Physics 301 28-Oct-1999 13-1 Superfluid Helium As We Mentioned Last Time, the Transition of 4He at 2.17 K at 1 Atm Is Believed T

Physics 301 28-Oct-1999 13-1

Superﬂuid Helium

As we mentioned last time, the transition of 4He at 2.17 K at 1 atm is believed to be the condensation of most of the helium atoms into the ground state—a Bose-Einstein condensation. That this does not occur at the calculated temperature of 3.1 K is believed to be due to the fact that there are interactions among helium atoms so that helium cannot really be described as a non-interacting boson gas! Above the transition temperature, helium is refered to as He I and below the transition, it’s called He II.

K&K present several reasons why thinking of liquid helium as a non-interacting gas is not totally oﬀ the wall. You should read them and also study the phase diagrams for both 4He and 3He (K&K ﬁgures 7.14 and 7.15).

The fact that something happens at 2.17 K is shown by the heat capacity versus temperature curve (figure 7.12 in K&K) which is similar to the heat capacity curve for a phase transition and also not all that different from the curve you’re going to calculate for the Bose-Einstein condensate in the homework problem (despite what the textbook problem actually says about a marked difference in the curves!). In addition to the heat capacity, the mechanical properties of 4He are markedly different below the transition temperature. The liquid helium becomes a superfluid which means it flows without viscosity (that is, friction).

The existence of a Bose-Einstein condensate does not necessarily imply the existence of superﬂuidity.

To understand this we need to examine the mechanics of friction on a microscopic level. Without worrying about the details, friction must be caused by molecular collisions which transfer energy from the average motion (represented by the bulk velocity) to the microscopic motion (internal energy).

If our Bose-Einstein condensate were really formed from a gas of non-interacting particles, then it would be possible to excite any molecule in the condensate out of the ground state and into the first excited state simply by providing the requisite energy (and momentum). In the last lecture, we calculated that under typical conditions, the energy 31 difference between the ground state and the first excited state was about 10− erg, an incredibly small amount of energy that would be very easy to provide given that thermal 16 energies are about 10− erg.

In order to have superfluid behavior, it must be that there are interactions among the molecules such that it’s not possible to excite just one molecule out of the condensate and into the first excited state. A better way to say this is that due to the molecular interactions, the single particle states are not discrete energy states of the superfluid. We need to consider the normal modes of the fluid—the longitudinal oscillations or the sound waves. In particular, we can consider travelling sound waves of wave vector k

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and frequency ω. (Rather than the standing waves which carry no net momentum.) A travelling wave carries energy in units of ¯hω and momentum in units ¯hk.Thenumberof units is determined by the number of phonons or quasiparticles in the wave.

Now imagine an object of mass M moving though a stationary superfluid with velocity Vi. In order for there to be a force on the object, there must be a momentum transfer to the superfluid. In order to do this, the object must create excitations in the fluid which contain momentum (the quasiparticles in the travelling waves we just discussed). Of course, if quasiparticles already exist, the object could “collide” with a quasiparticle and scatter it to a new state of energy and momentum. (This can also be viewed as the absorption of one quasiparticle and the emission of another.) We will assume that there are not very many existing quasiparticles and consider only the creation (emission) of a quasiparticle.

So, let’s consider this emission process. Before the event, the object has velocity Vi and afterwards it has velocity Vf . We must conserve both energy and momentum,

1 1 MV 2 = MV 2 +¯hω , 2 i 2 f and MVi = MVf +¯hk .

We can go through some algebra with the goal of solving for Vi k. The momentum equation can be rewritten as · MV ¯hk = MV , i − f squared and divided by 2M,

1 1 1 MV 2 ¯hV k + ¯h2k2 = MV 2 . 2 i − i · 2M 2 f Subtract from the energy equation to get

1 ¯hV k =¯hω + ¯h2k2 , i · 2M or k ¯hω ¯hk V = + . i · k ¯hk 2M

What we really want to do is place a lower limit on the magnitude of Vi. This means we can drop the term containing M on the right hand side. The smallest value will occur when Vi is parallel to k/k, the unit vector in the k direction. This corresponds to emission of the quasiparticle in the forward direction. Thus

¯hω V > . i ¯hk

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I’ve left the ¯h’s there in order to emphasize that the right hand side is the ratio of the energy to the momentum of an excitation,

Suppose the excitations are single particle states with momentumh ¯k and energy ¯h2k2/2m. This is the travelling wave analog to the standing wave particle in a box states we’ve discussed many times. Then the right hand side becomes ¯hk/2m which goes to zero as k 0. Thus, an object moving with an arbitrarily small velocity can produce an excitation→ and feel a drag force—there is no superﬂuid in this case. (Note: k must be bigger than 1/L,whereL is the size of the box containing the superﬂuid, but as we’ve already seen∼ the energies corresponding to this k are tiny compared to thermal energies.)

Suppose the excitations are sound waves (as we’ve been assuming) and the phase velocity is independent of k.Then ω V > = v , i k s

where vs is the phase velocity of sound in the fluid. This means that if an object flows through the fluid at less than the velocity of sound, the flow is without drag! That is, the fluid is a superfluid.

In fact, the vs is not independent of k and what sets the limit is the minimum phase velocity of any excitation that can be created by the object moving through the ﬂuid. 1 Figure 7.17 in K&K shows that this minimum is about 5000 cm s− for the low lying excitations in 4He. K&K point out that Helium ions have been observed to travel without 1 drag through He II at speeds up to about 5000 cm s− !

Comment 1: It appears that we’ve shown that any fluid should be a superfluid as long as we don’t move things through it faster than its speed of sound. In our derivation, we made an assumption that doesn’t apply in most cases. Can you figure out which assumption it was?

Comment 2: As a general rule, superﬂuidity or superconductivity requires the condensation of many particles into a single (ground) state and a threshold for creating excitations. The minimum velocity required to create an excitation is the threshold for non-viscous ﬂow.

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Heat and Work

Now we want to discuss the material covered in chapter 8 of K&K. This material might be considered to have a more classical thermodynamics rather than statistical mechanics ﬂavor. We’ve already discussed a lot of this material in bits and pieces throughout the term, so we will try to focus on the material not yet covered and just hit the highlights of the remaining material.

Heat and work occur during processes. They are energy transfers. Work is an energy transfer by macroscopic means and heat is an energy transfer by microscopic means. We’ve discussed reversible processes several times and we’ll assume reversible processes unless we explicitly state otherwise. When work is done to or by a system, the macroscopic parameters of the system are changed—for example changing the volume causes pdV work to be performed. Performing work changes the energy, U, of a system. But work does not change the entropy. Heat transfer changes the entropy as well as the energy:

dU =¯dQ = τdσ.

A very important activity in any modern society is the conversion of heat to work. This is why we have power plants and engines, etc. Basically all forms of mechanical or electrical energy that we use involve heat to work conversion. Not all of them involve fossil fuels, and in some cases it may be hard to see where the heat enters. For example, what about hydro-electric power? This is the storage of water behind a dam and then releasing the gravitational potential energy of the water to run an electric generator. Where is the heat supplied? Heat is supplied in the form of sunlight which keeps the weather going which provides water in the atmosphere to make rain to fill the lake behind the dam. Of course, the economics of this process are quite different from the economics of an oil fired electrical generating plant.

It was the steam engine (conversion of heat, obtained by burning coal, into work) that allowed the industrial revolution to proceed. The desire to make better steam engines produced thermodynamics!

With an irreversible process you can turn work completely into heat. Actually, this statement is not well defined. What we really mean to say is that with an irreversible process we can use work to increase the internal energy of a system and leave that system in a final configuration that would be exactly the same as if we had reversible heated the system. For example, consider a viscous fluid in an insulating container. Immersed in the fluid is a paddle wheel which is connected by a string running over various pulleys and whatever to a weight. The weight is allowed to fall under the influence of gravity. Because the fluid is so viscous, the weight drops at a constant slow speed. Once the weight reaches the end of its travel, we wait for the fluid to stop sloshing and the temperature and pressure in the fluid to become uniform. Thus essentially all of the mechanical gravitational

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potential energy is converted to internal energy of the fluid. We can take the fluid from the same initial state to the same final state by heating slowly (reversibly!) until we have the same temperature rise.

There is no known way to convert (reversibly or non-reversibly) heat (more properly, internal energy, U) entirely into work with no other change. This is one of the ways of stating the second law of thermodynamics.

It is certainly possible to convert heat into work. (I’m getting tired of trying to say it exactly correctly, so I’ll just use the vernacular and you know what I mean, right?) The constraints are that you can’t convert all of it to work or there must be some permanent change in the system or both. For example, suppose we reversibly add heat to an ideal gas while we keep the volume constant. Then we insulate the gas and allow it to reversibly expand until its temperature is the same as when we started. Then the internal energy of the gas is the same as when we started, so we have completely converted the heat into work, but the system is not the same as when we started. The gas now occupies a bigger volume and has a lower pressure.

The problem is that when we reversibly add heat to a system we add internal energy dU =¯dQ and we also add entropy dσ =¯dQ/τ, but when we use the system to perform work, we remove only the energy dU =¯dW and leave the entropy! If we want to continue using the system to convert heat to work, we have to remove the entropy as well as the energy, so there is no accumulation of entropy. The only way to remove entropy (reversibly) is to remove heat. We want to remove less heat than we added (so we have some energy left over for work) so we must remove the heat at a lower temperature than it was added in order to transfer the same amount of entropy.

To make this a little more quantitative, consider some time interval (perhaps a com- plete cycle of a cyclic engine) during which heat Qh is transfered into the system at temperature τ ,heat Q is transfered into the system at temperature τ , and energy W h − l l − in the form of work is transfered into the system. (So heat Ql > 0 leaves the system and work W>0 is performed on the outside world.) At the end of this time interval we want the system to be in the same state it was when we started. This means ∆U =0=Q Q W, h − l − and Q Q ∆σ =0= h l . τh − τl We ﬁnd Q τ h = h , Ql τl and W τl ηC = =1 . Qh − τh

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The ratio of the heat input and output is the same as the ratio of the temperatures of the input and output reservoirs. The energy conversion efficiency or just efficiency, η is defined as the work output over the heat input, W/Qinput. For the ideal engine we’ve been considering, the efficiency is ηC,theCarnot efficiency, and is the upper limit to the efficiency of any real (i.e. non-reversible) engine operating between temperature extremes τh and τl. Carnot might be called the father of thermodynamics. He worked in the early 1800’s and understood the second law. This was before heat was recognized as a form of energy!

Of course, this definition of efficiency is motivated by the fact that if you’re an electric power company, you can charge your customers based on W but you have to pay your suppliers based on Qh and you want to maximize profits!

We live on the surface of the Earth and any engine must dump its waste heat, Ql, at what amounts to room temperature, about 300 K. This is roughly the equilibrium temperature of the surface of the Earth and is set by radiation equilibrium between the Sun and Earth and between the Earth and space (T = 3 K). See problem 5 in chapter 4 of K&K. Aside: are you surprised that room temperature and the surface temperature of the Earth are about the same? Anyway, the waste heat goes into the environment and usually generates thermal pollution. There may come a time when a cost is associated with Ql. In this case it’s still desirable to maximize η, because that minimizes Ql.

Because waste heat must be dumped at room temperature, improving the Carnot eﬃciency requires increasing the high temperature, τh. But this is not so easy to do, especially in an economically viable power plant that’s supposed to last for many years,

Comment 1: no real engine is reversible, so all real engines operating between temperature extremes τh and τl will have an eﬃciency less than the Carnot eﬃciency.

Comment 2: many practical engines are designed in such a way that heat is exchanged at intermediate temperatures as well as the extremes. Such engines, even if perfectly reversible, frictionless, etc. have an efficiency less than the Carnot efficiency. However, if one assumes a perfect, reversible engine operating according to the specified design, one can calculate an efficiency (lower than the Carnot efficiency) which is the upper limit that can be achieved by that engine design.

A reversible refrigerator uses work supplied from the outside to remove heat from a low temperature reservoir and deposit that heat plus the work used as heat in a high temperature reservoir. Such a refrigerator is basically the reversible engine just discussed, but run backwards! The signs of Qh, Ql,andW all change but anything involving their ratios remains the same. (Note if you wanted to “derive” refrigerators you could start from the same idea we used with the engine—entropy is only transfered when there’s a heat transfer and entropy must not be allowed to accumulate in the refrigerator.) With refrigerators, one uses a coeﬃcient of performance, this is deﬁned as the ratio of the heat

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removed from the low temperature reservoir to the work required. This is

Q γ = l , W

and for a reversible refrigerator operating between temperatures τl and τh,theCarnot coeﬃcient of performance is τ γ = l , C τ τ h − l and this is an upper limit to the performance of any refrigerator operating between the same temperature extremes.

Aside: If you go to a department store and look at air conditioners, you will find something called an energy efficiency rating (EER) which is basically the coefficient of performance. But, I believe these are given in BTU per hour per watt. That is they have dimensions instead of being dimensionless! To convert to a dimensionless number you must multiply by 1055 Joules 1 Hour J Hour =0.29 . 1BTU · 3600 Seconds BTU s A typical EER you’ll find on an air conditioner is roughly 10, so the “real” γ is about 3!

Note that all reversible engines operating between the same two temperature reservoirs must have the same eﬃciency. Similarly, all reversible refrigerators operating between the same two temperature reservoirs must have the same coeﬃcients of performance.

To see this, suppose that one has two reversible engines operating between the same two temperature reservoirs but they have different efficiencies. Run the low efficiency engine for some time, taking heat Qh from the high temperature reservoir, producing work W , and dumping heat Q = Q W in the low temperature reservoir. Now run l h − the other engine in reverse (it’s reversible!) as a refrigerator removing heat Ql from the low temperature reservoir, so the low temperature reservoir is exactly the same as when we started. To do this, the refrigerator is supplied with work W 0 and it dumps heat Qh0 = Ql + W 0 to the high temperature reservoir. Since this is the more efficient of the two reversible engines, W 0

Similar kinds of arguments can be used to show that all reversible refrigerators operating between the same two reservoirs must have the same coefficients of performance, that reversible engines are more efficient than irreversible engines, and that reversible refrigerators have higher coefficients of performance than irreversible refrigerators.

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The Carnot Cycle

We’ve mentioned the Carnot eﬃciency and we’ve talked about heat engines, but how would one make a heat engine that (were it reversible) would actually have the Carnot eﬃciency? Simple, make an engine that uses the Carnot cycle.

The Carnot cycle is most conveniently plotted on a temperature-entropy diagram. We plot the entropy of the “working substance” in an engine on the horizontal axis and the temperature of the working substance on the vertical axis. The working substance might be an ideal gas. It’s whatever component actually receives the heat and undergoes changes in its entropy and internal energy and performs work on the outside world. There are four steps in a Carnot cycle. In step ab, the temperature is constant at τh while the entropy is increased from σ1 to σ2. This is the step in which the system is in contact with the high temperature reservoir and heat Qh = τh(σ2 σ1) is added to the system. If the system is an ideal gas, then it must expand to keep the− temperature constant, so it does work on the outside world. In step bc, the temperature is lowered at constant entropy. No heat is exchanged, the gas expands and does more work on the outside world. In step cd, entropy is removed at constant temperature by placing the system in contact with the low temperature reservoir. The heat removed is Ql = τl(σ2 σ1). In this step, the gas is compressed in order to maintain constant temperature, so the− outside world does work on the gas. In step da, the system is returned to the starting temperature, τh,by isentropic compression, so more work is done on the system. The hatched area within the path followed by the system is the total heat added to the system in one cycle of operation,

Q = Q Q = τdσ. h − l I Since the system returns to its starting conﬁguration (same U) this is also the work done in one cycle.

Whatever the working substance, a Carnot cycle al- ways looks like a rectangle on a τσ diagram. (It has two isothermal segments and two isentropic segments.) Given a working substance we can plot the Carnot cycle on a pV diagram. The ﬁgure shows the Carnot cycle for a monatomic ideal gas. The vertices abcd in this diagram are the same as the vertices abcd in the τσ diagram. So paths ab and cd are the constant temperature paths with τ = τh and τ = τl. Along these paths pV = const. Paths bc and da are the constant entropy paths with σ = σ2 and σ = σ1.

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Along these paths pV 5/3 = const. The work done on the outside world in one cycle is the hatched area within the path, W = pdV . I

The arrows on the paths indicate clockwise traversal. In this direction, the Carnot cycle is a Carnot engine producing work and waste heat from high temperature input heat. If the cycle is run in reverse—counterclockwise—one has a Carnot refrigerator using work to move heat to a higher temperature reservoir.

As an example of a non-Carnot cycle, suppose we have a monatomic ideal gas which is the working substance of a reversible engine and it follows the rectangu- larpathonthepV diagram shown in the ﬁgure. Along da, heat is added at constant volume V1. On ab, heat is added at constant pressure. p2, and work is performed. On bc heat is removed at constant volume, V2,andon cd, heat is removed at constant pressure, p1, while the outside world does some work on the system. As before the total work done on the outside world is the area within the path and in this case,

W =(p p )(V V ) . 2 − 1 2 − 1 The heat added is 3 5 Q = N(τ τ )+ N(τ τ ) , in 2 a − d 2 b − a 3 5 = (p V p V )+ (p V p V ) , 2 2 1 − 1 1 2 2 2 − 2 1 5 3 = p V p V p V . 2 2 2 − 2 1 − 2 1 1 The actual eﬃciency of this reversible engine is

W p2V2 p2V1 p1V2 + p1V1 η = = 5 − − 3 , Qin p V p V p V 2 2 2 − 2 1 − 2 1 1 while the Carnot eﬃciency is p1V1 ηC =1 . − p2V2 These formulae aren’t all that illuminating so let’s consider a numerical example: suppose p2 =2p1 and V2 =2V1. Then the temperature at the hottest (upper right) vertex is 4 times the temperature at the lowest (lower left) vertex, so the Carnot eﬃciency is 3 η = . C 4

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The actual efficiency is 2 η = . 13 If you’re actually trying to build a heat engine that operates on this cycle, then as you improve the engine by reducing friction, heat losses, etc., you will approach an efficiency of 2/13 and this should be your goal, not the Carnot efficiency.

If you want to approach the Carnot efficiency, you must redesign the cycle to be more like a Carnot cycle. In the cycle shown, the extreme temperatures are reached only at the upper right and lower left vertices. Most heat transfers are at less extreme temperatures and this is why the actual efficiency is so much less than the Carnot efficiency .

Other Thermodynamic Functions

We have concentrated on the internal energy, U, in the preceding discussion. If we consider a constant temperature process, then the work done on the system is the change in the Helmholtz free energy. This is because at constant temperature, d(τσ)=τdσ,so

dW¯ = dU d(τσ)=dF (constant temperature) . −

Many processes occur at constant pressure, such as all processes open to the atmosphere. If a process occurs at constant pressure, then we are letting the system adjust it’s volume “as necessary,” so we cannot really use or supply any pdV work performed by or on the system. The pdV work “just happens.” If the system can perform work in other ways, then we can divide the work into the pdV work and other work,

dW¯ =¯dWother +¯dWpV ,

and dW¯ = dU dW¯ dQ¯ , other − pV − = dU + pdV dQ¯ , − = dU + d(pV ) dQ¯ , − = dH dQ¯ (constant pressure), − where H = U + pV , is called the enthalpy. In any constant pressure process the heat added plus the non-pdV work done is the change in enthalpy. In particular, if there is no non-pdV work done, the change in enthalpy is just the heat added. Note thatdW ¯ other is what K&K call the “eﬀective” work in a constant pressure process.

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In the event that we have a reversible process that occurs at constant temperature and constant pressure, the Gibbs free energy is useful. This is deﬁned as G = F + pV = H τσ = U + pV τσ. − − It should be clear that

dW¯ other = dG (constant temperature and pressure) . Also, a system which is allowed to come to equilibrium at constant temperature and pressure will come to equilibrium at a minimum of the Gibbs free energy.

As an example of the use of the Gibbs free energy, consider a system (cell) of two noninteracting electrodes in an electrolyte consisting of sulfuric acid dissolved in water. The sulfuric acid becomes two hydrogen ions and one sulfate ion, + H SO 2H +SO−− . 2 4 ↔ 4 When current is forced through the system in the direction to supply electrons to the cathode, the reaction at the cathode is + 2H +2e− H , → 2 and the reaction at the anode is 1 SO−− +H O H SO + O +2e− , 4 2 → 2 4 2 2 and the net reaction is 1 H O H + O . 2 → 2 2 2 If the current is passed through the cell slowly and the cell is open to the atmosphere and kept at constant temperature, then the process occurs at constant τ and p. The “other” work is electrical work. 1 W = G(H O) G(H ) G(O ) , other 2 − 2 − 2 2 where the Gibbs free energies can be looked up in tables and it is found that the difference is 1 ∆G = 273, 000 J mol− . − The other work done is electrical work equal to the charge times the voltage. Since we have two electrons per mole, W = 2eN V , other − 0 0 or ∆G V0 = =1.229 Volts , −2N0e where N0 is Avogadro’s number and e isthechargeonanelectron. V0 is the voltage that is established with no current flowing. At higher voltage, current flows and the cell liberates hydrogen and oxygen (electrolysis). If these gases are kept in the cell and allowed to run the reactions in reverse, one obtains a fuel cell in which the reaction of hydrogen and oxygen to form water generates electricity “directly.”