The Gas Constant and Unit Systems

Every gas calculation eventually needs one number: the gas constant $R$ . It links pressure, volume, amount, and temperature in the ideal gas law. The trouble is that $R$ has many faces, because pressure and volume can each be expressed in several unit systems. A student who plugs $R = 0.0821$ into a calculation that needs joules will get a wrong answer by many orders of magnitude. This is not a conceptual error about gases. It is a dimensional bookkeeping error, and it is one of the most common mistakes in introductory chemistry. This article catalogs the numerical values of $R$ in the unit systems you will actually meet. It shows where each value comes from. It builds a habit of dimensional consistency that will serve you in every gas-law, thermodynamics, and kinetic-theory problem that follows.

Key takeaways

$R$ is a single physical constant; its numerical value changes only because our unit choices for $P$ , $V$ , and $n$ change.
The SI value is $R = 8.31446 \ \text{J mol}^{-1}\text{K}^{-1}$ , exact to the precision quoted since the 2019 SI redefinition fixed the Boltzmann constant.
Common alternate values: $0.082057 \ \text{L atm mol}^{-1}\text{K}^{-1}$ , $8.31446 \ \text{L kPa mol}^{-1}\text{K}^{-1}$ , $62.364 \ \text{L Torr mol}^{-1}\text{K}^{-1}$ , and $1.98720 \ \text{cal mol}^{-1}\text{K}^{-1}$ .
$R$ is exactly $N_A k_B$ , tying the macroscopic gas constant directly to the microscopic Boltzmann constant.
Dimensional consistency means every term in an equation must reduce to the same units. Always convert $P$ , $V$ , $T$ before multiplying.
Key idea: $R$ never changes physically; only its numerical clothing changes with the units chosen for pressure and volume, so always match $R$ 's units to your data before calculating.

The core insight

The gas constant $R$ is a fixed conversion factor between the macroscopic thermal energy scale ( $RT$ , in energy units) and the amount of substance (moles), and the many "different values" of $R$ quoted in textbooks are the same physical quantity expressed in different, mutually convertible pressure-volume-energy units.

The physical picture

The ideal gas law, $PV = nRT$ , is a statement of dimensional bookkeeping as much as physics. The product $PV$ has units of energy (pressure times volume is force per area times volume, which is force times length, i.e. energy). So $nRT$ must also carry units of energy. That means $R$ itself must carry units of energy per mole per kelvin. This is the entire reason $R$ has "so many values" in tables: energy can be measured in joules, liter-atmospheres, calories, or erg, and each choice forces a different numerical value of $R$ while the underlying physical constant is untouched. Think of $R$ as a exchange rate between temperature-scaled thermal energy and the mole. Just as £1 and $1.27 are "the same money" expressed in different currencies,$ 8.314 \ \text{J mol}^{-1}\text{K}^{-1} $and$ 0.08206 \ \text{L atm mol}^{-1}\text{K}^{-1} $are the same physical constant expressed in different unit currencies. The deep reason$ R $exists at all is statistical: it is Avogadro's number of Boltzmann's constant,$ R = N_A k_B $. Boltzmann's constant$ k_B $sets the energy scale per molecule ($ k_B T $), and multiplying by$ N_A $converts "per molecule" to "per mole." This connection foreshadows kinetic molecular theory, where$ k_B T$ appears directly in the distribution of molecular speeds and energies.

Figure

Same constant, different unit "currencies," all equal to R

Figure

Dimensional check: PV and nRT must match units term by term

The core relations

PV = nRT

where: $P$ = pressure, $V$ = volume, $n$ = amount in moles, $T$ = absolute temperature in kelvin, $R$ = gas constant matched in units to $P$ and $V$ .

R = N_A k_B

where: $N_A = 6.02214076 \times 10^{23} \ \text{mol}^{-1}$ (Avogadro constant, exact by SI definition), $k_B = 1.380649 \times 10^{-23} \ \text{J K}^{-1}$ (Boltzmann constant, exact by SI definition).

R = 8.31446 \ \text{J mol}^{-1}\text{K}^{-1} = 0.0820573 \ \text{L atm mol}^{-1}\text{K}^{-1} = 62.3637 \ \text{L Torr mol}^{-1}\text{K}^{-1} = 1.98720 \ \text{cal mol}^{-1}\text{K}^{-1}

where: each numerical value follows from converting the same physical constant into the pressure-volume-energy unit combination shown.

Derivation AFrom the SI value of R to the liter·atmosphere value

Goal. Show explicitly how $R = 8.31446 \ \text{J mol}^{-1}\text{K}^{-1}$ becomes $R = 0.0820573 \ \text{L atm mol}^{-1}\text{K}^{-1}$ , using only unit conversions—no new physics.

Step 1. Start from the exact SI definition. Since 2019, $k_B = 1.380649 \times 10^{-23} \ \text{J K}^{-1}$ and $N_A = 6.02214076 \times 10^{23} \ \text{mol}^{-1}$ are both exact defined constants. Their product is

R = (1.380649 \times 10^{-23})(6.02214076 \times 10^{23}) \ \text{J mol}^{-1}\text{K}^{-1} = 8.31446 \ \text{J mol}^{-1}\text{K}^{-1}

Step 2. Recall that 1 J = 1 Pa·m³, so we must convert to liter·atmosphere. We need two conversion factors: 1 atm = 101325 Pa (exact, by definition), and 1 m³ = 1000 L (exact).

Step 3. Convert pascal·cubic meter to atmosphere·liter.

1 \ \text{J} = 1 \ \text{Pa}\,\cdot\,\text{m}^3 = \frac{1}{101325} \ \text{atm} \times 1000 \ \text{L} = \frac{1000}{101325} \ \text{L}\,\cdot\,\text{atm} = 9.86923 \times 10^{-3} \ \text{L}\,\cdot\,\text{atm}

Step 4. Multiply the SI value of R by this conversion factor.

R = 8.31446 \ \text{J mol}^{-1}\text{K}^{-1} \times 9.86923 \times 10^{-3} \ \frac{\text{L}\,\cdot\,\text{atm}}{\text{J}} = 0.0820573 \ \text{L atm mol}^{-1}\text{K}^{-1}

Step 5. Sanity check against Torr. Since 1 atm = 760 Torr exactly, multiply by 760:

0.0820573 \times 760 = 62.3636 \ \text{L Torr mol}^{-1}\text{K}^{-1}

This matches the tabulated value to five significant figures.

Boxed result.

\boxed{R = 0.0820573 \ \text{L atm mol}^{-1}\text{K}^{-1}}

obtained purely by converting units on the exact SI value of $R$ , with no adjustable physics.

Worked examples

Example 1 — Foundational: volume of a mole of gas at STP. Find the molar volume of an ideal gas at $T = 273.15 \ \text{K}$ and $P = 1.000 \ \text{atm}$ . Rearrange the ideal gas law for $V$ :

V = \frac{nRT}{P}

Because pressure is in atmospheres, choose $R = 0.0820573 \ \text{L atm mol}^{-1}\text{K}^{-1}$ so units cancel cleanly. With $n = 1.000 \ \text{mol}$ :

V = \frac{(1.000 \ \text{mol})(0.0820573 \ \text{L atm mol}^{-1}\text{K}^{-1})(273.15 \ \text{K})}{1.000 \ \text{atm}}

V = 22.414 \ \text{L}

Sanity check: this is the textbook "22.4 L per mole at STP" figure, confirming both the arithmetic and the choice of $R$ . Note that IUPAC's current STP (0 °C, 1 bar) gives a slightly different value, 22.711 L, because 1 bar (100000 Pa) is slightly less than 1 atm (101325 Pa)—a reminder that "STP" itself is a unit-system convention, not a universal constant.

Example 2 — Applied: pressure inside a gas cylinder, unit systems collide. A rigid 40.0 L steel cylinder holds 3.20 kg of oxygen gas ( $\text{O}_2$ , molar mass $M = 32.00 \ \text{g mol}^{-1}$ ) at $27.0\ ^\circ\text{C}$ . Find the pressure in kilopascals, and separately in atmospheres, then verify both answers agree.

Step 1. Convert mass to moles.

n = \frac{3200 \ \text{g}}{32.00 \ \text{g mol}^{-1}} = 100.0 \ \text{mol}

Step 2. Convert temperature to kelvin.

T = 27.0 + 273.15 = 300.15 \ \text{K}

Step 3. Compute pressure in SI-compatible units (kPa, using L and kPa forms of R). Since $R = 8.31446 \ \text{J mol}^{-1}\text{K}^{-1} = 8.31446 \ \text{L kPa mol}^{-1}\text{K}^{-1}$ (because 1 J = 1 L·kPa exactly, as 1 L = $10^{-3}$ m³ and 1 kPa = $10^3$ Pa, and their product of scale factors is 1):

P = \frac{nRT}{V} = \frac{(100.0 \ \text{mol})(8.31446 \ \text{L kPa mol}^{-1}\text{K}^{-1})(300.15 \ \text{K})}{40.0 \ \text{L}}

P = \frac{249{,}615 \ \text{L}\,\cdot\,\text{kPa}}{40.0 \ \text{L}} = 6240 \ \text{kPa}

Step 4. Recompute in atmospheres directly, as an independent check.

P = \frac{(100.0 \ \text{mol})(0.0820573 \ \text{L atm mol}^{-1}\text{K}^{-1})(300.15 \ \text{K})}{40.0 \ \text{L}}

P = \frac{246.35 \ \text{L}\,\cdot\,\text{atm}}{40.0 \ \text{L}} = 61.6 \ \text{atm}

Step 5. Cross-check the two answers agree. Convert 6240 kPa to atm: $6240 \ \text{kPa} \times \frac{1 \ \text{atm}}{101.325 \ \text{kPa}} = 61.6 \ \text{atm}$ . The two independent unit paths agree to three significant figures, confirming the calculation is dimensionally sound. Both values are physically reasonable for a compressed gas cylinder (tens of atmospheres, not thousands or fractions).

Limiting cases & common traps

Situation	What goes wrong	Fix
Pressure given in atm, using $R = 8.314$	Answer is off by a factor of ~101	Convert P to Pa, or switch to $R = 0.08206 \ \text{L atm mol}^{-1}\text{K}^{-1}$
Temperature left in Celsius	Nonsensical or negative results	Always convert to kelvin: $T(\text{K}) = T(^\circ\text{C}) + 273.15$
Volume left in mL or cm³	Off by factor of 1000	Convert to liters (or m³) to match chosen R
Using cal-based R in a joule equation	Energy answer off by factor of 4.184	Match $R$ 's energy unit to the rest of the equation (1 cal = 4.184 J exactly)
Confusing STP conventions (1 atm vs 1 bar)	Molar volume quoted as 22.4 L vs 22.7 L, both "right"	State which STP convention is being used

The chemist behind it

The gas constant is a joint legacy of Émile Clapeyron, who first combined Boyle's and Gay-Lussac's laws into a single equation of state in 1834, and Dmitri Mendeleev, who in 1874 gave the constant its now-familiar symbol and generalized form $PV = RT$ per mole. But the deeper meaning of $R$ —as Avogadro's number of Boltzmann's constant—was only clarified once the molecular nature of gases was firmly established by kinetic theory in the late nineteenth century, linking a macroscopic measured quantity to a microscopic per-molecule energy scale.

Origins · 1 minLudwig Boltzmann and how this was discovered

Ludwig Boltzmann

1844–1906 · Statistical mechanics pioneer whose constant k_B, when multiplied by Avogadro's number, gives exactly R, unifying the macroscopic gas constant with molecular kinetic energy.

Ask the tutor: Want to see how R and $k_BT$ reappear when we derive the pressure of a gas from molecular collisions in kinetic molecular theory?

Why is R sometimes written with more or fewer significant figures?

The 2019 SI redefinition fixed $k_B$ and $N_A$ as exact numbers, making $R = 8.314462618...$ J mol⁻¹ K⁻¹ an exact (not measured) constant, in principle knowable to infinite precision. In practice, textbooks round to 8.314 or 8.3145 J mol⁻¹ K⁻¹ for everyday use. More digits appear in precision thermodynamic calculations; fewer are fine for typical stoichiometry and gas-law problems where other data limit your precision anyway.

How do I know which unit version of R to use in a problem?

Look at the units given for pressure and volume in the problem. If pressure is in atmospheres and volume in liters, use $R = 0.08206$ L atm mol⁻¹ K⁻¹. If everything is SI (pascals, cubic meters), use $R = 8.314$ J mol⁻¹ K⁻¹. The fastest check is dimensional: after multiplying, the units on both sides of $PV = nRT$ must match exactly, with no leftover conversion factors.

Is R the same constant that appears in the Arrhenius equation and Nernst equation?

Yes. The same $R$ appears throughout physical chemistry because it always plays the same role: converting a per-molecule energy scale ( $k_B T$ ) into a per-mole energy scale ( $RT$ ). In the Arrhenius equation, $RT$ sets the thermal energy scale against which the activation energy is compared. In the Nernst equation, $RT/F$ sets the voltage scale for electrochemical equilibria. It is the identical universal constant every time.

Why does STP give 22.4 L per mole in some books and 22.7 L in others?

This reflects two different official definitions of "standard conditions," not two different gas laws. The older, still widely taught convention uses 1 atm (101.325 kPa) and 0 °C, giving 22.414 L/mol. IUPAC's current recommended standard uses 1 bar (100 kPa) exactly, giving 22.711 L/mol. Both are internally consistent; always state which convention a problem assumes, especially in gas-density calculations.

Does R ever fail, i.e., break down for real gases?

The constant $R$ itself never fails—it is exact. What breaks down is the ideal gas law's assumption that molecules have no volume and no intermolecular attractions. For real gases, especially at high pressure or low temperature, you replace $PV = nRT$ with a corrected equation of state (such as van der Waals), but $R$ remains the same constant inside that corrected equation. The physics that fails is the "ideal" approximation, not the value of $R$ .