Does volume change as much as pressure changes in $pV=nRT$? If I have an ideal gas where  $pV=nRT$  applies, and I compress the gas to half the volume, does this mean that the pressure has necessarily doubled? And if it did, that should mean the temperature is unchanged?
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.
But how do these 3 variables play together? $P$, $V$ and $T$? Is there another formula that governs how they would change in relation to each other?
 A: 
If I have an ideal gas where PV=nRT applies, 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, V and T? 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.
A: There is indeed more information needed to figure out exactly how these three variables interact. This information can be phrased in several ways, but they all boil down to something like: how much heat escapes the gas during compression?

If you're doing this compression in a totally insulated chamber, where no heat can flow between the gas and its environment, then the compression is roughly* adiabatic. Alternatively, if you're in a chamber that's not completely insulated, but you do the compression quickly enough that heat doesn't have a chance to escape, that's also a roughly adiabatic process. The point is that no heat leaves the gas during this process. In an adiabatic compression, pressure and volume are related by the following formula:
$$P_iV_i^\gamma=P_fV_f^\gamma$$
where $P_i$ and $V_i$ are the initial pressure and volume, $P_f$ and $V_f$ are the final pressure and volume, and $\gamma=C_P/C_V$ is the ratio of specific heats at constant pressure and volume for the gas in question, which is also related to the number of degrees of freedom of the gas molecules. Combining this formula with the ideal gas law should give you enough information to determine the final state. Generally, in an adiabatic process, the temperature and pressure of the gas will both increase when it's compressed.

In contrast, if you're doing this compression in a chamber that's in contact with a heat bath, and you do the compression slowly enough to let the maximum amount of heat escape the gas as it's being compressed, then the compression is roughly* isothermal. In an isothermal compression, the temperature of the gas does not increase during the compression. Applying the ideal gas law will also tell you that the pressure will increase. This can happen because the heat that would have raised the gas's temperature during compression was allowed to escape to the environment.

And, of course, if you allow some heat to escape, but not the maximum amount, then you're doing something in between an adiabatic compression and an isothermal compression. In that situation, the temperature of the gas will increase, but not as much as in the adiabatic case. The pressure of the gas will also increase, by an amount greater than in the isothermal case but less than in the adiabatic case.

*"Roughly" here refers to the fact that these are all idealized descriptions, where the minimum possible entropy is generated. If you allow the gas to leave thermodynamic equilibrium as it's being compressed (which always happens to some extent in real life), then your results may not be quite the same as those above.
