Does Doubling Gas Molecules Affect RMS Speed?

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SUMMARY

Doubling the number of gas molecules in a constant volume container does not decrease the root mean square (RMS) speed of the gas molecules. Instead, the RMS speed remains constant if the pressure is also kept constant. Given an initial RMS speed of 1300 m/s, the final RMS speed will also be 1300 m/s. The relationship between temperature, pressure, and the number of molecules is governed by the ideal gas law, PV=nRT, where temperature must adjust to maintain constant pressure.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of root mean square speed formula (vrms = √(3RT/m))
  • Familiarity with concepts of pressure, volume, and temperature in gas behavior
  • Basic principles of thermodynamics related to gas laws
NEXT STEPS
  • Study the ideal gas law and its applications in thermodynamics
  • Learn about the relationship between temperature and pressure in gas systems
  • Explore the derivation and implications of the RMS speed formula
  • Investigate real gas behavior versus ideal gas assumptions
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone studying thermodynamics or gas behavior, particularly in the context of ideal gases and their properties.

ajmCane22
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Homework Statement



An ideal gas is kept in a container of constant volume. The pressure of the gas is also kept constant
(a) If the number of molecules in the gas is doubled, does the rms speed increase, decrease, or stay the same?
(b) If the initial rms speed is 1300 , what is the final rms speed?

Homework Equations



PV=nRT...I think

The Attempt at a Solution


For part I, I believe the rms speed decreases. So if the only thing changing is the number of molecules wouldn't doubling the molecules cause rms to decrease by half? I know the answer is no, because the computer said so, I'm just not sure why it's wrong. And how much does it decrease then?
 
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If the volume and the pressure must remain constant, the only way to get in more molecules is cooling the gas, and let in more gas to keep the pressure constant.

If the original tempererature was T, what is the new temperature?

After you've found it use

v_{rms} = \sqrt { \frac {3 R T} {m} }

where m is the mass of one mole of the gas molecule
 

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