R = 8.314 J K
^{-1} mol
^{-1} is the value in SI units. It works for:
- P in Pa, V in m^{3}, n in mol, T, in K
- P in kPa, V in L = dm^{3}, n in mol, T in K
This is the standard value which relates to
R = 0.08206 L atm K
^{-1} mol
^{-1} works for:
- P in atm, V in L, n in mol, T in K
If given a volume in some other unit (like mL or cm
^{3}), convert to L or m
^{3} as appropriate. If given a pressure in some other unit (ie. 747 mmHg), convert to a suitable pressure unit and apply above...
If given a temperature not in K, convert to K. The ideal gas law and underlying theory
requires a thermodynamic temperature with a zero that is physically meaningful and corresponding to absolute zero. You cannot have a temperature in Celsius and use PV = nRT because Charles' Law tells us that V is proportional to T only if T = 0 corresponds to the theoretical V = 0 limit for a gas... in other words, if T is in Celsius then
.
The ideal gas law can be modified to uses masses in kg rather than chemical amount in moles, but this also means the equation becomes
with the above R values, or a change to PV = mRT with R a gas constant with units of g or kg. (Note: SI would say mas must be in kg, but since we seek moles and M is usually in g mol
^{-1}, using m in g is usually appropriate. As with all cases, keeping track of units means making sure that they suit the situation rather than applying blanket rules blindly.)
You can have an R in bar or other units, but it is generally easier to change to P to atm or kPa or Pa and use the above Rs.
Addendum / Note: A link above gives R = 8.314 L kPa K
^{-1} mol
^{-1}. This is the same as R = 8.314 J K
^{-1} mol
^{-1} as, first:
- 1 J = 1 N m
- 1 m^{3} Pa = 1 ^{3} N m^{-2} = 1 N m
- So, 1 J = 1 m^{3} Pa
Now, looking at the units of R:
- 1 L = 1 dm^{3} = (0.1 m)^{3} = 10^{-3} m^{3}
- 1 kPa = 10^{3} Pa
- So, 1 L kPa = 10^{-3} . 10^{3} m^{3} Pa = 1 m^{3} Pa = 1 J
- And thus, 1 L kPa K^{-1} mol^{-1} = 1 J K^{-1} mol^{-1}
This is why I don't bother to list R with two sets of units (and nor do most data sheets) and just recognise that R will work with either the combination (P in kPa, V in L / dm
^{3}) or the SI option (P in Pa, V in m
^{3}).