JEE Mains 2025 Physics Adiabatic Compression question of Monoatomic Gas | 8th April Shift 2 Question Solved
In this post, we solve a key Physics question from JEE Mains 2025 – 8th April Shift 2. The problem involves adiabatic compression of a monoatomic gas, an important concept in thermodynamics. Follow along to understand the approach and final answer with full clarity.
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Question
A monoatomic gas having γ = 5/3 is stored in a thermally insulated container and the gas is suddenly compressed to (1/8)ᵗʰ of its initial volume. The ratio of final pressure and initial pressure is (γ is the ratio of specific heats of the gas at constant pressure and at constant volume):
- (1) 40
- (2) 28
- (3) 32
- (4) 16
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Detailed Step-by-Step Solution
Step 1: Identify the Type of Process
The gas is stored in a thermally insulated container, and compression happens suddenly. This indicates an adiabatic process, where no heat exchange occurs.
Step 2: Use the Adiabatic Equation
For adiabatic processes, the pressure and volume relation is given by:
P × Vγ = constant
From this, we derive:
P₂ / P₁ = (V₁ / V₂)γ
Step 3: Insert the Given Values
Given γ = 5/3 for a monoatomic gas and V₂ = (1/8) × V₁:
P₂ / P₁ = (V₁ / (1/8)V₁)5/3 = 85/3
Now, simplify:
85/3 = (2³)5/3 = 2⁵ = 32
Step 4: Final Answer
Correct Option: (3) 32
Conclusion
This JEE Mains 2025 Physics question tests your understanding of adiabatic processes and the application of thermodynamic formulas. Mastering such questions will help you score higher in the actual exam.
