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 Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0\times10-15 atm. at room temperature (300 K).

Taking R = 8.3 JK-1 mole-1, 1 atm=105 Pa and N Avogadro = 6 \times 1023 mole-1, the mean distance between molecules of gas in an evacuated vessel will be of the order of :

 

  • Option 1)

    0.2\: \mu m

  • Option 2)

    0.2\: m m

  • Option 3)

    0.2\: c m

  • Option 4)

    0.2\: n m

 

Answers (1)

best_answer

As we have learned

Formula for mean free path -

Y= \frac{KT}{\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 ^3P = \frac{1}{N_A }= R_T \: \: \: or \: \: D = \left ( \frac{3RT}{4\pi N_AP} \right )^{1/3}

Put P = 4 \times 10^{-10 } Pa , R = 83 , N_A = 6 \times 10^{23 } \\ T = 300 K

D = 0.2 mm 

 

 

 

 

 

 

 


Option 1)

0.2\: \mu m

Option 2)

0.2\: m m

Option 3)

0.2\: c m

Option 4)

0.2\: n m

Posted by

Avinash

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