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: Identify the Given Information
- We have 2 moles of gas 1 at temperature \( T_0 \) (let's denote this as \( n_1 = 2 \) and \( T_1 = T_0 \)).
- We have 4 moles of gas 2 at temperature \( 2T_0 \) (let's denote this as \( n_2 = 4 \) and \( T_2 = 2T_0 \)).

### Step 2: Write the Equation for Total Internal Energy
The total internal energy of an ideal gas can be expressed as:
\[ U = \frac{f}{2} nRT \]
where \( f \) is the degrees of freedom (for a monatomic gas, \( f = 3 \)), \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.

### Step 3: Set Up the Energy Conservation Equation
The total internal energy of the mixture should equal the sum of the internal energies of the two gases:
\[ U_{\text{mixture}} = U_1 + U_2 \]
Substituting the expressions for internal energy:
\[ \frac{3}{2} n_{\text{final}} R T = \frac{3}{2} n_1 R T_1 + \frac{3}{2} n_2 R T_2 \]

### Step 4: Simplify the Equation
The \( \frac{3}{2} R \) cancels out from both sides:
\[ n_{\text{final}} T = n_1 T_1 + n_2 T_2 \]

### Step 5: Calculate the Total Number of Moles
The total number of moles in the mixture:
\[ n_{\text{final}} = n_1 + n_2 = 2 + 4 = 6 \]

### Step 6: Substitute the Known Values
Now substituting the values into the equation:
\[ 6T = 2T_0 + 4(2T_0) \]
This simplifies to:
\[ 6T = 2T_0 + 8T_0 = 10T_0 \]

### Step 7: Solve for the Final Temperature \( T \)
Now, divide both sides by 6:
\[ T = \frac{10T_0}{6} = \frac{5T_0}{3} \]

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