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Universal Gas Constant R: Value, Units and Meaning

Physics · Kinetic Theory · NEET

The universal gas constant R is the constant in the ideal gas equation PV = nRT. Its value is R = 8.31 J/mol/K (also written 8.314 J/mol/K). It is called "universal" because it is the same for every ideal gas, no matter which gas you use. Memory hook: R links pressure-volume (energy) to "one mole, one kelvin" — so its units read as energy per mole per kelvin (J/mol/K).
Universal Gas Constant RPV = nRTideal gas equationR = 8.31 J/mol/KPer MOLER (moles): PV=nRTPer MOLECULEkB = R/N_A = 1.38e-23
R appears in PV = nRT and works per mole (value 8.31 J/mol/K). Divide R by Avogadro's number to get the Boltzmann constant kB, which works per molecule.

Your doubts, answered

What exactly is R and where does it come from?

R is the proportionality constant in the ideal gas law PV = nRT. When you combine Boyle's law, Charles's law and Avogadro's law, PV/T for a fixed amount of gas is constant. For one mole this constant is R. So R just packages that constant into a single number: R = PV/(nT). Its accepted value is 8.31 J/mol/K.

Why is R called "universal"?

Because it has the same value for every ideal gas — hydrogen, oxygen, argon, helium, any of them. Real gases differ in mass and size, but at low pressure they all obey PV = nRT with the SAME R. A quantity that does not change with the gas is called universal. (Compare: the specific gas constant r = R/M is different for each gas because it depends on molar mass M.)

What is the value of R in different units?

R = 8.31 J/mol/K (SI, most used in NEET). R = 8.314 J/mol/K (more precise). R = 2 cal/mol/K (approximately, since 1 cal = 4.18 J). R = 0.0821 atm L/mol/K (when pressure is in atm and volume in litres). R = 0.083 bar L/mol/K (used in NEET 2024). Always match the units of R to the units of P and V in the problem.

How is R related to the Boltzmann constant kB?

They describe the same physics at different scales. R works per mole; kB works per molecule. The link is R = N_A × kB, where N_A = 6.022 × 10^23 is Avogadro's number. So kB = R/N_A = 8.31 / 6.022×10^23 = 1.38 × 10^-23 J/K. Use R with PV = nRT (moles); use kB with average KE = (3/2)kBT (per molecule).

Why are the units of R energy per mole per kelvin?

From R = PV/(nT): P has units N/m^2 and V has units m^3, so PV has units N·m = joule (energy). Dividing by n (mol) and T (K) gives J/mol/K. That is why R carries units of energy, and why RT has units of energy per mole — it appears in energy expressions.

⚠️ The NEET trap
✗ Using R = 8.31 J/mol/K but leaving pressure in atm and volume in litres, or forgetting to convert temperature to kelvin.
✓ Match R to the units in the problem. If P is in atm and V in litres, use R = 0.0821 atm L/mol/K; if P is in bar and V in litres, use R = 0.083 bar L/mol/K; only use 8.31 J/mol/K when P is in pascal (N/m^2) and V is in m^3. Always convert temperature to kelvin (T = t°C + 273).
🧠 NEET often GIVES you the R value to use (e.g. R = 0.083 or R = 100/12) — that is a hint about which units to keep. Use the R they give, do not swap in 8.31.

Real NEET questions

2024

The volume occupied by 1.8 g of water vapour at 374 °C and 1 bar pressure will be: [Use R = 0.083 bar L K⁻¹ mol⁻¹]

A · A. 96.66 L
B · B. 55.87 L
C · C. 3.10 L
D · D. 5.37 L ✓
Solution: Step 1 - moles: n = mass/molar mass = 1.8 g / 18 g/mol = 0.1 mol. Step 2 - temperature in kelvin: T = 374 + 273 = 647 K. Step 3 - apply PV = nRT, so V = nRT/P. Step 4 - substitute with the GIVEN R = 0.083 bar L/mol/K and P = 1 bar: V = (0.1 × 0.083 × 647)/1 = 5.37 L. Answer D. Notice R = 0.083 is used because P is in bar and V in litres.
2025

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn, its pressure drops to 11 atm at 27 °C. The mass of oxygen withdrawn is nearly: [Given R = 100/12 J mol⁻¹ K⁻¹, molar mass of O₂ = 32, 1 atm = 1.01 × 10⁵ N m⁻²]

A · A. 0.116 kg
B · B. 0.156 kg ✓
C · C. 0.125 kg
D · D. 0.144 kg
Solution: Step 1 - final state uses PV = nRT, so final moles n' = PV/(RT). Step 2 - SI units: P = 11 × 1.01×10⁵ = 11.11×10⁵ N/m², V = 30 litre = 30×10⁻³ m³, T = 27 + 273 = 300 K, R = 100/12 J/mol/K. Step 3 - n' = (11.11×10⁵ × 30×10⁻³)/((100/12) × 300) = 33330/2500 ≈ 13.33 mol. Step 4 - moles withdrawn = 18.20 − 13.33 = 4.87 mol. Step 5 - mass = 4.87 × 32 ≈ 156 g = 0.156 kg. Answer B. The odd R = 100/12 is given only to make the arithmetic clean.

Solved Kinetic Theory NEET PYQs

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Frequently asked

What is the value of the universal gas constant R?

R = 8.31 J/mol/K in SI units (more precisely 8.314 J/mol/K). It equals about 2 cal/mol/K, 0.0821 atm L/mol/K, or 0.083 bar L/mol/K depending on the units of pressure and volume.

What are the SI units of R?

Joule per mole per kelvin, written J/mol/K or J mol⁻¹ K⁻¹. This comes from R = PV/(nT): PV has units of energy (joule), divided by mole and kelvin.

Is R the same for all gases?

Yes. The universal gas constant R is identical for every ideal gas. What changes from gas to gas is the specific gas constant r = R/M, which depends on the molar mass M of that gas.

What is the relation between R and Boltzmann constant kB?

R = N_A × kB, where N_A = 6.022 × 10^23 per mole. So kB = R/N_A = 1.38 × 10^-23 J/K. R is per mole; kB is per molecule.

When should I use 8.31 versus 0.0821 for R?

Use 8.31 J/mol/K when pressure is in pascal (N/m²) and volume in m³. Use 0.0821 atm L/mol/K when pressure is in atm and volume in litres. Always match R to the given units, and keep temperature in kelvin.

Related concepts:
Ideal Gas Equation PV=nRT: Meaning and UsesBoltzmann Constant kB: Meaning and ValueAvogadro Number and Mole ConceptMayer's Relation Cp - Cv = RSTP: Standard Temperature and Pressure
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Physics MCQs (free, chapter-wise)Kinetic Theory PYQs