RMS velocity of molecules in a mixture

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SUMMARY

The discussion centers on calculating the root mean square (RMS) velocity of molecules in a gas mixture. The correct formula for a single monoatomic gas is given as ##v_{rms} = \sqrt{\frac{3k_bT}{m}}##, where ##m## is the mass of a molecule and ##k_b## is the Boltzmann constant. Participants clarify that the average kinetic energy of a gas mixture is not simply the sum of the individual gases' energies but rather involves the average of their RMS velocities. The relationship between the RMS velocities of two gases at equal temperatures is established as ##\frac{v_{rms_1}}{v_{rms_2}} = 2##.

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Saptarshi Sarkar
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Homework Statement
A vessel at temperature T contains equal number of molecules of two gases whose masses are m and 4m respectively. What is the RMS velocity of molecules in the mixture?
Relevant Equations
##v_{rms} = \frac {\sqrt {3k_bT}} m##
I know that for a single monoatomic gas with RMS velocity ##v_{rms}## , $$\frac 1 2mv_{rms}^2 = \frac 3 2k_bT$$ where ##m## is mass of a single molecule, ##k_b## is Boltzmann constant and ##T## is temperature of the gas.

For a mixture of gas, I know that the average kinetic energy after mixing the gases will be equal to the sum of average kinetic energy of the two constituent gases before mixing.

##\frac 1 2mv_{rms_1}^2 + \frac 1 24mv_{rms_2}^2 = \frac 1 25mv_{rms}^2##

But, I have no idea how to use this to find the RMS velocity of the mix.
 
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Saptarshi Sarkar said:
the average kinetic energy after mixing the gases will be equal to the sum of average kinetic energy of the two constituent gases before mixing.
I don’t think you mean that.
It might be the average of the averages, which fits better with the equation you wrote, but not the sum.
What does your first equation tell you about the relationship between the two individual rms speeds?
 
haruspex said:
I don’t think you mean that.
It might be the average of the averages, which fits better with the equation you wrote, but not the sum.
What does your first equation tell you about the relationship between the two individual rms speeds?

I did a mistake and wrote the formula wrong, it should be ##v_{rms} = \sqrt \frac {3k_bT} m##

From this I can say that if the absolute temperature of the two gases be equal, then ##\frac {v_{rms_1}} {v_{rms_2}} = \sqrt \frac {4m} m = 2##
 
Saptarshi Sarkar said:
I did a mistake and wrote the formula wrong, it should be ##v_{rms} = \sqrt \frac {3k_bT} m##

From this I can say that if the absolute temperature of the two gases be equal, then ##\frac {v_{rms_1}} {v_{rms_2}} = \sqrt \frac {4m} m = 2##
Right, so consider one molecule of each. What is the RMS speed of the pair?
 
haruspex said:
Right, so consider one molecule of each. What is the RMS speed of the pair?

Should it be ##\frac 3 2v_{rms_1}##?
 
Saptarshi Sarkar said:
Should it be ##\frac 3 2v_{rms_1}##?
How do you find the RMS of two numbers?
 
haruspex said:
How do you find the RMS of two numbers?

By taking the root of the average of the square. So, will it be ##\sqrt \frac {5v_{rms_1}^2} 2##?
 
Saptarshi Sarkar said:
By taking the root of the average of the square. So, will it be ##\sqrt \frac {5v_{rms_1}^2} 2##?
Yes.
 
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