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 $H_2;O_2;C{H}_4$ would be

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

Answer 1

Answer: (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 $\left(C{H}_4\right)$ is given by
$V_{H_2}:V_{O_2}:V_{C_4}=n_{H_2}:n_{O_2}:n_{C{H}_4}$
$\Rightarrow V_{H_2}:V_{O_2}:V_{C_4}:=\frac{m_{H_2}}{M_{H_2}}:\frac{m_{O_2}}{M_{O_2}}:\frac{m_{C{H}_4}}{M_{C{H}_4}}$
Given, $m_{H_2}=m_{O_2}=m_{C_4}=m$
$\left[\because n=\frac{\text{ mass }}{\text{ molarmass }}\right]$
Thus, $V_{H_2}:V_{O_2}:V_{C{H}_4}:=\frac{m}{2}:\frac{m}{32}:\frac{m}{16}$
$=16:1:2$