The Gas Constant R: MDCAT Chemistry notes
The Gas Constant R for MDCAT: values of R in different units, how they are calculated from STP data, the physical meaning of R and why it is universal.
What is R?
$R$ is the proportionality constant in the ideal gas equation $PV=nRT$. It is called the general or universal gas constant because it has the same value for all gases. It does not depend on the nature or molar mass of the gas, nor on the pressure, volume or temperature chosen.
$$R=\frac{PV}{nT}$$
Calculating R from STP data
One mole of an ideal gas at STP ($P=1$ atm, $T=273.15$ K) occupies 22.414 dm³.
In atm dm³ units:
$$R=\frac{1\ \text{atm}\times22.414\ \text{dm}^3}{1\ \text{mol}\times273.15\ \text{K}}=0.0821\ \text{atm dm}^3\ \text{K}^{-1}\text{mol}^{-1}$$
In SI units ($P=101325$ N m$^{-2}$, $V=0.022414$ m³):
$$R=\frac{101325\times0.022414}{1\times273.15}=8.314\ \text{N m K}^{-1}\text{mol}^{-1}=8.314\ \text{J K}^{-1}\text{mol}^{-1}$$
since 1 N m = 1 J.
| Units | Value of R |
|---|---|
| atm dm³ K⁻¹ mol⁻¹ | 0.0821 |
| J K⁻¹ mol⁻¹ (SI) | 8.314 |
| cal K⁻¹ mol⁻¹ | 1.987 (about 2) |
| mm Hg (torr) dm³ K⁻¹ mol⁻¹ | 62.4 |
| mm Hg cm³ K⁻¹ mol⁻¹ | 62400 |
So 0.0821 atm dm³ and 8.314 J represent the same quantity of energy, since 1 atm dm³ = 101.3 J.
Physical meaning of R
$R$ has the units of energy per kelvin per mole. From $PV=nRT$, for 1 mol at constant pressure, raising the temperature by 1 K gives $P\Delta V=R\times1$. So R is the work done by one mole of an ideal gas when it expands against constant pressure on being heated through 1 K. Its value is 8.314 J, i.e. 0.0821 dm³ atm.
Note: this is the expansion work only. The total heat actually supplied to raise 1 mol of gas by 1 K at constant pressure is the molar heat capacity $C_p$, which is larger than $R$ (for a monatomic ideal gas $C_p=\frac{5}{2}R$) because the internal energy of the gas also rises.
Choosing the right value
- Pressure in atm and volume in dm³: use $R=0.0821$.
- Pressure in Pa (N m⁻²) and volume in m³, or energy in joules: use $R=8.314$.
- Pressure in mm Hg and volume in dm³: use $R=62.4$.
Worked example. Pressure of 2 mol of gas in 10 dm³ at 300 K: $P=\frac{2\times0.0821\times300}{10}=4.93$ atm.
Common MDCAT traps
- 0.0821 goes with atm dm³, not J or cal; 8.314 goes with J, not cal.
- R is the same for every gas; it does not depend on molar mass.
- "R at STP" is simply 0.0821 atm dm³ K⁻¹ mol⁻¹; R does not change with conditions.
- 0.821 and 8.324 are distractors.
Quick revision
- $R=0.0821$ atm dm³ K⁻¹ mol⁻¹.
- $R=8.314$ J K⁻¹ mol⁻¹.
- $R\approx2$ cal K⁻¹ mol⁻¹.
- R is universal for all ideal gases.
- R = expansion work of 1 mol gas per kelvin at constant pressure.