The internal energy of air in 4 m × 4 m × 3 m sized room at 1 atmospheric pressure will be___________________ × 10⁶ J (Consider air as diatomic molecule) Numerical answer type. Work it out, then open the answer.

Answer: 12. Diatomic air: U = 5/2nRT = 5/2PV. V = 4 × 4 × 3 = 48 m³, P = 1.013 × 10⁵ Pa U = 5/2 × 1.013 × 10⁵ × 48 ≈ 12.2 × 10⁶ J So the answer is 12 ( × 10⁶ J).

The internal energy of air in 4 m × 4 m × 3 m sized room at 1… | JEE Main 2025 (2 Apr, Shift 2)

Q8JEE Main 2025 · 2 APR · SHIFT 2Medium

Degrees of Freedom & Internal Energy

The internal energy of air in $4\text{ m} \times 4\text{ m} \times 3\text{ m}$ sized room at 1 atmospheric pressure will be___________________ $\times 10^{6}\text{ J}$

(Consider air as diatomic molecule)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 12

Diatomic air: $U = \frac{5}{2}nRT = \frac{5}{2}PV$ .
$V = 4 \times 4 \times 3 = 48\ \text{m}^3$ , $P = 1.013 \times 10^5$ Pa
$U = \frac{5}{2} \times 1.013 \times 10^5 \times 48 \approx 12.2 \times 10^6$ J
So the answer is $12$ ( $\times 10^6$ J).

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