The total kinetic energy of a mixture of 4 g of `H_2 and 4g` of He at 300 K is

Options

A. 1800 R
B. 1750 R
C. 1950 R
D. 2500 R

Answer

To find the total kinetic energy of a mixture of 4 g of \( H_2 \) and 4 g of He at 300 K, we can follow these steps:

### Step 1: Calculate the number of moles of \( H_2 \) and He
1. **For Hydrogen (\( H_2 \))**:
- Molar mass of \( H_2 \) = 2 g/mol
- Mass of \( H_2 \) = 4 g
- Number of moles of \( H_2 \) = \( \frac{\text{mass}}{\text{molar mass}} = \frac{4 \text{ g}}{2 \text{ g/mol}} = 2 \text{ moles} \)

2. **For Helium (He)**:
- Molar mass of He = 4 g/mol
- Mass of He = 4 g
- Number of moles of He = \( \frac{4 \text{ g}}{4 \text{ g/mol}} = 1 \text{ mole} \)

### Step 2: Calculate the internal energy for \( H_2 \)
- The formula for internal energy \( U \) for a diatomic gas is given by:
\[ U = \frac{F}{2} nRT \]
where \( F \) is the degrees of freedom (5 for diatomic gases), \( n \) is the number of moles, \( R \) is the gas constant (8.314 J/(mol·K)), and \( T \) is the temperature in Kelvin.

- For \( H_2 \):
\[ U_{H_2} = \frac{5}{2} \times n_{H_2} \times R \times T = \frac{5}{2} \times 2 \times 8.314 \times 300 \]
\[ U_{H_2} = \frac{5}{2} \times 2 \times 8.314 \times 300 = 5 \times 8.314 \times 300 = 12471 J \]

### Step 3: Calculate the internal energy for He
- The formula for internal energy \( U \) for a monatomic gas is given by:
\[ U = \frac{3}{2} nRT \]
where \( F \) is the degrees of freedom (3 for monatomic gases).

- For He:
\[ U_{He} = \frac{3}{2} \times n_{He} \times R \times T = \frac{3}{2} \times 1 \times 8.314 \times 300 \]
\[ U_{He} = \frac{3}{2} \times 8.314 \times 300 = 3 \times 8.314 \times 150 = 3741.3 J \]

### Step 4: Calculate the total kinetic energy
- The total kinetic energy of the mixture is the sum of the kinetic energies of both gases:
\[ KE_{total} = U_{H_2} + U_{He} = 12471 J + 3741.3 J = 16212.3 J \]

### Final Answer
The total kinetic energy of the mixture of 4 g of \( H_2 \) and 4 g of He at 300 K is approximately **16212.3 J**.

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