Answer to Question #184740 in Molecular Physics | Thermodynamics for Khamosh Ray

Question

c)

For a given gas the coefficient of viscosity is 1.9 x 10 Nsm and the diffusion

coefficient is 1.2 x 10-5 m?s-. Calculate the density and mean free path at average

molecular velocity of 380 ms-1

Answer 1

$\rho=\frac{\eta}{D}=\frac{1.9\cdot 10^{-5}}{1.2\cdot 10^{-5}}=1.58~\frac{kg}{m^3},$

$\lambda=\frac{3D}{v}=\frac{3\cdot 1.2\cdot 10^{-5}}{380}=94.7~nm.$