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 :

• A. $0.2 \,\mu m$
• B. $0.2 \,mm$
• C. $0.2\, cm$
• D. $0.2\, nm$

Answer 1

The Correct Option is B

$ \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 $