Thermal Physics: Kinetic Energy of Neon Molecules

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SUMMARY

The total random kinetic energy of one mole of neon at a temperature of 305 K is calculated to be 3801.83 J. To find the speed at which a mole of neon must move to match this kinetic energy, the equation (1/2)mv² = (3/2)kT is utilized. The root mean square speed is determined using the formula v(rms) = sqrt(3kT/m), where k is the Boltzmann constant and m is the molar mass of neon (20.2 x 10^-3 kg). The solution confirms the correct application of kinetic energy equations in thermal physics.

PREREQUISITES
  • Understanding of kinetic energy equations in physics
  • Familiarity with the ideal gas law
  • Knowledge of the Boltzmann constant (k)
  • Basic concepts of thermal energy and temperature
NEXT STEPS
  • Study the derivation of the kinetic energy formula (KE = 1/2 mv²)
  • Learn about the ideal gas law and its applications in thermal physics
  • Explore the concept of root mean square speed (v(rms)) in gas molecules
  • Investigate the relationship between temperature and kinetic energy in gases
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Students studying thermal physics, educators teaching kinetic theory, and anyone interested in the behavior of gases at varying temperatures.

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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
 

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