In the realm of kinetic theory of gases, the internal energy of gas is solely due to the sum of kinetic energies of all particles, since kinetic energy is frame dependent I was thinking whether a container having gas will heat up if I put it in a moving train, which is accelerating?
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asked Aug 17, 2019 at 9:13
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2$\begingroup$ Possible duplicate of Is temperature a Lorentz invariant in relativity? $\endgroup$– PM 2RingCommented Aug 17, 2019 at 9:34
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3$\begingroup$ Possible duplicate of Why isn't temperature frame dependent? $\endgroup$– BioPhysicistCommented Aug 17, 2019 at 9:34
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$\begingroup$ Thanks for pointing that out , that article would be sufficient $\endgroup$– Aditya PrakashCommented Aug 17, 2019 at 9:49
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$\begingroup$ By the way, if the container of gas did increase in temperature due tovelocity it would you would have a way to detect constant velocity in violation of the principle of relativity. $\endgroup$– Bob DCommented Aug 17, 2019 at 11:59
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1 Answer
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The total energy of the gas is the sum of its internal energy at the molecular level and its external energy at the macroscopic level.
The external kinetic energy is due to the velocity of the center of mass of the collection of molecules with respect to an external frame of reference. The internal kinetic energy does not depend on an external frame of reference.
The temperature of the container of gas will not increases due to the velocity of the container.
Hope this helps
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$\begingroup$ Possibly promoting posting questions without checking for duplicates first? $\endgroup$ Commented Aug 17, 2019 at 12:10
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$\begingroup$ @Aaron Stevens Wasn’t aware of that responsibility. Can I find that policy somewhere? $\endgroup$– Bob DCommented Aug 17, 2019 at 12:39
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$\begingroup$ I don't think its an actual policy. I guess its the same idea as to answering homework questions where the OP has not given any effort. Its promoting similar activity from other users who can say "Well this person's questions got an answer." $\endgroup$ Commented Aug 17, 2019 at 13:07
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1$\begingroup$ It's up to you. If you think your answer is still useful, then I don't think one down vote should matter. $\endgroup$ Commented Aug 17, 2019 at 13:14