How to Model The Following Question as an Adiabatic Process? I was recently trying to attempt this question which was on 2019 JEE Advanced Physics Paper 2.
The question is as follows-


A small particle of mass m moving inside a heavy, hollow and straight tube along the tube axis undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the piston from closed end is $$L=L_{0}$$ the particle speed is $$v=v_{0}$$
The piston is moved inward at a very low speed V such that $$V<<(dL/L)v_{0}$$
Where dL is the infinitesimal displacement of the piston.

The choices given are

a)The particle’s kinetic energy increases by a factor of 4 when the piston is moved inward from $L_{0}$ to $\frac{L_0}{2}$

b)After each collision with the piston, the particle speed increases by $2V$
c)If the piston moves inward by $dL$, the particle speed increases by $2V.\frac{dL}{L}$
d)The rate at which the particle strikes the piston is $\frac{v}{L}$

I was able to solve the question by using Newtonian Mechanics by calculating the collision frequency and using the coefficient of restitution etc. to find all the required answers however the question in itself seemed to indicate or resemble some sort of a Adiabatic process with a slow moving piston hence I decided to think if I can model it as an Adiabatic or in general some sort of thermodynamic process? I have tried a few attempts but nothing of use has come up.
 A: This is a quite neat question, but the condition
$$V\ll \frac{d L}{L} v_0$$
was kind of confusing to me. In retrospect I understand it, but I think this is just badly written and that makes the question hard. How I really solved the question was by thinking about the full problem and then asking how it could be simplified if we assumed that the particle was somehow moving much faster than the piston and then seeing where that takes me.
So let's get to it. The point of the adiabatic treatment is to imagine that before the piston moves even a tiny bit, the particle goes back and forth so many times one can essentially average over these collisions to describe the long-term evolution of the system. This is true if the time between the collisions is much shorter than the time it takes for the piston to hit the end of the tube. The time between collisions is
$$\Delta t_{col} \sim \frac{2L}{v}$$
where I put the approximate $\sim$ since $L$ is evolving and the precise computation of $t_{col}$ would be more involved.
The time for the piston to come to the end of the tube is
$$T_{pist} = \frac{L}{V}$$
The condition $t_{col}\ll T_{pist}$ then reduces to the requirement that $V \ll v$ at all times. We will see that $v$ grows in time, so it would actually be sufficient to require only $V \ll v_0$ without any enigmatic reference to quantities such as "$dL/L$".

Now let us take a look at the evolution of the system. At every collision of the particle with the piston, the particle velocity grows by $2V$. You can see this by transforming into the rest-frame of the piston and treating the collision only as the particle bouncing off a static wall and then transforming back. Using the approximate formula for the time between collisions we obtain
$$\frac{\Delta v_{coll}}{\Delta t_{coll}} \approx \frac{V}{L} v$$
Note that this formula is valid up to other terms of relative size $V/v \ll 1$. We can also relate this to the change of $L$, since during $\Delta t_{coll}$ the piston will move by $\Delta L_{coll} = -V \Delta t_{coll}$. We then get
$$\frac{\Delta v_{coll}}{\Delta L_{coll}} \approx -\frac{v}{L}$$
Given the assumptions above, one can then describe the approximate evolution of the velocity smoothed over many collisions by the differential equation
$$\frac{d v}{dL} = -\frac{v}{L}$$
This can be integrated to yield
$$\frac{v}{v_0} = \frac{L_0}{L}$$
I believe that this sketch of the derivation gives you a good idea of how to work out the details and answer the question.
