Equal masses of `H_2, O_2`, 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 `H_2 : O_2 : CH_4` would be

Options

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

Explanation

To solve the problem of finding the ratio of the volumes of gases \( H_2 : O_2 : CH_4 \) when equal masses of each gas are taken in a container at the same temperature and pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law states that \( PV = nRT \), where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Identify Given Conditions**:
We are given that the temperature is \( 27^\circ C \) (which is \( 300 \, K \)), and that equal masses of \( H_2 \), \( O_2 \), and \( CH_4 \) are taken. The pressure is assumed to be 1 atm.

3. **Relate Volume to Moles**:
Since the pressure and temperature are constant, the volume of a gas is directly proportional to the number of moles (\( V \propto n \)). Therefore, the ratio of the volumes of the gases will be the same as the ratio of the number of moles of the gases.

4. **Calculate Moles of Each Gas**:
The number of moles \( n \) of a gas can be calculated using the formula:
\[ n = \frac{m}{M} \]
where \( m \) is the mass of the gas and \( M \) is the molar mass of the gas.

5. **Define Molar Masses**:
- For \( H_2 \): Molar mass \( M_{H_2} = 2 \, g/mol \)
- For \( O_2 \): Molar mass \( M_{O_2} = 32 \, g/mol \)
- For \( CH_4 \): Molar mass \( M_{CH_4} = 16 \, g/mol \)

6. **Set Up the Ratios**:
Since equal masses are taken, let the mass of each gas be \( m \). The number of moles for each gas can be expressed as:
\[ n_{H_2} = \frac{m}{2}, \quad n_{O_2} = \frac{m}{32}, \quad n_{CH_4} = \frac{m}{16} \]

7. **Calculate the Volume Ratios**:
The volume ratio can be derived from the number of moles:
\[ V_{H_2} : V_{O_2} : V_{CH_4} = n_{H_2} : n_{O_2} : n_{CH_4} = \frac{m/2}{m/32} : \frac{m/32}{m/32} : \frac{m/16}{m/32} \]
Simplifying this gives:
\[ V_{H_2} : V_{O_2} : V_{CH_4} = \frac{32}{2} : \frac{32}{32} : \frac{32}{16} = 16 : 1 : 2 \]

8. **Final Ratio**:
The ratio of the volumes of the gases \( H_2 : O_2 : CH_4 \) is:
\[ 16 : 1 : 2 \]

### Conclusion:
The final ratio of the volumes of the gases \( H_2 : O_2 : CH_4 \) is \( 16 : 1 : 2 \).