1. Use the relation for ideal gases

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

$C_V = \frac{f}{2}R, \qquad C_P = C_V + R = \left(\frac{f}{2}+1\right)R$

Hence,

$\gamma = \frac{C_P}{C_V} = \frac{\frac{f}{2}+1}{\frac{f}{2}} = \frac{f+2}{f}$

2. Find $\gamma$ for each molecule type

(A) Monatomic

A monatomic gas has only 3 translational degrees of freedom.

$f=3$

So,

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

Thus,

$(A) \to (IV)$

(B) Diatomic rigid molecules

A rigid diatomic molecule has:

• 3 translational degrees
• 2 rotational degrees

So,

$f=5$

Therefore,

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

Thus,

$(B) \to (I)$

(C) Diatomic non-rigid molecules

A non-rigid diatomic molecule also has one vibrational mode active. A vibrational mode contributes 2 degrees of freedom (one kinetic + one potential).

So,

$f = 3 + 2 + 2 = 7$

Hence,

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

Thus,

$(C) \to (II)$

(D) Triatomic rigid molecules

For a rigid triatomic molecule (non-linear rigid molecule),

• 3 translational degrees
• 3 rotational degrees

So,

$f=6$

Hence,

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

Thus,

$(D) \to (III)$

3. Final matching

The correct matching is:

• $(A) \to (IV)$
• $(B) \to (I)$
• $(C) \to (II)$
• $(D) \to (III)$

This corresponds to Option C.

4. Comparison with stored answer

Stored correct answer: C

Derived answer: C

So, the derived answer agrees with the stored answer.