Question:
Modern vacuum pumps can evacuate a vessel down to a pressure of $4.0 \times 10^{-15}\, atm$ at room temperature $(300\, K)$. Taking $R = 8.3\, JK^{-1} \,mole^{-1}, 1\, atm = 10^5\, Pa$ and $N_{\text{Avogadro}} = 6 \times 10^{23} mole^{-1}$, the mean distance between molecules of gas in an evacuated vessel will be of the order of :
Modern vacuum pumps can evacuate a vessel down to a pressure of $4.0 \times 10^{-15}\, atm$ at room temperature $(300\, K)$. Taking $R = 8.3\, JK^{-1} \,mole^{-1}, 1\, atm = 10^5\, Pa$ and $N_{\text{Avogadro}} = 6 \times 10^{23} mole^{-1}$, the mean distance between molecules of gas in an evacuated vessel will be of the order of :
Updated On: Apr 19, 2026
- $0.2 \,\mu m$
- $0.2 \,mm$
- $0.2\, cm$
- $0.2\, nm$
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The Correct Option is B
Solution and Explanation
$ \lambda =\frac{ kT }{\sqrt{2} \pi d ^{2} P } $
$=\frac{1.38 \times 10^{-23} \times 300}{\sqrt{2} \pi \times 10^{-20} \times 4 \times 10^{-10}} $
$=\frac{1.38 \times 3}{\sqrt{2} \times 4 \pi} \times 10^{-9} $
$= 0.2\, nm $
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