The ratio of kinetic energies of 2gm of `H_2 `and 4gm of `CH_4` at a given temperature is 

Options

A. `4:1`
B. `2:32`
C. `1:4`
D. `16:2`

Answer

To solve the problem of finding the ratio of kinetic energies of 2g of \( H_2 \) and 4g of \( CH_4 \) at a given temperature, we will follow these steps:

### Step 1: Calculate the number of moles of \( H_2 \)

The formula to calculate the number of moles (\( n \)) is given by:

\[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

For hydrogen gas (\( H_2 \)):
- Given mass = 2g
- Molar mass of \( H_2 = 2 \, \text{g/mol} \)

Calculating the moles of \( H_2 \):

\[ n_{H_2} = \frac{2 \, \text{g}}{2 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 2: Calculate the number of moles of \( CH_4 \)

For methane gas (\( CH_4 \)):
- Given mass = 4g
- Molar mass of \( CH_4 = 12 \, \text{g/mol (C)} + 4 \, \text{g/mol (H)} = 16 \, \text{g/mol} \)

Calculating the moles of \( CH_4 \):

\[ n_{CH_4} = \frac{4 \, \text{g}}{16 \, \text{g/mol}} = 0.25 \, \text{mol} \]

### Step 3: Write the formula for kinetic energy

The kinetic energy (\( KE \)) of a gas is given by the formula:

\[ KE = \frac{3}{2} nRT \]

Where:
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature (assumed to be the same for both gases)

### Step 4: Calculate the kinetic energy for \( H_2 \)

Substituting the values for \( H_2 \):

\[ KE_{H_2} = \frac{3}{2} \times n_{H_2} \times R \times T = \frac{3}{2} \times 1 \times R \times T = \frac{3}{2} RT \]

### Step 5: Calculate the kinetic energy for \( CH_4 \)

Substituting the values for \( CH_4 \):

\[ KE_{CH_4} = \frac{3}{2} \times n_{CH_4} \times R \times T = \frac{3}{2} \times 0.25 \times R \times T = \frac{3}{8} RT \]

### Step 6: Find the ratio of kinetic energies

Now, we can find the ratio of the kinetic energies:

\[ \text{Ratio} = \frac{KE_{H_2}}{KE_{CH_4}} = \frac{\frac{3}{2} RT}{\frac{3}{8} RT} \]

The \( RT \) cancels out:

\[ \text{Ratio} = \frac{\frac{3}{2}}{\frac{3}{8}} = \frac{3}{2} \times \frac{8}{3} = \frac{8}{2} = 4 \]

Thus, the ratio of the kinetic energies of 2g of \( H_2 \) and 4g of \( CH_4 \) is:

\[ \text{Ratio} = 4:1 \]

### Final Answer:
The ratio of kinetic energies of 2g of \( H_2 \) and 4g of \( CH_4 \) at a given temperature is \( 4:1 \).

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