The Boltzmann constant kB = 1.38 x 10^-23 J/K is the gas constant R shared out over one single molecule, so kB = R / NA. It converts absolute temperature into energy per molecule: the average translational kinetic energy of one gas molecule is (3/2) kBT. Memory hook: R is the "per mole" money, kB is the "per molecule" coin you get after dividing by Avogadro's number NA.
The Boltzmann constant kB is the universal gas constant R divided by Avogadro's number NA. R describes one mole; dividing by NA scales it down to a single molecule, giving kB = 1.38 x 10^-23 J/K, which appears in PV = N kBT and average KE = (3/2) kBT.
Your doubts, answered
Is the Boltzmann constant the same as the universal gas constant R?
No. R = 8.314 J/mol/K works for one mole of gas (about 6.022 x 10^23 molecules). kB works for a single molecule. They are linked by kB = R / NA. Numerically 8.314 / (6.022 x 10^23) = 1.38 x 10^-23 J/K. So use R when you count in moles (PV = nRT) and use kB when you count individual molecules (PV = N kBT).
What is the exact value and unit of the Boltzmann constant?
kB = 1.38 x 10^-23 J/K (joule per kelvin). The unit J/K makes sense because kB multiplies a temperature (in kelvin) to give an energy (in joule): kBT has units of joule. NCERT gives this same SI value 1.38 x 10^-23 J K^-1.
Why is average kinetic energy of a molecule (3/2) kBT and not (3/2) RT?
(3/2) RT is the total translational kinetic energy of one whole mole of gas. To get the energy of just ONE molecule you divide by NA, and R/NA = kB. So per molecule the average translational KE = (3/2) kBT. Do not mix them: RT is per mole, kBT is per molecule.
How is the Boltzmann constant different from the Stefan-Boltzmann constant?
They are two totally different constants that share the name Boltzmann. The Boltzmann constant kB = 1.38 x 10^-23 J/K links temperature to molecular energy. The Stefan-Boltzmann constant sigma = 5.67 x 10^-8 W/m^2/K^4 appears in radiated power E = sigma T^4 (heat radiation). Different symbol, different value, different unit.
Where does kB come from in the ideal gas law?
Start from PV = nRT. Since n = N/NA (N = number of molecules) and R = NA kB, substitute to get PV = (N/NA)(NA kB)T = N kBT. This microscopic form PV = N kBT, or P = n kBT with n = number density, is exactly how NCERT introduces kB.
⚠️ The NEET trap ✗ Average kinetic energy of one molecule = (3/2) RT, using R = 8.314 J/mol/K. ✓ Average KE of one molecule = (3/2) kBT with kB = 1.38 x 10^-23 J/K. R is per mole; you must divide by NA to reach one molecule, and R/NA = kB. Using R here overshoots the answer by a factor of NA. 🧠 Per mole vs per molecule mix-up
Real NEET questions
2020
The average thermal energy for a mono-atomic gas is: (kB is the Boltzmann constant and T the absolute temperature)
A · (5/2) kB T
B · (7/2) kB T
C · (1/2) kB T
D · (3/2) kB T ✓
Solution: Step 1: Use the law of equipartition of energy. Each active degree of freedom carries average energy (1/2) kBT per molecule. Step 2: A monatomic gas (like He, Ar) has only 3 translational degrees of freedom, so f = 3. Step 3: Average thermal energy per molecule = f x (1/2) kBT = 3 x (1/2) kBT = (3/2) kBT. Note kB appears (not R) because this is per molecule. Answer: D.
2018
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (mass of an oxygen molecule m = 2.76 x 10^-26 kg, Boltzmann constant kB = 1.38 x 10^-23 J/K, escape speed = 11200 m/s)
A · 5.016 x 10^4 K
B · 8.360 x 10^4 K ✓
C · 2.508 x 10^4 K
D · 1.254 x 10^4 K
Solution: Step 1: The molecule escapes when v_rms = v_escape. rms speed uses kB: v_rms = sqrt(3 kB T / m). Step 2: Set sqrt(3 kB T / m) = v_escape and square: 3 kB T / m = v_escape^2. Step 3: Solve for T: T = m x v_escape^2 / (3 kB). Step 4: Substitute: T = (2.76 x 10^-26 x (11200)^2) / (3 x 1.38 x 10^-23). Numerator = 2.76 x 10^-26 x 1.2544 x 10^8 = 3.462 x 10^-18. Denominator = 4.14 x 10^-23. T = 3.462 x 10^-18 / 4.14 x 10^-23 = 8.36 x 10^4 K. Answer: B.
2016
A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. Which of the following gives the density of the gas? (k = Boltzmann constant)
A · P/(kT)
B · Pm/(kT) ✓
C · P/(kTV)
D · mkT
Solution: Step 1: Use the microscopic gas law PV = N kBT, where N is the number of molecules. Step 2: Number density n = N/V = P/(kBT). Step 3: Density = mass per unit volume = n x m = m P/(kBT). This uses kB (not R) because we count individual molecules. Dimensionally P/(kBT) is per m^3 and multiplying by m (mass per molecule) gives kg per m^3. Answer: B: Pm/(kT).
Solved Kinetic Theory NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the amount of energy, per molecule, that goes with each degree of temperature. Its value is kB = 1.38 x 10^-23 J/K. It is the bridge between temperature (which you feel) and molecular energy (which you cannot see).
What is the formula linking kB, R and NA?
kB = R / NA. With R = 8.314 J/mol/K and NA = 6.022 x 10^23 per mole, this gives kB = 1.38 x 10^-23 J/K.
What is the unit of the Boltzmann constant?
Joule per kelvin (J/K), written J K^-1 in SI. This is the same as an energy divided by a temperature.
Does the average kinetic energy of a molecule depend on the gas type?
No. The average translational kinetic energy is (3/2) kBT for every ideal gas, whether monatomic, diatomic or polyatomic. It depends only on temperature, not on which gas it is.
Why do we need kB when we already have R?
R is convenient for moles and lab-scale problems (PV = nRT). kB is needed when a problem talks about ONE molecule, such as average kinetic energy (3/2) kBT or rms speed sqrt(3 kB T / m).
