Hour 2-13 (1) PDF - Physics Past Paper

Hour 2 – Basic ideas of systems, states, changes, equations of state. Revision of ideal gases. Conventions. Units of measurements. Theories. Thermodynamics is the study of the transformations of energy and the relations between the bulk properties of matter. System...

Hour 2 – Basic ideas of systems, states, changes, equations of state. Revision of ideal gases. Conventions. Units of measurements. Theories. Thermodynamics is the study of the transformations of energy and the relations between the bulk properties of matter. System Surroundings Universe Figure 1.1 Schematic of the Universe, system and surroundings System – the part of the world in which we have a special interest; that part of the universe that we are studying. Surroundings – region outside the system (where we make our measurements and observations) The system can be: Open – exchanges matter and energy with the surroundings Closed – exchanges only energy with the surroundings Isolated – exchanges neither energy nor matter with the surroundings The states of matter: Solid – well-defined boundary, minimal movement of particles, does not take the shape of its container Liquid - well-defined boundary, moderate movement of particles, takes shape of the container Gas – no well-defined boundary, almost free movement of particles, expands to fill container We examine gases because it is the simplest and also most important state (we are surrounded by gases). 1 The states of gases Physical state refers to a specific condition of a sample of matter described by its physical form (solid, liquid or gas). Thermodynamic state refers to the set of variables required to completely describe the system; these are volume, pressure, temperature and number of moles. For a pure gas, the thermodynamic state is completely defined by the following thermodynamic variables: Volume, V; Amount, n; Pressure, P; Temperature, T These four variables are related by an equation known as the equation of state; if three of the variables are known, then the fourth one is fixed. Hence, an equation of state is an equation that relates the variables that define the state of a system to each other. 𝑃 = 𝑓(𝑇, 𝑉, 𝑛) This is the general equation of state. The equation of state for a perfect (ideal) gas is: 𝑃𝑉 = 𝑛𝑅𝑇 Where R is the gas constant (8.314 J K-1 mol-1). Pressure Pressure is defined as force divided by area over which the force is applied. 𝐹 𝑃= 𝐴 F is force, measured in Newtons (N) and A is area (m2). Recall that: 𝐹 = 𝑚𝑎 Where mass, m is measured in kg and acceleration is measured in ms-2 For a gas, the pressure arises from the gas molecule battering the walls of the container. The SI unit of pressure is the Pascal (Pa). 1Pa = 1kg m-1 s-2. A pressure of 105 Pa is the standard pressure for reporting data. Below are other units of pressure. 2  Table 1.1 Units of pressure Name Symbol Value Pascal Pa Nm ; kg m-1 s-2 -2 bar Bar 105 Pa atmosphere Atm 101, 325 Pa Torricelli Torr 133.32 Pa (101, 325/760) Pa millimetre of mm Hg 133.32 Pa mercury pound per square Psi 6.89 kPa inch Questions: (1) What happens in a vessel containing two regions of gas separated by a movable piston where one region is of high pressure and the other of low pressure? (2) Show that the pressure at the foot of a column of liquid is: 𝑷 = 𝒉𝝆𝒈 [Hint: volume of cross-sectional area = A x h; density =m/v; F = m/a = m/g where g is acceleration due to gravity. g = 9.81 ms-2 on earth] Temperature Temperature is defined as the property of an object that determines in which direction the energy will flow when the object is in contact with another object. If energy flows from A to B, then A has a higher temperature than B. If A and B have the same temperature, then they are in thermal equilibrium and there is no net flow of energy. Temperature is measured on the Celsius and Kelvin scales where: T (K) = 0 C +273.15 0 K = -273.15 C 3 Types of thermal boundaries: Diathermic boundary – one in which a change of state is observed when two objects of different temperatures are brought into contact; this boundary allows the transfer of heat energy. Adiabatic boundary – one in which no change occurs even though temperatures are different; this boundary does not allow the Zeroth Law of Thermodynamics If A and B are in thermal equilibrium, and B and C are in thermal equilibrium, then A and C are in thermal equilibrium. Amount of substance (mol) 12 One mole of substance = number of particles of substance as there are in 12 g of C. This number is Avogadro’s number (NA = 6.022 x 10 23) Molar mass = mass of 1 mol of substance Molar mass = mass divided by number of moles (g mol-1) The gas laws for perfect (ideal) gases PV = nRT is a combination of the gas laws. This relation is only valid at low pressures. (1) Boyle’s law At constant temperature, the pressure of a fixed amount of gas is inversely proportional to its volume. 4  (2) Charles’ law At constant pressure, the volume of a fixed amount of gas is directly proportional to temperature. (3) Avogadro’s principle At a given temperature and pressure, equal volumes of gas contain equal numbers of particles. 𝑉∝𝑛 Molar volume = volume of sample divided by amount of substance (in mol) 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑛𝑅 = 𝑇1 𝑇2 Alternatively: 𝑛𝑇 𝑅𝑛𝑇 𝑉∝ ;𝑉 = 𝑃 𝑃 5 Hour 3 & 4 – Kinetic theory of gases. Maxwell distribution. Kinetic theory of gases The model for the kinetic theory of gases is based on the following assumptions: (i) The gas consists of molecules of mass (m) in ceaseless, random motion. (ii) The size of the molecules is negligible; their diameters are much smaller than the average distance travelled between collisions. (iii) The molecules interact only through brief, infrequent and elastic collisions (hence, the interactions are negligible). An elastic collision is one in which the total translational kinetic energy of the molecules is conserved; that is, there is no loss of momentum. We can use the kinetic theory to derive a relationship between the pressure of a gas and the speed of its colliding particles. When a particle of mass (m) and velocity (v) collides with the wall of a container, the momemtum before and after collision is given by: ± 𝒎𝒗𝒙 mvx -mvx The modulus of the change in momentum of one particle is: 2𝑚𝑣𝑥 The number of collisions in time Δt depends on how many particles can reach the wall in that time. The distance from the centre of the box can be given by: Speed = distance / time Distance (l) = speed x time 𝑙 = 𝑣𝑥 ∆𝑡 Now the volume of the box is given by: A x l 6 Hence, the volume of particles within reach of the wall of the box is: V=Axl 𝑉 = 𝐴𝑣𝑥 ∆𝑡 The number density of particles is nNA/V where n is the total amount of particles in the container of volume V. Hence the number of particles in the volume V is given by: 𝑛𝑁𝐴 (𝐴𝑣𝑥∆𝑡 ) 𝑉 At any instant, half of the particles will move to the right and half will move to the left; so the average number of collision will be: 1 𝑛𝑁𝐴 ( (𝐴𝑣𝑥 ∆𝑡)) 2 𝑉 Now, the total momentum change is: Total momentum change = number of particles x momentum change = 1 𝑛𝑁𝐴 ( (𝐴𝑣𝑥 ∆𝑡)) ∗ 2𝑚𝑣𝑥 2 𝑉 Now the Rate of momentum change = total momentum change / time change = force on the walls: 𝑚𝑛𝑁𝐴 𝐴𝑣𝑥2 = ∆𝑡 𝑉∆𝑡 𝑚𝑛𝑁𝐴 𝐴𝑣𝑥2 = 𝑉 Since force = rate of change of momentum, and Pressure = Force/Area; The pressure (P) is given by: 𝑚𝑛𝑁𝐴 𝐴𝑣𝑥2 𝑚𝑛𝑁𝐴 𝑣𝑥2 𝑃= = 𝑉𝐴 𝑉 Where vx2 is the average velocity squared. However the actual speed of the particle is: 𝑐 = √𝑣𝑥2 7 But the particle moves along all coordinates, so: 𝑐 = √𝑣𝑥2 + 𝑣𝑦2 + 𝑣𝑧2 But since the motion is random, and does not go in any fixed direction, then the coordinates are equivalent and so: 𝑐 = √3𝑣𝑥2 Which gives: 1⁄ 𝑐 2 = 𝑣 2 3 𝑥 Substituting this value into the pressure equation we get: 1⁄ 𝑚𝑛𝑁 𝑐 2 𝑃= 3 𝐴 𝑉 But since molar mass, M = mNA we get: 1⁄ 𝑛𝑀𝑐 2 𝑃= 3 𝑉 Or 𝑷𝑽 = 𝟏⁄𝟑 𝒏𝑴𝒄𝟐 Assuming the kinetic model is accurate, then for an ideal gas (PV=nRT) we get: 𝒏𝑹𝑻 = 𝟏⁄𝟑 𝒏𝑴𝒄𝟐 Which rearranges to give: 3𝑅𝑇 𝑐=√ 𝑀 This relationship was proven experimentally (the Zartmann-Ko experiment). This relationship implies that the rms speed of molecules is proportional to square root of temperature and inversely proportional to square root of molecular mass. 8 This means that: At higher temperature, the molecules will move faster on average At fixed temperature, heavier molecules will travel on average more slowly This refers to average speeds (rms speeds). However, the individual molecules have widely ranging speeds. Some travel at moderate speeds, then accelerate on collision, while others may abruptly come to a halt. The distribution of molecular speeds will predict or describe the fraction of molecules travelling with a particular speed, once temperature and molecular mass are known. The precise form of this distribution is known as the Maxwell-Boltzmann distribution. The Maxwell-Boltzmann distribution Given that f is the fraction of molecules with speed v: 3⁄ 𝑀 2 𝑀𝑣 2 2 − 2𝑅𝑇 𝑓(𝑣) = 4𝜋 [ ] 𝑣 𝑒 2𝜋𝑅𝑇 Tip: In order to calculate the fraction of molecules in a narrow range between v and Δv, calculate f(v) and multiply by Δv. The message of this distribution is best understood with graphical interpretation: 9 Note that: (i) The fraction of molecules within a range of Δv increases with Δv. That means that for a larger range of speeds, you capture a larger fraction of molecules. (ii) The fraction of molecules with very high speeds will be very small. (iii) The fraction of molecules with high speed and high mass is very small; heavy molecules move more slowly. (iv) At high temperature, a greater fraction of molecules will move faster. (v) The fraction of molecules with very low speeds is very small. 10 Experimental determination of molecular speeds Molecular speeds have been determined by two methods; the molecular beam method and the Zartman-Ko experiment. Zartman-Ko Experiment The Zartman-Ko experiment was performed in the 1930s by I.F. Zartman and C.C. Ko utilising apparatus shown in the diagram below. The oven contains Bi (bismuth) vapour at about 800°C, some of this vapour escapes through a slit and is collimated by another slit a short distance away. Above the second slit is a drum that rotates about a horizontal axis at 6,000 rpm. At those instants when the slit in the drum faces the Bi beam, a burst of molecules enters the drum. These molecules reach the opposite face of the drum at various times, depending upon their speeds. There is a glass plate attached on the opposite side of the drum and because the drum is turning, the faster and slower molecules strike different parts of the plate. From the resulting distribution of deposited Bi on the plate it is possible to infer the distribution of speeds in the beam, and this distribution agrees with the prediction of Maxwell-Boltzmann statistics. 11 Molecular Beam Method (adapted from Singh, Das and Singh, 2007) The Molecular Beam Method utilises the apparatus shown by the diagram below. Gas molecules pass through a pinhole in the source S, they are collimated by the slits and then pass through one of the openings between the cogs in the cogwheel C1. The two cogwheels C1 and C2 are mounted on an axel which is rotated rapidly. The molecules which have passed through C1 will only pass through C2 if the time required for the molecules to travel from C1 to C2 is the same as the time required for the cogwheel C2 to rotate once (that is, when C2 is in exactly the same position as C1 was when the molecules passed through it). If υ is the frequency of rotation of the cogwheels, then the time required for one rotation is given by 1 𝑡= 𝜐 If the distance between the two cogwheels is given by l, then the speed of the molecules (u) which travels the distance l in time t is 𝑙 𝑢= = 𝑙𝑣 𝑡 The molecules with the speed determined above are admitted into the radiometer (R) where their number is determined from the radiometer deflection. The frequency of rotation of the cogwheels is changed so that the speeds of other molecules are determined. 12 Calculation of molecular speed Consider a molecular beam apparatus in which the distance between the two cogwheels is 30 cm and the rotation frequency of the cogwheels is 6,000 rpm, what is the speed of the molecules (in ms-1) that reach the radiometer under these conditions? 𝑢 = 𝑙𝑣 𝑙 = 30 cm = 30𝑐𝑚⁄100 𝑐𝑚 𝑚−1 = 0.30 𝑚 𝑣 = 6,000 𝑟𝑝𝑚 = 6000 𝑚𝑖𝑛−1⁄60 sec 𝑚𝑖𝑛−1 = 100 𝑠 −1 𝑢 = 0.30 𝑚 𝑥 100 𝑠 −1 𝒖 = 𝟑𝟎 𝒎𝒔−𝟏 13 Hour 5 – Real Gases. The Van der Waals equation. Other gas equations. Critical constants. Real Gases Real gases do not obey the ideal gas law 𝑃𝑉 = 𝑛𝑅𝑇 because they have intermolecular interactions which affect their behaviour. Ideal or perfect gases are assumed to have NO intermolecular interactions. However, real gases do experience: (i) Attractive intermolecular forces which assist compression; and (ii) Repulsive intermolecular forces which assist expansion. Consider the changes in intermolecular interactions with intermolecular distance. At intermediate intermolecular distances, van der Waals attractions are dominant; the gases are contracted. When the molecules are almost touching, repulsive intermolecular forces become dominant and the molecules spring apart. At very low pressures, gases behave close to ideality, because their molecules are significantly separated. At moderate pressures, deviations arise due to attractions (because the molecules are at intermediate intermolecular distances). At high pressures, deviations arise due to repulsions because the molecules are being forced too close together. The Compression factor (Z) The compression factor of a gas is the ratio of its molar volume (Vm) to the molar volume of a perfect gas (V0m) at the same temperature and pressure. 𝑆𝑖𝑛𝑐𝑒 𝑃𝑉 = 𝑛𝑅𝑇 𝑅𝑇 𝑉𝑜𝑚 = , 𝑠𝑖𝑛𝑐𝑒 𝑛 = 1 𝑃 𝑉𝑚𝑃 𝑁𝑜𝑤 𝑠𝑖𝑛𝑐𝑒 𝑍 = 𝑅𝑇 𝑇ℎ𝑖𝑠 𝑔𝑖𝑣𝑒𝑠 𝑷𝑽𝒎 = 𝑹𝑻𝒁 14 For ideal gases, Z = 1, so deviation from Z = 1 is a measure of deviation from ideal behaviour. At very high pressure, repulsion is dominant, so real volumes are larger than expected and Z > 1. At intermediate pressures, attractions are dominant so real volumes are smaller than expected and we see Z < 1. At low pressures, real gases show ideal behaviour and Z = 1. Real isotherm of carbon dioxide (CO2) For this segment, you will need to refer to the figure of the isotherm of CO2: 15 Along the isotherm at 20oC, the sample is a gas at point A. As the sample is compressed through B to C, the pressure increases (Boyle’s law). One way of compressing the gas is by pushing in a piston on a sealed container of gas. Further compression from C through D to E yields no pressure increase – we observe this as the piston moving in reducing the volume without any increase in pressure. This happens because the gas compresses to a liquid (with a visible surface boundary) immediately left of C. The volume of the liquid is much smaller than that of the gas. So the piston can move in freely until it touches the surface of the liquid at E. The pressure corresponding to the line CDE is called the vapour pressure. From E to F, the small reduction in volume accompanied by a large increase in pressure represents the effect of trying to compress a liquid. The further motion of the piston brings the molecules into closer contact resulting in repulsive interactions (this increases the pressure). Along the isotherm Tc, there is no abrupt change from the gaseous to the liquid state. At a particular temperature and pressure (called the critical point), the gaseous phase transforms continuously into the condensed phase without evidence of a surface boundary between them. The two phases are in equilibrium. The pressure, temperature and volume at the critical point are called critical constants, Pc, Tc and Vc. At Tc and above, a single form of matter exists at all stages of the compression. This state is defined as a gas but due to its large density is called a supercritical fluid. This means that: A gas can only be compressed to a liquid by the application of pressure if the temperature is at or below Tc. The liquid state does not form at and above Tc. Critical pressure is the pressure needed to liquefy a gas at the critical temperature. Critical volume of a gas is the volume occupied by 1 mol of gas at the critical temperature and critical pressure. 16 Real gas equations Virial coefficients Based on the ideal behaviour of gases at low temperature, it was proposed that the real gas law is an expansion beginning with the ideal gas law: 𝑃𝑉𝑚 = 𝑅𝑇[1 + 𝐵 ′ 𝑝 + 𝐵 ′ 𝑝2 + … ] or 𝐵 𝐶 𝑃𝑉𝑚 = 𝑅𝑇 [1 + + 2 + …] 𝑉𝑚 𝑉𝑚 This equation is called the Virial equation of state. The coefficients B and C are the second and third virial coefficients. Generally, each progressive term is much smaller (and therefore less important) than the preceding one, so many terms are seldom used. The van der Waals equation The virial equation of state is not easy to use. The van der Waals equation of state is an approximation to the virial equation. However, it shows how intermolecular interactions contribute to deviations from ideal gas behaviour. Van der Waals proposed that: (a) Repulsion between molecules prevents them from approaching closer than a certain distance. This reduces the volume in which they live from V to V – nb where n = moles of gas and b is a constant. Hence we see the following change in the ideal gas equation: 𝑃(𝑉 − 𝑛𝑏) = 𝑛𝑅𝑇 (b) Attractive interactions reduce the pressure that the gas exerts because they slow the molecules down and have two effects i. Molecules collide with the walls less frequently and ii. They collide with weaker impact. The reduction in pressure is modelled by: 𝑛 2 𝑎( ) 𝑉 17 where n/V represents the concentration of molecules. Hence the van der Waals equation becomes: 𝑛𝑅𝑇 𝑛 2 𝑃= −𝑎( ) (𝑉 − 𝑛𝑏) 𝑉 This is sometimes expressed as: 𝑛 2 [𝑃 + 𝑎 ( ) ] (𝑉 − 𝑛𝑏) = 𝑛𝑅𝑇 𝑉 where a and b are called van der Waals coefficients. We examine the reliability of the van der Waals equation by plotting isotherms. Examine the isotherms for real gas behaviour compared to the behaviour predicted by vdW equation. The calculated isotherms do resemble experimental isotherms except for oscillations below Tc. These oscillations are called van der Waals loops and are unrealistic as they predict that under some conditions, increasing pressure increases volume! We know that this is not physically possible. The loops are replaced by horizontal lines called the Maxwell construction. Van der Waals coefficients are found by comparing calculated isotherms with the experimental isotherms. The critical constants can be expressed in terms of a and b. They are: 𝑉𝑐 = 3𝑏 𝑎 𝑃𝑐 = 27𝑏 2 8𝑎 𝑇𝑐 = 27𝑅𝑏 NB. It is not possible for a single, simple expression to be the true equation of state of all substances. When the van der Waals equation fails, we either use one of the other proposed equations of state, invent a new one or go back to the virial equation. 18 Other proposed equations of state Berthelot equation: 𝑅𝑇 𝑎 𝑃= − 𝑉𝑚 − 𝑏 𝑇𝑉𝑚2 Dieterici equation: −𝑎 𝑅𝑇𝑒 𝑅𝑉𝑚 𝑇 𝑃= 𝑉𝑚 − 𝑏 Virial (Kamerlingh Onnes) equation: 𝑅𝑇 𝐵(𝑇) 𝐶(𝑇) 𝑃= [1 + + + …] 𝑉𝑚 𝑉𝑚 𝑉𝑚2 19 Hr 6 & 7 – Energy as Heat and Work; Expansion work WORK is the fundamental physical property in thermodynamics. Work is done when an object is moved against an opposing force: 𝒘=𝑭𝒙𝒅 For example, the expansion of a gas pushes a piston, raising a weight or a chemical reaction drives an electric current that operates a motor which raises a weight. ENERGY is defined as the capacity of a system to do work. First Law of Thermodynamics When work is done on an otherwise isolated system, e.g. compressing a gas, the system’s capacity to do work is increased, i.e the energy is increased. This is recorded by a + sign. When the system does work (piston moves out by gas expansion) the energy of the system is reduced (it has less capacity to do work). This is recorded by a – sign. The energy of a system may be changed by heat exchange. When the energy of a system changes as a result of temperature differences between system and surroundings, energy has been transferred from the hotter body to the cooler body as heat. The First Law of Thermodynamics states that the internal energy of an isolated system is constant. 20  𝜟𝑼 = 𝒒 + 𝒘 The internal energy of a system refers to the total energy within the system. It is the sum of all the kinetic energies and potential energies of the system. The internal energy of a system may be changed by doing work on it or by work done by it. If a system is isolated from its surroundings, no change in internal energy can take place, hence the First Law. This is a consequence of the law of conservation of energy. An exothermic process releases energy as heat e.g. combustion reactions An endothermic process absorbs energy as heat. When an endothermic process occurs in an adiabatic system the temperature falls (because the walls do not allow thermal equilibrium with surroundings); conversely if an exothermic process occurs, the temperature rises. When an endothermic process occurs in a diathermic container, energy (as heat) enters the system from the surroundings and the temperature of the system remains unchanged. If the process is exothermic, then energy (as heat) leaves the system and again the process is isothermal. The terms heat and work can also be distinguished at the molecular level. Work is the transfer of energy that makes use of the organized motion of molecules. When a weight is raised or lowered, its molecules move in an organized way; this organized motion of molecules is harnessed to do work on the system. Conversely, when the system does work, it causes atoms or electrons in the surroundings to move in an organized way. Heat is the transfer of energy that makes use of disorderly molecular motion (called thermal motion). The thermal motion of the particles in the hot surroundings stimulates the molecules in the cooler system to move more vigorously resulting in increased energy. The opposite happens when the system heats the surroundings. 21 Reversible changes A reversible change is one that can be reversed by an infinitesimal change of a variable. A system is said to be in equilibrium with its surroundings if an infinitesimal change in the conditions (in opposite directions) result in opposite changes of state. Let us now consider infinitesimal changes of state. Recall the First Law states ∆𝑈 = 𝑞 + 𝑤. This refers to a measureable change. An infinitesimal change in U is expressed as: 𝑑𝑈 = 𝑑𝑞 + 𝑑𝑤 Expansion work Expansion work is the work arising from a change in volume; e.g. work done by a gas as it expands and pushes back the atmosphere. Note that compression is referred to as negative expansion. The work required to move an object a distance dz against an opposing force F is expressed as1: 𝑑𝑤 = −𝐹𝑑𝑧 Consider that the force on the wall is: 𝐹 = 𝑝𝑒𝑥 𝐴 Where A is the area of the wall. 1 Note the negative sign corresponding to work done by the system. Later we will see how this sign convention synchronises with expansion work being –ve and compression being +ve. 22 Hence, work done is: 𝑑𝑤 = −𝑝𝑒𝑥 𝐴𝑑𝑧 But since volume is carved out by moving an area A through a distance dz, we see: 𝑑𝑉 = 𝐴𝑑𝑧 Hence: 𝑑𝑤 = −𝑝𝑒𝑥 𝑑𝑉 This is the general expression for expansion work in differential form. To obtain the total work done for volume change Vi to Vf, we integrate dV within these limits: 𝑉𝑓 𝑤 = ∫ −𝑝𝑒𝑥 𝑑𝑉 𝑉𝑖 Hence we get: 𝒘 = −𝒑𝒆𝒙 (𝑽𝒇 − 𝑽𝒊 ) 1. Free expansion This is expansion against zero opposing force, e.g. expansion in a vacuum. This means that 𝑝𝑒𝑥 = 0. This gives: 𝑑𝑤 = 0 ∗ 𝑑𝑉 = 0. Hence no work is done when a system expands freely. 2. Expansion against constant pressure For example, if a gas expands, pushing a piston out against the atmosphere, the pressure will be constant throughout the expansion. We express this as: 𝑤 = − ∫ 𝑝𝑒𝑥 𝑑𝑉 = −𝑝𝑒𝑥 (𝑉𝑓 − 𝑉𝑖 ) This can be expressed as 𝒘 = −𝒑𝒆𝒙 ∆𝑽 for a constant pressure expansion. 23 3. Reversible expansion If during the expansion, the external pressure pex is increased infinitesimally then the gas contracts slightly; on the other hand, if the pex is decreased infinitesimally the gas expands hence the change is reversible. To achieve reversibility, we set pex = p at each stage of the expansion, hence we cannot treat it as a constant anymore. Therefore our work expression becomes: 𝑑𝑤 = −𝑝𝑒𝑥 𝑑𝑉 = −𝑝𝑑𝑉 Note now that the pressure is integrable. We express this as: 𝒘 = − ∫ 𝒑 𝒅𝑽. We can only find the work done if we know how pressure p changes with volume for this system. 4. Isothermal reversible expansion An isothermal change is one that takes place at constant temperature. We will now apply the work expression for reversible expansion to an isothermal case. Consider a fixed number of moles of perfect gas, in a system in thermal contact with surroundings, such as a constant-temperature water bath. Since 𝑝𝑉 = 𝑛𝑅𝑇 then expressing pressure as a function of volume we get: 𝑛𝑅𝑇 𝑝=. We can now substitute for p in the work expression for reversible expansion: 𝑉 𝑛𝑅𝑇 𝑤 = − ∫ 𝑝 𝑑𝑉 = − ∫ 𝑑𝑉 𝑉 24 Since n is constant, the integral becomes: 1 𝑤 = −𝑛𝑅𝑇 ∫ 𝑑𝑉 𝑉 If integrating between the limits of initial and final volumes, Vi and Vf, the resulting work is expressed by: 𝑽𝒇 𝒘 = −𝒏𝑹𝑻𝒍𝒏 ( ) 𝑽𝒊 25 Hr 8 – Heat capacity and Enthalpy; Revision of Thermochemistry Heat Capacity The Heat capacity of a substance is the energy required to raise the temperature of that substance by 1 unit. C = q/∆T If the internal energy of a system at constant volume (e.g. sample of gas in a container of fixed volume) is plotted against temperature then a curve such as shown below is obtained. The slope of the tangent to the curve at any temperature gives the heat capacity of the system at that temperature. The heat capacity at constant volume is denoted Cv and is defined as the change in internal energy per unit change in temperature. Cv = (∂U/∂T)v Graph shows the variation of internal energy with temperature at constant volume. As temperature increases, the internal energy increases. For the system illustrated the heat capacity at B is greater than at A. NB. Heat capacities are extensive properties eg) 100 g of water has 100 times the heat capacity of 1 g of water and therefore requires 100 times the heat to bring about the same rise in temperature. The molar heat capacity at constant volume (Cv,m) is the energy required as heat to raise the temperature of 1 mol of substance by a unit of temperature. Generally heat capacities depend on the temperature and decrease at low temperatures. However, over small ranges of T, at and above room temperature, the variation is quite small and heat capacities can be treated as almost independent of T. 26 The heat capacity can be used to relate a change in internal energy to a change in temperature of a constant-volume system. dU = CvdT for infinitesimal changes and ∆U = Cv∆T for measurable changes Since at constant volume; w = 0 then ∆U = q and so q = Cv∆T Question: Explain why the heat capacity of a sample is infinite at the temperature of a phase transition. Enthalpy When the system is free to change its volume, the change in internal energy is not equal to the heat applied; some of this energy is used to do expansion work. ∆U = q + w However heat applied at constant pressure is equal to the change in enthalpy. The enthalpy, H is defined as H = U + pV Enthalpy is a state function and so the change in enthalpy, ∆H between Hi and Hf is independent of the path taken. In fact, ∆H = qp for a system at constant pressure that does no additional work. Justification for ΔH = qp For a general infinitesimal change in the state of the system: 𝑈 → 𝑈 + 𝑑𝑈; 𝑝 → 𝑝 + 𝑑𝑝; 𝑉 → 𝑉 + 𝑑𝑉 Therefore, the expression: 𝐻 = 𝑈 + 𝑝𝑉 Becomes: 𝐻 + 𝑑𝐻 = 𝑈 + 𝑑𝑈 + (𝑝 + 𝑑𝑃)(𝑉 + 𝑑𝑉) When brackets are cleared, this becomes: 𝐻 + 𝑑𝐻 = 𝑈 + 𝑑𝑈 + 𝑝𝑉 + 𝑝𝑑𝑉 + 𝑉𝑑𝑝 + 𝑑𝑉𝑑𝑝 dVdp is the product of two infinitesimally small values and can safely be ignored. 27 And since H = U+pV, we get: 𝐻 + 𝑑𝐻 = 𝐻 + 𝑑𝑈 + 𝑝𝑑𝑉 + 𝑉𝑑𝑝 We see that H drops on (being on RHS and LHS). Now substituting for dU = dq + dw we get: 𝑑𝐻 = 𝑑𝑞 + 𝑑𝑤 + 𝑝𝑑𝑉 + 𝑉𝑑𝑝 If the system does only expansion work, then dw = -pdV and the equation becomes: 𝑑𝐻 = 𝑑𝑞 − 𝑝𝑑𝑉 + 𝑝𝑑𝑉 + 𝑉𝑑𝑝 By definition of enthalpy, heating occurs at constant pressure, so dp = 0 which gives: 𝒅𝑯 = 𝒅𝒒 at constant pressure and no additional work. Moving from the infinitesimal change to the macroscopic change we get: ∆𝑯 = 𝒒𝒑 For an ideal gas; enthalpy is related to internal energy in the following way; H = U + nRT from pV = nRT Therefore for a reaction involving gases; ∆H = ∆U + ∆ng RT where ∆ng is the change in the amount of gas molecules in the reaction. Variation of enthalpy with temperature The enthalpy of a substance increases with increasing temperature. This relation differs for constant pressure and constant volume systems. The slope of the tangent to a plot of enthalpy against T at constant pressure is called the heat capacity at constant pressure, Cp. Cp = (∂H/∂T)p The heat capacity at constant pressure is used to relate the change in enthalpy to a change in temperature: dH = CpdT for infinitesimal changes and ∆H = Cp∆T for measurable changes. At constant p; ∆H = qp therefore qp = Cp∆T 28 Hours 9 & 10 – Exact, inexact and total differentials. Changes in internal energy. Thermal expansivity, isothermal compressibility. Joule-Thomson effect. Exact, inexact and total differentials Exact differentials are infinitesimal quantities which, when integrated, gives a result that is independent of the path between the initial and final stages. This means that infinitesimal changes in state functions are exact differentials. For instance, since U is a state function: 𝑓 ∆𝑈 = ∫ 𝑑𝑈 𝑖 this means that dU is an exact differential. Inexact differentials are infinitesimal quantities which, when integrated, gives a result that depends on the path between initial and final stages. This means that infinitesimal changes in path functions are inexact differentials. For instance, since q is a path function: 𝑓 𝑞=∫ 𝑑𝑞 𝑖,𝑝𝑎𝑡ℎ dq is an inexact differential. Partial vs total differentials Partial derivatives are the derivatives we get when we hold constant all except ONE of the independent variables in a function and differentiate with respect to that one alone. Whereas the total derivative (full derivative) of a function f, of several variables, e.g. t, x, and y, with respect to one of its input variables, e.g., t, is different from its partial derivative. Calculation of the total derivative of f with respect to t does not assume that the other independent variables are constant while t varies; instead, it allows the other independent variables to depend on t. 29 The total derivative adds in these indirect dependencies to find the overall dependence of f on t. For example, the total derivative of f(x, y, t) with respect to t is : 𝑑𝑓 𝜕𝑓 𝑑𝑡 𝜕𝑓 𝑑𝑥 𝜕𝑓 𝑑𝑦 = + + 𝑑𝑡 𝜕𝑡 𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 Simplifying and multiplying both sides by dt we get: 𝜕𝑓 𝜕𝑓 𝜕𝑓 𝑑𝑓 = 𝑑𝑡 + 𝑑𝑥 + 𝑑𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑦 If we know that the function f is dependent on x, y and z, then we can have three partial derivatives: 𝑆𝑖𝑛𝑐𝑒 𝑓(𝑥, 𝑦, 𝑧) then the partial derivatives are: 𝜕𝑓 𝜕𝑓 𝜕𝑓 𝑎𝑛𝑑 𝜕𝑥 𝜕𝑦 𝜕𝑧 Also if x, y and z are functions of t, then we see that: 𝑑𝑓 𝑑𝑓 𝑑𝑥 = ∗ 𝑑𝑡 𝑑𝑥 𝑑𝑡 For a function z = z(x, y): 𝜕𝑦 𝜕𝑥 𝜕𝑧 ( ) ∗ ( ) ∗ ( ) = −1 𝜕𝑥 𝑧 𝜕𝑧 𝑦 𝜕𝑦 𝑥 Also: 𝜕𝑦 1 ( ) = 𝜕𝑥 𝑧 (𝜕𝑥 ) 𝜕𝑦 𝑧 Note that skilful use of partial derivatives can often transform an unfamiliar quantity into a quantity that can be recognised, interpreted or measured. These manipulations will be observed throughout our study of thermodynamics. 30 Questions: 1. Show that the following function has an exact differential: 𝑥 2 𝑦 + 3𝑦 2. 2. What is the total differential of 𝑧 = 𝑥 2 + 2𝑦 2 − 2𝑥𝑦 + 2𝑥 − 4𝑦 − 8? 𝜕2𝑧 𝜕2𝑧 3. Show that 𝜕𝑦𝜕𝑥 = 𝜕𝑥𝜕𝑦 for the function above. Changes in internal energy When the volume and temperature of a system are changed, the internal energy U changes. We will consider an adiabatic and non-adiabatic path will be discussed. These two paths correspond to different values of q and w but they correspond to the same value of ΔU. Consider a closed system of constant composition. The system is at constant internal pressure, and we know that U is a function of V, T and p. However, p is fixed if V and T are known. So U becomes a function of V and T only. As 𝑉 → 𝑉 + 𝑑𝑉 (a small change in V) we see a corresponding change in U (note the use of the partial derivative): 𝜕𝑈 𝑈 → 𝑈 ′ = 𝑈 + ( ) 𝑑𝑉 𝜕𝑉 𝑇 Similarly for a small change in temperature: 𝐴𝑠 𝑇 → 𝑇 + 𝑑𝑇 𝜕𝑈 𝑈 → 𝑈′ = 𝑈 + ( ) 𝑑𝑇 𝜕𝑇 𝑉 If both V and T change, then: 𝜕𝑈 𝜕𝑈 𝑈 ′ = 𝑈 + ( ) 𝑑𝑉 + ( ) 𝑑𝑇 𝜕𝑉 𝑇 𝜕𝑇 𝑉 This rearranges to: 𝑈 ′ = 𝑈 + 𝑑𝑈 = 𝑈 + (𝑆𝑢𝑚 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠) Which implies that: 𝜕𝑈 𝜕𝑈 𝑑𝑈 = ( ) 𝑑𝑉 + ( ) 𝑑𝑇 𝜕𝑉 𝑇 𝜕𝑇 𝑉 31 Note that this is the total derivative. However, we saw earlier that: 𝜕𝑈 ( ) = 𝐶𝑉 𝜕𝑇 𝑉 And now we define a new term, the internal pressure, πT (which has the same dimensions as pressure2): 𝜕𝑈 𝜋𝑇 = ( ) 𝜕𝑉 𝑇 This property measures the change in internal energy U as volume changes. Combining these two definitions we see that: 𝑑𝑈 = 𝜋𝑇 𝑑𝑉 + 𝐶𝑉 𝑑𝑇 Internal pressure is a measure of cohesive forces within the sample; for an ideal gas, πT = 0. As a gas expands at constant temperature, work done by the system lowers U; initially T falls, but heat enters the system, raising the temperature and returning the system to its original state, thereby restoring U. So the overall change in internal energy with volume is zero. So for a 𝜕𝑈 perfect gas, internal pressure is zero. (𝜕𝑉 ) = 0 𝑇 Thermal expansivity We now seek to see how internal energy U changes with temperature at constant pressure; we want to observe the thermal expansion. Starting with the relationship for changes in internal energy: 𝑑𝑈 = 𝜋𝑇 𝑑𝑉 + 𝐶𝑉 𝑑𝑇 2 Show that internal pressure has units of pressure. 32 Dividing both sides by dT, at constant pressure we get [Eq 7.4-1]: 𝜕𝑈 𝜕𝑉 𝑑𝑇 ( ) = 𝜋𝑇 ( ) + 𝐶𝑉 𝜕𝑇 𝑃 𝜕𝑇 𝑃 𝑑𝑇 𝜕𝑉 Since (𝜕𝑇 ) is the slope of a plot of V against T, it is related to the expansion coefficient, α. 𝑃 The expansion coefficient (α) is defined as: 1 𝜕𝑉 𝛼= ( ) 𝑉 𝜕𝑇 𝑃 Or 𝜕𝑉 𝛼𝑉 = ( ) 𝜕𝑇 𝑃 This relationship means that when α is large, the volume of a sample responds strongly to temperature changes. Substituting αV into Eq 7.4-1 we get: 𝜕𝑈 ( ) = 𝜋𝑇 𝛼𝑉 + 𝐶𝑉 𝜕𝑇 𝑃 Recall that for a perfect gas, πT = 0. Hence for a perfect gas: 𝜕𝑈 ( ) = 𝐶𝑉 𝜕𝑇 𝑃 𝜕𝑈 Recall that by definition 𝐶𝑉 = ( 𝜕𝑇 ) by definition. This therefore means that for a perfect gas: 𝑉 𝜕𝑈 𝜕𝑈 ( ) =( ) 𝜕𝑇 𝑃 𝜕𝑇 𝑉 Question: 1. Show that the expansion coefficient of a perfect gas is α = 1/T 33 Isothermal compressibility (temperature dependence of enthalpy) Remember that when values are given for ΔH, T is stated. This is because ΔH is different at different temperatures. We will now explore how ΔH changes with temperature. 𝜕𝐻 Recall that ( 𝜕𝑇 ) = 𝐶𝑃 and ΔH =𝑞𝑝 𝑃 Now, since H is a state function, it can be defined by the state variables T and P: 𝜕𝐻 𝜕𝐻 𝑑𝐻 = ( ) 𝑑𝑃 + ( ) 𝑑𝑇 𝜕𝑃 𝑇 𝜕𝑇 𝑃 Substituting for the second term: 𝜕𝐻 𝑑𝐻 = ( ) + 𝐶𝑃 𝜕𝑃 𝑇 Through a similar but longer manipulation as for dU, it can be shown that: 𝜕𝐻 𝛼𝜇 ( ) = [1 − ] ∗ 𝐶𝑃 𝜕𝑇 𝑉 𝜅𝑇 Where κT, the isothermal compressibility is defined as: 1 𝜕𝑉 𝜅𝑇 = − ( ) 𝑉 𝜕𝑃 𝑇 And μ is the Joule-Thomson coefficient, defined as: 𝜕𝑇 𝜇=( ) 𝜕𝑃 𝐻 A large value of κT indicates that for a small change in pressure, large changes in volume results. The negative sign ensures that κT is always positive since a positive change in pressure brings about a negative change in volume. Question: 1 Show that the isothermal compressibility of a perfect gas is 𝜅𝑇 = 𝑃 34 Joule-Thomson effect A gas can be cooled by adiabatic expansion if the process is isenthalpic (constant enthalpy). A gas expands through a porous barrier from one constant pressure to the next, and a temperature difference was observed (since work was done). Insulation of the system made the process adiabatic. A lower temperature was observed on the low P side (larger volume) and ΔT was proportional to ΔP. This effect is used to cool gases, by allowing the gas to expand adiabatically. As the gas expands, the molecules move away from each other, struggling against the attractive forces of their nearest neighbours. In order to do so, some ke has been converted to pe and so they travel more slowly. Since their average speed will then decrease, the temperature drops and the gas is cooler than before expansion. When a gas is cooled by expansion through a narrow opening, this is called the Joule-Thomson effect. 35 Hour 11 – Relations between Cp and Cv. Work of adiabatic expansion. Ideal gas adiabats. Relation between Cv and Cp The difference between CV and CP arises by the work needed to change the volume of the system at constant pressure. At constant pressure, some of the energy supplied to a system escapes back to the surroundings as work done via expansion. Hence the temperature of the system rises less than when the heating occurs at constant volume. At constant pressure, a smaller increase in temperature implies a larger heat capacity. So we can say that in most cases, CP is greater than CV. Using the definitions of these heat capacities (with manipulation for Cv), we see [Eq. 8.1-1]: ∂H ∂U CP − CV = ( ) − ( ) ∂T P ∂T P But H = U + PV which and since PV = nRT H = U + nRT Substituting for H we get: ∂U ∂U CP − CV = [( ) + nR] − ( ) ∂T P ∂T P So, for a perfect gas [Eq. 8.1-2]: 𝐂𝐏 − 𝐂𝐕 = 𝐧𝐑 For a general case, using the original definition of Cv [Eq. 8.1-3]: ∂H ∂U CP − CV = ( ) −( ) ∂T P ∂T V ∂U ∂(PV) ∂U =( ) +( ) −( ) ∂T P ∂T P ∂T V ∂U Recall that:(∂T) = απT V + CV P ∂U and CV = (∂T) V 36 So substituting in Eq. 8.1-3 we get: ∂U ∂U ( ) − ( ) = απT V + CV − CV ∂T P ∂T V = απT V Therefore: ∂(PV) CP − CV = απT V + ( ) ∂T P Since P is constant: ∂(PV) ∂V ( ) = P [ ] = PαV ∂T P ∂T P 1 ∂V since α = V (∂T). P This leaves us with: CP − CV = απT V + PαV Where the term απTV corresponds to work due to stretching bonds of the material and PαV corresponds to expansion work. On grouping terms we get: 𝐂𝐏 − 𝐂𝐕 = 𝛂𝐕(𝐏 + 𝛑𝐓 ) Work of adiabatic expansion When an ideal gas expands adiabatically, work is done so the kinetic energy (k.e.) of molecules falls, causing their average speed to decrease and hence their temperature falls. Since temperature difference is easy to measure, we can use it to measure the work done by adiabatic expansion. We will now derive an expression to link the measurable temperature difference with the change in volume which we can easily relate to the work done during adiabatic expansion. Consider a reversible adiabatic expansion (P = Pex at all stages of the expansion) for which we know that dw = −PdV. Recall that dU = dw + dq and that for a perfect gas dU = CV dT 37 For an adiabatic change, 𝐝𝐪 = 𝟎 This implies that dU = dw = −PdV = CV dT This gives us: CV dT = −PdV For a perfect gas PV = nRT so substituting for P: nRT CV dT = − dV V Dividing both sides by T we get: dT dV CV = −nR T V Assuming Cv is independent of T and integrating between initial and final states: Tf Vf dT dV CV ∫ = −nR ∫ Ti T Vi V dx Recall ∫ = lnx + c so the integration gives: x Tf Vf CV ln = −nRln Ti Vi Which rearranges to: CV Tf Vi ln ( ) = ln ( ) nR Ti Vf C Let c = nRV ; we get: Tf c Vi ln ( ) = ln ( ) Ti Vf Taking anti natural log of both sides we get [Eq 8.2-1]: 𝐓𝐟 𝐜 𝐕𝐢 ( ) = 𝐓𝐢 𝐕𝐟 38 Why is this expression useful? 1. It helps us to find the work done during adiabatic expansion. Recall dU = dw = −PdV = CV dT Which gives 𝐝𝐰𝐚𝐝 = 𝐂𝐕 𝐝𝐓 for infinitesimal changes and 𝐰𝐚𝐝 = 𝐂𝐕 𝚫𝐓 for measurable changes. So in order to find the work done by adiabatic expansion, we can use Eq 8.2-1 to find the final temperature and use: 𝐰𝐚𝐝 = 𝐂𝐕 (𝐓𝐟 − 𝐓𝐢 ) 2. It helps us predict how the temperature of an ideal gas will fall, when it is expanded adiabatically. The temperature falls most steeply for gases with low molar heat capacity. Questions: 1. Calculate the work done as 0.020 mol Ar, initially at 250C expands adiabatically and reversibly from 0.50 L to 1.00 L. The molar heat capacity of argon at constant volume is 12.48 J K-1 mol-1. 2. Calculate the final temperature, the work done, and the change of internal energy when 0.020 mol ammonia is used in a reversible adiabatic expansion from 0.5 L to 2.00 L, with initial temperature 25 0C. 39 Ideal gas adiabats An adiabat shows how pressure varies with volume when a gas expands adiabatically. When compared with an isotherm, we see that the pressure falls more rapidly (curve is steeper) with increasing volume because the temperature decreases in adiabatic expansion. In order to see this relationship mathematically, we need to derive an expression for the change in pressure with volume upon adiabatic expansion. For a perfect gas, PV = nRT regardless of the stages. Hence: Pi Vi Pf Vf = Ti Tf On rearranging we get [Eq 8.3-1]: Pi Vi Ti = Pf Vf Tf However, from Eq 8.2-1: Ti Vf 1/c Vf γ−1 =( ) =( ) Tf Vi Vi R Where γ is defined as CP,m/CV,m (the ratio of the molar heat capacities) and γ = 1 + C which V,m implies that γ-1 = 1/c 40  Ti Vf γ Vi =( ) ( ) Tf Vi Vf Therefore: Pi Vi Ti Vf γ Vi = =( ) ( ) Pf Vf Tf Vi Vf Which simplifies to: 𝐏𝐢 𝐕𝐟 𝛄 =( ) 𝐏𝐟 𝐕𝐢 Usually expressed as: 𝐏𝐕 𝛄 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 Illustration: 41 Hours 12 &13 Entropy: Purpose, definition and measurement. Second and Third laws. A spontaneous change is a change that can occur without work being done to bring it about. It has a natural tendency to occur. A non-spontaneous change is a change that can only occur by doing work; it has no natural tendency to occur. The apparent driving force for spontaneous change is the tendency for matter and energy to become disordered. Hence, spontaneous changes are those which lead to a more disorderly dispersal of the total energy of the isolated system. The measure of disorder of the matter and energy in a system is called the entropy, S. S has units of J K –1mol –1 or J K –1. Recall that the total energy of an isolated system is constant so only a redistribution of energy is possible. When matter and energy become disordered, entropy is increased. Second Law of Thermodynamics The energy of an isolated system increases in the course of a spontaneous change. 𝛥𝑆𝑡𝑜𝑡 > 0 Where ΔStot is the entropy change is the system and surroundings. The second law can also be worded: The entropy of the universe tends to increase. In order to measure entropy quantitatively, we use the thermodynamic definition of entropy: 𝑑𝑞𝑟𝑒𝑣 𝑑𝑆 = for an infinitesimal change, and 𝑇 𝑓 𝑑𝑞𝑟𝑒𝑣 𝛥𝑆 = ∫𝑖 for a measurable change. 𝑇 𝑑𝑞𝑟𝑒𝑣 Interpretation of the equation 𝑑𝑆 = : 𝑇 (1) The change in entropy is directly related to heat (q) since q is a result of molecular disorder. 42  (2) The change in entropy is inversely proportional to temperature (T); this implies that heat supplied to a hot body will not cause as much change in entropy as if the same heat were supplied to a cold body. Eg. The difference between sneezing on a busy street vs sneezing in a quiet library. Calculation of ΔS for specific changes (1) Phase transition enthalpy Consider this direction of phase changes: 𝑠𝑜𝑙𝑖𝑑 → 𝑙𝑖𝑞𝑢𝑖𝑑 → 𝑔𝑎𝑠 ΔS > 0 for these changes. Let Ttrs be the normal transition temperature for the phase change. Recall that ΔH = qp so: 𝑞𝑟𝑒𝑣,𝑝 𝛥𝐻𝑡𝑟𝑠 𝛥𝑆 = = = 𝛥𝑆𝑡𝑟𝑠 𝑇𝑡𝑟𝑠 𝑇𝑡𝑟𝑠 NB. When ΔHtrs < 0 (exothermic process), ΔStrs < 0 (non-spontaneous change, such as freezing and condensation. However, when ΔHtrs > 0 (endothermic process), ΔStrs > 0 (spontaneous change such as melting and evaporation). Trouton’s rule: a wide range of liquids give approximately the same ΔSvap ≈ 85 J K –1mol –1. Large deviations from Trouton’s rule indicate the presence of partial order in the liquid. For instance, hydrogen bonding in H2O gives a more orderly structure than is found in H2S. (2) Entropy change on expansion of a perfect gas 𝑞𝑟𝑒𝑣 We know 𝛥𝑆 =. Since the process is isothermal and reversible, then ΔU = 0 hence we 𝑇 𝑉 deduce that: 𝑞𝑟𝑒𝑣 = −𝑤𝑟𝑒𝑣 = −(−𝑛𝑅𝑇𝑙𝑛 ( 𝑉𝑓 ). Therefore: 𝑖 𝑉𝑓 𝛥𝑆 = 𝑛𝑅𝑙𝑛 ( ) 𝑉𝑖 (3) Variation of entropy with temperature Generally: 𝑓 𝑞𝑟𝑒𝑣 𝑆(𝑇𝑓 ) = 𝑆(𝑇𝑖 ) + ∫ 𝑖 𝑇 43  𝑑𝑞 Recall that 𝐶𝑃 = 𝑑𝑇 therefore 𝑑𝑞𝑟𝑒𝑣 = 𝐶𝑃 𝑑𝑇 at constant P, therefore we get: 𝑓 𝑑𝑇 𝑆(𝑇𝑓 ) = 𝑆(𝑇𝑖 ) + 𝐶𝑃 ∫ 𝑖 𝑇 Which gives: 𝑇𝑓 𝑆(𝑇𝑓 ) = 𝑆(𝑇𝑖 ) + 𝐶𝑃 𝑙𝑛 ( ) 𝑇𝑖 This relationship can be used to relate the entropy of a substance at T = 0 to its entropy at some temperature T (be sure to include the entropy of each phase transition). That is, the entropy of a system at T, is related to its entropy at T = 0 by measuring CP at different temperatures and using: 𝑇𝑓𝑢𝑠 𝑇𝑏 𝑇 𝑑𝑇 𝛥𝐻𝑓𝑢𝑠 𝑑𝑇 𝛥𝐻𝑣𝑎𝑝 𝑑𝑇 𝑆(𝑇) = 𝑆(0) + ∫ 𝐶𝑃 (𝑠) + + ∫ 𝐶𝑃 (𝑙) + + ∫ 𝐶𝑃 (𝑔) 0 𝑇 𝑇𝑓𝑢𝑠 𝑇𝑓𝑢𝑠 𝑇 𝑇𝑣𝑎𝑝 𝑇𝑏 𝑇 However, it is difficult to determine S at T = 0; in order to evaluate this, we apply the Debye approximation. 44 Carnot cycle Evidence is provided here to support S being a state function. The total entropy change around a closed cycle called the Carnot cycle is zero. A Carnot cycle, which is named after the French engineer Sadi Carnot, consists of four reversible stages 1. Reversible isothermal expansion from A to B at Th; the entropy change is qh/Th, where qh is the energy supplied to the system as heat from the hot source. 2. Reversible adiabatic expansion from B to C. No energy leaves the system as heat, so the change in entropy is zero. In the course of this expansion, the temperature falls from Th to Tc, the temperature of the cold sink. 3. Reversible isothermal compression from C to D at Tc. Energy is released as heat to the cold sink; the change in entropy of the system is qc /Tc; in this expression qc is negative. 4. Reversible adiabatic compression from D to A. No energy enters the system as heat, so the change in entropy is zero. The temperature rises from Tc to Th. 𝑞 𝑞𝑐 The total change in entropy around the cycle is dS = 𝑇ℎ + ℎ 𝑇𝑐 45 However, we can show that, for a perfect gas: 𝑞ℎ 𝑇ℎ = - (See justification on page 83 of Physical Chemistry by Atkins) 𝑞𝑐 𝑇𝑐 Substitution of this relation into the preceding equation gives zero on the right, which Means that around the closed cycle dS = 0. This is sufficient proof that S (entropy) is a state function. 46