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Suppose I place a cylinder filled with ideal gas is placed on a train, which starts moving. Does the kinetic energy of the gas molecules increase? Will the temperature of the gas rise as a consequence?

Further details: This question arises from a book I was reading, which states that

The motion of molecules is truly random. In other words, the center of mass of the gas is assumed to be at rest and any rotation about the center of mass is assumed to be absent. Any systematic motion of a gas sample has no effect on temperature.


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  • $\begingroup$ Could you please tell me which book it was $\endgroup$ – Shashaank Mar 3 '17 at 7:29

Kinetic energy does indeed depend on overall motion, as is apparent from the formula $K=mv^2/2$. Naively looking at the definition of temperature, $3kT/2=mv^2/2$, where $v$ is the rms velocity of the gas molecules, one could (erroneously) be convinced that temperature behaves in the same manner.

However, there is a distinction between kinetic energy and temperature: temperature is only definable by the above formula under strict conditions; namely, the distribution of velocities within the gas must be isotropic. On a moving train, the velocities of all the molecules of the gas are boosted along the direction of motion. The velocity distribution is no longer isotropic, and therefore the temperature of the gas cannot be defined as above. Only in the frame where the gas velocity distribution is isotropic (i.e. as viewed by an observer standing on the train) can the temperature be defined. By these arguments, temperature should be a frame-independent quantity, since the conditions for its measurement require an inertial frame relative to the gas.


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