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  1. If I have a Number of gas molecules in a container each having some velocity $v(i)$, what would be the $K.E$ of the gas?

    Will it be the sum of kinetic energies of the individual particles or will it be $\frac{1}{2}mv^2$ where $v$ is the velocity of the center of mass? My book says we have to use the center of mass velocity but according to me the center of mass should stay at rest since there are no external forces (inter molecular attraction if exists is an internal force ) $$$$

  2. What if the container is moving too? What will be the kinetic energy of the gas?

    The sum of K.E's of particles (their velocity somehow calculated by adding the velocity of the container) or by using the velocity of the center of mass. If in the second case both the ways give the same answer please explain me how. And should we include the macroscopic K.E also in the Internal Energy or not $$$$

  3. Please explain which way is the correct and also explain why the other method is wrong.

Edit This is after @sammygerbil last comment Please refer to my last comment in which I quote Feynman

Feynman writes enter image description here

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E2n
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    I think you are failing to distinguish between temperature and kinetic energy. Temperature is related to the random component of KE. See also Do temperature and kinetic energy depend on overall motion? and How does temperature relate to the kinetic energy of molecules? – sammy gerbil Feb 18 '17 at 16:08
  • @sammygerbil Ok , I sort of understand you. But if we just look at the above question seeing the mechanics point of view and neglecting the thermodynamic one. If atoms are replaced with many big balls then what is the K.E of the system ? Is it the addition of the k.e of different balls ( using their separate velocities) or is K.E of the sysmtem found out by using Center of Mass Velocity because I think the center of mass should remain at rest ( imagine the box to be very very large so that balls can move considerable distances keeping com at rest since no external force)? – E2n Feb 18 '17 at 16:18
  • @sammygerbil And if we keep this all mechanics point of view what will be the answer to 2nd question ? – E2n Feb 18 '17 at 16:18
  • You tagged the question as thermodynamics, which implies you are asking about thermal motion and temperature. Outside of that context you have to be clear what you are asking, and which frame of reference you are measuring from. If you are only interested in bulk motion you use the motion of the CM (rest frame). If only interested in internal energy you use the random motion excluding CM motion (CM frame). If you want total KE it is the sum of the two, which is the total KE of all particles (rest frame). There is no standard definition which says one is correct and the others are not. – sammy gerbil Feb 18 '17 at 16:30
  • @sammygerbil Actually I couldn't understand wholly what you are saying.If I just want the total K.E of the system ( total energy of the system) , I need to sum the kinetic energies of all the parts ( if I am seeing only the mechanics part ) , because the COM as it is stays at rest . Is that right I guess it should be right because that is why we have Moment of Inertia ! ? – E2n Feb 18 '17 at 16:48
  • If the container is moving, the CM is not at rest. – sammy gerbil Feb 18 '17 at 16:50
  • @sammygerbil My above comment was for the container at rest. Is it right then ? – E2n Feb 18 '17 at 16:51
  • If the container is at rest then the only motion is random motion. Whether or not the box moves, the total KE is the sum of KE for all particles in whatever frame you are measuring from. KE depends on frame of reference. – sammy gerbil Feb 18 '17 at 16:55
  • @sammygerbil Exactly the K.E has to be the sum of the K.E of all the particles . Now taking the classical atoms of an ideal gas in such a container ( the example i have given in the above question ) , Feynman quotes " That the kinetic energy of the center of mass motion is all the energy there is " Now how can this be ? Feynman's example is the same as mine with the container at rest . Now the COM cannot move since there are no external forces and hence COM cannot have a kinetic energy . All the energy is again in the speed of the atoms . But why does he say it is due to COM motion. How ? – E2n Feb 18 '17 at 17:05
  • Sorry, this site is not a forum for discussion. When you have a rep of 50 you will be able to discuss in the chat room. For now, please update your question to reflect what you are asking. It would be helpful if you quote or post an image of the whole of the paragaph(s) in Feynman's book. You might be taking his words out of context. – sammy gerbil Feb 18 '17 at 17:12
  • Let us continue this discussion in chat. – E2n Feb 18 '17 at 17:30
  • You misquoted Feynman : "We suppose that the internal motion can be discarded, therefore FOR THIS PURPOSE the KE of the CM motion is all the energy there is." – sammy gerbil Feb 19 '17 at 01:21
  • @sammygerbil Sorry , I didn't understand what he meant . My English is not good ! I thought that he was saying that we have to use the the COM velocity to calculate the internal K.E. That was what was confusing ! Please remove 2 doubts - 1) If the box is at rest and the atoms jiggle and move inside then the COM has to stay at rest ( IS this right ) and then the K.E is the sum of the K.E of each. I think you said this in your comments. When the box also moves with will the total K.E of the system be the sum of K.E of each atom plus the K.E of the box due to its velocity ? Sorry for troubles ! – E2n Feb 21 '17 at 08:43

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