FOR DECELERATIONCould you please explain this using mathematical notations just like you did in post#12 .
$$\Delta U + \Delta (KE)=-W$$where W is the positive work that the gas does on its surroundings (the container) and ##\Delta (KE)## is the change in kinetic energy (negative). So,
$$\Delta U=-(\Delta (KE)+W)\geq0$$with the = sign applying to very gradual deceleration and the ##\geq## sign applying to very rapid deceleration. Alternate and equivalently, $$W\leq-\Delta(KE)$$This equation says that, because of viscous dissipation of mechanical energy, the amount of work that the gas does on its surroundings (the container) is less than its decrease in kinetic energy. (Not all the kinetic energy can be converted to work).
$$\Delta U + \Delta (KE)=-W$$
where, in this case, W is negative, since the surroundings are doing work on the system, and ##\Delta (KE)## is positive. So,
$$\Delta U=-(\Delta (KE)+W)\geq0$$with the = sign applying to very gradual acceleration and the ##\geq## sign applying to very rapid acceleration. Alternate and equivalently, $$-W\geq\Delta (KE)$$This equation says that, because of viscous dissipation of mechanical energy, the amount of work that the surroundings needs to do to accelerate the gas (-W) is greater than its increase in kinetic energy. (Additional work is required to increase the kinetic energy).
So we see that, both in rapid acceleration and rapid deceleration, the internal energy increases (actually, as we shall see shortly, by exactly the same amount). If we were to use the same rationale that we had used in the original analysis of the problem to determine the change in internal energy for acceleration, we would have written that ##\Delta U=-\Delta (KE)## and we would have concluded that, for acceleration (##\Delta (KE)>0##), ##\Delta U## is negative.
Actually, we will now show that for the same magnitude of the change in kinetic energy, the change in internal energy is exactly the same for extremely rapid acceleration as for extremely rapid deceleration. We know from Galileo that the laws of physics do not change when a system is observed from two different inertial frames of reference. So, consider deceleration as observed from the following two inertial frames of reference:
Frame 1. The laboratory frame that has already been used to analyze the deceleration problem (where the observer is stationary on the laboratory floor)
Frame 2. An inertial frame in which the observer is moving with the original velocity of the container (in the positive x direction) forever (relative to to the laboratory frame).
As reckoned by an observer in inertial frame 2, the container is standing still, and the barrier that is going to stop the motion of the container when it hits appears to be moving toward the container in the negative x direction with velocity v. Once the container hits the barrier, both the container and the barrier are moving with the velocity v in the negative x direction. So, as far as the observer in frame 2 is concerned, the container and gas start out standing still, and, after the barrier hits them, they are instantly accelerated in the negative x direction to a velocity of magnitude v. So, as reckoned from frame 2, the container has experienced an acceleration exactly equal to the deceleration reckoned by an observer in the laboratory frame (frame 1). This shows that the change in internal energy for the acceleration and deceleration situations must be exactly the same for the same magnitude of the change in kinetic energy. The two situations are indistinguishable physically (since neither inertial observer can prove that it is his frame that is stationary and it is the other frame that is moving).