What Temperature Gives Nitrogen Molecules an RMS Speed of 1.5 m/s?

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SUMMARY

The discussion focuses on calculating the temperature at which nitrogen molecules achieve an RMS speed of 1.5 m/s using the equation V = (3*K_b*T/m)^0.5. The user correctly identifies the mass of a nitrogen molecule as approximately 4.677E-26 kg, derived from the combined mass of protons, neutrons, and electrons. Upon rearranging the equation to solve for temperature, the user arrives at a temperature of 0.0025 Kelvin, confirming the accuracy of their calculations. The discussion highlights the importance of consulting a periodic table for elemental masses to streamline such calculations.

PREREQUISITES
  • Understanding of the ideal gas law and kinetic theory of gases
  • Familiarity with Boltzmann's constant (K_b)
  • Knowledge of molecular mass calculations
  • Basic algebra for rearranging equations
NEXT STEPS
  • Study the ideal gas law and its applications in thermodynamics
  • Learn more about Boltzmann's constant and its significance in statistical mechanics
  • Explore molecular mass determination using the periodic table
  • Investigate the relationship between temperature and molecular speed in different gases
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Students in physics or chemistry, educators teaching thermodynamics, and anyone interested in molecular behavior and kinetic theory.

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Homework Statement



1.5 is a typical walking speed.
At what temperature would nitrogen molecules have an rms speed of 1.5 ?


Homework Equations



V=(3*K_b*T/m)^.5
V is the rms speed
K_b is Boltzmann's constant
T is temp in kelvin
m is mass


The Attempt at a Solution



I am going to make the foolish assumption that when they mean mass, they mean the mass of one nitrogen molecule. So the mass of a proton and a neutron is 1.67E-27kg and the mass of an electron is 9.11E-31kg. One nitrogen atom has 7 protons, 7 neutrons and 7 electrons and multiply that by two because of the second nitrogen atom. So ((1.67E-27)(14)+(7)(9.11E-32))(2)=4.677E-26 kg.

Now I rearrange my equation to solve for T and plug in.

mV^2/(3*K_b)=T

(4.677E-26)(1.5)^2/(3*K_b) = .0025 kelvin

Am I close?

Thanks!
 
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Never mind. My answer is correct.
 
Good job. For future reference, it may be easier and quicker to look up the mass of nitrogen or other elements in a periodic table.
 

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