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JEE Main 2024 (29 Jan, Shift 2) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2024 (29 Jan, Shift 2)

Q12JEE Main 2024 · 29 JAN · SHIFT 2Medium

Degrees of Freedom & Internal Energy

NN moles of a polyatomic gas (f=6)(f = 6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of NN is:

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Answer: (c)

  1. Mixture acts diatomic, so Cv,mix=52RC_{v,mix} = \frac{5}{2}R.
  2. Cv,mix=N(3R)+2(32R)N+2=52RC_{v,mix} = \frac{N(3R) + 2\left(\frac{3}{2}R\right)}{N + 2} = \frac{5}{2}R
  3. 3N+3=52N+5⇒N2=2⇒N=43N + 3 = \frac{5}{2}N + 5 \Rightarrow \frac{N}{2} = 2 \Rightarrow N = 4
  4. So the answer is (c) 4.
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