PPT1 PDF - The First Law: Energy Is Conserved

Chapter 2 The First Law: Energy Is Conserved The First Law: you can’t win. The Second Law: you can’t break even. The Third Law: you can’t come close to breaking even, unless it’s really cold. — A common scientific joke, with many variations. Concepts A scientific law is a regularity observed in na...

Chapter 2 The First Law: Energy Is Conserved The First Law: you can’t win. The Second Law: you can’t break even. The Third Law: you can’t come close to breaking even, unless it’s really cold. — A common scientific joke, with many variations. Concepts A scientific law is a regularity observed in nature and formulated after a large number of observations. Because they are solidly grounded in experimental observations, scientific laws may be modified after further experience, but they are rarely refuted. The branch of science we call thermodynamics deals with the exchange of energy. Scientific observations of how the exchange of energy happens ultimately lead to laws that govern the direction of all natural processes. The First Law of Thermodynamics states that energy is conserved; that is, in any process, energy can neither be created nor destroyed. The law is based on experiments carried out in the early 19th century. In one such experiment, a falling weight, via a pulley, turned a paddle that stirred a bucket of water. As the weight fell and the paddle turned, the temperature of the water increased. This experiment showed that the potential energy of the weight could be converted into internal energy manifested in the increased temperature of the water, just as could be achieved by applying heat to the water. The mechanical motion of the apparatus churning the water is described as work. Classical thermodynamics centers on work and heat, but other forms of energy such as electrical energy have since been incorporated into the First Law. Since Einstein proposed his famous equation E = mc 2 , we have known that the individual laws requiring the conservation of mass and energy are incomplete, and that what is really conserved is a combined quantity called mass-energy. However, absent nuclear reactions and radioactivity, the conversion of energy into mass, or mass into energy, is not observed, and so in most of everyday life, the original formulation of the First Law is perfectly valid. The Second Law of Thermodynamics, which was formulated at about the same time, deals not with the quantity of energy but with its flow. Scientists observing steam engines, in which wood or coal are burned to generate heat that is converted into work, noticed that the conversion of heat into work was very inefficient. Pursuing the theory of these heat engines, they showed that no engine, no matter how well designed, can convert heat into work with perfect efficiency. Some of the heat must be discharged into a colder 133 14 Chapter 2 | The First Law: Energy Is Conserved environment in order to drive the conversion of the rest into work. In fact, the observation that heat always spontaneously flows from objects at high temperature into objects at lower temperature has been shown to be equivalent to the Second Law. This law therefore dictates the direction of all spontaneous processes and not simply the flow of heat. Being able to determine the direction of any process that occurs spontaneously is incredibly useful. As we will see in chapter 3, the Second Law introduces the concept of entropy. Entropy is the quantity that increases in any spontaneous process. The Third Law of Thermodynamics is the most recent. It was only clearly stated in the 1920s. It shows, within limitations, that the entropy of systems goes to zero as the temperature approaches zero Kelvin. The Third Law is important in the biological sciences because it gives us an absolute measure of the entropy. But, of course, biological systems never naturally approach zero Kelvin. For that reason, we will give it limited attention in this book. Everything in biology, from nearly instantaneous chemical reactions to the incredibly slow process of evolution, is governed by the three laws of thermodynamics. The laws explain the hybridization of DNA, the folding of proteins, the pumping of ions across cell membranes, and the metabolism of foodstuffs. Understanding these laws is crucial to understanding life itself. Applications Thermodynamics applies to everything from black holes, so massive that even light cannot escape from them, to massless neutrino particles. Thermodynamics can answer these and many other questions: • How do you calculate the work done when a muscle contracts or stretches? • How can chemical reactions be used to do work or to produce heat? • How much heat can be generated by burning 1 gram (g) of sugar or eating and digesting 1 g of sugar? Many examples of the applications of thermodynamics will be illustrated in this and the subsequent three chapters. Energy Conversion and Conservation A large set of experiments, done over a period of many years, has shown that energy can be converted from one form to another, but that the total amount of energy remains constant. We will eventually discuss this quantitatively, but a couple of examples will make the idea clearer. Consider the cellphone that you have carelessly left on the window ledge of the fifth floor of your apartment building. Perched precariously, high above the sidewalk, it possesses gravitational potential energy. Knocked off the ledge, it accelerates towards its untimely end. First the potential energy is converted to kinetic energy, and then the friction of the air begins to convert the kinetic energy into heat. Once it hits the hard concrete, some of the kinetic energy is retained in the splintered fragments of the ‘phone as they fly in every direction. Some of the kinetic energy is transferred to planet Earth, which is jogged an imperceptibly tiny amount in its orbit by the impact. But many new forms of energy also appear. Some heat is produced. There may be some light energy in the form of sparks. Some energy is used to break the chemical bonds in the metal, glass, and plastic components as they shatter. Some sound energy is certainly produced. A more interesting question: how much energy arrives from the Sun and how is it transformed? Sunlight hitting the desert or a solar collector is mainly transformed into heat. However, some of the light energy can be converted into electrical energy using solar cells, and sunlight striking a green leaf is partly transformed into useful chemical energy through photosynthesis. It is vitally important that we know and understand the various kinds of energy involved in biochemical processes and what limits, if at all, their interconversion. Energy Exchanges | 15 Systems and Surroundings Clear definitions are perhaps more important in thermodynamics than in any other branch of science. But our first pair of definitions is easy. The system is defined as the specific part of the universe on which we choose to focus; in other words, the system is whatever we say it is. It might be the Sun, Earth, a person, a mammalian liver, a single living cell, or a mole of liquid water. Everything else in the universe we call the surroundings, and what separates the system from the surroundings is termed the boundary (figure 2.1). Thermodynamics deals with three idealized kinds of systems, illustrated in figure 2.1. • Isolated systems have no exchange of any kind — neither matter nor energy — with the surroundings. • Closed systems can exchange energy but not matter with the surroundings. • Open systems exchange both matter and energy with the surroundings. Isolated systems are the most difficult to construct, because it is hard to completely cut off energy transfer to a system. However, the contents of a sealed and thermally insulated flask come very close to an isolated system, especially over a short period of time with negligible heat flow in and out. An isolated system is really an idealized concept, which can only be approached in the limit. Such concepts are common in thermodynamics. They are often easy to theoretically analyze but can usually only be realized in practice by extrapolation. A closed system can simply be one within a physical enclosure. Many chemical reactions are performed in a closed system, a sealed flask being an example. The chemicals stay inside the flask, but heat can come in or out. The most difficult type of system to consider (and also the most common) is the open system. An open system can exchange both matter and energy with the surroundings. A fertilized egg being hatched by a hen is an example: oxygen goes in and carbon dioxide comes out of the egg; heat is also exchanged between the egg and its surroundings. We should emphasize that it is entirely up to us to specify our system and to define real or imaginary boundaries that separate it from its surroundings. If we specify as our system 10 mol of H2O (180 g) poured into a beaker and left to evaporate, then the liquid water remaining in the beaker [(180 - x) g] plus the water that has evaporated from it (x g), constitutes a closed system. As another example, if a solution containing the enzyme catalase is added to an open beaker containing a hydrogen peroxide solution, the enzyme will accelerate the reaction 2 H2 O2 S 2 H2 O + O2 and oxygen gas will come out of the beaker. If we choose the liquid content in the beaker as our system, we have an open system. But if we choose the liquid content plus the oxygen evolved as our system, we have a closed system. We can choose the system according to our interests, objectives, and convenience. We must, however, always define the chosen system clearly, to avoid confusion. Energy Exchanges Now that we have defined a system, we can focus on how its energy can be changed. In thermodynamics, we are interested in changes in energy, more than the absolute value of energy, which is in any case difficult to define. We can add energy to a system in several ways. Adding matter to an open system, for example, increases the chemical energy of the system because the matter can undergo various physical and/or chemical reactions. We do not have to think about the large amount of energy potentially available from nuclear reactions unless such reactions are actually occurring in the system; that is, usually we do not have to include the E = mc2 energy term because this term does not change significantly in an ordinary reaction. It is convenient to divide energy exchange between system and surroundings into different types. Two of the most common types of energy exchanges are work and heat. Open system heat matter Closed system heat Isolated system FIGURE 2.1 The three kinds of thermodynamic systems. The black box is the system boundary. In an open system, matter and energy can pass through the boundary between system and surroundings. In a closed system, only energy can pass through the boundary. In an isolated system, nothing can pass through the boundary. 16 Chapter 2 | The First Law: Energy Is Conserved Work (a) x0 In classical mechanics, work is defined as the product of a force times a distance. In thermodynamics, the system can do work on the surroundings, or the surroundings can do work on the system. To calculate the mechanical work done by the system or on the system, multiply the external force on the system by the distance moved — the displacement: work = external force * displacement x (b) Fex (c) x We must be consistent about the sign of the work to keep proper accounting of the energy exchanges between the system and the surroundings. Physical chemists and biochemists generally follow the convention that the work is positive if the surroundings are doing work on the system, and negative if the system is doing work on the surroundings. So when we use Eq. 2.1, we need to keep a watchful eye to make sure that the sign of work, which depends on the proper choice of signs for the force and the displacement, is always consistent with this convention. Some physicists and engineers define work done by the system as positive, and so one has to be careful not to blindly apply formulas from textbooks in those disciplines without checking the sign convention used. Sign conventions can be a nuisance, but as long we specify these clearly, the nuisance can be kept to a minimum. Work of Extending a Spring Any spring has an equilibrium length along its axis equal to x 0 . This is its length when not subject to an external force. We can apply an external force Fex either to extend or to compress the spring to a new length x, as shown in figure 2.2. At this new length, the stretched or compressed spring itself exerts a spring force Fsp equal in magnitude, and opposite in sign, to the external force. According to Hooke’s Law, the spring force Fsp that balances Fex exactly is directly proportional to the change in the length of the spring, and has a sign such that it opposes the change in length away from equilibrium (i.e., opposite to x - x 0) Fsp = -Fex = -k(x - x 0), Fex FIGURE 2.2 (a) A spring under no external force, with an equilibrium length x0. (b) a spring subjected to a compressive (negative) external force Fex; x 6 x0 (c) a spring subjected to a extensive (positive) external force Fex; x 7 x0. (2.1) (2.2) where k is a constant for a given spring (if we do not extend it beyond its elastic limits). The negative sign on the right-hand side reflects the spring force acting in the opposite direction to the change in length. The magnitude of k will be different for different springs. To calculate the work done in changing the length of the spring, we choose an x-axis with one end of the spring fixed at x = 0. The other end is free to move along this axis; its position when there are no forces acting on it is x 0; in general, its position is at x. As we change the length of the spring, the spring force changes. This means we cannot simply multiply the force by the change in length; we must integrate the force multiplied by the differential of the displacement. w = 3 Fex(x)dx = 3 k(x - x0)dx (2.3) In the preceding equation w is positive if the external force Fex and the displacement dx are of the same sign, and negative if they are of opposite sign. This is consistent with our sign convention for w: if the direction of the external force is the same as the direction of the displacement, the external force is doing work on the system (the spring), and w is positive. The work done on the spring when its length is changed from x 1 to x 2 is: x2 w = 3 k(x - x0)dx x1 Changing variables: x⬘ = x - x 0 and so dx⬘ = dx . (2.4) Energy Exchanges | 17 x2 - x0 w = 3 kx⬘ dx⬘ = 兩 1> 2 kx⬘ 2 兩 xx21 -- xx00 x1 - x0 (2.5) = 1>2 k c ( x 2 - x 0)2 - ( x 1 - x 0 ) 2 d x0 This integration is depicted in figure 2.3. The green line graphs the external force as a function of displacement x. The area under the green line — in other words, its integral — is the work done. The right angled triangle whose apex is at x 0 and whose right edge is the vertical line at x = x 1 has width equal to x 1 - x 0 and vertical height equal to k(x1 − x0). Its area is therefore 1>2 k(x 1 - x 0)2 . Likewise, the triangle bounded by a vertex at x 0 and the vertical line at x = x 2 has an area 1>2 k(x2 - x0)2 . The difference between these two triangles is an area of the light green polygon between values x = x 1 and x = x 2 . This is the work done stretching the spring from x 1 to x 2 . A graph of this sort — with force on the y axis and displacement on the x axis — is called a force-extension curve. We have thus calculated the work done on the system by extending the spring. If the spring was originally extended, the system would do work on the surroundings when the spring returned to its equilibrium position. A muscle fiber does work in a similar fashion. It can do work by stretching or contracting against an external force. According to the Système International (SI), the unit of work, and indeed of any form of energy, is the joule (J). One joule is 1 newton * meter. The unit of force is the newton (N), and the spring constant k must therefore have units of N/m or N m−1. One newton is the magnitude of the force that will cause an acceleration of 1 m s−2 when applied to a mass of 1 kilogram (kg): 1 N = 1 kg m s−2. Older units for energy are the erg: 1 erg = 10 −7 J; and the calorie: 1 cal = 4.184 J. Additional energy conversion tables can be found in table A.3 in the appendix. Most international scientific organizations now insist on SI units which we will use throughout this book. The basic SI units of length, mass, and time are meter (m), kilogram (kg), and second (s). While other units still survive, particularly in older published works, their use is discouraged. Work Done Against a Constant Force The force due to the gravitation attraction of the Earth on mass m is given by Fex = mg, where g is the acceleration due to gravity on the Earth’s surface. This force is directed down towards the Earth’s center, and it depends on the inverse square of the distance. On the Earth’s surface, approximately 6370 km from the center of the planet, the acceleration due to gravity is approximately 44 times that on a TV satellite in geosynchronous orbit, 42,000 km from the center. However, since almost all of us live and work within 10 km of the Earth’s surface, the variation in our distance from the center of the Earth is negligible. Even at the Earth’s surface, the acceleration due to gravity varies from a sea level value of 9.832 m s−2 at the poles to 9.780 m s−2 at the equator, mostly because the centrifugal acceleration caused by the Earth’s rotation reduces the pull to the center to the Earth at the equator. Other effects such as altitude and variations in density of the Earth’s crust also affect local gravity. For this reason, scientists have adopted a standard gravity; approximately the mean gravity at a latitude of 45°N. Standard g is called g0: g0 = 9.80665 m s−2. If a force does not change significantly with distance, it can be treated as a constant for the purpose of the integration in Eq. 2.2 w = 3 Fex (x) dx = 3 mg dx ⬇ mg 3 dx. Fex (2.6) x1 x2 x FIGURE 2.3 The force-extension curve for a spring obeying Hooke’s Law. On extending the spring from x1 to x2, the work done on the system is the area of the light green polygon. 18 Chapter 2 | The First Law: Energy Is Conserved Integrating this between two values of x, x 1 and x 2, gives θ laser x2 r tip FIGURE 2.4 Molecular force microscopy; one end of a biomolecule is attached to a surface and the other to an atomic force microscope tip. As the tip is pulled back from the surface, the force bends the cantilever. Force (pN) (2.7) x1 surface A B C Extension (nm) FIGURE 2.5 Force-extension curve for single stranded DNA. At low levels of extension, the apparatus straightens the DNA, which takes little force (A). However, once the DNA is fully extended (C), the force required to further stretch it is large. Force (pN) w = mg 3 dx = mg ( x2 - x1 ) = mgh , cantilever A B C 65 pN Extension (nm) FIGURE 2.6 Force-extension curve for double stranded DNA. At low levels of extension, the apparatus straightens the DNA with little force (A). Once the DNA is fully extended (B), the force increases to 65 pN, and begins to unzip the DNA (C), converting a gently pitched Watson-Crick double helix to an extended single-stranded conformation. When this process is complete the DNA behaves as if single stranded. where h = x 2 - x 1. Molecular Force Microscopy Using technology that became available in the last twenty years, starting with the invention of the scanning tunneling microscope, we can reproduce Robert Hooke’s experiments on springs, but now using biological molecules. The essence of these experiments is simple; we tether one end of a biological molecule to a surface, using a chemical cross-linking agent or a very tight noncovalent binder, and the other end to a movable object, such as a polymer bead or an atomic force microscope (AFM) tip. We then move the object away from the surface, either the bead with a pair of tightly focused counter-propagating laser beams called ‘laser tweezers’ or the AFM tip mechanically. This applies an external force, stretching the molecule. Figure 2.4 shows the basics of a molecular force microscope. A linear biopolymer is attached chemically to a surface and also to the tip of an atomic force microscope; the tip is glued on the end of a flexible cantilever — a very narrow, flexible wafer of silicon oxide or nitride. As the cantilever is withdrawn from the surface, the distance by which it is displaced can be measured very precisely using a laser interferometer. The force on the tip also bends the cantilever, and the angle by which it bends can be measured by the deflection in the direction of the laser and directly converted into a force. So, measuring these two quantities, the distance the cantilever moves from the surface and the bending angle, allows us to construct a force-extension curve for the biomolecule. Few, if any, biomolecules obey Hooke’s Law, but some, like single-stranded DNA (ss-DNA), have a comparatively simple force extension curve. Unstretched ss-DNA can be considered a random coil (figure 2.5A), or worm-like chain, and as you might expect, as you stretch it, it uncoils rather easily (figure 2.5B). However, once the coils are gone and the chain becomes nearly straight, further stretching requires elongating covalent bonds beyond their ‘natural length’, or opening bond angles more than the nearly-tetrahedral angle most would like to have, or moving torsion angles into sterically unfavorable ‘eclipsed’ conformations. This requires considerable force, and so the force-extension curve bends dramatically upwards (point C). However, with double-stranded DNA, one can convert it to a stretched form by unwinding the strands and breaking the hydrogen bonds that hold together the base pairs in the double helix; this takes much less force than needed to snap a covalent bond — 65 pN, in fact, for DNA from the double-stranded virus phage l. At that force, the double helix begins to ‘unwind’ (figure 2.6C), and the stretched chain increases in length by 70%, as shown by the flat plateau in figure 2.6. The detailed structure of the stretched form of DNA is not known. Different structures can form depending on whether one or both DNA strands are attached to the pulling surfaces, and whether ‘nicks’ occur in the strands. Reducing the force on the ends of the DNA allows the DNA strands to re-anneal and re-form the coiled form of B-DNA. E X A M P L E 2 .1 B-DNA (ds-DNA) has a length of 0.34 nm per base pair. The stretching of l phage DNA occurs at a force of 65 pN, while that of the artificial DNA poly(dAdT) occurs at 35 pN. Compute the work done in kJ/mol in stretching a base-pair length of DNA by 70%. Energy Exchanges | 19 SOLUTION The change in length of a nucleotide length of DNA on stretching is ( 0.70 * 0.34 ) nm = 0.24 nm or 2.4 * 10−10 m . For l DNA, the force required is 65 pN = 6.5 * 10−11 N . The work done in Joules is therefore 2.4 * 10−10 m * 6.5 * 10−11 N = 1.6 * 10−20 J . For one mole of base pairs, the work done is that for a single base pair, multiplied by Avogadro’s number: w = 1.6 * 10−20 J * 6.022 * 1023 mol−1 = 1.0 * 104 J mol−1, or 10 kJ mol−1 . For poly(dAdT), the work done is w = 2.4 * 10−10 m × 3.5 * 10−11 N * 6.022 * 1023 mol−1 * 10−3 kJ−1 J = 5.1 kJ mol−1. In practice, integrating curves such as those shown in figure 2.5 and figure 2.6 requires either fitting the curve to a function (based on theory or a curve-fitting procedure), or else using computer numerical integration. Work Increasing or Decreasing the Volume of a Gas Consider a system such as a gas or liquid enclosed in a container with a movable wall or piston (figure 2.7). The system can expand and do work on the surroundings if the external pressure pex is less than the pressure of the system p, or the surroundings can do work on the system by compressing it if pex is greater than p. By definition, pressure is just force per unit area, and so the external pressure pex is related to the external force Fex by pex = Fex >A , where A is the cross-sectional area of the piston. We can then write Fex dx = Fex >A * A dx = -pex dV , (2.8) since the product A dx = -dV is the change in volume of the gas container. Why the negative sign? If the change in the x coordinate has the same direction and therefore the same sign as the external force, the volume of the container decreases, and so dV is negative if dx is positive. Thus, from the equation above, force times displacement is equivalent to pressure times the negative of the change of volume. The work done on the system is therefore: w = 3 Fex dx = - 3 pex dV (2.9) Work associated with a change in the volume of the system is often termed pressure–volume work, or pV work. The SI unit of pressure is the pascal, or Pa; 1 Pa = 1 N m - 2 . The SI unit of volume is of course the cubic meter, or m3. 1 m3 is a larger volume than most chemists and biochemists use in practice, and so we commonly substitute the liter(L); 1 L = 10 - 3 m3. Similarly, 1 pascal is a much lower pressure than ambient, and so we often work in bars; 1 bar = 105 Pa. A bar is approximately the pressure exerted by the atmosphere at sea level. However, we have to be careful with non-SI units, and we advise you to convert all units to SI before doing any calculations. SI carries with it a sort of warranty; if you use the correct formula, and all units are SI units, the result will be in an SI unit. Thus, for example, if we multiply the SI unit of pressure (1 Pa) by the SI unit of volume (1 m3), the result is the SI unit of energy (1 J). Fex x pex A dx dV=–Adx EXAMPLE 2.2 Calculate the pV work done when a system containing a gas expands from 1.0 L to 2.0 L against a constant external pressure of 10 bar. SOLUTION First convert to SI units! pex = 10 bar * 105 Pa bar - 1 = 106 Pa; V 1 = 1 L * 10 - 3 m3 L - 1 = 10 - 3 m3; V 2 = 2 * 10 - 3 m3 FIGURE 2.7 Relationship between force and pressure. As the gas is compressed under an external force acting on a piston of area A, it creates a pressure pex. The change in displacement along the direction of the force, dx, reduces the volume of the system by an amount dV = ⫺Adx. 20 Chapter 2 | The First Law: Energy Is Conserved The external pressure is constant, and so it can be taken outside the integral: V2 V2 w = - 3 pex dV = -pex 3 dV = -pex ( V2 -V1 ) V1 V1 = -106 Pa * (2 * 10 - 3 m3 -10 - 3 m3) = -103 J The system does work; the sign of w is negative. Friction Friction — specifically, kinetic friction — is the force that acts on a surface in motion and in contact with a solid, liquid, or gas to retard the motion. The frictional force is approximately independent of the velocity and the contact area but proportional to the load on the surface. Because friction involves a force applied over a distance, an object moving against a frictional force does work on the surroundings. This work is quite different from the work done compressing a piston or extending a spring. When we extend the spring or compress the piston we store potential energy that can be released when the spring is allowed to contract or the gas to expand. The work we do on the system can be returned to us as work done by the system on the surroundings. These processes are considered conservative processes. In contrast, frictional work is converted to heat, not to potential energy, and cannot easily be reconverted back to work. As we will see in chapter 3, a heat engine is needed to convert heat energy back into work. This is an intrinsically inefficient process. Therefore, frictional work is called a dissipative process rather than a conservative process. It is essential to understand, however, that, whether the process is conservative or dissipative, energy is always conserved. A conservative process stores work as potential energy whereas a dissipative process converts it to heat. Heat When two bodies are in contact with each other, their temperatures tend to become equal. Energy is being exchanged. The hot body will lose energy and cool down; the cold body will gain energy and warm up. This energy exchange is said to occur by heat transfer, and the energy that passes through the system-surroundings boundaries because of a temperature difference between the two is called heat. The sign convention for heat is the same as that for work: heat is positive if it flows into the system and negative if it flows out of the system. For a closed system, the transfer of an infinitesimal quantity of heat dq will result in a change dT in its temperature. The ratio dq/dT is called the heat capacity of the system and is given the symbol C. The heat capacity for every material is the quantity of heat that is needed to raise its temperature by 1⬚C or 1 K. dq = C or dq = CdT dt (2.10) C is a distinct property of a given material, and will also tend to vary with temperature. Thus, the heat q gained when the temperature of the system changes from T1 to T2 should be evaluated from the integral T2 q = 3 C dT . (2.10a) T1 If C is constant in the temperature range T1 to T2, then it can be removed from the integral, and q = C( T2 - T1 ) . (2.10b) Energy Exchanges | 21 For a hot body in contact with a cold body, the heat capacity of both bodies must be known, and Eqs. 2.10 –2.10b must be applied to each body individually. The heat capacity C is an extensive quantity; that is, it increases with the amount of material in the body. An intensive quantity is one that remains unchanged when a system is subdivided. If 100 g of water at a uniform temperature is divided into two portions, the temperature of each part is unchanged. Thus, temperature is an intensive property. Similarly, the pressure of a system is also an intensive property. For a pure chemical substance, we can convert C into an intensive quantity — one that depends only on the properties of the substance, and not its amount — by dividing by the number of moles of substance present: Cm = C>n, where n is the number of moles and Cm is the molar heat capacity. In general, when any property has an equivalent molar property (such as how volume has molar volume) the property is extensive and the molar property is intensive. Energy is extensive, but energy mol−1 is intensive; mass is extensive, but density or mass volume−1 is intensive. A heat capacity is usually determined by measuring the change in temperature when a known quantity of electrical energy is dissipated in the material. The SI unit of heat capacity is J K−1; that of molar heat capacity is J mol−1 K−1 . In older books, heat capacities are often given in cal K−1 . Until it was discovered that the heat capacity of liquid water depends slightly on temperature, one calorie or cal was the quantity of heat needed to raise the temperature of 1 g of water by 1 K. Later it was simply defined as 1 cal = 4.184 J. We also occasionally encounter the term specific heat capacity, or c (lower case), which is the heat capacity divided by the mass: c = C>m . It has the SI unit J kg−1 K−1, but it is often given in J g−1 K−1 . EXAMPLE 2.3 Calculate the heat, in joules, necessary to increase the temperature of 100.0 g of liquid water from 25.0°C to 75.0°C at ambient (1 bar) pressure. The molar heat capacity of liquid water is 75.4 J mol−1 K−1 and is nearly independent of temperature. SOLUTION ( 100.0 g ) > ( 18.015 g>mol ) = 5.551 mol ( 75.0⬚C - 25.0⬚C ) = 50 K 5.551 mol * 50.0 K * 75.4 J mol - 1K-1 = 2.09 * 104 J. Internal Energy You are probably already familiar with the idea that when we transfer heat energy to a body, raising its temperature, the energy goes into motion of its constituent atoms or molecules. The field that studies how that energy is distributed among the countless modes of motion possible in a substance is called statistical mechanics. We will consider statistical mechanics in more detail in chapter 5; here we are only concerned with what happens when we transfer heat to a substance. The simplest instance of statistical mechanics is the kinetic theory of gases, which was the first successful attempt to explain the properties of a class of materials (ideal gases, which obey the ideal gas equation of state pV = nRT ) in terms to the motion of atoms and molecules. The theory was constructed by the German physicist Rudolf Clausius, on two premises: • Gases are made up of molecules in ceaseless random motion, colliding with each other and with the container. • Gases have no attractive forces between them, and their collisions are elastic (they neither gain or lose energy). Our analysis starts with a rectangular-sided container bounded by edges of length a, b, and c, corresponding to the x, y, and z directions. Let’s consider a single gas molecule in that 22 Chapter 2 | vy -vz The First Law: Energy Is Conserved vy vz c b FIGURE 2.8 A particle collides with the z wall of a box, reversing the sign of the z component of its momentum, while leaving the x and y components unchanged. container, of mass m, moving with a velocity vector v. The momentum of that molecule is given by: p = mv (2.11) It is unfortunate that the symbol for momentum, p, is similar to that for pressure, p. However, momentum is a vector, and so will either be in bold font, or, if we are discussing one component of the momentum, it will have a subscript corresponding to a Cartesian direction (e.g., px). Pressure, in contrast, is always a scalar. For the moment, let’s assume the molecule is a point particle, and therefore, even if there are other molecules present, it will not collide with them. Since we have forbidden collisions except with the wall of the container, the particle will simply bounce ceaselessly off the walls, like a frictionless billiard ball. We show this behavior for a two-dimensional box in figure 2.8. Each time the ball hits a wall, its momentum in the direction normal to the wall will change sign. Let’s consider a collision with the z wall (the wall perpendicular to the z direction). If the component of the momentum along the z direction is pz,1 = mv z before the collision, it is pz,2 = -mv z after the collision. The momentum along the other two directions is unchanged by the collision (see figure 2.8). The change in the z component of the momentum per collision with the z wall is therefore ⌬pz = pz, 2 - pz, 1 = -2mv z . (2.12) Momentum, like energy, is a conserved quantity. If momentum is lost by the particle, it must be gained by the wall, which recoils very slightly every time a particle collides with it. So the gain in momentum in the wall is: ⌬pz, wall = - ⌬pz = 2mv z (2.13) The particle typically collides with the z wall hundreds of times per second. The exact time interval t between collisions is obtained by dividing the distance traveled in the z direction between collisions, which is twice the length of the box in that direction or 2c, divided by the magnitude of the velocity in the z direction, vz. t = 2c>v z (2.14) If there are t seconds between collisions, in 1 second there will be 1/t collisions. The number of collisions per second is therefore 1/t = v z /2c. In a time ⌬t there are v z ⌬t/2c collisions, and in this increment of time ⌬t the total momentum transferred to the wall by this single particle is ⌬pz, total = vz ⌬t/2c * ⌬pz, wall , (2.15) which from Eq. 2.13 becomes ⌬pz, total = v z ⌬t/2c * 2mv z = mv 2z ⌬t/c. (2.16) This momentum transfer occurs in tiny bursts. However, if we ‘smooth out’ the collisions, we can treat the transfer of momentum as a continuous process. Think of it this way: in a building with a corrugated iron roof, at the beginning of a rain shower, we can hear the raindrops collide individually with the roof. However, as the rain gets heavier, and there are hundreds of drops hitting the roof per second, the individual impacts merge into a continuous roar. The mathematical operation of going from the discrete to the continuous is accomplished by replacing the differences in Eq. 2.16 by differentials. ⌬pz, total/⌬t _ dpz, total/dt = mv 2z /c (2.17) Energy Exchanges | 23 The derivative of momentum is force. So the time-averaged force on the z wall due to this single particle is: Fz = mv2z >c (2.18) Pressure is force per unit area. The area of the z wall is ab. So the pressure on the z wall is p = Fz >ab = mv2z >abc = mv2z >V, (2.19) since the product abc is the volume V of the container. We’ve gone as far as we possibly can with a single particle. A real gas is composed of a large number of individual particles, each with its own z velocity. We’ll assume the gas is pure and that therefore all of the masses of the particles are the same. If we label the particles 1,2,3… the total pressure is therefore the sum of the pressure of particles 1, 2, 3… p = mv2z,1 V + mv2z,2 V + mv2z,3 V N Á = a m b v2z, i V ia =1 (2.20) The mean square velocity in the z direction is just the sum of all squared velocities, divided by the number of particles N. It is therefore: 1 N 8 v 2z 9 = a b a v 2z, i N (2.21) i= 1 But there is nothing special about the z direction. By symmetry, the mean square velocities in the x, y and z directions are the same. Furthermore, by Pythagoras’ theorem, the total velocity squared is v 2 = v 2x + v 2y + v 2z = 3v 2z . So combining this with Eqs. 2.20 and 2.21 we get: mN p = a b 8v2 9 (2.22) 3V Multiplying both sides by V: pV = 1>3 mN 8 v 2 9 (2.23) N, the total number of molecules, is related to the total number of moles n, by n = N/N A, where NA is Avogadro’s constant, the number of particles in a mole. The best current value of Avogadro’s constant is (6.0221418 { 0.0000003) * 1023. The mass of a mole M is equal to the mass m of the particle times the same constant: M = m * N A. So the product mN is the same as the product nM. Replacing this in Eq. 2.18, and then comparing the result with the Ideal Gas Law, we get: pV = 1>3 nM 8 v 2 9 = nRT (2.24) Canceling the number of moles, and multiplying across by 3/2, we get > 2 M 8 v 2 9 = 3>2 RT . 1 (2.25) But the left-hand side of this equation is the total kinetic energy of all the gas particles. We call this total energy the molar internal energy Um = 1> 2 M 6 v 2 7, an intensive quantity. Similarly, if we multiply both sides of Eq. 2.25 by n, we get U, the internal energy, an extensive quantity. This equation is remarkable for what it is missing: p and V! It says that the internal energy of an ideal gas depends only on the temperature, and is entirely independent of the volume and the pressure. This makes intuitive sense; since the theory assumes no interactions between the gas molecules, reducing or increasing the distance between them should make no difference. 24 Chapter 2 | The First Law: Energy Is Conserved Constant Volume Heat Capacity What happens when we heat the container? In classical thermodynamics, we express the First Law by the equation ⌬U = w + q. (2.26) That is, there are two ways we can add energy to a system; we can heat it, or do work on it. As long as we keep the volume fixed, we do no work: by Eq. 2.9, doing work requires changing the volume. Therefore, at constant volume, ⌬U = qV . (2.27) In thermodynamics, a subscripted variable means we keep that variable constant. If we now take the partial differential with respect to temperature, and include Eq. 2.25, we get a 0q 0U b = a b = 3>2 nR . 0T V 0T V (2.28) But dq/dT is the heat capacity (Eq. 2.10) so we call (0q/0T)V the constant volume heat capacity. We can once again divide across by the number of moles, n, to make everything intensive. CV,m = a Gas CV,m/R Helium Neon Argon Krypton Xenon 1.5000 1.5000 1.5000 1.5000 1.5000 0qm 0Um b = a b = 3>2 R 0T V 0T V (2.29) This is a remarkable result; it says that the heat capacities of all ideal gases that can be regarded as point particles should be the same! How well does it work? Quite excellently, it appears, as we can see by the table on the right! Amazingly, 127 g of xenon has the same heat capacity and the same internal energy as 4 g of helium. Theory predicts CV,m /R for an ideal gas to be 3/2; experiment shows it is 1.5000; and, therefore, agrees with theory to five significant digits. In fact, modern theoretical chemistry can predict most thermodynamic properties of simple gases more accurately than they can be measured. Constant Volume Heat Capacity of Diatomic Gases Gas CV,m/R y (cm−1) H2 2.467 4395 N2 2.503 2360 CO 2.506 2170 O2 2.535 1580 F2 2.770 892 Since we’ve succeeded with the heat capacities of monatomic gases, let’s see if we can equally elegantly explain the molar constant volume heat capacities of diatomic gases at 1 bar and 298 K. While they’re nowhere nearly as similar to each other as the monatomics, it is clear the first four diatomic gases in the table are very close to a value of CV,m/R = 5/2, with fluorine a little bit larger. Why is this? A diatomic gas, in addition to moving translationally in three dimensions like monatomic gases, can also rotate. However, only rotations about two dimensions are meaningful; rotation about its long axis leaves the molecule in an identical state. These two additional degrees of freedom or modes can also absorb energy, just as the three translational modes absorb energy. Because the molecule now has five active modes, CV,m/R has a value of approximately 5/2. The principle that energy will be shared out equally among all active modes is called the equipartition theorem. What about fluorine? One additional mode available to a diatomic is the vibrational mode along the bond axis. Why is this mode inactive in the first four diatomics and only partly active in fluorine? The answer lies in quantum mechanics, which we will treat in more detail in chapter 11. But briefly, the energy required to raise a diatomic molecule into its first vibrational excited state is often quite high. Vibrational energies are typically given in wavenumbers, and are listed in the table. It can be seen that fluorine, and to a lesser extent oxygen, have comparatively low vibrational frequencies, and CV,m/R values slightly higher than 5/2, because a small but significant fraction of the molecules can be thermally excited to a higher vibrational state at room temperature. Constant Volume Heat Capacity of Monatomic Solids | 25 Moving away from gases and towards ‘condensed matter’ — liquids and solids — let’s now consider the heat capacities of the simplest solids: metals, which are composed of crystals with isolated atoms in a simple lattice. The molar heat capacities of metals at room temperature are all close to 3R; this phenomenon is so general it has been given the name Dulong and Petit’s Law. Why do atoms in a crystal lattice have molar heat capacities of 3R, while those in an ideal gas have CV,m values of (3/2)R? The answer, again, lies in the number of modes available. An atom in a crystal lattice interacts with its neighbors, and as it moves away from its equilibrium position in the lattice, its potential energy increases. Atoms in an ideal gas do not interact with each other at all, and so have no potential energy. So atoms in a lattice can put RT/2 units of internal energy into kinetic energy along a particular direction, and RT/2 units of energy into potential energy; thus, per mole, their total internal energy is double that of the same atom in an ideal gas at the same temperature. Vibrational modes always have kinetic and potential energy components, and so can always absorb RT units of energy per mode, when thermally accessible. Rotational modes, like translational modes, can absorb RT units of energy in a crystal lattice, but only RT/2 units in an ideal gas, where there is no potential energy associated with rotation. We can gain additional insight from the molar heat capacities of a couple of metals as a function of temperature. These are shown in figure 2.9. The heat capacities of both sodium and aluminum drop to zero at 0 K; this is a consequence of quantum mechanics and is true of all substances. They rise rapidly above a temperature of about 10 K and eventually reach a plateau near 3R = 24.94 J mol - 1K - 1; as can be seen, aluminum, a hard metal where the lattice vibrations are spaced quite far apart, rises more slowly and has not quite reached the plateau at room temperature. In sodium, a softer metal, it is easier to displace atoms from their equilibrium positions, and so the heat capacity increases faster as a function of temperature, but still reaches approximately the same plateau. CV, m (J mol–1 K–1) Constant Volume Heat Capacity of Monatomic Solids Metal CV,m/R Mg Al Fe Pt Pb 2.91 2.78 2.96 2.97 2.98 25 20 Na 15 Al 10 5 50 100 150 200 250 300 T (K) FIGURE 2.9 The constant volume heat capacities of sodium and aluminum, as a function of temperature. Heat Capacity of Molecular Solids and Liquids 125 Cp, m (J mol–1 K–1) In figure 2.10, we show the temperature dependence of the constant pressure molar heat capacities of glycine and alanine. We use these because constant volume heat capacities are very difficult to obtain for most solids. At low temperature, the two curves resemble each other and those of metals, except that in addition to translations, restricted rotations of the molecules within the lattice (called librations) also contribute 3R to the heat capacities. At higher temperatures, instead of reaching a plateau at 6R, as the amino acids seem to be trying to do at 100 K, the heat capacities continue to increase, as more and more vibrational modes become thermally accessible and contribute to Cp,m. Alanine, a larger molecule with more low-frequency vibrational modes, diverges appreciably from glycine at these higher temperatures. Finally, we consider the heat capacities of more complex molecular solids and liquids; a variety of these are given in table 2.1. All of them are at 298 K except where specified. There are some general trends: • With the exception of water ice, all of the heat capacities exceed 6R, which is the heat capacity due to lattice vibrations and hindered rotations alone. Vibrations play an important role in the heat capacities of most molecules at ambient temperatures. • As can be seen from the last three rows, the molar heat capacities increase with molecular mass M, because increasing the number of atoms increases the number of low-energy vibrations that absorb heat. 100 75 50 ala gly 25 50 100 150 200 250 300 T (K) FIGURE 2.10 The constant pressure heat capacities of the amino acids glycine and alanine, as a function of temperature. 26 Chapter 2 | The First Law: Energy Is Conserved TABLE 2.1 Heat capacities of molecules Compound H2O (l, 273 K) Cp. m J mol−1K−1 M g mol−1 75.4 18.0 H2O (s, 273 K) 38.1 18.0 potassium iodide 52.8 166.0 a-D-glucose 219.2 180.2 L-tyrosine 216.4 181.2 anthracene 211.7 178.2 decanoic acid 375.6 172.3 dodecanoic acid 404.3 200.3 tetradecanoic acid 432.0 228.4 • For molecules of similar mass (rows 3–8), those that are more flexible (which usually means softer solids, such as decanoic acid) have more low frequency vibrations and therefore higher heat capacities than harder materials composed of more rigid molecules, molecules linked together by hydrogen bonds, or ionic solids. • Liquids have higher heat capacities at the melting points than the corresponding solids; an example is water and water ice. For the same reasons, we find that macromolecules in random coils, such as single-stranded nucleic acids and denatured proteins, have higher heat capacities than the corresponding folded proteins and double-stranded DNA and RNA. We shall see later that this difference in heat capacities is important in understanding the stability of native proteins, and to a lesser extent, nucleic acids. State and Path Variables Consider a closed system in the absence of all external fields. This statement describes a system that we can only approximate on Earth. We cannot turn off gravity, for example. However, for many practical purposes we can ignore the effects of gravity. If the system consists of a pure liquid, specifying the pressure p, volume V, and temperature T of the liquid is sufficient to specify many other properties of the liquid, such as its density, surface tension, refractive index, and so forth. These other properties of the system, and p, V, and T, are called variables of state (or state variables). The variables of state depend only on the state of the system, not on how the system arrived at that state. The useful characteristic of variables of state is that when a few ( p, V, T, chemical composition, etc.) are used to specify a system, all other variables of state are determined implicitly. Any property of the system that depends only on the variables of state must itself be a variable of state. As we have seen, the internal energy U of an ideal monatomic gas is a function only of the temperature, a state variable, and so it is itself a state variable. As we have seen, we can raise the temperature and thus increase the internal energy of an ideal, monatomic gas by heating it. However, we could also raise its temperature by doing frictional work. This might seem like an odd idea — gases don’t exert much friction on objects moving through them — but recall that a space vehicle or a meteorite entering the Earth’s atmosphere glows red hot from friction. Move a surface fast enough through a gas, and you will certainly heat it up! So while the internal energy of a system is called a state variable, depending only on the state of the system and not how it arrived where it is, heat and work are path variables, State and Path Variables | 27 which depend on the route taken to the final state of the system. Properties of the system that we can measure, such as pressure, volume, and temperature, are state variables. When we measure the temperature of a beaker of water, we don’t care whether it warmed up by sitting in strong sunlight or on a hot plate. Once we specify the path, however, a path variable becomes a state variable. For example, the heat capacity, dq/dT is clearly a path variable, because it depends on q, itself a path variable. But if we specify a constant volume path, then CV , the heat capacity at constant volume, is a state variable, equal to dU/dT. Because state variables do not depend on the path taken, but only on the state of the system, the change in a state variable as a result of a physical process depends only on the initial and final state, and not how we get there. For this reason, state variables are intimately linked to conservation laws. If we take the system from state 1 to state 2, and then back to state 1 — a cyclic path — all the state variables must return to their initial values. This means that internal energy cannot be reduced or increased over a cyclic path; if it could, we could generate internal energy in a system by taking it around a cyclic path which would violate the First Law. Because work and heat are not state variables, we can create or destroy them in a cyclic path, and in fact we will see in chapter 3 how we can use a cyclic path to turn heat into work. However, the internal energy, the sum of heat and work, must be unchanged, and therefore whatever heat we consume must be turned into work. Reversible Paths and Reversible Processes The external pressure pex and the pressure of the system p are not equal when the system is undergoing compression or expansion; if they were, nothing would happen. Unfortunately, if pex is very different from p, then the work will not simply change the volume of the gas, but impart velocity to it, and thence generate frictional heat. If pex is kept nearly equal to p during the process (to be precise, in the limit where pex - p goes to zero), the process is said to be reversible. A very small change in pressure will reverse the process from an expansion to a compression, or vice versa. The sign of the work will change from positive to negative, depending on an infinitesimal change in the pressure. If pex and p differ significantly, irreversible expansion or compression will occur, depending on whether pex is smaller or greater than p. To visualize a reversible compression or expansion, consider the expansion of an ideal gas in a cylinder with a frictionless and weightless piston, as illustrated in figure 2.11a. Suppose that the pressure outside the cylinder is always at 1 bar and the pressure inside the cylinder is initially at 2 bar. If we remove the stops that hold the piston in position, the gas will expand irreversibly until a final state is reached at which its pressure becomes 1 bar. During the course of expansion, the pressure of the system is always significantly greater than that of the surroundings (that is why the expansion is called irreversible). The two become the same only at the end of the expansion. We can carry out the expansion in a different way. Instead of holding the piston in position with stops at the beginning, we put many small weights on top of the piston to make up for the pressure difference, as shown in figure 2.11(b). The expansion is then carried out in a stepwise manner by removing one weight at a time. If the number of weights is very large (and the weight of each very small), the pressure of the system is always almost the same as that of the surroundings during the course of expansion. Furthermore, if we add rather than remove a weight, the process will be reversed. A reversible path is one in which the process can be reversed by an infinitesimal change of the variable that controls the process. The path illustrated in figure 2.11(a) is not a reversible one; the path illustrated in figure 2.11(b) becomes a reversible one when the weights are infinitesimally small and their number approaches infinity. 28 Chapter 2 | The First Law: Energy Is Conserved FIGURE 2.11 Comparison of (a) an irreversible expansion and (b) a reversible expansion. In a reversible expansion, the internal pressure is always nearly equal to the external pressure. (a) Irreversible expansion by removal of stops pex stop p pex stop p (b) Reversible expansion by one-at-a-time removal of weights pex pex p p In the special case when a system is expanding irreversibly against a vacuum, pex is zero and no work is done. Reversible heating or cooling is similar to reversible expansion or compression. Here, the temperature difference between the system and the surroundings must be very small— infinitesimal. Heat flows in or out, depending on very small changes in temperature. If the temperatures of two bodies differ significantly, irreversible heat exchange occurs when the two come into contact. Most processes considered by classical thermodynamics assume reversibility. Equations of State A few variables of state are usually sufficient to specify all others. This means that equations exist that can relate the variables of state. Such equations are termed equations of state. The simplest and most frequently used equations of state link p, V, and T. State and Path Variables | 29 V = m/r (2.30) At a higher level of accuracy, the volume of a solid or liquid does change somewhat with T and p. Experimental data for 1 mol of liquid water are shown in figures 2.12 and 2.13. The molar volume is plotted versus temperature at constant pressure in figure 2.12, and versus pressure at constant temperature in figure 2.13. Equations can be obtained for V as a function of p and T for liquid water from these data. If we want to substantially improve on the primitive equation of state V = V ⬚, we can introduce a linear dependence on temperature and pressure. This gives the following intensive equation of state, which can be made extensive by multiplying both sides by n. V m = V m⬚[1 + a(T - T⬚)][1 - k(p - p⬚)] (2.31) Vm (m3 × 10–6) 19.0 18.8 18.6 18.4 18.2 18.0 18.2 18.0 17.8 17.6 17.4 In this expression, a is the coefficient of volume expansion, defined by 1 0V a = a b V 0T p k = - 1 0V a b V 0p T 200 400 600 8001000 p (bar) (2.32) and k (the Greek letter kappa) is the isothermal compressibility… (2.33) 20 40 60 80 100 T (°C) FIGURE 2.12 The molar volume of liquid water as a function of temperature at a constant pressure of 1 bar. Vm (m3 × 10–6) Solids or Liquids The volume of a solid or a liquid does not change very much with either pressure or temperature. For the time being, we will not consider changes from solid to liquid or any other change except p, V, and T. Therefore, a first approximation for the equation of state of a solid or liquid is that the volume is constant, or V = V ⬚, where V ⬚ is the volume under standard conditions (298 K, 1 bar). This means that calculating the volume of a solid or liquid simply requires finding the density or specific volume at one temperature and using that value for any temperature and pressure. The volume is related to the density r by FIGURE 2.13 The molar volume of liquid water as a function of pressure at a constant temperature of 298 K. a is the fraction by which the volume of the system increases per degree temperature increase, while k is the fraction by which it decreases per Pa of increased pressure. T⬚ and p⬚ are the standard temperature and pressure, respectively. Isothermal compressibilities and coefficients of expansion are tabulated in handbooks of chemical and physical data. The dependence of the volume of liquid water on temperature (figure 2.12) is more complicated. The plot is not a straight line, and the molar volume of water actually has a minimum value at 277 K (4°C). Most of the time, we use the simplest approximation, that V is independent of T and p for a solid or liquid. 30 20 Vm (L) Gases The volume of a gas varies greatly with T and p but is nearly independent of the type of gas. The simplest approximate equation of state for gases is the Ideal Gas Law, ( pV = nRT ), which has already been introduced in Eq. 2.24. A plot of how the volume of 1 mol of an ideal gas depends on pressure is shown in figure 2.14. Note that, although 1000 bar is necessary to change the volume of liquid water by 3%, a change from 1 bar to 1000 bar for an ideal gas will cause a change in volume by a factor of 1/1000. The ideal gas equation has the great advantage that it contains no constants applying to individual gases; it applies to all gases if the pressure is low enough. It is an exact limiting equation for all gases as p approaches zero. For higher pressures, it is an approximation for real gases. The answers obtained using Eq. 2.24 are usually accurate within a percent for most gases near room temperature and atmospheric pressure. 500 K 10 273 K 0 2 4 6 p (bar) 8 FIGURE 2.14 Molar volume of an ideal gas as a function of temperature and pressure. 10 30 Chapter 2 | The First Law: Energy Is Conserved There are several more accurate equations of state for gases; these correct the ideal gas equation with parameters specific to individual gases. One example is the van der Waals gas equation: a p+ n2a b (V - nb) = nRT , V2 (2.34) where a and b are parameters specific for each individual gas. The van der Waals a constant is an approximate measure of the attractive forces between molecules, and the b constant is an approximate measure of the intrinsic volume of the gas molecules. When both of these are zero, Eq. 2.34 simplifies to the ideal gas equation. The equations of state that we have been discussing have all been applied to systems containing only one component. For mixtures, the number of grams or moles of each component must be specified, and the equation of state will depend on