The gas constant R links energy and temperature in PV = nRT, and its value depends on your units: 8.314 J/(mol K), 0.08206 L atm/(mol K), 62.36 L torr/(mol K), 1.987 cal/(mol K), or 8.314 L kPa/(mol K). Pick the R whose pressure and volume units match yours, and always use kelvin for T.
Why One Constant Has Many Values
Beginners often ask which R is “the real one.” All of them are. R is a single physical constant, the same way a distance is the same whether you measure it in miles or kilometers. The number changes because the units change.
R connects the macroscopic world of pressure and volume to the microscopic world of molecular energy. In SI units it is 8.314 joules per mole per kelvin, which tells you that each mole of gas gains 8.314 joules of energy for every kelvin of temperature rise. Chemists usually work in liters and atmospheres instead of cubic meters and pascals, so they use 0.08206. Same constant, different ruler.
This is why unit matching is the whole game. The equation PV = nRT only balances if the units inside R cancel the units on P and V, leaving a clean number. Pick the R whose pressure unit matches your P and whose volume unit matches your V, and the arithmetic takes care of itself.
R Values in Every Common Unit
| R value | Use when pressure is in | Use when volume is in | Common in |
|---|---|---|---|
| 8.314 J / (mol K) | pascals (Pa) | cubic meters (m3) | Physics, SI problems |
| 0.08206 L atm / (mol K) | atmospheres (atm) | liters (L) | General chemistry |
| 62.36 L torr / (mol K) | torr (same as mmHg) | liters (L) | Manometer problems |
| 1.987 cal / (mol K) | Energy-form problems (no P, V) | Thermochemistry | |
| 8.314 L kPa / (mol K) | kilopascals (kPa) | liters (L) | Engineering, metric labs |
Two facts make this table less intimidating than it looks. First, 1 J equals 1 kPa L exactly, which is why the first and last rows share the number 8.314. Second, the chemistry row, 0.08206, handles the large majority of textbook problems.
The Golden Rule: Match R to Your Units
Example A: A problem gives P = 250 kPa and V = 4.0 L at 300 K. The pressure is in kilopascals and the volume in liters, so reach for R = 8.314 L kPa / (mol K):
n = PV / RT = (250 x 4.0) / (8.314 x 300) = 1000 / 2494.2 = 0.401 mol
Example B: A problem gives P = 2.5 atm and V = 1500 mL at 25 C. Two conversions come first: 1500 mL = 1.5 L, and 25 C = 298 K. Now atm and liters pair with R = 0.08206:
n = (2.5 x 1.5) / (0.08206 x 298) = 3.75 / 24.45 = 0.153 mol
If Example B had kept milliliters, the answer would have come out 1000 times too large. Converting mL to L before substituting is a two-second step that saves the whole problem.
Where Does 0.08206 Come From?
The chemistry value of R is not arbitrary. It falls out of a single measurement: at 0 C and 1 atm, one mole of an ideal gas occupies 22.414 L. Rearrange the gas law to R = PV / nT and plug in:
R = (1 atm x 22.414 L) / (1 mol x 273.15 K) = 22.414 / 273.15 = 0.08206 L atm / (mol K)
Every other R value is the same idea in different units. Multiply 0.08206 by 101.325 kPa per atm and you get 8.314 L kPa / (mol K). Multiply by 760 torr per atm and you get 62.36 L torr / (mol K). You can verify any of these by solving a known case, for instance with an ideal gas law calculator set to the matching units, and watching the same physical answer appear.
Why Temperature Must Always Be in Kelvin
Every R value ends in “per kelvin,” and that is not a suggestion. The gas law relates pressure and volume to absolute temperature, the scale on which zero means zero molecular motion.
See what goes wrong otherwise. A gas at 20 C heated to 40 C has not doubled in “hotness”: 20 C is 293 K and 40 C is 313 K, a ratio of 1.07, not 2.0. Plug Celsius straight into PV = nRT and a 0 C sample divides by zero and breaks the universe. Convert first, always: K = C + 273.15.
Fahrenheit needs two steps: K = (F – 32) x 5/9 + 273.15. So 68 F becomes 20 C, which becomes 293.15 K. Tedious by hand, instant in a calculator, and non-negotiable either way.
Quick-Reference: Which R Do I Pick?
- atm and L: 0.08206 (the default for chemistry class)
- kPa and L: 8.314
- Pa and m3: 8.314 (same number, SI form)
- mmHg or torr, with L: 62.36
- psi and L: convert psi to atm first (divide by 14.6959), then use 0.08206
- Energy problems with no P or V: 8.314 J / (mol K) or 1.987 cal / (mol K)
Frequently Asked Questions
What is the value of the gas constant R?
R = 0.08206 L atm / (mol K) for pressure in atmospheres and volume in liters, or 8.314 J / (mol K) in SI units. Other common forms are 62.36 L torr / (mol K) and 1.987 cal / (mol K).
Why are there different values of R?
Because R is measured in units built from pressure and volume, and those come in many systems. It is one constant wearing different unit outfits: 0.08206 L atm / (mol K) and 8.314 J / (mol K) describe the same physics.
Which R value do I use for kPa?
Use R = 8.314 L kPa / (mol K) when pressure is in kilopascals and volume is in liters. Alternatively, divide kPa by 101.325 to get atm and use 0.08206.
Do I use 0.08206 or 8.314?
Use 0.08206 when P is in atm and V is in liters. Use 8.314 when working in SI units (pascals and cubic meters) or with kPa and liters. Match R to your given units.
Is R the same for all gases?
Yes. R is universal, which is why it is called the universal gas constant. The ideal gas law treats every gas identically; differences between real gases only appear when the ideal assumptions break down.
What are the units of R?
Energy per mole per kelvin, written to match the problem: L atm / (mol K), J / (mol K), L torr / (mol K), or cal / (mol K). The “per kelvin” is constant across every form because temperature must always be absolute.
