Why does adiabatic contraction bring Carnot engine back to initial state? (and other questions)
I have many questions (all in bold) around the Carnot cycle. I know I'm not the first one to ask a bunch of questions in the middle of a description of my own understanding, but having read other answers, I really think this is a rather individual process.
I'm using the pebble model : pebbles are resting on a piston which moves vertically under the varying pressure of an ideal gas.
At point A, the gas is at the same temperature as the hot reservoir and its pressure equals $P_\text{pebbles+atm}$.
* Isothermic expansion : remove a pebble. $P_\text{pebbles+atm}$ decreases. Gas pressure is now greater than $P_\text{pebbles+atm}$, therefore the gas expands until pressures are equal. Gas has worked and therefore cooled. Hot reservoir transfers heat to the gas. Gas will heat, therefore pressure will once again exceed $P_\text{pebbles+atm}$ and the piston will move up a little more, and so on. Until I wrote this, my understanding was that one needed to remove pebbles one by one, and I've just realised my description only involves removing one (infinitesimally heavy) pebble, and gas expands until I cease heat exchange. Is there a mistake ?
At point B, heat exchange ceases. 749
Gas is at hot reservoir temperature, with pressure slightly greater than $P_\text{pebbles+atm}$, therefore expanding an extra (infinitesimal) step.
* Adiabatic expansion : I remove more pebbles (can I remove them in bulk here ?), $P_\text{pebbles+atm}$ decreases therefore the gas expands frictionlessly and cools a good deal. I'm assuming the gas needs to cool to a temperature close to the cool reservoir's temperature in order to initiate reversible contraction. If so, how do I know this will actually happen ?
At point C, gas is at the same temperature as cool reservoir.
* Isothermic contraction : I add a pebble. Does it have to be the same weight as the one I removed to initiate isothermic expansion ? $P_\text{pebbles+atm}$ increases and gas pressure is now less than $P_\text{pebbles+atm}$, therefore gas contracts and receives work from pebbles+atm. It heats a little, gives away this heat to cool reservoir, pressure decreases, gas contracts, and so on.
At point D, heat exchange ceasses. Gas contracts infinitesimally after the last release of heat.
* Adiabatic contraction : I add more pebbles (as many as I removed to initiate adiabatic expansion ?). $P_\text{pebbles+atm}$ increases again, gas contracts until pressure increases to match $P_\text{pebbles+atm}$.
If all removed pebbles were eventually repositioned on the piston, then $P_\text{pebbles+atm}$ obviously has identical initial and final values, and so does gas pressure. Further, having proven that $U=\frac{3}{2}nRT$, one can prove that the volume ratios during adiabatic transitions satisfy $(\frac{T_{cool}}{T_{hot}})^\frac{3}{2}=\frac{V_B}{V_C}=\frac{V_{final}}{V_D}$, therefore, if I want $V_{final}=V_{initial}$, I need to ensure that $V_D=\frac{V_CV_{initial}}{V_B}$. In other words, I can't initiate adiabatic contraction at any random moment, but I need do it when isothermic contraction reaches that precise volume. Is this correct ? I've never seen this precaution mentioned. 749