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JEE Main 2022 (26 Jun, Shift 1) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2022 (26 Jun, Shift 1)

Q22JEE Main 2022 · 26 JUN · SHIFT 1Medium

Degrees of Freedom & Internal Energy

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel, temperature of the gas increases by:

(R = universal gas constant)

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Answer: (b)

  1. γ=1.4⇒Cv=Rγ−1=52R\gamma = 1.4 \Rightarrow C_v = \frac{R}{\gamma-1} = \frac{5}{2}R.
  2. Per mole: 12Mv2=CvΔT\frac{1}{2}Mv^2 = C_v\Delta T.
  3. ΔT=Mv22⋅52R=Mv25R\Delta T = \frac{Mv^2}{2\cdot\frac{5}{2}R} = \frac{Mv^2}{5R}
  4. So the answer is (b) Mv25R\frac{Mv^2}{5R}.
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