Which of the following will have maximum total kinetic energy at temperature 300 K

Options

A. `1 kg, H_(2)`
B. `1 kg, He`
C. `1/2 kg H_(2)+1/2 kg He`
D. `1/4 kg H_(2)+3/4 kg He`

Explanation

To determine which of the given options has the maximum total kinetic energy at a temperature of 300 K, we will use the formula for the total kinetic energy of a gas, which is given by:

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

where:
- \( KE \) is the total kinetic energy,
- \( n \) is the number of moles of the gas,
- \( R \) is the universal gas constant (approximately 8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin.

Now, let's analyze each option step by step.

### Step 1: Calculate the total kinetic energy for each option.

#### Option A: 1 kg of Hydrogen (H₂)
1. **Calculate the number of moles (n)**:
- Molecular mass of H₂ = 2 g/mol
- \( n = \frac{1000 \text{ g}}{2 \text{ g/mol}} = 500 \text{ moles} \)

2. **Calculate the kinetic energy**:
\[ KE_A = \frac{3}{2} nRT = \frac{3}{2} \times 500 \times R \times 300 \]
\[ KE_A = 750R \times 300 = 750R \times 300 \]

#### Option B: 1 kg of Helium (He)
1. **Calculate the number of moles (n)**:
- Molecular mass of He = 4 g/mol
- \( n = \frac{1000 \text{ g}}{4 \text{ g/mol}} = 250 \text{ moles} \)

2. **Calculate the kinetic energy**:
\[ KE_B = \frac{3}{2} nRT = \frac{3}{2} \times 250 \times R \times 300 \]
\[ KE_B = 375R \times 300 = 375R \times 300 \]

#### Option C: Mixture of 0.5 kg Hydrogen and 0.5 kg Helium
1. **Calculate the number of moles for Hydrogen**:
- \( n_{H_2} = \frac{500 \text{ g}}{2 \text{ g/mol}} = 250 \text{ moles} \)

2. **Calculate the number of moles for Helium**:
- \( n_{He} = \frac{500 \text{ g}}{4 \text{ g/mol}} = 125 \text{ moles} \)

3. **Calculate the total kinetic energy**:
\[ KE_C = \frac{3}{2} n_{H_2}RT + \frac{3}{2} n_{He}RT \]
\[ KE_C = \frac{3}{2} \times 250 \times R \times 300 + \frac{3}{2} \times 125 \times R \times 300 \]
\[ KE_C = 375R \times 300 + 187.5R \times 300 = 562.5R \times 300 \]

#### Option D: 0.25 kg Hydrogen and 0.75 kg Helium
1. **Calculate the number of moles for Hydrogen**:
- \( n_{H_2} = \frac{250 \text{ g}}{2 \text{ g/mol}} = 125 \text{ moles} \)

2. **Calculate the number of moles for Helium**:
- \( n_{He} = \frac{750 \text{ g}}{4 \text{ g/mol}} = 187.5 \text{ moles} \)

3. **Calculate the total kinetic energy**:
\[ KE_D = \frac{3}{2} n_{H_2}RT + \frac{3}{2} n_{He}RT \]
\[ KE_D = \frac{3}{2} \times 125 \times R \times 300 + \frac{3}{2} \times 187.5 \times R \times 300 \]
\[ KE_D = 187.5R \times 300 + 281.25R \times 300 = 468.75R \times 300 \]

### Step 2: Compare the total kinetic energies
- \( KE_A = 750R \times 300 \)
- \( KE_B = 375R \times 300 \)
- \( KE_C = 562.5R \times 300 \)
- \( KE_D = 468.75R \times 300 \)

### Conclusion
The maximum total kinetic energy is from Option A: **1 kg of Hydrogen**.