Systematic changes do not affect thermodynamic equilibrium.

What does this mean? And what kind of systematic changes are allowed?

The container with gas is stationary till some time then it's given a constant velocity and the final temperature is asked; the answer says that systematic changes don't affect thermodynamic equilibrium and temperature remains constant.

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    $\begingroup$ The internal energy of a gas is measured from a reference frame at rest with respect to the center of mass of the system. $\endgroup$ – jm22b Mar 23 '16 at 15:57
  • $\begingroup$ So,unless the container is accelerating the internal energy doesnt change ? $\endgroup$ – Apoorva Uplap Mar 23 '16 at 16:04
  • $\begingroup$ @Jacobadtr or orderly motion of particles does not count. :) $\endgroup$ – Yashas Feb 18 '17 at 17:08

Consider two molecules moving in opposite direction in a container. enter image description here

Case 1: Container's velocity=0

Both molecules rebound with their initial velocities and net KE remains constant.

Case 2: Container's velocity=V

For molecule moving along V, if initial velocity is v, rebound velocity=v-2V and for the molecule in the left, rebound velocity=v+2V

Final KE of system=0.5m((v+2V)^2+(v-2V)^2)=0.5m(2v^2+8V^2).

Since V is very less than v, the final KE=0.5m(2v^2)=Initial KE of system.

Since, the system consists of gas molecules in all random directions, net KE remains almost the same and change in T isn't noticeable.

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  • $\begingroup$ "Since v is very less than V" ??? In my books the *V^2 would dominate then. $\endgroup$ – anna v Feb 18 '17 at 19:48
  • $\begingroup$ Sorry, confused V with v. Corrected! $\endgroup$ – Deechit Poudel Feb 19 '17 at 4:40

It is simpler to think of the thermodynamic definition of temperature, to start with, the one on the left:


where N is the number of molecules, n the number of moles, R the gas constant, and k the Boltzmann constant.

For the expression on the right, randomness is implicit, Brownian motion after all led the molecular model, and it is basically random.

The identification of temperature as the same quantity in both expressions holds when the conditions for both expressions hold.


Moving a fixed volume at a fixed pressure does not change its temperature , it is a definite macrostate for which the microstates on the right are used to average out the molecular velocities.

If randomness does not hold , as with a moving system, the fact that we do not observe a rise in temperature in moving systems leads to an answer by Deechit Poudel, the rms of the velocity of gas molecules is large, and the contribution of the motion is small for every day motions. This can be verified with this calculator:

For 20C in air the most probable speed of the molecules is 400m/second, Larger than the velocity of sound.

For supersonic velocities and velocities commensurate with the speed of sound, the relationship demands as an axiom the randomness, imo.

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Where are you measuring the temperature?

If you measure it using a thermometer that is inside chamber, travelling with the gas, then you will not observe any difference relative to when the chamber is at rest (obviously, we are assuming some classical, global reference frame here...).

However, if you measure from outside the chamber, in the global rest frame, then you will find that it is hotter.

To see how the gas can have a different temperature depending on where you measure it from, you have to consider that temperature is just a measure of the momentum change the gas particles undergo when they hit the thermometer. If the thermometer is travelling along with the gas, then it records only their thermal motion in the moving frame. If the thermometer is at rest and the gas whizzes past and hits it, you get the thermal motion plus the relative motion.

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This means that changes that happen to the whole of the gas won't affect its temperature. This is because the temperature of a gas is proportional to the average kinetic energy of the particles.

$ T \alpha KE_{average}$

Since in an ideal gas all particles move in straight random paths, an increase in velocity in one direction for all particles will not actually change their average kinetic energies.

Therefore the thermodynamic equilibrium remains unchanged.

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  • $\begingroup$ So,an impulse/constant force,etc also wont change anything if its giving all the molecules of gas an additional velocity in the same direction ? $\endgroup$ – Apoorva Uplap Mar 23 '16 at 16:08
  • $\begingroup$ The part about the average kinetic energy not changing is just wrong. The average kinetic energy is ${}\propto\langle v^2 \rangle$, if we now translate the gas by velocity $\vec u$, this changes, as $\langle v'^2 \rangle = \frac 1 N \sum_{i=1}^N (\vec v_i - \vec u)^2 = \frac 1 N \sum_{i=1}^N \left(v^2 - 2\vec v_i \cdot \vec u + u^2 \right) = \langle v^2 \rangle +u^2 - 2 \vec u \cdot \langle \vec v \rangle = \langle v^2 \rangle + u^2$ (as the average velocity is zero). $\endgroup$ – Sebastian Riese Mar 23 '16 at 16:50
  • $\begingroup$ @SebastianRiese umm no it doesn't. If all components are shifted by the same value the mean remains constant. Basic Statistics. $\endgroup$ – Jaywalker Mar 23 '16 at 19:05

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