1. Use the relation between degrees of freedom and heat capacity ratio

For an ideal gas with degrees of freedom $f$,

$$C_v = \frac{f}{2}R, \qquad C_p = C_v + R = \frac{f+2}{2}R$$

Hence,

$$\gamma = \frac{C_p}{C_v} = \frac{f+2}{f}$$

2. Find $\gamma$ for each gas type

(C) Monoatomic gas

A monoatomic gas has $f=3$ So,

$$\gamma = \frac{3+2}{3} = \frac{5}{3}$$

Thus,

$$(C) \to (I)$$

(D) Diatomic rigid gas

A rigid diatomic gas has 3 translational and 2 rotational degrees of freedom: $f=5$ So,

$$\gamma = \frac{5+2}{5} = \frac{7}{5}$$

Thus,

$$(D) \to (II)$$

(B) Diatomic non-rigid gas

For non-rigid diatomic gas, vibrational mode is also active. Then total degrees of freedom are: $f=7$ (3 translational + 2 rotational + 2 vibrational)

So,

$$\gamma = \frac{7+2}{7} = \frac{9}{7}$$

Thus,

$$(B) \to (IV)$$

(A) Triatomic rigid gas

For a rigid triatomic gas, commonly taken degrees of freedom: $f=6$ So,

$$\gamma = \frac{6+2}{6} = \frac{4}{3}$$

Thus,

$$(A) \to (III)$$

3. Final matching

$$A \to III, \quad B \to IV, \quad C \to I, \quad D \to II$$

4. Compare with options This corresponds to Option A.

5. Comparison with stored correct answer Stored correct answer is A, which matches our derived answer.