`n_1` mole of a monoatomic gas is mixed with `n_2` mole of a diatomic gas. (Assume gases are ideal). If for the mixture adiabatic contant is found to be 1.5, then `n_1 : n_2` is :

Options

A. `1 : 1`
B. `1 : 3`
C. `2 :3`
D. `3 : 2`

Explanation

To solve the problem, we need to find the ratio \( n_1 : n_2 \) given that \( n_1 \) moles of a monoatomic gas are mixed with \( n_2 \) moles of a diatomic gas, and the adiabatic constant \( \gamma \) of the mixture is 1.5.

### Step-by-Step Solution:

1. **Identify Degrees of Freedom**:
- For a monoatomic gas, the degrees of freedom \( F_1 = 3 \).
- For a diatomic gas, the degrees of freedom \( F_2 = 5 \).

2. **Calculate Molar Specific Heats**:
- The molar specific heat at constant volume for the monoatomic gas:
\[ C_{V1} = \frac{F_1}{2} R = \frac{3}{2} R \]
- The molar specific heat at constant volume for the diatomic gas:
\[ C_{V2} = \frac{F_2}{2} R = \frac{5}{2} R \]

3. **Calculate Molar Specific Heats at Constant Pressure**:
- The molar specific heat at constant pressure for the monoatomic gas:
\[ C_{P1} = C_{V1} + R = \frac{3}{2} R + R = \frac{5}{2} R \]
- The molar specific heat at constant pressure for the diatomic gas:
\[ C_{P2} = C_{V2} + R = \frac{5}{2} R + R = \frac{7}{2} R \]

4. **Use the Formula for the Adiabatic Constant**:
- The formula for the adiabatic constant \( \gamma \) of the mixture is given by:
\[ \gamma = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 C_{V1} + n_2 C_{V2}} \]
- Substituting the values we have:
\[ 1.5 = \frac{n_1 \left(\frac{5}{2} R\right) + n_2 \left(\frac{7}{2} R\right)}{n_1 \left(\frac{3}{2} R\right) + n_2 \left(\frac{5}{2} R\right)} \]

5. **Cancel Out Common Terms**:
- Since \( R \) is common in both the numerator and denominator, we can cancel it out:
\[ 1.5 = \frac{n_1 \left(\frac{5}{2}\right) + n_2 \left(\frac{7}{2}\right)}{n_1 \left(\frac{3}{2}\right) + n_2 \left(\frac{5}{2}\right)} \]

6. **Cross Multiply**:
- Cross multiplying gives:
\[ 1.5 \left(n_1 \left(\frac{3}{2}\right) + n_2 \left(\frac{5}{2}\right)\right) = n_1 \left(\frac{5}{2}\right) + n_2 \left(\frac{7}{2}\right) \]
- This simplifies to:
\[ 4.5 n_1 + 7.5 n_2 = 5 n_1 + 7 n_2 \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ 4.5 n_1 - 5 n_1 + 7.5 n_2 - 7 n_2 = 0 \]
- This simplifies to:
\[ -0.5 n_1 + 0.5 n_2 = 0 \]
- Thus, we have:
\[ n_1 = n_2 \]

8. **Finding the Ratio**:
- Therefore, the ratio \( n_1 : n_2 = 1 : 1 \).

### Final Answer:
The ratio \( n_1 : n_2 \) is \( 1 : 1 \).