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 solve the problem of finding 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: Determine the number of moles of each gas.
- For \(H_2\):
- Molar mass of \(H_2\) = 2 g/mol
- Number of moles of \(H_2\) = \(\frac{4 \text{ g}}{2 \text{ g/mol}} = 2 \text{ moles}\)

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

### Step 2: Calculate the kinetic energy of \(H_2\).
- The formula for kinetic energy (U) is given by:
\[ U = \frac{f}{2} nRT \]
where \(f\) is the degrees of freedom, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

- For \(H_2\) (a diatomic gas), the degrees of freedom \(f = 5\).
- Thus, the kinetic energy of \(H_2\) is:
\[ U_{H_2} = \frac{5}{2} \times 2 \text{ moles} \times R \times 300 \text{ K} \]
\[ U_{H_2} = 5 \times R \times 300 = 1500R \]

### Step 3: Calculate the kinetic energy of He.
- For He (a monoatomic gas), the degrees of freedom \(f = 3\).
- Thus, the kinetic energy of He is:
\[ U_{He} = \frac{3}{2} \times 1 \text{ mole} \times R \times 300 \text{ K} \]
\[ U_{He} = \frac{3}{2} \times R \times 300 = 450R \]

### Step 4: Calculate the total kinetic energy of the mixture.
- The total kinetic energy \(U_{total}\) is the sum of the kinetic energies of \(H_2\) and He:
\[ U_{total} = U_{H_2} + U_{He} = 1500R + 450R = 1950R \]

### Final Answer:
The total kinetic energy of the mixture of 4 g of \(H_2\) and 4 g of He at 300 K is \(1950R\).

---