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I came across a question in which speed distribution of molecules of an ideal gas trapped in a vessel and the temperature of the gas was given. Then it was given that the molecules of a particular speed escape from the vessel and we had to determine the new temperature of the gas. I just by trial and error figured out that the new temperature could be obtained if I assumed that the root mean square speed of the gas divided by the root of temperature that is $$v_{rms}/(T)^{0.5}$$ remained constant which by formula is equal to $(3R/M)^{0.5}$ and it makes sense.

But I don’t understand that why is it only the root mean square speed of the gas that remains constant why does say average velocity or the most probable speed not remain constant?

For a clearer idea of what I am asking, I am also attaching the question : Speed distribution of molecules of an ideal mono atomic gas trapped in a vessel is given in the following table.

Speed of molecules Percentage of molecules

100 m/s                     10
200 m/s                     20
600 m/s                     40
800 m/s                     20
1000 m/s                    10

Temperature of this gas is 324 K. If all the molecules of speed 800 m/s escape from the vessel, by what amount will the temperature of the gas change?

Any help would be greatly appreciated!

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Because the heat energy remains the same and there is no expansion work done (expansion of gas while applying a force on a surface via pressure). This means that total energy is conserved. The energy of a molecule is its kinetic energy, 1/2mv^2, and the total energy is the sum of the kinetic energy of every molecule. To keep this sum from changing, the root mean square must be constant.

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From the maxwell's distibution we can derive three velocities:

  1. The rms speed as you rightly mentioned is $\sqrt{\frac{3RT}{M}}$

  2. The most probable speed: $\sqrt{\frac{2RT}{M}}$

and finally the

  1. The average speed: $\sqrt{\frac{8RT}{\pi M}}$

As you can see $\frac{𝑣}{{𝑇}^{0.5}}$ is constant for all velocities mentioned. Rms isnt special (but it is the most important.

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  • $\begingroup$ But I calculated new and initial v rms and used that as $v$ in equating the initial and final ratio of v/T^{0.5} which gives me the right answer. So I am confused why should v rms be only used to find the final temperature. Considering other velocities in equating the ratio doesn’t give the correct answer. And thanks! $\endgroup$
    – Nil
    Jul 19 at 18:21
  • $\begingroup$ I can't quite follow what you did in this comment. I suggest that you edit your original post and include what you are trying to say in the comment, but with more detail. (Note that you had to add more information in the comment, which means the original post is not complete or detailed enough.) $\endgroup$
    – garyp
    Jul 19 at 19:53
  • $\begingroup$ @Nil can you tell me what is the question, I can help you. $\endgroup$ Jul 20 at 6:01
  • $\begingroup$ @AbhiramCherukupalli I have edited the original post to now include the question I was referring to. I hope that now it will be easier to comprehend what I mean to say and ask. $\endgroup$
    – Nil
    Jul 21 at 5:40

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