I was trying to evaporate water at a lower temperature than its boiling point.So i tried the experiment in a vacuum chamber that was actually meant for sputtering.I placed a small plastic beaker of water in the chamber and tried to vacuum the chamber.To my surprise the water became very cold in less than two mins.can anyone please explain the reason..or am I doing something wrong with the procedure.??
I am interested to your question (similar to my case).and will explain in detail soon. Bit hurry for this time, please consult me.
at normal temperature, the process of water evaporation occurs very slowly. While decreasing the pressure the liquid boiling point is decreasing and evaporation becomes more intensive. So by decreasing the temperature water starts actively evaporate (can start boiling if not overheated), and as a result the whole system loose energy and temperature go down. Also see Raoult's law.
Same reason that when you spill a volatile liquid on yourself it feels cold. The more volatile the liquid, the colder it feels. Water feels colder than room temperature and alcohol and acetone feel colder than water because the phase change from liquid from gas requires energy (is endothermic), it absorbs the energy from surroundings and thus cools you or whatever it is in contact with. AND as you lower the surroundinug pressure, you increase the evaporation rate. Note when "trying to evaporate water at a lower temperature than its boiling point" you should be aware that the boiling point (bp) depends on the pressure and is defined as the temperature at which the vapor pressure equals the surrounding pressure. At normal atmospheric pressure, the bp is 100 deg C. Here is a nice chart of temp vs vapor pressure http://intro.chem.okstate.edu/1515sp01/database/vpwater.html - this is equivalent to bp vs pressure. At 1 deg C, the pressure is 4.9 mmHg, so water's bp is 1 deg C when the pressure is 4.9 mmHg. The rapid evaporation causes rapid cooling.
I agree with the previous responses. An important experimental note is that you should be careful with your vacuum pumps when pumping on a vessel containing significant quantities of condensable vapors, like water or alcohol. Oil-sealed mechanical rotary-vane pumps can have problems with vapors condensing in the oil and reducing the performance, and ultimately the lifetime, of the pump. This is often a problem when metal hoses are cleaned with solvents and not fully dried. This can be mitigated by opening the ballast valve on the pump until the vapors are gone, and by more frequent oil changes in case of heavy condensate loads.
With a name like Boyle, he should know about gasses. I focused on answering the original poster's question, but Dennis raises important practical points. If pumping significant amounts of water or other volatiles, change the pump oil much more frequently than usual or put in a vapor trap (e.g., l-N2 or dry ice cooled for water vapor) or use a telfon sealed pump.
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.
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