How can I calculate the ideal gas law equations?

1 Answer
May 15, 2014

There are two versions of the ideal gas equation:
Molar version: #pV=nRT# (n is no. of moles & R is molar gas constant).
Molecular version: #pV=NkT# (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 #PV = nRT#.

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 #(atmL) / (molK)#.

The constant value then becomes 0.0821 #(atmL)/(molK)#

If you choose not to have your students work in standard pressure unit factor, you may also use: 8.31 #(kPaL)/(molK)#.

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 #PM = dRT#.

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 (atm L) / (mol K)#.

In SI #R=8.31 J K^(-1) mol^(-1)#

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

#PV = nRT#

P = 3.00 atm
V = ?
n = 2.0 mol
R = 0.0821 #(atmL)/(molK)#
T = 20 + 273 = 293 K

#V = (nRT)/P#

#V = (2.0 mol(0.0821(atmL)/(molK))293K)/(3.00 atm)#

#V = 16.0L#

I hope this is helpful.