Thermal Physics: Kinetic Energy of Neon Molecules

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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.
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[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:
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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 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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