A rectangular (simplified) container with rigid surfaces, has a certain mass of ideal gas within it, and it accelerates in free space, undergoing rectilinear motion. There are no dissipative forces. Now, since the container moves, its kinetic energy increases, and since the temperature of the enclosed gas is dependent on the kinetic energy, their temperature should consequently increase, which does not occur. What's the reason behind it?
I try to solve it in the following manner:
Lets consider a rectangular container and lets label its vertical edges A and B. Assuming that it moves along BA ,after a certain time the edge A moves to A', B to B', and thus, the effective space for the molecules remains always unchanged. Now considering the force parameter in the problem, here the forcing agent is only the edge B colliding with the molecules in the vicinity(let vicinity be the region BC, C being sufficiently close to B) of B. Now, I argue that the impulsive force imparted by B on the molecules near it, modifies the initial random motion direction of these nearby molecules, and forces them to move along BA direction. But after BC, i.e. in the region CA, the molecules don't attain a perfectly horizontal motion, and the alignment along BA deteriorates as one moves from B to A, in a disciplined manner, and yet not changing the Maxwellian distribution. I consider layers within the container, at the not-so-clearly defined C junction, there's a continuous yet slightly increasing variation from BA direction, although the resultant velocity of these molecules may be in NW/SW directions. These molecules collide with those in the next layer, and these latter molecules undergo further variation from BA (not to forget that in all these layers, some molecules might even move in NE/SE/E(AB) directions, the fraction of which is relatively low, but increases, and becomes relatively higher while moving along BA). Thus this variation considerably increases in the vicinity of the A edge, and as such, magnitudes of the velocities of these molecules eventually re-organize themselves (due to the changing angle-factor at each layer of molecules) in such a way that the final energy-distribution curve varies too little from the initial one, to be of any significance at a practical level, whence the temperature of the enclosed gas remains unchanged.
Is this a correct approach, as far as an explanation is concerned?
P.S. also provide me with an alternative simpler explanation, which I am sure there is...
Answer
The equation $$ T = \frac{m\langle v^2 \rangle}{3 k_B} $$ for $\langle v^2 \rangle$ the average speed of a particle in the gas does not hold in frames other than the rest frame of the gas, where the rest frame is the one in which $\langle v \rangle = 0$ for all gas particles. This is because the derivation of this law assumes that the movement of the particles is "random", in particular, that there is no preferred direction, and "no preferred direction" is equivalent to $\langle v \rangle = 0$.
Hence, there is nothing to explain here, your statement
since the temperature of the enclosed gas is dependent on the kinetic-energy, their temperature must consequently increase
is simply unfounded.
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