RE-NEET 2026 Physics Adiabatic Processes MCQ Question with Solution

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (γ=5/3) decreases from 60K to 50K. The work done by the gas in the process is : (Take the universal gas constant as R=8.3 J mol⁻¹ K⁻¹)

A

83 J

B

124.5 J

C

166 J

D

41.5 J

Correct Answer

Option B

Detailed Explanation

NCERT Topic: Thermodynamics → First Law of Thermodynamics → Adiabatic Process

Difficulty: 🟡 Medium

✅ Ans: 2 → 124.5 J

For an adiabatic process:

Q=0

By first law,

Q=\Delta U+W \Rightarrow W=-\Delta U

For monatomic gas:

C_V=\frac32R

W=nC_V(T_1-T_2)

=1\times\frac32(8.3)(60-50)

\boxed{W=124.5\,J}

❌ Why others are wrong?

83 J ❌ → uses $R\Delta T$ , ignoring $C_V=\frac32R$
166 J ❌ → incorrect heat-capacity factor
41.5 J ❌ → incorrect numerical factor

📌 NCERT: “In an adiabatic process, there is no exchange of heat between the system and its surroundings.”

🧠 NEET Trick: Adiabatic → $Q=0$ → $W=-\Delta U=nC_V(T_1-T_2)$ .

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