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MHT CET 2026 (15 May, Shift 2) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from MHT CET 2026 (15 May, Shift 2)

Q1MHT CET 2026 · 15 MAY · SHIFT 2Medium

Degrees of Freedom & Internal Energy

The internal energy of a diatomic molecule (air) in 4 m×4 m×3 m4\text{ m} \times 4\text{ m} \times 3\text{ m} sized room at 10510^{5} N/m2^{2} atmospheric pressure will be x×106x \times 10^{6} J. The value of xx is

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Answer: (b)

  1. For a diatomic gas U=52nRT=52PVU = \frac{5}{2}nRT = \frac{5}{2}PV.
  2. V=4×4×3=48 m3V = 4 \times 4 \times 3 = 48\ \text{m}^3
  3. U=52×105×48=12×106U = \frac{5}{2} \times 10^5 \times 48 = 12 \times 10^6 J
  4. So the answer is (b) x=12x = 12.
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