B

To find the value of \( \gamma \) for the mixture of a monatomic gas and a diatomic gas, we can use the following formula:
\( \gamma = \frac{C_p}{C_v} \), where \( C_p \) is the molar specific heat at constant pressure and \( C_v \) is the molar specific heat at constant volume.
For the monatomic gas (1 mole):
- \( C_{v1} = \frac{3}{2} R \)
- \( C_{p1} = C_{v1} + R = \frac{3}{2}R + R = \frac{5}{2}R \)
For the diatomic gas (1 mole):
- \( C_{v2} = \frac{5}{2} R \)
- \( C_{p2} = C_{v2} + R = \frac{5}{2}R + R = \frac{7}{2}R \)
Now, the total molar specific heats for the mixture:
\( C_v = C_{v1} + C_{v2} = \frac{3}{2}R + \frac{5}{2}R = 4R \)
\( C_p = C_{p1} + C_{p2} = \frac{5}{2}R + \frac{7}{2}R = 6R \)
Now calculate the mixture \( \gamma \):
\( \gamma = \frac{C_p}{C_v} = \frac{6R}{4R} = \frac{6}{4} = 1.5 \)
Thus, the value of \( \gamma \) for the mixture is 1.50.
Therefore, the correct answer is option B.