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 need to find the value of \( T_0 \) given that the final temperature of the gas mixture is \( 10 T_0 \). We will use the concept of internal energy and the fact that the total internal energy of the system is conserved.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Moles of gas 1, \( n_1 = 0.1 \) moles, Temperature \( T_1 = 200 \) K
- Moles of gas 2, \( n_2 = 0.05 \) moles, Temperature \( T_2 = 400 \) K
- Final temperature \( T_f = 10 T_0 \)

2. **Calculate the Internal Energy of Each Gas:**
The internal energy \( U \) of an ideal gas can be expressed as:
\[ U = \frac{F}{2} n R T \]
where \( F \) is the degrees of freedom (we will assume \( F = 3 \) for a monoatomic gas), \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.

- For gas 1:
\[ U_1 = \frac{F}{2} n_1 R T_1 = \frac{3}{2} \times 0.1 \times R \times 200 = 30R \]

- For gas 2:
\[ U_2 = \frac{F}{2} n_2 R T_2 = \frac{3}{2} \times 0.05 \times R \times 400 = 30R \]

3. **Total Internal Energy Before Mixing:**
The total internal energy before mixing is:
\[ U_{initial} = U_1 + U_2 = 30R + 30R = 60R \]

4. **Total Number of Moles After Mixing:**
The total number of moles after mixing is:
\[ n_{total} = n_1 + n_2 = 0.1 + 0.05 = 0.15 \text{ moles} \]

5. **Internal Energy After Mixing:**
The internal energy after mixing at final temperature \( T_f \) is:
\[ U_{final} = \frac{F}{2} n_{total} R T_f = \frac{3}{2} \times 0.15 \times R \times (10 T_0) = 2.25 R T_0 \]

6. **Set Up the Energy Conservation Equation:**
Since the internal energy is conserved, we have:
\[ U_{initial} = U_{final} \]
Therefore:
\[ 60R = 2.25 R T_0 \]

7. **Solve for \( T_0 \):**
Cancel \( R \) from both sides:
\[ 60 = 2.25 T_0 \]
Now, solve for \( T_0 \):
\[ T_0 = \frac{60}{2.25} = \frac{6000}{225} = \frac{80}{3} \approx 26.67 \text{ K} \]

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

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