Does adiabatic process actually happen in real life?

I think everyone is familiar with the equation $pv^{\gamma}=constant$ and how it was derived.

But does everyone actually know $\frac{du}{dv}=-p$ which the p here actually mean instantaneous pressure exert by the gas system to the surrounding?

This can be prove by consider the mechanism change in internal energy which is $\frac{1}{\gamma-1}pv$ is equal to $-p$

consider a gas system that occupied volume v and pressure p, $p(v_0)=p_0$ undergo expansion of volume from $v_0$ to $v_0+h$ where h is very small increase and obtain pressure $p_1$ so

$\frac{1}{\gamma-1}p_0v_0-p_0(h)=\frac{1}{\gamma-1}p_1(v_0+h)$

By using binomial approximation, one can obtain

$p_1\approx p_0-h\gamma\frac{p}{v}$

While from $p_0v_0^{\gamma}=constant$

$p_0v_0^{\gamma}=p_1(v_0+h)^{\gamma}$

$p_1=p_0(\frac{1}{1+\frac{h}{v_0}})^{\gamma}=p_0(1+\frac{h}{v_0})^{-\gamma}\approx p_0-h\gamma\frac{p}{v}$

This hold even for -h

$\frac{1}{\gamma-1}p_0v_0-p_0(-h)=\frac{1}{\gamma-1}p_1(v_0-h)$

By using binomial approximation, one can obtain

$p_1\approx p_0+h\gamma\frac{p}{v}$

While from $p_0v_0^{\gamma}=constant$

$p_0v_0^{\gamma}=p_1(v_0-h)^{\gamma}$

$p_1=p_0(\frac{1}{1-\frac{h}{v_0}})^{\gamma}=p_0(1-\frac{h}{v_0})^{-\gamma}\approx p_0+h\gamma\frac{p}{v}$

In adiabatic free expansion,$\frac{du}{dv}=0$

Hence for +h

$\frac{1}{\gamma-1}p_0v_0-0(+h)=\frac{1}{\gamma-1}p_1(v_0+h)$

while for -h

$\frac{1}{\gamma-1}p_0v_0-0(-h)=\frac{1}{\gamma-1}p_1(v_0-h)$

which implies pv=constant

I think there a physics question in physics book long ago with the question an insulated cylinder contain gas system on left and right which is isolated by a partition, the pressure on the left side is higher than pressure on right, need to find the pressure when both side come to the same pressure. The book actually solve it using conservation of internal energy and taking adiabatic expansion on left side and compression on right side, I think this is not true? First for left partition to expand adiabatically, it must do a work which is same as its instantaneous pressure p but on the right side is different pressure than $p$, so it is not possible right? Also if both side happen adiabatically, the change of internal energy at the left partition will not be the same as right partition?

The question as below

A cylinder is close at both ends and has insulating walls. It is divided into two compartments by a perfectly insulating partition that is perpendicular to the axis of the cylinder. Each compartment contains 1.00 mol of oxygen, which behave as an ideal gas with $\gamma=1.4$. Initially the two compartments have equals volumes, and their temperatures are 550k and 250k. The partition is then allowed to move slowly until the pressure on its two sides is equal. Find the final temperatures in the two compartments.

This the solution in the book

Let

$p_1$ be the initial pressure and $p_2$ be final pressure in 1st compartment

$p_3$ be the initial pressure and $p_4$ be final pressure in 2nd compartment

$v_1$ be the initial volume and $v_2$ be final volume in 1st compartment

$v_3$ be the initial volume and $v_4$ be final volume in 2nd compartment

$T_1$ be the initial temperature and $T_2$ be final temperature in 2nd compartment

$T_3$ be the initial temperature and $T_4$ be final temperature in 2nd compartment

solving the equation

$p_1v_1^{\gamma}=p_2v_2^{\gamma}$

$p_3v_3^{\gamma}=p_4v_4^{\gamma}$

$p_2=p_4$ and $v_1=v_3$ solve the equation and using PV=nRT obtain $v_2=1.756v_4$ and $T_2=1.756T_4$

Since work done by the adiabatically expanding gas is equal and opposite to the work done by the adiabatically compress gas.

$\frac{nR}{\gamma-1}(T_1-T_3)=-\frac{nR}{\gamma-1}(T_4-T_2)$

$T_2+T_4=T_1+T_3=800k$ and using ratio above get $T_2=510K$ and $T_4=290K$

I was thinking is this correct? The gas pressure at 1st compartment should be using to counter the pressure on right compartment? As I tried to demonstrate above when the gas start to expand, in order to follow the curve $pv^{\gamma}$= constant, the equation should be

$\frac{1}{\gamma-1}p_1v_1-p_1(h)=\frac{1}{\gamma-1}p_2(v_1+h)$ and not

$\frac{1}{\gamma-1}p_0v_0-p_3(h)=\frac{1}{\gamma-1}p_1(v_0+h)$

Will it be more accurate if consider the approach below?

The right compartment undergo adiabatic compression

consider the length of cylinder be X and surface area A while y is the unknown

$p_4=p_3.(\frac{0.5XA}{(0.5-y)XA})$

According to conservation of energy

$2.5p_1v_1-(2.5p_4v_4-2.5p_3v_3)=2.5p_2v_2$

$p_1(0.5XA)-(p_3(\frac{0.5}{0.5-y})^{\gamma}(0.5-y)XA)-2.5p_3(0.5XA)=2.5p_2(0.5+y)XA$

Noticed "XA=Volume of cylinder" can be cancel out

Getting $p_2=\frac{0.5p_1-[p_3(0.5)^{\gamma}(0.5-y)^{1-\gamma}-0.5p_3]}{0.5+y}$

But $p_2=p_4$ at same pressure

$p_3.(\frac{0.5}{(0.5-y)})=\frac{0.5p_1-[p_3(0.5)^{\gamma}(0.5-y)^{1-\gamma}-0.5p_3]}{0.5+y}$

Using $p_1v_1=nR(550)$ and $p_3v_3=nR(250)$, $v_1=v_2$

Obtain $p_1=2.2p_3$ substitute in the equation

$p_3.(\frac{0.5}{(0.5-y)})=\frac{1.1p_3-[p_3(0.5)^{\gamma}(0.5-y)^{1-\gamma}-0.5p_3]}{0.5+y}$

$p_3$ can be cancel out and left y to be found.