause: formula misuse
Correction
v_rms = √(3RT/M), so v_rms ∝ √T. To double v_rms, T must quadruple (factor of 4). Common error: assume doubling T doubles v_rms.
Past Year Questions
6 questions from NEET 2020, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse
How NEET usually asks this
4 recurring patterns from past papers — click to collapse
Average translational KE per molecule = (3/2)kT (monoatomic). Total via DoF.
Common distractors
uses 3 2 RT not 3 2 kT
Confuses per-mole RT with per-molecule kT
PV = nRT. Given two of (P, V, T, n) and changes, find missing.
Common distractors
forgets temperature conversion
Mixes °C with K
Mean free path scaling with n, d. lambda = 1/(sqrt(2)*pi*n*d^2).
Common distractors
ignores d squared
Treats d linearly
Find new T given v_rms scales by factor k. v_rms ∝ sqrt(T), so T_new = T*k^2.
Common distractors
uses linear scaling
Treats v_rms ∝ T not sqrt(T)
Test yourself on this topic with real past-paper questions:
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