Does Doubling Gas Molecules Affect RMS Speed?

AI Thread Summary
Doubling the number of gas molecules in a constant volume and pressure scenario does not affect the root mean square (RMS) speed of the gas molecules. The RMS speed is dependent on temperature, not the number of molecules. If the number of molecules is increased while maintaining constant pressure and volume, the gas must cool down, resulting in a decrease in temperature. The final RMS speed can be calculated using the equation v_{rms} = √(3RT/m), where R is the gas constant and m is the mass of one mole of gas. Thus, the RMS speed remains unchanged if the temperature is adjusted accordingly.
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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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