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.

  • 4
    $\begingroup$ 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
    Commented Mar 23, 2016 at 15:57
  • $\begingroup$ So,unless the container is accelerating the internal energy doesnt change ? $\endgroup$ Commented Mar 23, 2016 at 16:04
  • $\begingroup$ @Jacobadtr or orderly motion of particles does not count. :) $\endgroup$
    – Yashas
    Commented Feb 18, 2017 at 17:08

6 Answers 6


Consider two gas molecules in a container, moving in opposite directions with speed $v_{g}$.

enter image description here Case 1: When the container is at rest.

Both gas molecules rebound with their initial velocities and net KE of the gaseous system remains constant.

The KE of the gaseous system is $$KE = 2. \frac {1}{2}mv_{g}^2 = mv_{g}^2$$

Case 2: When the container sets in motion with a velocity $v_{c}$.

The molecule moving along $v_{c}$ rebounds with a velocity $v_{g}-2v_{c}$, while the molecule moving opposite to $v_{c}$ rebounds with a velocity $v_{g}+2v_{c}$.

Final KE of the gaseous system is $$KE= \frac {1} {2} m[(v_{g}+2v_{c})^2+(v_{g}-2v_{c})^2] = \frac {1} {2} m(2v_{g}^2+8v_{c}^2)$$

Since $v_{c}$ is very less when compared to $v_{g}$, the final KE can be simplified to; $$KE = \frac {1}{2} m(2v_{g}^2) = mv_{g}^2$$ which is equal to the KE of the system before the container sets in motion.

This is an oversimplification of the actual model consisting of millions of gas molecules. However, since the system should consist of molecules in all random directions, you can always pair a molecule with another moving just opposite to it. Hence, the net KE of the system remains almost the same and the change in temperature isn't noticeable.

  • $\begingroup$ "Since v is very less than V" ??? In my books the *V^2 would dominate then. $\endgroup$
    – anna v
    Commented Feb 18, 2017 at 19:48
  • $\begingroup$ To be exact the kinetic energies according to you aren't equal? $\endgroup$
    – Kashmiri
    Commented Aug 19, 2020 at 4:20

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.


To add to the other already good answers that discuss the physics behind the phenomenon, I'd like to also point at the basic reason for the apparence of temperature from the human perspective.

First, remember how we experience pain and the sense of touch. If some molecules are ripped from your skin, then the brain receives a pain signal and you feel the sensation that we interpret as pain. If all molecules moved exactly aligned, then uoi wouldn't have felt it. In other words, what you feel is not the rapid and sudden and violent motion of these skin particles but rather their motion relative to the their neighbour skin particles.

Now to temperature. When you touch something hot, then the first set of particles and molecules of your skin get energy transferred which converts into kinetic energy at the microscale in the form of vibrations and fluctuations. They move randomly about and are not aligned.

That is why you feel temperature - you feel the relative differences in speeds, accelerations and positions of particles and moleces. Had they all moved equally in an aligned manner, then you wouldn't have detected any temperature. But naturally in its very nature, random vibrations and fluctuations are not aligned, and this is thus what we in more scientific ways have designed to define as temperature.

And now to your question. If all particles are moved simultaneously - if the gas bottle is moved on a truck or if your body is moving in a car, then this motion is aligned for all particles of the respective bodies and objects. There is no relative motion at the microscale for us to feel. Our fingertips haven't detected any particles moving differently than the rest.

This so why uniform motion doesn't cause a temperature rise. Temperature is strictly defined as microscale relative motion for this very reason - it doesn't make sense to define in any macroscale manner because we don't have any sense of feeling that relatives to hotness or coldness at this scale.


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.


One way to think about this is the consequences of temperature. For example: ionization into a plasma.

In a gas a temperatures above around 10,000K or so (depends somewhat on pressure), the molecules of gas will start to ionize, so we will see:

neutral atom => ion + free electron

Now, temperature relates to the average kinetic energy of particles in a gas. In the extreme case of a gas so low pressure that we just have a single neutral atom, we can imagine that single atom zipping through emptiness at high speed. Could that ionize, if it is traveling fast enough?


In order to ionize from thermal energy, it is not enough for one atom to be traveling at high speed. Since its ionizing, converting kinetic energy into ionization energy, would violate the laws of conservation of momentum and/or conservation of energy (depending on how you try setting the resulting velocities of the ion and free electron after the collision).

Two particles have to collide. And what matters is the relative velocity of the two atoms. If the relative velocity is high enough, then one of them can ionize, when they collide, and no laws of conservation of energy or momentum will be violated.

If you have a jar of low temperature gas moving at high speed, then the particles of the gas are not moving at high speed relative to each other, so no ionization takes place. In this sense, we can see that the temperature relates to the relative speed of the particles of the gas, not the macroscopic speed of the jar of gas itself.


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.

  • $\begingroup$ 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 ? $\endgroup$ Commented Mar 23, 2016 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). $\endgroup$ Commented Mar 23, 2016 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
    Commented Mar 23, 2016 at 19:05

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