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Physics · Thermodynamics

JEE Main 2025 — 2 April, Evening Shift — Question 66

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 \qquad ×106 J\times 10^{6} \mathrm{~J}. (Consider air as diatomic molecule)

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

U=nCvTU=n C_{v} T

=n×5R2T=52(nRT)=52PVU=52×105×4×4×3=12×106 J\begin{aligned} \\ & =n \times \frac{5 R}{2} T \\ & =\frac{5}{2}(n R T)=\frac{5}{2} P V U \\ & =\frac{5}{2} \times 10^{5} \times 4 \times 4 \times 3 \\ & =12 \times 10^{6} \mathrm{~J} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
2 April, Evening Shift
Subject
Physics
Chapter
Thermodynamics
Topic
Calculation of Work, Heat and Internal Energy

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