If 2 mol of an ideal monatomic gas at temperature `T_(0)` are mixed with 4 mol of another ideal monatoic gas at temperature `2 T_(0)` then the temperature of the mixture is

Options

A. `(5)/(3)T_(0)`
B. `(3)/(2)T_(0)`
C. `(4)/(3)T_(0)`
D. `(5)/(4)T_(0)`

Answer

To find the temperature of the mixture of two ideal monatomic gases, we can use the principle of conservation of energy. Here's a step-by-step solution:

### Step 1: Understand the given data
We have:
- Gas 1: 2 moles at temperature \( T_0 \)
- Gas 2: 4 moles at temperature \( 2T_0 \)

### Step 2: Write the formula for internal energy
The internal energy \( U \) of an ideal gas is given by:
\[ U = \frac{f}{2} n R T \]
where:
- \( f \) is the degrees of freedom (for monatomic gas, \( f = 3 \)),
- \( n \) is the number of moles,
- \( R \) is the universal gas constant,
- \( T \) is the temperature.

### Step 3: Calculate the internal energy of each gas
For Gas 1:
\[ U_1 = \frac{3}{2} \cdot n_1 \cdot R \cdot T_1 = \frac{3}{2} \cdot 2 \cdot R \cdot T_0 = 3 R T_0 \]

For Gas 2:
\[ U_2 = \frac{3}{2} \cdot n_2 \cdot R \cdot T_2 = \frac{3}{2} \cdot 4 \cdot R \cdot (2T_0) = 12 R T_0 \]

### Step 4: Write the total internal energy of the mixture
The total internal energy \( U \) of the mixture is:
\[ U = U_1 + U_2 = 3 R T_0 + 12 R T_0 = 15 R T_0 \]

### Step 5: Set up the equation for the mixture
For the mixture of gases, the total internal energy can also be expressed as:
\[ U = \frac{3}{2} (n_1 + n_2) R T \]
where \( n_1 + n_2 = 2 + 4 = 6 \) moles.

### Step 6: Equate the two expressions for internal energy
Now we equate the two expressions for \( U \):
\[ 15 R T_0 = \frac{3}{2} \cdot 6 \cdot R \cdot T \]

### Step 7: Simplify the equation
\[ 15 R T_0 = 9 R T \]
Dividing both sides by \( R \) (assuming \( R \neq 0 \)):
\[ 15 T_0 = 9 T \]

### Step 8: Solve for \( T \)
\[ T = \frac{15 T_0}{9} = \frac{5}{3} T_0 \]

### Conclusion
The temperature of the mixture is:
\[ T = \frac{5}{3} T_0 \]