Effect of Uniform Motion on Internal Energy of an Ideal Gas in an Isolated container [duplicate]

Question

“The internal energy of a gas contained in a beaker is U, if the box starts moving with velocity V, then internal energy of the gas will be...”

Elaboration

So, further explaining the situation, we have an Insulating Container which contains gas with Internal Energy U at rest. Now, if we put the container in an Uniform Motion (with velocity V), what will be the impact on its Internal Energy, will it Increase/Decrease/Remain Constant.

Important Considerations

1. The Container is considered to be insulated, therefore, there will be no exchange in heat energy.
2. The gas filled inside the container is considered to be an Ideal Gas.
3. No external force is acting on the container when it is in motion, that is, it is not coming from rest to velocity V but is kept under the uniform motion with constant velocity V, therefore, there will be no Compression in the gas molecules towards the wall opposite to the direction in motion and hence, no change in pressure.

My Thoughts

I took this situation in two different point of views and tried to prove the answer in two approaches.

Approach #I

First of all, we know that when the container is at rest, the gas molecules with still be in motion and thus have a Kinetic Energy due to their Random Motion, Let's Call this KE_r.

When the container is observed in an uniform motion, its new Kinetic Energy will be the sum of its Kinetic Energy due to its Random Motion within the container and the Translatory Motion of the Container itself (KE_x). Therefore, the resultant Kinetic Energy is given by: KE_motion = KE_r + KE_x (KE_r <<< KE_x)

Clearly, KE_rest < KE_motion

Now, Internal Energy of a system is defined as:

Internal energy (U) is defined as the total energy of a closed system. Internal energy is the sum of potential energy of the system and the system's kinetic energy.

By the above definition, we can conclude that U ∝ KE ∴ U_rest < U_motion

In other words, due to the increase in Kinetic Energy of every individual gas molecule in the container, the Internal Energy will increase.

Approach #II

By First Law of Thermodynamics,

ΔQ = ΔU + W

Since we're considering the beaker to be insulated, ΔQ = 0. Therefore, ΔU = -W. Since the Pressure and Volume of the Gas within Container is not changing, we can say:
PΔV = 0.
∴ W = 0
∴ ΔU = 0

Note: W = 0 can also be proved by the fact that W = ∫F·dx, however in uniform motion, F = 0. Therefore, W = 0.
Therefore, the Internal Energy of the Gas Molecules in the container should remain the same.

Problem

As you can observed above, both of my conclusions are contradicting each other. I would like to know where my assumptions and theory went wrong on the paper.

Additional Request

If possible, can you also explain the given situation without the 3rd assumption, which is:

No external force is acting on the container when it is in motion, that is, it is not coming from rest to velocity V but is kept under the uniform motion with constant velocity V, therefore, there will be no Compression in the gas molecules towards the wall opposite to the direction in motion and hence, no change in pressure.

For example, if the container is starting from rest (v = 0) to some velocity (v = V) with some acceleration a under some Force F?
I know both situations are quite different to each other but it would be very helpful to understand both the situations to come to a final conclusion.