One way I think to intuitively understand the reduction of temperature
is that the total energy inside the system remains constant due to no
exchange of heat but the volume increases due to expansion.
You're not thinking about it correctly. The total energy of the system, which in the case of an ideal gas is purely kinetic energy, decreases as a result of the expansion. This is due to the fact that the system does work and expends some of its internal energy in the process of doing so. From the first law,
$$\Delta U=Q-W$$
Where $Q$ is heat and is positive if heat transfers to the system, and W is work and is positive if done by the system. For the adiabatic expansion, $Q=0$ and therefore $\Delta U=-W$. In this case $W$ is positive when the system does work, which decreases internal energy. For an ideal gas, any process, $\Delta U=mC_{v}\Delta T$. So a decrease in internal energy results in a decrease in temperature.
My question is: whether there is any form of exchange of energy
between the system and the surroundings in a form other than heat due
to the movement of the piston as the gas expands or the reduction of
temperature is purely due to increase in the volume?
Yes, the other form of energy exchange (transfer) besides heat is work. As explained above, that's why the internal energy and temperature decreases when the gas expands adiabatically. When the gas expands it transfers internal energy to its surroundings if no heat transfer to the system is possible to supply energy for doing work.
I am confused whether the U in this equation is the total internal
energy or internal energy density?
$U$ in these equations is the total internal energy. The specific internal energy, or energy density as you call it, is the total internal energy divided by the mass. It is usually shown as a lower case $u$. Then in the first law equation you would also have lower case $q$ and lower case $w$, meaning heat and work transfer per unit mass of the system. In the equation $\Delta U=mC_{v}\Delta T$ if you divide by the mass, you get the specific internal energy, or $\Delta u=C_{v}\Delta T$.
Secondly, even if so energy is transferred out of the system in the
form of work done by the gas, the increase in volume also must
contribute to the decrease in temperature from a purely intuitive
point of view since the temperature depends on the internal energy
density rather than just the total internal energy. Is my reasoning
correct?
No that is not correct because temperature does not depend on the internal energy density rather than the total internal energy. It's the same for both.
Temperature is what we call an intensive thermodynamic property, meaning it does not depend on the amount of mass or volume of the system. Pressure is another intensive thermodynamic property.
Let's say you have a room full of air and its temperature is 20 $^0$C. If you divide the room in half each half has half the volume and half the mass of air of the whole room, but the temperature of each half of the room is still 20 $^0$C.
If we take the equation
$$\Delta U=mC_{v}\Delta T$$
where $\Delta U$ is the change in total internal energy and divide it by the mass, we get
$$\Delta u=C_{v}\Delta T$$
Where now $\Delta u$ is the specific internal energy (energy density). But $\Delta T$ is the same in both equations.
Follow up question: doesn't halving the room also halves the total
internal energy? But the temperature remains the same.
Yes each half has half the total internal energy because each half has half the mass of the total but the energy density of each half is the same and the temperature of each half is the same. That's because internal energy is an extensive property, meaning it depends on the mass. But temperature is an intensive property and does not depend on the mass. That's my point.
Here’s another perspective. The temperature of the room full of an ideal gas is a measure of the average kinetic energy per molecule of the gas molecules in the room. If you divide the room in half each half has half of the total kinetic energy (half the total number of molecules and thus half the internal energy), but the average kinetic energy per molecule in each half is the same so the temperature is the same.
Hope this helps.
