Assume temperature of 27oC p = 105 Pa, n = 1 mole diameter of air molecules as 10-10m, which of the following is correct?

λ = v 2 π d 2 n N a = R T 2 π d 2 p N a = 2 5 3 × 3 0 0 2 × π × 1 0 − 2 0 × 1 0 5 × 6 × 1 0 2 3 x 3 = v n N a = ( 2 5 3 × 3 0 0 1 0 5 × 6 × 1 0 2 3 ) 1 / 3

Assume temperature of 27oC p = 105 Pa, n = 1 mole diameter of air molecules as 10-10m, which of the following is correct?

Mean free path > interatomic spacing

Mean free path = interatomic spacing

Mean free path < interatomic spacing

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physics ncert solutions class 11th Physics Class 11th Chemistry Chemical Kinetics

Assume temperature of 27^oC p = 10⁵ Pa, n = 1 mole diameter of air molecules as 10^-10m, which of the following is correct?

Option 1 -

Mean free path > interatomic spacing

Option 2 -

Mean free path = interatomic spacing

Option 3 -

Mean free path < interatomic spacing

Option 4 -

Data insufficient

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

$λ = \frac{v}{\sqrt[]{2} π d^{2} n N_{a}} = \frac{R T}{\sqrt[]{2} π d^{2} p N_{a}}$

= $\frac{\frac{25}{3} \times 300}{\sqrt[]{2} \times π \times 1 0^{- 20} \times 1 0^{5} \times 6 \times 1 0^{23}}$

$x^{3} = \frac{v}{n N_{a}}$

$= {(\frac{\frac{25}{3} \times 300}{1 0^{5} \times 6 \times 1 0^{23}})}^{1 / 3}$

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