Small dust particles suspended in air seem to dance randomly about, a phenomenon

Question

Small dust particles suspended in air seem to dance randomly about, a phenomenon called Brownian motion.

For this problem you will need to know Boltzmann's constant: k_{ m B}=1.38 imes10^{-23};{ m J/K}.

A) What would you expect the mean translational kinetic energy K of such particles to be if they are in air at a temperature of 290 K?
Express the mean translational kinetic energy numerically, in joules, to two significant figures.


B) Find an expression for the rms (root-mean-square) speed v_rms of these particles, assuming them to be spheres of diameter d and density rho.

Explanation / Answer

SOLVE: A) the mean translational kinetic energy K =(3/2) kT where Boltzmann constant k =1.38*10^-23 J/K Temperature T =290 K substitute the given data in above equation ,we get the mean translational kinetic energy K =(3/2)[ (1.38*10^-23 J/K)(290 K)] = 600.3*10^-23 J = 60.03*10^-22 J B) the root-mean-square speed v_rms =square root of(3kT/m) ..... (1) volume of the sphere V =4/3 pi (d/2)^3 where d= diameter of the sphere d/2=radius of the sphere the density of the sphere rho = mass/volume rho =m/(4/3) pi (d/2)^3 m =rho*(4/3) pi (d/2)^3 ..... (2) substitute the equation (2) in equation (1) ,we get the rms (root-mean-square) speed is v_rms =square root of(3kT/rho*(4/3) pi (d/2)^3)

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