The Physics of Energy, page 35
(8.60)
The resulting partition function per particle is
(8.61)
The probability that any two particles are in the same state is negligibly small (Problem 8.16), so the system is not degenerate and its partition function is given by eq. (8.51),
(8.62)
where, following the discussion after eq. (8.50), we have included the factor of required when the gas particles are indistinguishable.
As in the previous examples, the internal energy and entropy follow from eqs. (8.37) and (8.38),
(8.63)
(8.64)
where we have replaced by V. This result for the internal energy agrees with the equipartition theorem (§5.1.3), and confirms that the definition of temperature we have used throughout this chapter, , coincides with the phenomenological definition (5.6) of temperature used in §5.
Note that the Gibbs factor of in does not affect the internal energy, since it appears as an additive constant in and is removed by the T derivative in the computation of U (see eq. (8.37)). The does, however, contribute a term to the entropy. can be simplified by using Stirling's approximation (B.73),
(8.65)
where is the number density of particles. The importance of the extra N-dependence introduced by the factor of is explored in Problem 8.15. This equation for is known as the Sackur–Tetrode equation.
Example 8.4 Internal Energy and Entropy of Helium
The Sackur–Tetrode equation (8.65) allows us to explore the way that the entropy of a simple monatomic gas such as helium changes with temperature, pressure, and volume.
First, consider heating a sample of helium from to at constant volume. We expect the entropy to increase because we are adding disorder to the system by increasing the average energy of the particles. According to eq. (8.65), , in agreement with the computation in Example 8.2.
Next, consider heating a sample of helium from to at constant pressure. Because the gas expands – according to the ideal gas law, – the increase in information needed to specify the microstate increases even more thanin the constant volume case. So we expect a greater increase in entropy in this case. From eq. (8.65) we find .We can get the same result by following the heat flow using and conservation of energy, , or , where H is the enthalpy (see §5). Then , which agrees with eq. (8.65) because for helium.
Next, consider doubling the pressure on a sample of helium at constant temperature. This halves its volume,reducing the set of accessible states for the gas molecules, and should decrease the total entropy. According to eq. (8.65), indeed, when the volume is halved.
Finally, note that the entropy per particle , which is a local property of an ideal gas, depends only on thedensity () and temperature, which are both also local properties, and not on V and N separately. The factor inthe partition function is needed for this form of the dependence, as seen in eq. (8.65) .
displays the volume dependence of the entropy that we anticipated earlier in this chapter and derived in Example 8.2: it increases by if the volume of the system is doubled (while keeping N fixed). The temperature dependence of reflects the relation to the heat capacity (see Example 8.2). One flaw in the expression for is that it does not vanish as ; instead it seems to go to . This is because we ignored the quantization of the energy levels when we approximated the sum over n by an integral. This is not a physically relevant issue, however, since all gases condense to a liquid or solid at temperatures well above those where this approximation breaks down.
8.8Spontaneous Processes and Free Energy
The laws of thermodynamics determine which physical (and chemical) processes can occur spontaneously and which cannot. In many cases the time scales for these processes are long enough that the initial and final states can be regarded as states of thermal equilibrium, characterized by state functions such as entropy. In these cases the laws of thermodynamics can be applied straightforwardly to determine which processes will occur spontaneously. This approach is particularly important in the study of chemical changes and changes of phase. Iron, for example, will rust – forming hydrated iron oxide FeO(HO) when exposed to moisture and oxygen. The time scale for this reaction is long enough that it is reasonable to consider iron, water, oxygen, and iron oxide all to be in thermal equilibrium in fixed proportions at any given time. With this assumption, it is possible to compute the entropy of these ingredients and the surrounding environment and to determine whether the 2 Law allows this process to proceed or not.
The 1 Law requires that energy is conserved, though internal energy (including energy stored in chemical bonds – see §9) can be transformed into heat or vice versa in chemical and phase changes. Thus, for example, when iron rusts, heat is given off into the environment as chemical bonds between oxygen and iron are formed. Such a reaction is exothermic. In contrast, when water evaporates, heat from the environment is transformed into latent heat of vaporization and the surface from which the water evaporates is cooled. This is an endothermic reaction. Both exothermic and endothermic reactions can conserve energy, and both can occur spontaneously if they satisfy the 2 Law.
Before turning to the conditions for spontaneity imposed by the laws of thermodynamics, we stress that these are necessary, but not sufficient conditions for a process to occur. Activation barriers may prevent a process from proceeding even though it is allowed by both the first and second laws of thermodynamics. Such barriers depend upon the details of the reaction and can be studied with the tools of chemistry. A good example is the combustion of carbon (e.g. coal), . The laws of thermodynamics allow carbon to undergo a combustion reaction in the presence of oxygen at room temperature. Nevertheless, coal does not spontaneously ignite because carbon atoms and oxygen molecules repel one another – an example of an activation barrier. Instead, for combustion to occur the coal–oxygen mixture must be heated so that the kinetic energy of the oxygen molecules overcomes the activation barrier, allowing them to get close enough to the carbon atoms to react.
Keeping the possible presence of activation barriers in mind, we investigate the implications of the laws of thermodynamics. The role of the first law is relatively simple: the process must conserve energy. We assume henceforth that this is the case, and turn to the more complex question of the effects of the 2nd Law.
8.8.1 Spontaneous Processes at Constant Volume or Constant Pressure
Assuming that energy is conserved, entropy alone determines in which direction an isolated system can evolve. Most processes of interest to us, however, are not isolated, but rather occur in the presence of thermal reservoirs, such as the ambient environment or the interiors of heated reactors. Under these conditions it is necessary to include not only the change in entropy of the system, but also the entropy that can be taken up from or dumped into the environment. In such circumstances, the concept of free energy is extremely useful. Free energy can be used to determine, among other things, whether a given process can occur spontaneously if a system is kept in contact with a heat reservoir at constant temperature. There are two versions of free energy; these apply to processes that take place at constant volume (Helmholtz free energy) or constant pressure (Gibbs free energy). We encountered Helmholtz free energy in eq. (8.39). Most processes of interest in energy physics take place at fixed pressure, so the Gibbs free energy is of greater interest to us. Therefore we focus here on the constant pressure case and quote the constant volume result at the end.
When a system undergoes a transformation such as a chemical reaction (e.g. silver tarnishing) or a change in phase (e.g. ice melting), while kept at constant T and p, the entropy of the system changes by an amount . In addition, an amount of heat equal (but opposite in sign) to the change in enthalpy of the system is given off to the surroundings. This transfer of heat also adds entropy to the environment. The reaction can only proceed if the total change in entropy is positive. Signs are important here: if the system transforms to a lower enthalpy state (), then heat is given off, so is positive. The total change in entropy, including both the system and its surroundings, is
(8.66)
where denotes the fact that this result holds at constant p and T. The condition that a reaction can occur spontaneously, , is then
(8.67)
To make use of eq. (8.67), we define a new state function, the Gibbs free energy
(8.68)
At constant p and T, a transformation can occur spontaneously only if
(8.69)
Gibbs and Helmholtz Free Energy
Gibbs free energy is a state function defined by . A reaction can occur spontaneously at constant temperature and pressure if the change in G is negative, . Helmholtz free energy, , plays the same role at constant temperature and volume.
The change in enthalpy and free energy have been tabulated for many physical and chemical reactions including chemical reactions, mixing of solids, liquids, and gases, and dissolving of solids in water and other liquids. This data makes it possible to analyze the implications of thermodynamics for these processes.
The implications of eq. (8.69) can be quite counter-intuitive. Even an endothermic reaction – one in which the reactants absorb enthalpy from their surroundings () – can occur if the entropy of the system increases enough. A commercially available cold pack is a familiar example: it consists of ammonium nitrate (NH4 NO3) and water, initially isolated from one another. When the package is crushed, the NH4 NO3 mixes with the water, dissolves and absorbs heat from the environment. The reaction proceeds because the change in entropy as the NH4 NO3 dissolves is large and positive, so even though is positive, .
If the process of interest is restricted to occur at constant volume, then everywhere that enthalpy appeared in the preceding discussion, internal energy should be substituted instead. The criterion for a spontaneous reaction at constant temperature and volume is
(8.70)
From the definition (8.39) of Helmholtz free energy, , at constant V and T this criterion is
(8.71)
In §8.6.2 we found that . Because of the simple relation between F and Z, if we wish to compute G, it is often easier to compute F from fundamental principles and then use to obtain G.
We return to the subject of free energy and use it extensively when we discuss energy in matter in §9. Free energy plays a role in energy systems ranging from batteries (§37.3.1) to carbon sequestration (§35.4.2).
Discussion/Investigation Questions
8.1 In the game of ScrabbleTM, the letters of the English alphabet are inscribed on tiles and a prescribed number of tiles are provided for each letter. Consider an ensemble of ScrabbleTM tiles with a probability distribution defined by the frequency of tiles in the box. Explain how to calculate the information entropy of this ensemble. It is not necessary to actually compute the entropy, but you can if you wish.
8.2 Explain why the statements “No cyclic device can transform heat into work without expelling heat to the environment” and “No device can move thermal energy from a low-temperature reservoir to a high-temperature reservoir without doing work” follow from the laws of thermodynamics.
8.3 Living systems are highly ordered and therefore relatively low entropy compared to their surroundings; for example, a kilogram of wood from a pine tree has less entropy then the corresponding masses of water and carbon dioxide placed in a container in equilibrium at room temperature. When living systems grow they expand these highly ordered (relatively low entropy) domains. How can this be reconciled with the second law of thermodynamics?
8.4 An inventor claims to have constructed a device that removes CO2 from the air. It consists of a box with an opening through which the ambient air wafts. The CO2 is separated out by a complex system of tubes and membranes, and ends up in a cylinder, ready for disposal. The device consumes no energy. Do you think that this is possible?
8.5 According to the Sackur–Tetrode formula (8.65), the entropy of an ideal monatomic gas grows with the mass of the atoms. Thus the entropy of a volume of argon is much greater than the same volume of helium (at the same T and p). Can you explain this in terms of information entropy?
Problems
8.1 What is the information entropy of the results of flipping a biased coin 1000 times, if the coin comes up tails with probability 5/6 and heads with probability 1/6? For this weighted coin, can you find an encoding for sequential pairs of coin flips (e.g. HH, HT, etc.) in terms of sequences of bits (possibly different numbers of bits for different possible pairs) that takes fewer than 2 bits on average for each pair of flips? How many bits on average does your encoding take, and how close is it to the information-theoretic limit?
8.2 What is the information entropy of the results of spinning a roulette wheel with 38 possible equally likely outcomes (36 numbers, 0 and 00) 10 times? How many fair coins would you need to flip to get this much information entropy?
8.3 Consider an amount of helium gas at atmospheric pressure and room temperature (you can use K) enclosed in a 1-liter partition within a cubic meter. The volume outside the partition containing the helium is evacuated. The helium gas is released suddenly by opening a door on the partition, and flows out to fill the volume. What is the increase in entropy of the gas?
8.4 [T] Prove analytically that the entropy of a two-state system (8.12) is maximized when .
8.5 Consider air to be a mixture of 78% nitrogen, 21% oxygen, and 1% argon. Estimate the minimum amount of energy that it takes to separate a cubic meter of air into its constituents at STP, by computing the change in entropy after the separation and using the 2 Law.
8.6 A heat engine operates between a temperature and the ambient environment at temperature 298 K. How large must be for the engine to have a possible efficiency of 90%? Can you find some materials that have melting points above this temperature, out of which you might build such an engine?
8.7 A nuclear power plant operates at a maximum temperature of 375℃ and produces 1 GW of electric power. If the waste heat is dumped into river water at a temperature of 20℃ and environmental standards limit the effluent water to 30℃, what is the minimum amount of water (liters per second) that must feed through the plant to absorb the waste heat?
8.8 In a typical step in the CPU of a computer, a 32-bit register is overwritten with the result of an operation. The original values of the 32 bits are lost. Since the laws of physics are reversible, this information is added to the environment as entropy. For a computer running at 300 K that carries out operations per second, give a lower bound on the power requirement using the second law of thermodynamics.
8.9 [T] A simple system in which to study entropy and the Boltzmann distribution consists of N independent two-state subsystems coupled thermally to a reservoir at temperature T. The two states in each subsystem have energies (by definition) and ε. Find the partition function, the internal energy, and the entropy of the total system as a function of temperature. Explain why the internal energy approaches as . Is this what you expected? It may help if you explain why the entropy approaches in the same limit.
8.10 Repeat the calculations of Example 8.3 for the H molecule, with K. In particular, compute the probabilities that a given molecule is in the ground state or first excited state at temperatures , and K, compute the vibrational contribution to the heat capacity at each of these temperatures, and compare your results with Figure 5.7.
8.11 Consider a simple quantum system consisting of twenty independent simple harmonic oscillators each with frequency ω. The energy of this system is just the sum of the energies of the 20 oscillators. When the energy of this system is , show that its entropy is . What is the entropy of the system in thermal equilibrium for , 2, and 3? Assume that , corresponding to the vibrational excitation mode of O. Beginning in the state with , by how much does the entropy increase if the energy is increased to ? Use this result to estimate the temperature of the system by . Repeat for the shift from to , and estimate the heat capacity by comparing the change in U to the change in estimated T. Compare the heat capacity to to compute the effective number of degrees of freedom. Do your results agree qualitatively with the analysis of Example 8.3?
8.12 [T] Check the results quoted in eqs. (8.42)–(8.44) for a single harmonic oscillator. Show that as , the entropy and the heat capacity both vanish. Show that .
8.13 [T] Show that in general the condition that and the requirement that the entropy be finite at all temperatures (we cannot be infinitely ignorant about a finite system), require that as for all systems.
8.14 [T] Use the Sackur–Tetrode equation (8.64) to show that for an ideal monatomic gas. Likewise show that , verifying eq. (8.18) and the assertion of footnote 6.
8.15 [T] In §8.7.2 it was asserted that in order for entropy to be an extensive quantity, the classical partition function of a system of indistinguishable particles must be divided by . Consider a box evenly partitioned into two halves, each with volume V, and each containing N atoms of a monatomic ideal gas at a temperature T. Compute the entropy from eq. (8.64) omitting the last term. Now remove the partition, and compute the entropy of the resulting system of particles in a volume again from eq. (8.64) with the last term omitted. We have not lost any information about the system, but the entropy seems to have increased if we compute it without the Gibbs factor! Next, restore the factor and repeat the analysis.
8.16 [T] Compute the partition function per particle (8.61) for helium gas at NTP confined in a cube of side 1 cm. What quantum state has the highest probability of being occupied? What is the probability that a given helium atom is in this state? Given the number of atoms in the box, what is the expected average occupation number of this state?
