Modern vacuum pumps can evacuate a vessel down to a pressure of 4… - Vidyadip

MCQ

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

A
$0.2\,\mu m$
✓
$0.2\,mm$
C
$0.2\,cm$
D
$0.2\,nm$

✓

Answer

Correct option: B.

$0.2\,mm$

b
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 is only one

$\frac{4 \pi}{3} D^{3} P=\frac{1}{N_{A}}=R_{T} \quad$ or $\quad D=\left(\frac{3 R T}{4 \pi N_{A} P}\right)^{1 / 3}$

$\begin{aligned} \text { Put } P=& 4 \times 10^{-10} P a, R=83, N_{A}=6 \times 10^{23} \\ T=& 300 \mathrm{K} \end{aligned}$

$D=0.2 \mathrm{mm}$

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