At what temperature will the total kinetic energy of 0.30 mol of helium be same as the total kinetic energy of 0.40 mol of argon at 500 K.

Options

A. 375 K
B. 666.7 K
C. 573 K
D. 500 K

Explanation

To find the temperature at which the total kinetic energy of 0.30 mol of helium is the same as the total kinetic energy of 0.40 mol of argon at 500 K, we can use the formula for the kinetic energy of an ideal gas:

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

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

### Step-by-Step Solution:

1. **Write the kinetic energy equation for helium**:
\[ KE_{He} = \frac{3}{2} n_{He} R T_{He} \]
For helium, \( n_{He} = 0.30 \) mol, so:
\[ KE_{He} = \frac{3}{2} \times 0.30 \times R \times T_{He} \]

2. **Write the kinetic energy equation for argon**:
\[ KE_{Ar} = \frac{3}{2} n_{Ar} R T_{Ar} \]
For argon, \( n_{Ar} = 0.40 \) mol and \( T_{Ar} = 500 \) K, so:
\[ KE_{Ar} = \frac{3}{2} \times 0.40 \times R \times 500 \]

3. **Set the kinetic energies equal to each other**:
\[ KE_{He} = KE_{Ar} \]
Therefore:
\[ \frac{3}{2} \times 0.30 \times R \times T_{He} = \frac{3}{2} \times 0.40 \times R \times 500 \]

4. **Cancel out common terms**:
Since \( \frac{3}{2} \) and \( R \) are on both sides, we can simplify:
\[ 0.30 \times T_{He} = 0.40 \times 500 \]

5. **Calculate the right side**:
\[ 0.40 \times 500 = 200 \]
So we have:
\[ 0.30 \times T_{He} = 200 \]

6. **Solve for \( T_{He} \)**:
\[ T_{He} = \frac{200}{0.30} = \frac{2000}{3} \approx 666.67 \, K \]

### Final Answer:
The temperature at which the total kinetic energy of 0.30 mol of helium is the same as that of 0.40 mol of argon at 500 K is approximately **666.67 K**.