Thermal Physics: Kinetic Energy of Neon Molecules

AI Thread Summary
The total random kinetic energy of one mole of neon at 305 K is calculated to be 3801.83 J. To find the speed at which a mole of neon must move for its kinetic energy to equal this total, the relevant equations include (1/2)mv^2 and v(rms)=sqrt(3kT/m). The user initially expressed uncertainty about the correct approach for part b but later confirmed they figured it out using the kinetic energy formula. The discussion emphasizes the application of thermal physics principles to solve kinetic energy problems. Overall, the thread highlights the importance of understanding molecular motion in relation to temperature.
sun
Messages
39
Reaction score
0
[SOLVED] thermal physics

Homework Statement


a) What is the total random kinetic energy of all the molecules in one mole of neon at a temperature of 305 K? I got 3801.83J

(b) With what speed would a mole of neon have to move so that the kinetic energy of the mass as a whole would be equal to the total random kinetic energy of its molecules?

Homework Equations



(1/2)mv^2=(3/2)kT
v(rms)=sqrt(3kT/m)

The Attempt at a Solution


I'm not sure what to do in order to solve for part b. Not to sure if I'm even looking at the correct equations. Any suggestions?

Thanks
 
Last edited:
Physics news on Phys.org
any suggestions. I would love the help and greatly appreciate it. :)
 
I figured it out:

KE=1/2(mv^2)

KE=3801.83
m=20.2*10^-3
 
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 '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