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This question already has an answer here:

According to thermodynamics we were taught that temperature of a body is the measure of avg kinetic energy of the molecules which make up the body . Now if we consider an insulating box full of gas at stationary w.r.t an inertial frame of reference let s and tempertaure be t.Now if we change our frame in such a way that the box is in motion ,than the kinectic energy of the molecules also increase hence the temperature appear to be different but it is not so.(I know i somewhere made a mistake in defination of temperarure )

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marked as duplicate by By Symmetry, Jon Custer, sammy gerbil, Kyle Kanos, stafusa Jul 19 '18 at 23:14

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Yes you made a mistake in the definition of temperature. You need to differentiate the average translational kinetic energy (measured by temperature) of the contents of the system (the gas molecules themselves), and the translational kinetic energy of the center of mass of the system as a whole which has nothing to do with the temperature of the gas. The long version of the first law for a closed system (no mass transfer) is:

Q – W = ΔE = ΔU + ΔKE + ΔPE

Where

ΔE is now the total energy change of the system, which is the sum of change in internal energy and changes in energy with respect to external frames of reference.

ΔKE is the change in kinetic energy of the system as a whole. This relates to the velocity of the box full of gas.

ΔPE = The change in potential energy of the system as a whole, such as due to a change in elevation in a gravitational field.

Q, W, and ΔU are as always. The temperature of the gas only relates to ΔU which is the change in internal energy of the gas.

Hope this helps

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Temperature is related to our lack of complete information about the system (see the history of the definition of entropy).

An increase in temperature produces an increase in fluctuation or spread in momentum. Indeed, if you look at the Maxwell-Boltzmann distribution, which describes the particle speeds in an ideal gas, it has the form

$f(v) = \left(\frac{m}{2\pi k_B T}\right)^{3/2} e^{-\frac{mv^2}{2 k_B T}}$

You see that the distribution spreads wider and wider and temperature $T$ increases. Of course, the mean of the distribution is also a function of $T$ but it's the variance that captures the physical meaning of temperature.

For a speed distribution that does NOT follow Maxwell-Boltzmann distribution, you could have particles flying at high speed, but very little variance in the speed. This is what happens when you generate a supersonic atomic or molecular beam out of a nozzle (look up "supersonic beam sources"), or laser-cool atoms to a non-zero velocity (look up "VSCPT"). These two examples are known to create cold atoms (but moving at high velocity).

So to answer your question, since a frame transformation does not change the variance of the speed distribution, the temperature of your gas stays the same.

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The question can also be answered in the context of the principle of relativity (Einstein 1905)

“The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of co-ordinates in uniform translatory motion”

Thus the temperature of the gas should be the same in all inertial frames of reference, be it the reference frame s in which the container of gas it is at rest, in another inertial frame with respect to which the box is in uniform translational motion.

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