Question

Equal masses of ${H_2},{O_2}$ and methane have been taken in a container of volume $V$ at temperature ${27^\circ }C$ in identical conditions. What would be the ratio of the volumes of gases ${H_2}:{O_2}:methane$?
A) $8:16:1$
B) $16:8:1$
C) $16:1:2$
D) $8:1:2$

Answer 1

Hint:As we know that according to Avogadro’s law that states that the volume of gases is directly proportional to the number of moles of gases and number of moles is equivalent to the given masses of the solute to the molecular masses of the solvent.

Complete answer:
We already know that volume of gases is directly proportional to the number of moles of gases and the number of moles is equivalent to the given masses of the solute to the molecular masses of the solvent.

Let us assume that all the gases are present in $X\;g$ of the total gases. We also know that the molecular mass of hydrogen is $2g$, the molecular mass of the oxygen is $32g$ and the molecular mass of the methane is $16g$.

According to Avogadro’s law ‘equal volume of all gases under similar conditions of temperature and pressure contain equal numbers of molecules’. So we can use this law and determine the ratio of the given gases as:
Molar ratio of volumes of ${H_2}:{O_2}:C{H_4} = \dfrac{X}{2}:\dfrac{X}{{32}}:\dfrac{X}{{16}}$
Ratio will be: $ \Rightarrow \dfrac{1}{1}:\dfrac{1}{{16}}:\dfrac{1}{8} = 16:1:2$

Therefore the correct answer is option (C).

Note:Remember that Avogadro’s law is directly related to ideal gas equation because it link the temperature, pressure and volume as well as the number of moles of given substances or gases but it has limitations because it only provides the relationship for real gases because ideal gases behaves at low temperature and high pressure. Gases having low molecular masses like helium and hydrogen obeys Avogadro’s law to a higher extent than heavier molecules.