Category : JEE Main & Advanced

(1) According to this law, “Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules.”

Thus, $V\propto n$              (at constant T and P)

or $V=Kn$            (where K is constant)

or $\frac{{{V}_{1}}}{{{n}_{1}}}=\frac{{{V}_{2}}}{{{n}_{2}}}=.......=K$

Example,                 $\underset{1n\,litre}{\mathop{\underset{1\,litre}{\mathop{\underset{2\,litres}{\mathop{\underset{2\,volumes}{\mathop{\underset{2\,moles}{\mathop{2{{H}_{2}}(g)}}\,}}\,}}\,}}\,}}\,+\underset{1/2n\,litre}{\mathop{\underset{1/2\,litre}{\mathop{\underset{1\,litre}{\mathop{\underset{1\,volume}{\mathop{\underset{1\,mole}{\mathop{{{O}_{2}}(g)}}\,}}\,}}\,}}\,}}\,\xrightarrow{{}}\underset{1n\,litre}{\mathop{\underset{1\,litre}{\mathop{\underset{2\,litres}{\mathop{\underset{2\,volumes}{\mathop{\underset{2\,moles}{\mathop{2{{H}_{2}}O(g)}}\,}}\,}}\,}}\,}}\,$

(2) One mole of any gas contains the same number of molecules (Avogadro's number $=6.02\times {{10}^{23}}$) and by this law must occupy the same volume at a given temperature and pressure. The volume of one mole of a gas is called molar volume, Vm which is 22.4 L $mo{{l}^{-1}}$ at S.T.P. or N.T.P.

(3) This law can also express as, “The molar gas volume at a given temperature and pressure is a specific constant independent of the nature of the gas”.

Thus, ${{V}_{m}}=$ specific constant $=22.4\,L\,mo{{l}^{-1}}$ at S.T.P. or N.T.P.

You need to login to perform this action.
You will be redirected in 3 sec