Match the $$\frac{C_p}{C_v}$$ ratio for ideal gases with different type of molecules:
Molecule Type                                    $$C_p/C_v$$
(A) Monoatomic                                 (I) 7/5
(B) Diatomic rigid molecules             (II) 9/7
(C) Diatomic non-rigid molecules      (III) 4/3
(D) Triatomic rigid molecules            (IV) 5/3

For an ideal gas, the relation between the molar specific heats at constant pressure and volume is written first:

$$C_p = C_v + R$$

Here $$R$$ is the universal gas constant. To convert this into the required ratio, we recall the expression of $$C_v$$ in terms of the number of degrees of freedom $$f$$ of a single molecule:

$$C_v=\dfrac{f}{2}\,R$$

Substituting this value of $$C_v$$ in the earlier relation gives

$$C_p = \dfrac{f}{2}R + R = \left(\dfrac{f}{2}+1\right)R = \dfrac{f+2}{2}\,R$$

Now we build the desired ratio $$\dfrac{C_p}{C_v}$$, usually denoted by $$\gamma$$:

$$\gamma \;=\; \dfrac{C_p}{C_v} \;=\; \dfrac{\dfrac{f+2}{2}R}{\dfrac{f}{2}R} \;=\; \dfrac{f+2}{f} \;=\; 1 + \dfrac{2}{f}$$

Therefore, once we know the degrees of freedom $$f$$ of each type of molecule, we can obtain its $$\gamma$$ value directly.

We list the customary degrees of freedom for various molecular models:

• A monoatomic gas possesses only three translational degrees of freedom, so $$f=3$$.
• A diatomic rigid molecule (rotations allowed about two perpendicular axes) has five degrees of freedom, so $$f=5$$.
• A diatomic non-rigid molecule includes two additional vibrational degrees of freedom, making $$f=7$$.
• A triatomic rigid molecule (linear or planar but counted rigid) enjoys six degrees of freedom, so $$f=6$$.

Now we place each value of $$f$$ into $$\gamma = 1+\dfrac{2}{f}$$ and simplify step by step.

Monoatomic gas ($$f=3$$)

$$\gamma_{\text{mono}} = 1 + \dfrac{2}{3} = \dfrac{3}{3} + \dfrac{2}{3} = \dfrac{5}{3}$$

Diatomic rigid gas ($$f=5$$)

$$\gamma_{\text{dia, rigid}} = 1 + \dfrac{2}{5} = \dfrac{5}{5} + \dfrac{2}{5} = \dfrac{7}{5}$$

Diatomic non-rigid gas ($$f=7$$)

$$\gamma_{\text{dia, non-rigid}} = 1 + \dfrac{2}{7} = \dfrac{7}{7} + \dfrac{2}{7} = \dfrac{9}{7}$$

Triatomic rigid gas ($$f=6$$)

$$\gamma_{\text{tria, rigid}} = 1 + \dfrac{2}{6} = 1 + \dfrac{1}{3} = \dfrac{4}{3}$$

We now match each molecule category with its calculated $$\gamma$$ ratio:

$$\begin{aligned} \text{Monoatomic} &\longrightarrow \dfrac{5}{3}\;(\text{IV})\\[4pt] \text{Diatomic rigid} &\longrightarrow \dfrac{7}{5}\;(\text{I})\\[4pt] \text{Diatomic non-rigid} &\longrightarrow \dfrac{9}{7}\;(\text{II})\\[4pt] \text{Triatomic rigid} &\longrightarrow \dfrac{4}{3}\;(\text{III}) \end{aligned}$$

Reading these links against the options supplied, we obtain the correspondence

(A) - (IV), (B) - (I), (C) - (II), (D) - (III), which is exactly the content of Option C in the list.

Hence, the correct answer is Option 3.