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 $H_2:O_2:$ methane 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 $16:1:2$

According to the ideal gas equation,

$PV=nRT=\frac{W}{M}RT$

$V\propto \frac{1}{M}$

Here, V is the volume and M is the molecular weight of a gas.

$V_{H_2}:V_{O_2}:V_{C{H}_4}=\frac{1}{M_{H_2}}:\frac{1}{M_{O_2}}:\frac{1}{M_{C{H}_4}}$

$V_{H_2}:V_{O_2}:V_{C{H}_4}=\frac{1}{2}:\frac{1}{32}:\frac{1}{16}$

$V_{H_2}:V_{O_2}:V_{C{H}_4}=1:\frac{1}{16}:\frac{1}{8}$

On multiplying each ratio by 16, we get-

$V_{H_2}:V_{O_2}:V_{C{H}_4}=16:1:2$

So, the correct option is $C$.