One mole of a monoatomic gas at 300K is mixed with two moles of diatomic gas (degree of freedom = 5) at 600K. The temperature of the mixture will be

Options

A. 456 K
B. 531 K
C. 495 K
D. 501 K

Answer

To find the temperature of the mixture of one mole of a monoatomic gas at 300K and two moles of a diatomic gas at 600K, we will use the principle of conservation of energy, specifically the change in internal energy.

### Step-by-Step Solution:

1. **Identify the Gases and Their Properties:**
- Monoatomic gas: \( n_1 = 1 \) mole, \( T_1 = 300 \) K, \( C_{V1} = \frac{3}{2} R \)
- Diatomic gas: \( n_2 = 2 \) moles, \( T_2 = 600 \) K, \( C_{V2} = \frac{5}{2} R \)

2. **Write the Change in Internal Energy:**
The change in internal energy for each gas can be expressed as:
\[ \Delta U_1 = n_1 C_{V1} (T_f - T_1) \]
\[ \Delta U_2 = n_2 C_{V2} (T_f - T_2) \]

3. **Set Up the Equation:**
Since there is no heat exchange with the surroundings, the change in internal energy of the monoatomic gas will equal the negative change in internal energy of the diatomic gas:
\[ \Delta U_1 + \Delta U_2 = 0 \]
Substituting the expressions for \(\Delta U_1\) and \(\Delta U_2\):
\[ n_1 C_{V1} (T_f - T_1) + n_2 C_{V2} (T_f - T_2) = 0 \]

4. **Substitute the Values:**
\[ 1 \cdot \frac{3}{2} R (T_f - 300) + 2 \cdot \frac{5}{2} R (T_f - 600) = 0 \]
Simplifying this:
\[ \frac{3}{2} R (T_f - 300) + 5 R (T_f - 600) = 0 \]

5. **Factor Out the Common Term \( R \):**
Since \( R \) is a common factor, we can cancel it out:
\[ \frac{3}{2} (T_f - 300) + 5 (T_f - 600) = 0 \]

6. **Distribute and Combine Like Terms:**
\[ \frac{3}{2} T_f - 450 + 5 T_f - 3000 = 0 \]
Combine the terms:
\[ \left(\frac{3}{2} + 5\right) T_f - 3450 = 0 \]
\[ \frac{13}{2} T_f = 3450 \]

7. **Solve for \( T_f \):**
\[ T_f = \frac{3450 \cdot 2}{13} = \frac{6900}{13} \approx 530.76 \text{ K} \]

8. **Final Answer:**
Rounding to the nearest whole number gives us:
\[ T_f \approx 531 \text{ K} \]

### Final Result:
The temperature of the mixture will be approximately **531 K**.