Temperature change in a gas tank on a car There is a beautiful question I ran into:

A gas tank filled with gas at temperature $T$. The gas is at rest first. It is accelerated to a constant velocity of $V$. Assume the process is adiabatic.
Will the temperature of the gas change? (if yes - will it increase or decrease?)

I think there is a nice discussion we can have about how different reference frames measure temperature or how uniform velocity effects the MB distribution.
Thanks ahead for your answers and time :)
 A: As pointed out by @Chet Miller, during the acceleration phase part of the gas gets compressed, but at the same time part of the gas gets expanded, so that the net effect upon reaching constant velocity should be no change in temperature.
The following explains why the motion of the gas at constant velocity should not cause a change in temperature of the gas.
You need to differentiate the average translational kinetic energy (measured by temperature) of the contents of the system (the random motion of the gas molecules themselves), and the translational kinetic energy of the center of mass of the system as a whole, the tank of gas, which has nothing to do with the temperature of the gas.
In addition, if you were able to measure a change in the temperature of the gas in the tank moving at constant velocity, it would violate the special theory of relativity which states that the laws of physics are the same in all inertial frames (frames moving at constant velocity). Which is the same as saying the results of any experiment should be the same in all inertial frames, including the measurement of temperature.
If you were able to detect a change in temperature for the gas moving at constant velocity, you would have a means to detect absolute motion, in violation of relativity.

Thanks! At first, I thought the same way. But then I was thinking
about the energies this way:
$$U_{i}=U_{f} +\frac{(Vρ)v^2}{2}$$
where  is the internal energy apart from the kinetic energy of the
COM. this leads us to (using
$$U=\frac{PV}{γ-1}$$
and the law of ideal gases):
$$\Delta T≈v^2$$
The adiabatic index, the mass of gas molecule and Boltzmon constant
are not given and not of interest in this case. Please, where am I
wrong?

I believe where you went wrong is assuming there is a change in internal energy $U$.  There is  no change in the internal energy of the gas, i.e., $U_{f}=U_{i}$. The tank is adiabatic so $Q=0$. The tank is rigid so there is no boundary work done on the gas, or $W=0$. From first law, closed system, $\Delta U=Q-W$ and therefore $\Delta U=0$. If the gas is an ideal gas, $\Delta U=0$ means $\Delta T=0$ because for an ideal gas $\Delta U=nC_{V}\Delta T$.
The kinetic energy of the COM is not part of the internal energy. It's the energy of the system relative to an external (to the system) frame of reference. I like to refer to it as the "external" kinetic energy of the system (though some don't like the term.)
Hope this helps.
A: I'll give this a go!  My answer is that, no, the temperature of the gas will not increase.  Qualitatively, as you accelerate the chamber (let us say from right to left), the right wall of the chamber that is normal to the acceleration vector, will strike numerous gas particles.
Gas pressure can be related to the change in momentum of a gas particle striking a wall at rest.  Now the wall is moving and so the change in momentum will be greater.
Edit: However, the opposite wall (left) is moving away from the gas particles, so their change in momentum will decrease. These pressure changes should compensate each other and the temperature should remain the same.  With pV = nRT, if the pressure can logically be assumed not to change, and we know V does not change, therefore T does not change.
(At first I thought T would increase, and then realized I was only considering one wall!)
A: I'm going to play the devil's advocate, and argue that the temperature must increase.
First, I should take a moment to make the observation the question doesn't ask about a fluid that is in the gas phase, because the title of the question has "gas tank in a car" which implies we are talking about gasoline.
It is important to recognize that the internal energy of molecules is distributed among vibrational and rotational modes as well as translational.
As it turns out, it doesn't matter what phase the fluid is in as long as each molecule therein has at least rotational modes of motion accessible.
Without loss of rigor, we can make the following assumptions:
(0) The tank is a cube, which begins with velocity V0=[0,0,0] in [x,y,z], and ends with velocity V=[V,0,0].

(1) Arguments previously proposed based on symmetry of translational
modes are correct; that is, the compression on the "back" of the
tank cancels the expansion on the "front."

(2) The tank sustains near infinite acceleration (~V/dt) at a moment
when some number of GAS molecules, N, are impacting the walls
parallel to motion.

(3) Those molecules impacting the walls parallel to motion are
distributed uniformly among the four parallel walls, with N/4
molecules impacting per wall.

(4) The molecules in the tank share the three rotational degrees of
freedom (DoF) equally, *and* the molecules *impacting* the four walls
parallel to direction of motion at the moment of acceleration share 
rotational DoF equally among walls; thus N/(4*3) per degree of rotation
per wall.

(5) ALL the molecules in the tank initially rotate in only one of the
three available rotational DoF; i.e. either up-down, left-right,
forward-back.

(6) Molecules impacting have perfect contact; i.e. shear is
transferred to rotational motion only.

The three rotational DoF can be called [rx,ry,rz] where rx identifies rotation about the x axis, and so forth.
So, (6) shear is transferred to ry and rz.
Now, although an argument of symmetry might be made, analogous to those that have been made for translation, to those molecules already having ry and rz modes of rotation, there is no equivalent argument of symmetry for those molecules which begin with ry=rz=0, having only rx non-zero, to which ry or rz angular momentum is transferred. In fact, I don't believe there is an argument of symmetry for the other DoF either, since temperature is indifferent to rotational direction.
That is, those molecules for which previously only one degree of rotation (rx) was accessible now have two degrees accessible (rx and ry or rx and rz). With increasing accessible states (entropy), there will be an increase in temperature. There is no corresponding decrease in accessible states that does not already have a symmetric counterpart.
I should mention that because of the condition of "perfect contact" and the condition of adiabatic (energy cannot be transferred to the wall) one may argue that upon contact with walls parallel to the direction of motion, the molecule's rotation in any other direction must be converted to translational motion (not necessarily entirely). But, this conversion doesn't have importance with respect to the proposed argument.
Also, we can further predict the evolution of the system which may allow us to consider only translational motion in the definition of temperature. Once rotational energy in ry or rz has been transferred to the target molecule, the energy may be converted by collision with molecules in the system to translational energy of colliding molecules. If two molecules collide in the same instant on opposing sides of the molecule which has just received rotational energy, and these two molecules had exactly the same relative motion in the x direction with respect to the target molecule - zero - then the rotation would push one molecule in one direction, and the other in the opposite direction, each with equal and opposite momentum imparted, and the target molecule being left with zero rotational energy. This is not compensated or canceled but rather additive to the energy, and therefore temperature, of the system.
