Question on calculating RMS speed

[Eiano]

molecular motion is invisible in itself. when a small particle is suspended in a fluid, bombardment by molecules makes the particle jitter about at random.Robert Brown discovered this motion in 1827 while studying plant fertilization. Albert Winstein analyzed it in 1905 and Jean Perrin used it for an early measurement of Avogadro's number. The visible particle's average kinetic energy can be taken as 3/2kbT, the same as that of a molecule in an ideal gas. Consider a spherical particle of density 1000 kg/m^3 in water at 20 degree Celsius.

a)For a particle of diameter 3.00um, evaluate the rms speed.

b) The particle's actual motion is a random walk, but imagine that it moves with constant velocity equal in magnitude to its rms speed. In what time interval would it move by a distance equal to its own diameter?

c)Repeart parts (a) and (b) for a particle of mass 70.0kg, modeling your own body.d) Find the diameter of a particle whose rms speed is equal to its own diameter divided by 1 s.



So far I think that i have to use an equation like m=Pv and 1/2mv^2

(3/2)kT=(1/2)mv^2

anyone care to let me know if I am on the right track?
Thanks!

 

[Valhalla]

I think your on the right track for part number 1. Now look you have the diameter of a perfect sphere particle. Find the volume of just 1 particle (using the volume of a sphere) and you should be able to come up with the mass of one particle.

Then you should be able to use your equation.

 

[quicknote]

Yes, you are on the right track. To calculate the RMS speed, we can use the equation v = √(3kT/m), where k is the Boltzmann constant, T is the temperature in Kelvin, and m is the mass of the particle.

a) For a particle of diameter 3.00um (3x10^-6 m), we can calculate its mass using the density of water at 20°C, which is 1000 kg/m^3. The volume of the particle is given by V = (4/3)πr^3, where r is the radius. So, the mass of the particle is m = ρV = (1000 kg/m^3)(4/3)π(1.5x10^-6 m)^3 = 14.14x10^-15 kg. Plugging this into the equation, we get v = √[(3)(1.38x10^-23 J/K)(293 K)/(14.14x10^-15 kg)] = 0.964 m/s.

b) If we assume constant velocity equal to the RMS speed, then in one second, the particle would move a distance of 0.964 m. We can calculate the time it would take to move a distance equal to its own diameter (3.00x10^-6 m) using the equation d = vt, where d is the distance and t is the time. So, t = d/v = (3.00x10^-6 m)/(0.964 m/s) = 3.11x10^-6 s.

c) For a particle with a mass of 70.0 kg (similar to the mass of a human body), its RMS speed would be v = √[(3)(1.38x10^-23 J/K)(293 K)/(70.0 kg)] = 1.15x10^-10 m/s. If it moves at this speed for 1 second, it would only travel a distance of 1.15x10^-10 m, which is much smaller than its own diameter.

d) For a particle with an RMS speed equal to its own diameter divided by 1 second, we can rearrange the equation v = √(3kT/m) to solve for the diameter, which gives us d = √(3kT/v). Plugging in the