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JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases

Modern vacuum pumps can evacuate a vessel down to a pressure of \(4.0 \times {10^{ - 15}}\, atm\) at room temperature \((300\, K)\). Taking \(R = 8.0\, JK^{-1}\, mole^{-1}\) , \(1\, atm = 10^5\, Pa\) and \(N_ {Avogadro} = 6 \times 10^{23}\, mole^{-1}\) , the mean distance between molecules of gas in an evacuated vessel will be of the order of

  1. A \(0.2\,\mu m\)
  2. B \(0.2\,mm\)
  3. C \(0.2\,cm\)
  4. D \(0.2\,nm\)
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Answer & Solution

Correct Answer

(B) \(0.2\,mm\)

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Detailed explanation

Formula for mean free path- \(Y=\frac{K T}{\sqrt{2} \pi \sigma^{2} p}\) -wherein \(\sigma=\) Diameter of the molecule \(p=\) pressure of the gas \(T=\) temperature \(K=\) Boltzmann's Constant Let intermolecular distance be \(D\) then in a volume \(\frac{4 \pi}{3} D^{3}\) there…
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