Answer to Question #178366 in Molecular Physics | Thermodynamics for roohi

Question

For a given gas the coefficient of viscosity is 1.9 × 105 N s m2 and the diffusion coefficient is 1.2 × 105 m 2 s 1 . Calculate the density and mean free path at average molecular velocity of 380 m s1 . 

Answer 1

First we recall all the terms and their calculation formula to drieve our eqn..

We know, that,

1)Mean free path $=\boxed{\lambda={RT \over \sqrt 2 \pi d^2 N_A P}}$

2)And also given by

$\boxed{\lambda={{\mu \over P}{ \sqrt {\pi k T \over 2 m}}}}$

$​$

 $\mu=1.9\times 19^5$ N s m2

p = pressure

T = temperature

m = molecular mass

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And 

3)Average velocity is given by

$V=\sqrt{kT\over m \pi}$

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4)And Diffusion coffecient is calculated by

$D={kT\over 6\pi \mu R_o}$

Given:

$\bigstar$ coefficient of viscosity is 1.9 × 10^-5 N s m²

and

$\bigstar$ diffusion coefficient is 1.2 × 10^-5 m² s^-1

$\bigstar$ average molecular velocity is 380 ms^-1

now after manipulations we get,

$${6D\over K}={T\over \pi \mu}={\pi d^2 N_A\over R}$$

on comparing we get density,

$${\sqrt {6DR\over K N_A \pi}}={ d }$$

$\boxed{d=1.55\times10^{-6}}$

and we also get mean free path ..

$\boxed{\lambda =6.623 \times 10^{-3}}$