# How can I calculate the ideal gas law equations?

May 15, 2014

There are two versions of the ideal gas equation:
Molar version: $p V = n R T$ (n is no. of moles & R is molar gas constant).
Molecular version: $p V = N k T$ (N is no. of molecules & k is Boltzmann constant – i.e molecular gas constant).

Where the pressure - P, is in atmospheres (atm)* the volume - V, is in liters (L) the moles -n, are in moles (mol) and Temperature -T is in Kelvin (K) as in all gas law calculations.

*NB SI units for pressure is Pa and for volume is ${m}^{3}$.

The value and unit of molar gas constant, $R$, is derived from equation $P V = n R T$.

When we do the algebraic reconfiguration we end up with Pressure and Volume being decided by moles and Temperature, giving us a combined unit of $\frac{a t m L}{m o l K}$.

The constant value then becomes 0.0821 $\frac{a t m L}{m o l K}$

If you choose not to have your students work in standard pressure unit factor, you may also use: 8.31 $\frac{k P a L}{m o l K}$.

Temperature must always be in Kelvin (K) to avoid using 0 ºC and getting no solution when students divide.

There is a variation of the ideal gas law that uses the density of the gas with the equation $P M = \mathrm{dR} T$.

Where M is the Molar Mass in g/mol and d is the Density of the gas in g/L.

Pressure and Temperature must remain in the units atm and K and the Gas Law Constant remains $R = 0.0821 \frac{a t m L}{m o l K}$.

In SI $R = 8.31 J {K}^{- 1} m o {l}^{- 1}$

Let's take an example of a 2.0 moles of nitrogen at 20 ºC and 3.00 atm and find the volume.

$P V = n R T$

P = 3.00 atm
V = ?
n = 2.0 mol
R = 0.0821 $\frac{a t m L}{m o l K}$
T = 20 + 273 = 293 K

$V = \frac{n R T}{P}$

$V = \frac{2.0 m o l \left(0.0821 \frac{a t m L}{m o l K}\right) 293 K}{3.00 a t m}$

$V = 16.0 L$