A gaseos mixture consists of equal number of moles of two ideal gases having adiabatic exponents `gamma_(1)` and `gamma_(2)` and molar speific heats at constant volume `C_(v_(1))` and `C_(v_(2))` respectively. Which of the following statements is/are correct?

Options

A. Adiabatic exponent for gaseous mixture is equal to `(gamma_(1)+gamma_(2))/(2)`
B. Molar specific heat at constant volume for gaseous mixture is equal to `(C_(v_(1))+C_(v_(2)))/(2)`
C. Molar specific heat at constant pressure for gaseous mixture is equal to `(C_(v_(1))+C_(v_(2))+R)/(2)`
D. Adiabatic exponent for gaseous mixture is `1 +(2R)/(C_(v_(1))+C_(v_(2)))`

Explanation

To solve the problem regarding the gaseous mixture of two ideal gases, we need to analyze the given information and derive the necessary equations step by step.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have two ideal gases with equal number of moles.
- The adiabatic exponents are denoted as \( \gamma_1 \) and \( \gamma_2 \).
- The molar specific heats at constant volume are \( C_{v_1} \) and \( C_{v_2} \).

2. **Using the Relation between \( C_p \) and \( C_v \)**:
- The relationship between the specific heats is given by:
\[ C_p - C_v = R \]
- For the first gas:
\[ C_{p_1} = C_{v_1} + R \]
- For the second gas:
\[ C_{p_2} = C_{v_2} + R \]

3. **Calculating the Molar Heat Capacity at Constant Volume for the Mixture**:
- Since the number of moles of both gases is equal, let \( n_1 = n_2 = n \).
- The molar heat capacity at constant volume for the mixture can be calculated as:
\[ C_{v_{\text{mix}}} = \frac{n C_{v_1} + n C_{v_2}}{n + n} = \frac{C_{v_1} + C_{v_2}}{2} \]

4. **Calculating the Molar Heat Capacity at Constant Pressure for the Mixture**:
- Using the previously calculated \( C_{v_{\text{mix}}} \):
\[ C_{p_{\text{mix}}} = C_{v_{\text{mix}}} + R = \frac{C_{v_1} + C_{v_2}}{2} + R \]

5. **Calculating the Adiabatic Exponent for the Mixture**:
- The adiabatic exponent \( \gamma_{\text{mix}} \) is given by:
\[ \gamma_{\text{mix}} = \frac{C_{p_{\text{mix}}}}{C_{v_{\text{mix}}}} \]
- Substituting the values:
\[ \gamma_{\text{mix}} = \frac{\frac{C_{v_1} + C_{v_2}}{2} + R}{\frac{C_{v_1} + C_{v_2}}{2}} = 1 + \frac{2R}{C_{v_1} + C_{v_2}} \]

6. **Expressing \( C_{v_1} \) and \( C_{v_2} \) in terms of \( \gamma_1 \) and \( \gamma_2 \)**:
- From the relation \( C_v = \frac{R}{\gamma - 1} \):
\[ C_{v_1} = \frac{R}{\gamma_1 - 1}, \quad C_{v_2} = \frac{R}{\gamma_2 - 1} \]

7. **Substituting \( C_{v_1} \) and \( C_{v_2} \) into \( \gamma_{\text{mix}} \)**:
- Substitute these into the expression for \( \gamma_{\text{mix}} \):
\[ \gamma_{\text{mix}} = 1 + \frac{2R}{\frac{R}{\gamma_1 - 1} + \frac{R}{\gamma_2 - 1}} = 1 + \frac{2(\gamma_1 - 1)(\gamma_2 - 1)}{(\gamma_1 + \gamma_2 - 2)} \]

8. **Final Expressions**:
- The final expressions for \( C_{v_{\text{mix}}} \), \( C_{p_{\text{mix}}} \), and \( \gamma_{\text{mix}} \) are:
- \( C_{v_{\text{mix}}} = \frac{C_{v_1} + C_{v_2}}{2} \)
- \( C_{p_{\text{mix}}} = \frac{C_{v_1} + C_{v_2}}{2} + R \)
- \( \gamma_{\text{mix}} = 1 + \frac{2R}{C_{v_1} + C_{v_2}} \)

### Conclusion:
After calculating the values, we can check the correctness of the statements given in the question.