So far, we have seen how p, V and T are related. It is possible to write a single equation relating these quantities that takes into account the amount of gas being considered.
We can write the equation in the following form:

$pV = nRT$

where n is the amount (number of moles) of an ideal gas.
Or in the form:

$pV = NkT$

where N is the number of molecules and k is the Boltzmann constant described later in topic 20.8.
This equation is called the equation of state for an ideal gas (or the ideal gas equation). It relates all four of the variable quantities discussed at the beginning of this chapter. The constant of proportionality R is called the universal molar gas constant. Its experimental value is:

$R = 8.31\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$

Note that it doesn’t matter what gas we are considering–it could be a very ‘light’ gas like hydrogen, or a much ‘heavier’ one like carbon dioxide. So long as it is behaving as an ideal gas, we can use the same equation of state with the same constant R.

Calculating the number n of moles

Instead of knowing the mass of one molecule in unified atomic mass units, sometimes we may be given the molar mass (the mass of one mole) and the mass of gas we are concerned with, to find how many moles are present. To do this, we use the relationship:

$number\,of\,moles = \frac{{mass(g)}}{{molar\,mass\,(g\,mo{l^{ - 1}})}}$

For example: How many moles are there in $1.6 kg$ of oxygen?

$molar\,mass\,of\,oxygen - 16 = 32\,g\,mo{l^{ - 1}}$
$\begin{array}{l}
number\,of\,moles = \frac{{1600g}}{{32\,g\,mo{l^{ - 1}}}}\\
 = 50\,mol
\end{array}$

(Note that this tells us that there are 50 moles of oxygen molecules in $1.6 kg$ of oxygen. An oxygen molecule consists of two oxygen atoms – its formula is ${O_2} - so\,\,1.6\,kg$ of oxygen contains 100 moles of oxygen atoms.)
Now look at Worked examples 2 and 3.