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

To solve this problem, we need to understand the concept of molar volume and Avogadro's Law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

We are given equal masses of three gases: hydrogen (H_2), oxygen (O_2), and methane (CH_4), in a container of volume V at 27^\circ C. We need to find the ratio of the volumes of these gases.

1. First, calculate the molar mass of each gas:

   • Hydrogen, H_2: Molar mass = 2 g/mol
   • Oxygen, O_2: Molar mass = 32 g/mol
   • Methane, CH_4: Molar mass = 16 g/mol

2. Assume each gas has an equal mass, say m grams. Calculate the number of moles of each gas:

   • n_{H_2} = \frac{m}{2}
   • n_{O_2} = \frac{m}{32}
   • n_{CH_4} = \frac{m}{16}

3. Using Avogadro's Law, which states equal moles occupy equal volumes at the same temperature and pressure, the volume ratio will be directly proportional to the number of moles. Thus, calculate the ratio of moles:

   • Ratio of H_2 : O_2 : CH_4 = \frac{m}{2} : \frac{m}{32} : \frac{m}{16}
   • Clearing the common factor, the ratio becomes 16 : 1 : 2

4. Therefore, the ratio of the volumes of these gases under identical conditions is 16 : 1 : 2, which matches the given option.

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