The van der Waal's constant 'b' of a gas is 4 π × 1 0 - 4 L/mol . How near can the centres of the two molecules approach each other ? [Use : N A = 6 × 10 23 ]

b = 4 × 4 3 × π r 3 × N A ; We know that, 1 L = 1000 cm 3 4 × π × 1 0 - 4 × 1 0 0 0 = 4 × 4 3 × π × r 3 × 6 × 1 0 2 3 r = 5 × 1 0 - 9 cm Distance of closest approach = 2 r = 1 0 - 8 cm or 1 0 - 10 m

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ChemistryReal GasMiscellaneous/MixedEasy2 minPYQ_2020

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The van der Waal's constant 'b' of a gas is $4 π \times 1 0^{- 4} L/mol$ . How near can the centres of the two molecules approach each other ? [Use : $N_{A} = 6 \times 10^{23}$ ]

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Answer:

Solution:

$b = 4 \times \frac{4}{3} \times π r^{3} \times N_{A}$ ;

We know that, $1 L = 1000 {cm}^{3}$

$4 \times π \times 1 0^{- 4} \times 1000 = 4 \times \frac{4}{3} \times π \times r^{3} \times 6 \times 1 0^{23}$

$r = 5 \times 1 0^{- 9} cm$

Distance of closest approach $= 2 r$

$= 1 0^{- 8} cm$ or $1 0^{- 10} m$

Stream:NTA_ABHYASSubject:ChemistryTopic:Real GasSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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