Two perfect gases having masses `m_(1)andm_(2)` at temperatures `T_(1)andT_(2)` respectively are mixed without any loss of energy. If the molecular weights of the gas are `M_(1)andM_(2)` respectively, then the final temperature of the mixture is

Options

A. `((m_(1)T_(1)+m_(2)T_(2)))/((m_(1)+m_(2)))`
B. `((M_(1)T_(1)+M_(2)T_(2)))/((M_(1)+M_(2)))`
C. `([((m_(1)T_(1))/(M_(1)))+((m_(2)T_(2))/(M_(2)))])/([((m_(1))/(M_(1)))+((m_(2))/(M_(2)))])`
D. `([(M_(1)T_(1))/(m_(1))+(M_(2)T_(2))/(m_(2))])/([((M_(1))/(m_(1)))+((M_(2))/(m_(2)))])`

Explanation

To find the final temperature of the mixture of two perfect gases, we can use the principle of conservation of energy, which states that the total energy before mixing equals the total energy after mixing. Here’s a step-by-step solution:

### Step 1: Determine the number of moles of each gas
The number of moles \( n_1 \) of gas 1 and \( n_2 \) of gas 2 can be calculated using the formula:
\[ n_1 = \frac{m_1}{M_1} \]
\[ n_2 = \frac{m_2}{M_2} \]

### Step 2: Set up the equation for final temperature
The final temperature \( T_f \) of the mixture can be derived from the conservation of energy, which gives us:
\[ n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_f \]

### Step 3: Rearranging the equation
We can rearrange the equation to solve for \( T_f \):
\[ T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2} \]

### Step 4: Substitute the values of \( n_1 \) and \( n_2 \)
Substituting the expressions for \( n_1 \) and \( n_2 \) into the equation:
\[ T_f = \frac{\left(\frac{m_1}{M_1}\right) T_1 + \left(\frac{m_2}{M_2}\right) T_2}{\left(\frac{m_1}{M_1}\right) + \left(\frac{m_2}{M_2}\right)} \]

### Final Result
Thus, the final temperature \( T_f \) of the mixture is given by:
\[ T_f = \frac{\frac{m_1 T_1}{M_1} + \frac{m_2 T_2}{M_2}}{\frac{m_1}{M_1} + \frac{m_2}{M_2}} \]

---