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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2025 · 2 April · Evening Shift · Physics · Thermodynamics

The internal energy of air in 4 ~m × 4 ~m × 3 ~m sized room at 1...

The internal energy of air in 4 m×4 m×3 m4 \mathrm{~m} \times 4 \mathrm{~m} \times 3 \mathrm{~m} sized room at 1 atmospheric pressure will be _______ ×106 J\times 10^6 \mathrm{~J}. (Consider air as diatomic molecule)

Correct answer

0

Step-by-step solution

To find the internal energy of gas in the room.
<br/>U=nCv T=n5RT2=52PV=52×105×48=12×106 J<br/>\begin{aligned}<br/>& \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \mathrm{~T}=\mathrm{n} \frac{5 \mathrm{RT}}{2} \\ & =\frac{5}{2} \mathrm{PV}=\frac{5}{2} \times 10^5 \times 48=12 \times 10^6 \mathrm{~J}<br/>\end{aligned}
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