Most probable speed of aluminium atoms

AI Thread Summary
To calculate the most probable speed of aluminium atoms at 1173K, the formula vmp = √(2kT/m) is used. The molar mass of aluminium is 0.0270 kg/mol, but the calculation initially used this value incorrectly with the Boltzmann constant. The correct approach involves using the mass of an individual aluminium atom, which can be found by dividing the molar mass by Avogadro's number. Alternatively, one can use the gas constant instead of the Boltzmann constant while keeping the molar mass. This clarification resolves the confusion regarding the calculated speed.
Bugsy23
Messages
24
Reaction score
0

Homework Statement



Aluminium atoms at 1173K enter a vacuum chamber. Calculate the most probable speed for the atoms.

Homework Equations



vmp=\sqrt{}(2kT/m)

The Attempt at a Solution



The molar mass of aluminium is 0.0270kg mol^-1, so
vmp=\sqrt{}(2*1.381*10^-23JK^-1)*1173K/0.0270kg mol^-1=1.1*10^-9ms^-1

Which is a really tiny quantity and it seems unrealistic that particles at such a high temperature would move that slowly. I've used SI units though so I can't see where I would have gone wrong. Any help would be appreciated. Thanks
 
Physics news on Phys.org
Bugsy23 said:
The molar mass of aluminium is 0.0270kg mol^-1, so
vmp=\sqrt{}(2*1.381*10^-23JK^-1)*1173K/0.0270kg mol^-1=1.1*10^-9ms^-1
You're using the molar mass combined with Boltzmann constant. But that doesn't make sense. You need the mass of an individual molecule (in this case an individual aluminum atom) if you want to use Boltzmann constant. (You can do this by dividing the molar mass by Avogadro constant NA to find the mass per individual molecule [in this case atom].)

Alternatively, you can stick with the molar mass and substitute Boltzmann constant k with the gas constant R, since k = R/NA.
 
Last edited:
Thanks, I've got it now
 
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 hanged 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...

Similar threads

Back
Top