Kinetic Theory

JEE Main level, single correct, medium

18 of 168

Question 18 of 168

Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0×10−15\text{4.0}\times 1{0}^{-15} 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}^{23}{mole}^{-1}, the mean distance between molecules of gas in an evacuated vessel will be of the order of :

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Answer: (B) 0.2mm0.2mm

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=0.2 mm

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