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. The ratio of the volumes of gases ${H_2 : O_2 :CH_4}$ would be

• A. 8 : 16 : 1
• B. 16 : 8 : 1
• C. 16 : 1 : 2
• D. 8 : 1 : 2

Answer 1

The Correct Option is C

According to Avogadro's hypothesis,
Volume of a gas (V) $\propto$number of moles (.n)
Therefore, the ratio of the volumes of gases can be
determined in terms of their moles.
$\therefore $ The ratio of volumes of $H_2 : O_2 : $ methane $ (CH_4)$ is given by
$\, \, \, \, \, \, \, \, \, \, \, V_{H_2} : V_{O_2} : V_{CH_4}=n_{H_2} : n_{O_2} : n_{CH_4}$
$\Rightarrow V_{H_2} : V_{O_2} : V_{CH_4} : = \frac{m_{H_2}}{M_{H_2}} : \frac{m_{O_2}}{M_{O_2}} : \frac{m_{CH_4}}{M_{CH_4}}$
Given, $\, \, \, \, \, \, \, \, \, \, m_{H_2}= m_{O_2}=m_{CH_4}=m$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Bigg[\because n=\frac{mass}{molar\, mass}\Bigg]$
Thus, $V_{H_2} : V_{O_2} : V_{CH_4} : =\frac{m}{2} : \frac{m}{32} : \frac{m}{16}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $ = 16:1:2