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
 
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...
Back
Top