Equal masses of `H_(2),N_(2)` and methane taken in a container of volume V at temperature `27^(@)C` in identical conditions. The ratio of volumes of gases `H_(2):N_(2):` methane would be :-

Options

A. `56:4:7`
B. `7:04:56 ` AM
C. `2:28:16` AM
D. `8:14:01` AM

Explanation

To solve the problem of finding the ratio of volumes of gases \( H_2 : N_2 : \text{methane} \) when equal masses of each gas are taken in a container at the same temperature and pressure, we can follow these steps:

### Step 1: Understand the conditions
We are given equal masses of three gases: hydrogen (\( H_2 \)), nitrogen (\( N_2 \)), and methane (\( CH_4 \)). The temperature is \( 27^\circ C \) and the volume of the container is \( V \).

### Step 2: Use the Ideal Gas Law
According to the Ideal Gas Law, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of that gas. Therefore, we can express the volume of each gas in terms of its number of moles.

### Step 3: Calculate the number of moles for 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.

- For \( H_2 \): Molar mass \( M_{H_2} = 2 \, \text{g/mol} \)
- For \( N_2 \): Molar mass \( M_{N_2} = 28 \, \text{g/mol} \)
- For \( CH_4 \): Molar mass \( M_{CH_4} = 16 \, \text{g/mol} \)

Let’s assume we have \( m \) grams of each gas.

- Moles of \( H_2 \):
\[ n_{H_2} = \frac{m}{2} \]
- Moles of \( N_2 \):
\[ n_{N_2} = \frac{m}{28} \]
- Moles of \( CH_4 \):
\[ n_{CH_4} = \frac{m}{16} \]

### Step 4: Determine the ratio of moles
Now we can find the ratio of the number of moles:
\[ \text{Ratio} = n_{H_2} : n_{N_2} : n_{CH_4} = \frac{m}{2} : \frac{m}{28} : \frac{m}{16} \]

### Step 5: Simplify the ratio
To simplify, we can eliminate \( m \) from the ratio:
\[ \text{Ratio} = \frac{1}{2} : \frac{1}{28} : \frac{1}{16} \]

Now, to eliminate the fractions, we can find a common denominator. The least common multiple of \( 2, 28, \) and \( 16 \) is \( 112 \).

- Convert each term:
\[ \frac{1}{2} = \frac{56}{112}, \quad \frac{1}{28} = \frac{4}{112}, \quad \frac{1}{16} = \frac{7}{112} \]

Thus, the ratio becomes:
\[ 56 : 4 : 7 \]

### Step 6: Simplify further
Now, we can divide each term by 4 to simplify:
\[ 14 : 1 : \frac{7}{4} \]

### Final Ratio
To express this in whole numbers, we can multiply through by 4:
\[ 56 : 4 : 7 \]

Thus, the final ratio of volumes of gases \( H_2 : N_2 : CH_4 \) is:
\[ \text{Final Ratio} = 56 : 4 : 7 \]