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JEE Main 2021 (16 Mar, Shift 1) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2021 (16 Mar, Shift 1)

Q25JEE Main 2021 · 16 MAR · SHIFT 1Medium

Degrees of Freedom & Internal Energy

The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is:

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Answer: (d)

  1. Moles: O2=1632=0.5\text{O}_2 = \frac{16}{32} = 0.5, N2=2828=1\text{N}_2 = \frac{28}{28} = 1, CO2=4444=1\text{CO}_2 = \frac{44}{44} = 1; total n=2.5n = 2.5.
  2. Dalton's law: P=nRTV=2.5RTVP = \frac{nRT}{V} = \frac{2.5RT}{V}
  3. So the answer is (d) 52RTV\frac{5}{2}\frac{RT}{V}.
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