Question

Equal masses of H₂, O₂ and methane have been taken in a container of volume V at temperature 27°C in identical conditions. The ratio of the volumes of gases H₂ : O₂ : methane would be :

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

Answer 1

The Correct Option is B

To solve this problem, we need to determine the ratio of volumes for equal masses of hydrogen (H₂), oxygen (O₂), and methane (CH₄) when placed under identical conditions of volume, temperature, and pressure.

We know that the number of moles n of a gas is given by:

n = \frac{m}{M}

where m is the mass and M is the molar mass.

Since equal masses are taken for each gas, let the mass of each gas be m.

1. For hydrogen (H₂), its molar mass M_{\text{H}_2} = 2 \text{ g/mol}.
   • Number of moles n_{\text{H}_2} = \frac{m}{2}
2. For oxygen (O₂), its molar mass M_{\text{O}_2} = 32 \text{ g/mol}.
   • Number of moles n_{\text{O}_2} = \frac{m}{32}
3. For methane (CH₄), its molar mass M_{\text{CH}_4} = 16 \text{ g/mol}.
   • Number of moles n_{\text{CH}_4} = \frac{m}{16}

At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles (Avogadro's Law):

V \propto n

Therefore, the ratio of the volumes for H₂, O₂, and CH₄ will be the same as the ratio of the number of moles:

The ratio:

n_{\text{H}_2} : n_{\text{O}_2} : n_{\text{CH}_4} = \frac{m}{2} : \frac{m}{32} : \frac{m}{16} = \frac{1}{2} : \frac{1}{32} : \frac{1}{16}

To simplify this, multiply through by 32 (the least common multiple):

16 : 1 : 2

Thus, the correct answer is 16: 1 : 2.

This calculation shows that for equal masses, the volume taken by hydrogen is much larger due to its low molar mass, while oxygen takes the smallest volume.