NEET 2015 Physics Specific Heats MCQ Question

🧠 Explanation

For a gas having $n$ degrees of freedom, the molar specific heat at constant volume is:

$$C_v=\frac{n}{2}R$$

Using Mayer’s relation:

$$C_p-C_v=R$$

Therefore,

$$C_p=\frac{n}{2}R+R =\frac{n+2}{2}R$$

Hence,

$$\gamma=\frac{C_p}{C_v} =\frac{(n+2)/2}{n/2} =\boxed{1+\frac{2}{n}}$$

✅ Answer: (C) $\displaystyle 1+\frac{2}{n}$

⚡ Quick Revision:

$$\boxed{\gamma=1+\frac{2}{n}}$$

• Monoatomic: $n=3\Rightarrow\gamma=\frac53$
• Diatomic: $n=5\Rightarrow\gamma=\frac75$