Physics 301 24-Nov-2003 25-1Cooling by Expans ionNow we’re going to change the s ubject and consider the techniques used to get reallycold temperatures. Of course, the best way to learn about these techniques is to work ina lab where experiments in the micro-Kelvin to milli-Kel vin range are being d one. Oneof the thi ngs you mig ht learn, that you won’t find in K&K is that cryostats have tobe mechanically isolated from their surroundings. Any energy that is tr ansmitted to theinterior of the cryostat winds up as thermal energy that the cooling appa ratus must removefrom the system.To get to really cold temperatures requires s everal stages. The first stage is usuallythe cooling of h el ium to just above it’s liquefaction temperature. This is followed byliquefaction and pumping on the helium to get to a few tenths of a Kelvin, the heliu mdilution refrigerator to get to milli-Kelvins , and adiabatic demagnetization to get to mi cro-Kelvin. These days, one can use laser cooling and evaporative cooling of trapped atoms toget to na no-Kelvins. (Recall the Physics Today articles on the Bose-Einstein condensate,hand ed out with lecture 17.) These techniques are not discussed in K&K!So, how does one d o cooling by expansion? The basic scheme involves a workingsubstance (perhaps helium in the l a b, a banned CFC i n an old refrigerator?). The refrig-erator operates in a cycle. The working substance (called the gas from now on) travels thefollowing path. From the cold volume (where it picked up some thermal energy) it go es toa heat exchanger (where it picks up more thermal energy from the h ot gas, and then to acompressor. The compressor compresses the ga s, increasing its press ure and temperature.The hot, high temperature gas is allowed to cool in contact with the outsid e world (roomtemperatur e). The gas then flows through the heat exchanger where it gives up heat tothe cold gas on the way to the compressor. Then it goes through an expansion device,where it expands, does work on the outside world, and cools. Finally, the cold gas is sentback to the cold volume where it extracts heat.The expansion device is often a turbine which allows the gas to expand approximatelyisentropically. If a gi ven amou nt of gas has energy U1, volume V1and pressure p1on theinput side of the expansion device and U2, V2, and p2on the output side, then the totalenergy going in is U1+ p1V1= H1. (The pressure p1pushes the volume V1through thedevice.) Similarly, the energy coming out is U2+ p2V2= H2. Then the work done on theexpansion device isW = (U1+ p1V1) − (U2+ p2V2) = H1− H2=52N(τ1− τ2) ,where we have assumed an id eal monatomic gas so U = 3Nτ /2 and pV = Nτ and theenthalpy is H = 5Nτ /2. Note that the heat exchanger oper ates at con stant pressure, nomechanical work is done by the heat exchanger, and the energy transfered in the form ofheat appears as the change in enthalpy of the gas.Copyrightc 2003, P rincet on University Physics Department, Edward J. GrothPhysics 301 24-Nov-2003 25-2In a typical refrigerator for operation with helium , the gas is not liquefied in thisstep. There may several stages of expansion cooling required to get the helium close to itsliquefaction temperature. In househo ld refrigerator s and a ir conditioners the gas may beallowed to condense into a liquid. In this case the expansion device is often a porous plugwhich acts as the expansion valve in a Joule-Thomson device to be described shortly.Throttling ProcessesConsider a device with two insulated cham-bers. The chambers are separated by an insulatingpartition, with a small hole. The chambers havepistons which can be used to adjust the pressurein each chamber. Start with som e gas (or what-ever) in the left-han d chamber with volume V1andpressur e p1. Since the partition has a sma ll holein it, some o f the gas will get through the hole into the right-hand ch amber. We adjustboth pistons continuously, so the gas in the left hand chamb er stays at p ressure p1and thegas in the right-hand chamber stays at pressure p2< p1. Since p2< p1, the gas g raduallyflows from the left hand chamber to the right hand chamber. No heat is exchanged andno wo rk is done by the hole or the partition. The only work don e is the work done by thepistons. This is an ir reversi ble process. Changing the forces on the pistons by an infin ites-imal amount will not cause the process to go in the opposite direction. T here need not befriction or viscosity at the hole; in fact we assume there isn’t. There is a pressure drop atthe hole simply because it is small and molecules only run into it (and get through to theother side) every so often. This process is called a throttling process. We also assume thatthe speeds of the pistons are slow enough that we can ignore any kinetic energy of the gas.So, we start with some amount of gas containing internal energy U1, with pres sur ep1, occupying volume V1on the left-hand side of the partition. We en d up with thesame amount of gas containing internal energy U2, with pressure p2, occupying volume V2on the right-hand side of the partition. The work done on the gas by the left pi ston isW1= +p1V1and the work done o n the gas by the right piston is W2= −p2V2. As we’vealready mentioned, there is n o heat transfer. Therefore the change in interna l energy ofthe gas is equal to the work done on the ga s,U2− U1= W1+ W2= p1V1− p2V2,orU2+ p2V2= U1+ p1V1,orH2= H1.Copyrightc 2003, P rincet on University Physics Department, Edwa rd J. Grot hPhysics 301 24-Nov-2003 25-3The single small hole in our partition is simply to il lustrate the point. In practice, thesmall hole may be replaced by a partition containing many small holes, by a valve, by aporous pl ug, or even by a wad of cotton! What’s needed is a way to r estrict (throttle) theflow and produce a pressure drop without extracting heat or work.Note that in the refrigerator discussed in the previous section, the pressure drop wasnot prod uced by a throttling process. Instead, there was a turbine which extracted thework. That’s why that process was not a constant enthalpy process.Note also that since a throttling process is irreversible, the state of the gas as a wholeis not an equilib rium state (part of it’s at p1and part of it’s at p2). So we cannot reallyplot a throttling process on a p-V diagram.Some more comments on enthalpy:RecalldU = dQ − p dVand i n a constant volume (isochoric) process the change in energy is just the