Ratio of RMS Speed of Oxygen to Nitrogen Molecules

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SUMMARY

The ratio of the root mean square (RMS) speed of oxygen molecules to that of nitrogen molecules is calculated using their respective molar masses, which are 32 g/mol for oxygen and 28 g/mol for nitrogen. The derived ratio is approximately 0.935. The RMS speed can be expressed through the equation v_rms = sqrt(3RT/M), where R is the universal gas constant, T is the temperature in Kelvin, and M is the molar mass. This calculation does not require the number of molecules of each gas, simplifying the analysis.

PREREQUISITES
  • Understanding of root mean square (RMS) speed in kinetic theory
  • Familiarity with molar mass and its significance in gas calculations
  • Basic knowledge of the ideal gas law
  • Ability to manipulate algebraic equations for physical quantities
NEXT STEPS
  • Study the derivation of the RMS speed formula in kinetic theory
  • Learn about the ideal gas law and its applications in gas mixtures
  • Explore the impact of temperature on molecular speeds
  • Investigate the concept of molecular distribution and its relation to RMS speed
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Students studying thermodynamics, physicists analyzing gas behavior, and educators teaching kinetic theory concepts.

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Homework Statement


A container is filled with a mixture of nitrogen and oxygen. What is the ratio of the rms speed of oxygen molecules to that of nitrogen molecules? (Molar mass of oxygen=32gmol-1; molar mass of nitrogen=28gmol-1).


Homework Equations


v2=(v12+v22+...+vN2)/N


The Attempt at a Solution


Tried using momentum mv=mv but doesn't make sense to do that (even more because the question asks for rms speed) Answer is 0.935. Can't think of sensible ways to use rms speed equation provided above. The question doesn't even specify number of molecules of each so very confused. Please help...
 
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You can express the RMS speed in terms of different quantities.

RMS speed, see the first formula.
 
Thanks
 

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