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JEE Main 2009 (AIEEE) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2009 (AIEEE)

Q40JEE Main 2009 · AIEEEMedium

Degrees of Freedom & Internal Energy

One kg\text{kg} of a diatomic gas is at a pressure of 8×104 N/m2.8 \times 10^{4}\text{ N/m}^{2}. The density of the gas is 4 kg/m34\text{ kg/m}^{3}. What is the energy of the gas due to its thermal motion?

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Answer: (a)

  1. V=mρ=14=0.25 m3V = \frac{m}{\rho} = \frac{1}{4} = 0.25\ \text{m}^3
  2. Diatomic: U=52PV=52×8×104×0.25U = \frac{5}{2}PV = \frac{5}{2} \times 8 \times 10^4 \times 0.25
  3. U=5×104U = 5 \times 10^4 J
  4. So the answer is (a) 5×1045 \times 10^4 J.
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