0.1 mole of gas at 200K is mixed with 0.05 mole of same gas at 400K. If final temperature is equal to `10T_0` then find `T_0`

Answer

To solve the problem, we will use the principle of conservation of energy. The internal energy of the gas before mixing will be equal to the internal energy after mixing.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Number of moles of gas 1, \( n_1 = 0.1 \, \text{mol} \)
- Temperature of gas 1, \( T_1 = 200 \, \text{K} \)
- Number of moles of gas 2, \( n_2 = 0.05 \, \text{mol} \)
- Temperature of gas 2, \( T_2 = 400 \, \text{K} \)
- Final temperature, \( T = 10T_0 \)

2. **Use the Conservation of Energy:**
The total internal energy before mixing is equal to the total internal energy after mixing. The equation can be expressed as:
\[ n_1 T_1 + n_2 T_2 = (n_1 + n_2) T \]

3. **Substituting the Values:**
Substitute the known values into the equation:
\[ 0.1 \times 200 + 0.05 \times 400 = (0.1 + 0.05) T \]

4. **Calculate the Left Side:**
Calculate the left-hand side:
\[ 20 + 20 = 0.15 T \]
Thus, we have:
\[ 40 = 0.15 T \]

5. **Solve for T:**
Rearranging the equation to find \( T \):
\[ T = \frac{40}{0.15} = \frac{4000}{15} \approx 266.67 \, \text{K} \]

6. **Relate T to T0:**
We know from the problem statement that \( T = 10T_0 \). Therefore:
\[ 10T_0 = 266.67 \]

7. **Solve for T0:**
Dividing both sides by 10 gives:
\[ T_0 = \frac{266.67}{10} \approx 26.67 \, \text{K} \]

### Final Answer:
\[ T_0 \approx 26.67 \, \text{K} \]