JEE Main Thermodynamics: Cp/Cv Matching Concept 💡

 

❓ Question

Match the following:

LIST–I

A. Triatomic rigid gas
B. Diatomic non-rigid gas
C. Monoatomic gas
D. Diatomic rigid gas

LIST–II

1. $C_p/C_v = 5/3$
   

2. $C_p/C_v = 7/5$
   

3. $C_p/C_v = 4/3$
   

4. $C_p/C_v = 9/7$
   

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🖼️ Question Image

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✍️ Short Solution

This is a pure degrees of freedom question.
No formulas to memorize separately — just remember:

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

Where $f$= degrees of freedom 🔥

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🔹 Step 1 — Formula to Remember (MOST IMPORTANT 💯)**

For ideal gas:

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

So:

$$\boxed{\gamma = \frac{f+2}{f}}$$

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🔹 Step 2 — Monoatomic Gas

Monoatomic gas has:

$$f = 3$$

So:

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

✔️ Matches option 1

So:

$$\boxed{C \rightarrow 1}$$

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🔹 Step 3 — Diatomic Rigid Gas

Rigid diatomic gas:

• 3 translational

• 2 rotational

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

✔️ Matches option 2

$$\boxed{D \rightarrow 2}$$

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🔹 Step 4 — Triatomic Rigid Gas

Rigid triatomic (non-linear):

• 3 translational

• 3 rotational

$$f = 6$$
$$\gamma = \frac{6+2}{6} = \frac{8}{6} = \frac{4}{3}$$

✔️ Matches option 3

$$\boxed{A \rightarrow 3}$$

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🔹 Step 5 — Diatomic Non-Rigid Gas

Non-rigid diatomic:

• Vibrational mode active

• 3 translational

• 2 rotational

• 2 vibrational

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

✔️ Matches option 4

$$\boxed{B \rightarrow 4}$$

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✅ Final Matching

$$\boxed{ A \rightarrow 3,\quad B \rightarrow 4,\quad C \rightarrow 1,\quad D \rightarrow 2 }$$

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⭐ Golden JEE Insight

Just remember these 4 values:

• Monoatomic → $f=3$ → $5/3$
  

• Diatomic rigid → $f=5$ → $7/5$
  

• Triatomic rigid → $f=6$→ $4/3$
  

• Diatomic non-rigid → $f=7$ → $9/7$
  

🧠 One-line memory trick:

Gamma decreases as degrees of freedom increase