Particle physics / ideal gas problem

AI Thread Summary
Dust particles, with a diameter of 11.0 micrometers and a density of 2500 kg/m^3, are analyzed as an ideal gas to determine their root mean square (rms) speed at 23.0 degrees Celsius. The relevant formula for rms speed is v_{rms} = √(3k_{B}T/m), where m represents the mass of a single particle. The main challenge discussed involves converting the particle's density and size into a molecular mass to use in the equation. Participants suggest focusing on the mass of a single particle rather than the total number of particles for calculations. Overall, the discussion emphasizes the importance of understanding the relationship between temperature, kinetic energy, and particle mass in solving the problem.
greenskyy
Messages
17
Reaction score
0

Homework Statement


Dust particles are approximately 11.0 micrometers in diameter. They are pulverized rock, with density 2500 kg/m^3. If you treat dust as an ideal gas, what is the rms speed of a dust particle at 23.0 degrees C?

Homework Equations


I have no idea.
The only one I would bet money on is:

v_{rms}=\sqrt{\frac{3k_{B}T}{m}}

The Attempt at a Solution


I made a couple of far-reached guesses at this problem and came to the wrong answer. One method was to use the ideal gas law with pressure = 101325 Pa, and getting the number of particles to be 2.48*10^25. My main issue here is going from density and radius of a particle to a molecular mass or a correct number of particles. I am completely stumped by the problem and getting nowhere. If anyone could offer help, I would greatly appreciate it.
 
Physics news on Phys.org
Well, show us your work for what you've done so far and maybe we'll be able to offer more help! I'll give you a couple of hints though: The absolute temperature of an ideal gas is proportional to its random translational kinetic energy. How might this help you solve the problem?
 
Your starting formula is OK,

<br /> v_{rms}=\sqrt{\frac{3k_{B}T}{m}}<br />

but why do you want the number of he particles? In the formula, m means the mass of a single particle. Given the diameter and density, it is easy to get.

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top