A stationary vessel contains a mixture of equal moles of Nitrogen and Helium gas. The percentage of the total kinetic energy of the mixture that is possessed by the molecules of Nitrogen is ___________

Explanation

To find the percentage of the total kinetic energy of a mixture of equal moles of Nitrogen (N₂) and Helium (He) gas that is possessed by the molecules of Nitrogen, we can follow these steps:

### Step 1: Understand the Kinetic Energy Formula
The kinetic energy (KE) of a gas can be expressed as:
\[ KE = \frac{f}{2} nRT \]
where:
- \( f \) is the degrees of freedom of the gas,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Identify Degrees of Freedom
For Nitrogen (N₂), which is a diatomic gas, the degrees of freedom \( f \) is 5 (3 translational + 2 rotational).
For Helium (He), which is a monoatomic gas, the degrees of freedom \( f \) is 3 (3 translational).

### Step 3: Write the Kinetic Energy for Each Gas
- Kinetic energy of Nitrogen:
\[ KE_{N_2} = \frac{5}{2} nRT \]
- Kinetic energy of Helium:
\[ KE_{He} = \frac{3}{2} nRT \]

### Step 4: Calculate Total Kinetic Energy of the Mixture
The total kinetic energy of the mixture is the sum of the kinetic energies of both gases:
\[ KE_{total} = KE_{N_2} + KE_{He} = \frac{5}{2} nRT + \frac{3}{2} nRT = \frac{8}{2} nRT = 4nRT \]

### Step 5: Find the Percentage of Kinetic Energy from Nitrogen
To find the percentage of the total kinetic energy that is due to Nitrogen, we use the formula:
\[ \text{Percentage of } KE_{N_2} = \left( \frac{KE_{N_2}}{KE_{total}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage of } KE_{N_2} = \left( \frac{\frac{5}{2} nRT}{4nRT} \right) \times 100 \]

### Step 6: Simplify the Expression
The \( nRT \) terms cancel out:
\[ \text{Percentage of } KE_{N_2} = \left( \frac{5/2}{4} \right) \times 100 = \left( \frac{5}{8} \right) \times 100 = 62.5\% \]

### Final Answer
The percentage of the total kinetic energy of the mixture that is possessed by the molecules of Nitrogen is **62.5%**.

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