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 volume 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

Hint:
First find out the number of moles related to ${H_2}$, ${O_2}$ and methane when they have equal given masses. From the Ideal Gas Law we can say that volume is proportional to moles when the temperature and pressure are constant. So, to find the ratio of volumes of these gases we can find out the ratio of their moles.
Formula used:
-Ideal Gas Law: PV = nRT ……. (1)
Where, P = pressure;
              V = volume;
             n = moles;
             R = gas constant;
             T = temperature.

Complete answer:
-Here we have been given the question that the given masses of ${H_2}$, ${O_2}$ and methane are equal. So, let it be = x. The question also says that temperature is ${27^ \circ }C$ and conditions do not change.
We will use the Ideal Gas Law here to find the ratio of the volumes of ${H_2}$, ${O_2}$ and methane. According to equation (1):
                                                PV = nRT
Since in the question for all the 3 gases P, R and T are same; we can say that:
                                              V α n
                                            V α $\dfrac{W}{M}$ (2)
Where, W = given weight and M = molecular weight.
-So, now to calculate the ratio of volumes of ${H_2}$, ${O_2}$ and methane we will use equation (2):
${V_{{H_2}}}:{V_{{O_2}}}:{V_{C{H_4}}}$ = ${\left( {\dfrac{W}{M}} \right)_{{H_2}}}:{\left( {\dfrac{W}{M}} \right)_{{O_2}}}:{\left( {\dfrac{W}{M}} \right)_{C{H_4}}}$ (3)
Where, ${\left( {\dfrac{W}{M}} \right)_{{H_2}}} = \dfrac{x}{2}$ ; ${\left( {\dfrac{W}{M}} \right)_{{O_2}}} = \left( {\dfrac{x}{{32}}} \right)$ and ${\left( {\dfrac{W}{M}} \right)_{C{H_4}}} = \left( {\dfrac{x}{{16}}} \right)$. Now putting these values in equation (3):
  ${V_{{H_2}}}:{V_{{O_2}}}:{V_{C{H_4}}}$ = $\dfrac{x}{2}:\dfrac{x}{{32}}:\dfrac{x}{{16}}$
                                                                          = 16 : 1 : 2
So, the ratio of the volume of ${H_2}$, ${O_2}$ and methane is 16 : 1 : 2

The correct option will be: (C) 16 : 1 : 2

Note:
The ideal gas law basically states that at constant temperature and pressure n moles of gas will occupy the same volume. This law is valid only for ideal gases. Ideal gases are considered to be gases where the molecules behave as rigid spheres and are constantly moving in random directions. The collisions in these gases are elastic collisions.