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JEE Main 2020 (6 Sep, Shift 1) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2020 (6 Sep, Shift 1)

Q32JEE Main 2020 · 6 SEP · SHIFT 1Difficult

Degrees of Freedom & Internal Energy

Initially a gas of diatomic molecules is contained in a cylinder of volume V1 at a pressure P1 and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume 2V1 is given by P2. The ratio P2P1\frac{P_2}{P_1} is ________.

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

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Answer: 5

  1. 25% dissociate: moles become 0.75n+2(0.25n)=1.25n0.75n + 2(0.25n) = 1.25n.
  2. PV=nRT⇒P2P1=n2n1⋅T2T1⋅V1V2PV = nRT \Rightarrow \frac{P_2}{P_1} = \frac{n_2}{n_1}\cdot\frac{T_2}{T_1}\cdot\frac{V_1}{V_2}
  3. P2P1=1.25×2000250×12=1.25×8×0.5\frac{P_2}{P_1} = 1.25 \times \frac{2000}{250} \times \frac{1}{2} = 1.25 \times 8 \times 0.5
  4. So P2P1=5\frac{P_2}{P_1} = 5.
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