The van der Waals' constant 'b' of a gas is 4π×10−4L/mol. How near can the centeres of the two molecules approach each other? [Use :NA=6×1023]

Answer: bSolution: (b) b=4×34×πr3×NA, 4×π×10−4×1000 4×34×π×r3×6×1023 r=5×10−9cm Distance of closest approach =2r =10−8cm or 10−10m

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The van der Waals' constant 'b' of a gas is {4}{10}^{{-{4}}}{L

Question

The van der Waals' constant 'b' of a gas is $4 π \times 10^{- 4} L / m o l .$ How near can the centeres of the two molecules approach each other? [Use : $N_{A} = 6 \times 10^{23}$ ]

10^{- 7}

10^{- 10}

5 \times 10^{- 11}

5 \times 10^{- 9}

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Sebastian Discussed

The van der Waals' constant 'b' of a gas is

4 π \times 10^{- 4} L / m o l .

How near can the centeres of the two molecules approach each other? [Use :

N_{A} = 6 \times 10^{23}

]

14 mins ago

Discuss this question LIVE

14 mins ago

Answer: bSolution: (b)

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

4 \times π \times 10^{- 4} \times 1000

4 \times \frac{4}{3} \times π \times r^{3} \times 6 \times 10^{23}

r = 5 \times 10^{- 9} c m

Distance of closest approach

= 2 r

= 10^{- 8} c m or 10^{- 10} m

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Question Text	The van der Waals' constant 'b' of a gas is $4 π \times 10^{- 4} L / m o l .$ How near can the centeres of the two molecules approach each other? [Use : $N_{A} = 6 \times 10^{23}$ ]
Answer Type	Text solution:1
Upvotes	150