A closed vessel contains 0.1 mole of a monatomic ideal gas at 200k . If 0.05 mole of the same gas at 400 K is added to it , the final equilibrium temperature (in K ) of the gas in the vessel will be close to ...............

Answer

To solve the problem, we need to find the final equilibrium temperature of a closed vessel containing two different amounts of a monatomic ideal gas at different temperatures. Here’s the step-by-step solution:

### Step 1: Identify the given data
- Number of moles of the first gas, \( n_1 = 0.1 \) moles
- Temperature of the first gas, \( T_1 = 200 \) K
- Number of moles of the second gas, \( n_2 = 0.05 \) moles
- Temperature of the second gas, \( T_2 = 400 \) K

### Step 2: Calculate the total number of moles after mixing
The total number of moles after mixing the two gases is:
\[ n = n_1 + n_2 = 0.1 + 0.05 = 0.15 \text{ moles} \]

### Step 3: Use the principle of conservation of energy
For an ideal gas, the internal energy \( U \) is given by:
\[ U = \frac{f}{2} nRT \]
where \( f \) is the degrees of freedom (for a monatomic gas, \( f = 3 \)).

### Step 4: Write the internal energy for both gases
1. For the first gas:
\[ U_1 = \frac{3}{2} n_1 R T_1 = \frac{3}{2} \times 0.1 \times R \times 200 \]
2. For the second gas:
\[ U_2 = \frac{3}{2} n_2 R T_2 = \frac{3}{2} \times 0.05 \times R \times 400 \]

### Step 5: Calculate \( U_1 \) and \( U_2 \)
Calculating \( U_1 \):
\[ U_1 = \frac{3}{2} \times 0.1 \times R \times 200 = 30R \]

Calculating \( U_2 \):
\[ U_2 = \frac{3}{2} \times 0.05 \times R \times 400 = 30R \]

### Step 6: Write the total internal energy after mixing
The total internal energy after mixing is:
\[ U = U_1 + U_2 = 30R + 30R = 60R \]

### Step 7: Write the expression for the final internal energy
The final internal energy can also be expressed as:
\[ U = \frac{3}{2} n R T_f = \frac{3}{2} \times 0.15 \times R \times T_f \]

### Step 8: Set the two expressions for internal energy equal to each other
\[ 60R = \frac{3}{2} \times 0.15 \times R \times T_f \]

### Step 9: Simplify and solve for \( T_f \)
Cancel \( R \) from both sides:
\[ 60 = \frac{3}{2} \times 0.15 \times T_f \]
\[ 60 = 0.225 T_f \]
\[ T_f = \frac{60}{0.225} = \frac{6000}{225} = \frac{800}{3} \approx 266.67 \text{ K} \]

### Final Answer
The final equilibrium temperature \( T_f \) is approximately \( 267 \) K.

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