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JEE Main 20149 AprilOnline
Exam
JEE Main
Year
2014
Date
9 April
Mode
Online
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2014 · 9 April · Online · Physics · Kinetic Theory of Gases

Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0...

Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0×10−15\text{4.0} \times 1 {0}^{- 1 5} atm. at room temperature (300 K). Taking R = 8.3 JK-1 mole-1, 1 atm = 105 Pa and NAvogadro=6×1023mole−1{\text{N}}_{\text{Avogadro}} = 6 \times 1 {0}^{2 3} {mole}^{- 1}, the mean distance between molecules of gas in an evacuated vessel will be of the order of :

Options

  1. A0.2μm0 . 2 \mu m
  2. B0.2mm0 . 2mm
  3. C0.2cm0 . 2cm
  4. D0.2nm0 . 2nm

Correct answer

B. 0.2mm0 . 2mm

Step-by-step solution

Let the inter molecular distance be D
Then in a volume of 43πD3\dfrac{4}{3} \pi{D}^{3} there is only one molecule.

43πD3P=(1NA)RT\dfrac{4}{3} \pi{D}^{3} P = ( \dfrac{1}{{N}_{A}} )RT

D=(3RT4πNAP)13D = {( \dfrac{3RT}{4\pi{N}_{A} P} )}^{\dfrac{1}{3}}

Substituting P=4×10−10Pa,R=8.3,NA=6×1023,T=300KP = 4 \times {10}^{- 10} Pa, R = 8.3, {N}_{A} = 6 \times {10}^{23} , T = 300K

D=0.2mmD = 0.2mm

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