If I have an ideal gas where
PV=nRTapplies, and I compress the gas to half the volume, does this mean that the pressure has necessarily doubled?
Not necessarily. It depends on the process. If you do a reversible isothermal compression of the gas, then $PV$=constant and the temperature does not change. In that case halving the volume will double the pressure because there is no change in temperature. If, on the other hand, you do a reversible adiabatic compression ($Q=0$) on an ideal gas (such as with a reversible adiabatic compressor) then the process is $PV^{k}$= constant, and halving the volume will not result in doubling the pressure and there will be an increase in temperature per the ideal gas equation.
I am confused by this formula as the explanation I have always heard as to how an AC compressor works is that it heats up the freon gas by compressing (decreasing volume) thus increasing the pressure.
The ideal gas equation only provides the relationship between pressure, volume and temperature at the two equilibrium states that the process connects. So it tells you nothing about the process, or path, that connects the two states.
The ideal reversible compressor, in which there is no heat transfer, performs an adiabatic compression, the pressure increases with decreasing volume, but not by the same degree as I stated above.
But how do these 3 variables play together?
P,VandT? Is there another formula that governs how they would change in relation to each other?
Again, the variables in the ideal gas equation apply to the equilibrium states and do not define the process that connects the equilibrium states. If the volume doubles you have to be told that the temperature does not change in going from state 1 to state 2 in order for you apply the equation and to be able to say that the pressure doubles.
Hope this helps.