If your vacuum chamber is thermally isolated, then the temperature of the water decreases because it is expanding adiabatically into the gas state. This occurs because you are vacuuming the gas out at a sufficiently fast rate and you are not transferring heat into your vacuum chamber as you do so.
Without using integral calculus, the mathematical explanation is as follows: let the initial volume of the liquid be Vi and its initial temperature Ti.Once you have vacuumed out the atmospheric gas in the chamber, the water may now occupy the entire volume of the chamber, which means it is no longer confined to the beaker and that there are no other gases to occupy volume. Let this volume be Vf and it has a temperature Tf. In an adiabatic process, TVγ-1 = constant, where γ is a number > 1, which gives:
TiViγ-1 = TfVfγ-1 => Tf = Ti(Vi/Vf)γ−1
Because γ > 1 and Vf > Vi, Tf must be < Ti. This tells us that the water cools down as it expands.
Consider the First Law of Thermodynamics:
ΔU = Q + W = Q - PΔV = (7/2)NRΔT = (7/2)NR[(2/3)kB-1ΔKE]
where U = internal energy, Q = heat, W = work, which in thermal situations is Pressure-Volume work (PΔV), N = the number of molecules in the gas, R = the gas constant, T = temperature in kelvins, kB = Boltzmann's constant and KE = random translation kinetic energy of the system. Because gas expansion is always endothermic, the water molecules absorb energy from inside of the chamber to break the intermolecular forces that are holding them together in the liquid state. In doing so, some of the random translational kinetic energy is converted into uniform translational kinetic energy. The loss of random translational kinetic energy to uniform translational kinetic energy equates to a decrease in the temperature of the system. Also, the water molecules are doing (-)PV Work when they are expanding. Therefore, the temperature of the system decreases as the water expands into a gas because you are not supplying it with more energy in the form of heat to offset these losses.
