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JEE Main 2020 (5 Sep, Shift 1) · Pressure & Kinetic Energy of a Gas

Pressure & Kinetic Energy of a Gas question from JEE Main 2020 (5 Sep, Shift 1)

Q35JEE Main 2020 · 5 SEP · SHIFT 1Medium

Pressure & Kinetic Energy of a Gas

Number of molecules in a volume of 4 cm3 of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to?

(Given, mean kinetic energy of a molecule

(at T) is 4 ×\times 10–14 erg, g = 980 cm/s2, density of

mercury = 13.6 g/cm3)

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Answer: (d)

  1. P=hρg=2×13.6×980=26656P = h\rho g = 2 \times 13.6 \times 980 = 26656 dyne/cm2^2
  2. PV=23NEˉPV = \frac{2}{3}N\bar{E}, so N=3PV2EˉN = \frac{3PV}{2\bar{E}}
  3. N=3×26656×42×4×10−14≈4.0×1018N = \frac{3 \times 26656 \times 4}{2 \times 4 \times 10^{-14}} \approx 4.0 \times 10^{18}
  4. So the answer is (d) 4.0×10184.0 \times 10^{18}.
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