Calculate the mean free path of nitrogen molecule at $27 ° C$ when pressure is $1.0 a t m$ . Given, diameter of nitrogen molecule $= 1.5 \overset{\circ}{A}$ , $k_{B} = 1.38 x 10^{- 23} J K^{- 1}$ . If the average speed of nitrogen molecule is $675 m s^{- 1}$ . The time taken by the molecule between two successive collisions is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$0.4 n s$

$0.8 n s$

$0.3 n s$

$0.6 n s$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Here, $T = 27 ° C = 27 + 273 = 300 K$

$P = 1 a t m$ $= 1.01 \times 10^{5} N m^{- 2}$ , $d = 1.5 \overset{\circ}{A} = 1.5 \times 10^{- 10} m,$ $K_{B} = 1.38 \times 10^{- 23} J K^{- 1}$ $, λ = ?$

From $λ = \frac{k_{B} T}{\sqrt{2} π d^{2} p}$

$= \frac{1.38 \times 10^{- 23} \times 300}{1.414 \times 3.14 {(1.5 \times 10^{- 10})}^{2} \times 1.01 \times 10^{5}} = 4.1 \times 10^{- 7} m$

Time interval between two successive collisions

$t = \frac{d i s \tan c e}{s p e e d} = \frac{λ}{v} = \frac{4.1 \times 10^{- 7}}{675} = 0.6 \times 10^{- 9} s$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!