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JEE Main 2021 (26 Aug, Shift 2) · Gas Laws & Ideal Gas Equation

Gas Laws & Ideal Gas Equation question from JEE Main 2021 (26 Aug, Shift 2)

Q61JEE Main 2021 · 26 AUG · SHIFT 2Medium

Gas Laws & Ideal Gas Equation

A cylindrical container of volume 4.0 ×\times 10−-3 m3 contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is:

[Take gas constant as 8.3 J mol−1{}^{-1} K−1{}^{-1}]

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Answer: (c)

  1. Total moles n=1+2=3n = 1 + 2 = 3 (Dalton's law: pressures add).
  2. P=nRTV=3×8.3×4004.0×10−3=2.49×106 PaP = \frac{nRT}{V} = \frac{3 \times 8.3 \times 400}{4.0 \times 10^{-3}} = 2.49 \times 10^6\text{ Pa}.
  3. So the answer is (c) 24.9×10524.9 \times 10^5 Pa.
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