PHY 204 Thermal Physics Lecture Notes PDF
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Dr Y.A. Tanko
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These lecture notes cover thermal physics, focusing on thermodynamics, kinetic theory, and statistical physics. They detail the zeroth, first, second, and third laws of thermodynamics, along with concepts such as temperature, work, heat, and internal energy.
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LECTURE NOTES PHY 204 (THERMAL PHYSICS) BY DR Y.A. TANKO COURSE OUTLINE The foundations of classical thermodynamics, Zeroth law and concept of temperature The 1st of thermodynamic...
LECTURE NOTES PHY 204 (THERMAL PHYSICS) BY DR Y.A. TANKO COURSE OUTLINE The foundations of classical thermodynamics, Zeroth law and concept of temperature The 1st of thermodynamics, work, heat and internal energy Carnot cycle and 2nd law of thermodynamics Entropy and irreversibility Thermodynamic potentials and the Maxwell relation Qualitative discussion of phase transitions 3rd law of thermodynamics Ideal and real gases Elementary kinetic theory of gases, Boltzmann counting, Maxwell-Boltzmann law of distribution of velocities, simple application. References 1. Fundamentals of Physics 2nd Edition, by M.M. Kashimbila. 2. Thermal Physics 2nd Edition, by Philip M. Morse. 3. An Introduction to Thermal Physics, by Daniel V. Schroeder. 4. Thermal Physics Concepts and Practice, by Allen L. Wasserman. 1 INTRODUCTION Thermal physics can be divided into three, Thermodynamics, Kinetic theory and Statistical Physics. Thermodynamics: is the branch of physics that deals with heat and work, and those properties of matter that relate to heat and work. Temperature (T), is the degree of hotness or coldness of a body. The zeroth law of thermodynamics implies the existence of a property of a system which we shall call temperature. LAWS OF THERMODYNAMICS: The laws of thermodynamics include the zeroth law, the first law, the second law and the third law of thermodynamics. Zeroth law: it states that if two bodies A and B are in thermodynamic equilibrium and Body B and C are in thermodynamic equilibrium, then body A and C are also in thermodynamic equilibrium. Two bodies are said to be in thermodynamic equilibrium if they have the same temperature. ⟹ A = B and B = C, then A = C. The definition of zeroth law enables the use of the use of a thermometer as a measurement device. A scale however needs to be defined. The old metric temperature scale, Celsius (℃) was defined so that: 0 ℃ is the freezing point of water and 100 ℃ is the boiling point of water. These quantities vary with pressure however, so that different values would be obtained on top of mountain versus down in the valley, and so this is not a good standard. The modern Celsius scale is defined to be nearly the same but has 0.01 ℃ as the so-called triple point of water and -273.15 ℃ as absolute zero in Kelvin (K). The triple point of water is defined as the state where three phases of water (solid, liquid and gas) are observed to co-exist. The transformation between the absolute Kelvin scale and the Celsius is given by TK = TC + 273.15 (1A) TC = TK 273.15 (1B) A Kelvin is the same size as a degree Celsius, but kelvin temperatures are measured up from absolute zero instead of from the freezing point of water. In round numbers, room temperature is approximately 300 K. 2 DEFINATIONS Thermodynamic system: a quantity of fixed mass under investigation. Surroundings: everything external to the system. System boundary: interface separating system and surrounding. Universe: combination of system and surroundings. Isolated system: a system which is not influenced by its surroundings. Control volume: fixed volume over which mass can pass in and out of its boundary. Control surface: boundary of the control volume. Phase: a quantity of matter that is homogeneous throughout. Phase boundaries: interface between different phases. State: condition described by observable macroscopic properties i.e. pressure, temperature, volume (thermodynamic variables). Property: quantity which only depends on the state of the system and is independent of the history of the system. Equilibrium: states in which no spontaneous changes are observed with respect to time. Mechanical equilibrium: characterised by equal pressure. Thermal equilibrium: characterised by equal temperature. THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics is based on the principle of the conservation of energy and it states that “the total internal energy of a system is the sum of all the potential and kinetic energies of all the particles comprising the system”. The internal energy of a system can be changed through the supply (loss) of heat and by doing work. Heat is energy in transit from a high temperature system to a low temperature system. Internal energy of a system increases if it gains heat and decreases if it loses heat. Heat is taken to be positive if it is supplied and negative if it is lost. Work is transfer of energy by the compression or expansion of a system. Internal energy of a system increases when the system compresses and decreases when the system expands. We say work is done on a system when a system compresses and work is done by the system when the system expands. Work is taken to be positive when the system expands and negative when the system compresses. 3 WORK DONE DURING VOLUME CHANGES Consider a mass of gas enclosed in a cylinder by a frictionless piston whose cross-sectional area is A. The system is in equilibrium under the action of an external force F acting to the left and a force due to the pressure P of the gas acting to the right (Figure 1). For equilibrium, F = PA When the gas expands it moves the piston outwards through an infinitesimal distance dx. The work done by this force is dW = Fdx dW = PAdx (2) But Adx = dV is the infinitesimal change of volume of the system. Hence, the above equation becomes dW = PdV (3) Figure 1: A gas system is confined in a cylinder with a movable piston. The gas is maintained at the temperature T by the surroundings heat bath. In a finite change of volume from V1 to V2 , the total work done by the gas is obtained by integrating equation (3) V2 W= PdV V1 (4) In general, the pressure of the system may vary during the volume change. So, to evaluate the integral of equation (4), we use the ideal gas law: nRT P= (5) V Substitute (5) into (4), we have 4 V2 dV W = nRT V1 V V2 W = nRT In (6) V1 Since the gas is maintained at constant temperature, equation (6) is the expression for the work done under isothermal expansion. Also, when T is constant P1V1 = P2 V2 V2 P1 = (7) V1 P2 Using (7), (6) becomes P1 W = nRT In (8) P2 A state function: is a function whose change depends only on the initial and final states of the system irrespective of the process that brought about the change. Thus, a change in a state function can be expressed as the difference between the values of the function at the final state and the initial state of the system. If Z is an arbitrary state function, then ΔZ = Zf - Zi The variables that determines the state of a gas are its temperature, volume and pressure. This means, a state function is a function that depends on these variables only. The total internal energy (U) of a system is a state function. Its change is equal to the difference between its values at the final and initial states. ΔU = U f - U i Work and heat are not state functions. They depend on the process that brought about the change and not on the initial and final state. But the difference between heat and work depends on the initial and final states of the system only. The first law of thermodynamics state that the difference between heat (Q) and work (W) is a state function and is equal to the change in the total internal energy of the system. The first law is mathematically written as statement ΔU = Q W (9) Using the expression for work in terms of pressure and volume, equation (9) can be written as ΔU = Q PΔV (10) Note: SI units Pa for pressure and m3 for volume should be used to give joules. If litre is used for volume then kpa should be used for pressure to get joules. 5 Example 1: The volume of a gas decrease from 50 litres to 45 litres under the influence of a pressure of 20 kpa. Calculate the work done on the system. Solution: P = 20 kpa, Vi = 50 L and Vf = 45 L, W =? W = P∆V = P (Vf – Vi) = 20 (45 – 50) = -100 J Example 2: One mole of an ideal monoatomic gas is carried by a quasi-static isothermal process at 400 K to twice its original volume (a) How much work W12 was done by the gas? (b) How much heat Q12 was supplied to the gas? (c) What is the pressure ratio P1 to P2? Solution: (a) Let the initial volume be V, hence V1 = V and V2 = 2V, R = 8.314 J/mol. K V W12 = nRT In 2 V1 2V = 18.314 400 In = 2305J V (b) For an isothermal process, there is no increase in temperature and so there is increase in the internal energy of the system. i.e. dU = 0. Hence, Q12 = dU + W12 = 0 + 2305 = 2305J (c) Using the ideal gas law, with T1 = T2 we have P2 V1 1 = = P1 V2 2 Class work 1 1. A system does 50 J of energy on its environment while at the same time losing 20 J of energy to its environment. Calculate the change in its internal energy 2. A system gains 40 J of heat energy while the environment is doing a work of 10 J on the system. Calculate the change in internal energy of the system. 6 A cyclic process: is a process where the initial and the final state are the same. For a cyclic process, U f = U i and ∆U = 0 Q = W PV DIAGRAMS A PV diagram for a certain process is the graph of pressure against volume for the process. Work can be obtained from a PV diagram as area enclosed between pressure against volume curve and the volume axis. It is positive on intervals where the volume is increasing and negative on intervals where the volume is decreasing. For a cyclic process, the volume will increase on one interval and decrease on the other interval. The work done will be the difference between the works on these intervals. Thus, for a cyclic process the absolute value of the work done is equal to the area of the closed curve. It is positive if the process is clockwise and negative if the process is anticlockwise. Assignment 1 The following is a PV diagram for a certain process. Its internal energy at the states (2 L, 2 kPa) and (6 L, 6 kPa) are respectively 10 J and 20 J (a) Calculate the work done in taking the system from (2 L, 2 kPa) state to the (6 L, 6 kPa) state along the legs of the right-angled triangle. 7 (b) Calculate the amount of heat gained or lost when it is taken from the state (2 L, 2 kPa) to the state (6 L, 6 kPa) along the legs of the right-angled triangle. (c) Calculate the work done in taking the system from the (2 L, 2 kPa) state to the (6 L, 6 kPa) along the straight line joining the two points (i.e. along the hypotenuse of the right- angled triangle) and compare with the work done when the system is taken along the legs of the right-angled triangle. (d) Calculate the heat gained or lost when the system is taken from the state (2 L, 2 kPa) to the state (6 L, 6 kPa) along the line joining the two states (the hypotenuse) and compare with the heat obtained when the system is taken along the legs of the triangle. (e) Calculate the work and heat for the cyclic process that starts at the state (2 L, 2 kPa) and ends up at (2 L, 2 kPa) in a clockwise and anticlockwise direction. HEAT Heat is energy in transit from an object at higher temperature to an object at lower temperature. The SI unit of measurement for heat is the Joule. Another common unit of heat measurement is the Calorie (Cal). A Calorie is defined to be equal to the amount of heat energy required to raise the temperature of one gram of water by one ℃. It is equivalent to 4.18 J. HEAT AND CHANGE OF TEMPERATURE For a given substance, amount of heat supplied is directly proportional to the change of its temperature. For a given change in temperature, amount of heat supplied is directly proportional to mass Q = mcΔT (11) where Q is the quantity of heat supplied or lost, c is the specific heat capacity of the substance. Equation 11 can also be applied to the heat added to, or removed from a gas. It is more convenient to express the mass m of a gas in terms of the number of moles n of the gas. The total mass m of the gas is the sum of the masses of all the molecules of the gas. That is, m is equal to the mass of one molecule times the total number of molecules in one mole of the substance, times the total number of moles. That is m = m0 N A n (12) where m 0 is the mass of one molecule; N A is Avogadro’s number, the number of molecules in one mole of a substance; and n is the number of moles of the gas. The product of the mass 8 of one molecule times Avogadro’s number is called the molecular mass of the substance M, that is, M = m0 N A (13) The molecular mass is thus the mass of one mole of the gas. Substituting equation 13 into equation 12 gives for the mass of the gas m = nM (14) Equation 14 says that the mass of the gas is equal to the number of moles of the gas times the molecular mass of the gas. Substituting equation 14 for the mass m of the gas into equation 11, gives Q = nMcT (15) The product Mc is defined as the molar specific heat of the gas, or molar heat capacity, and is represented by the capital letter C. Hence, C = Mc (16) The heat absorbed or lost by a gas undergoing a thermodynamic process is found by substituting equation 16 into equation 15. Thus, Q = nCT (17) HEAT TRANSFER The transfer of thermal energy has historically been called heat transfer. Thermal energy can be transferred from one body to another by any or all the following mechanisms: 1. Convection 2. Conduction 3. Radiation Convection is the transfer of thermal energy by the actual motion of the medium itself. The medium in motion is usually a gas or a liquid. Convection is the most important heat transfer process for liquids and gases. When molecules at the bottom are heated their density decreases and they rise. The denser molecules at the top come down taking their place. This way heat is transferred from the bottom to the top by the actual movement of molecules. For example, when water in a dish is heated, first the molecules at the bottom of the dish are heated. Their density decreases and they rise up and the denser molecules at the top fall down resulting in the transfer of energy from the bottom to the top. 9 Conduction is the transfer of thermal energy by molecular action, without any motion of the medium. Conduction can occur in solids, liquids, and gases, but it is usually most important in solids. For example, when one end of a metal is heated, the molecules at that end will vibrate vigorously. This energy is transferred to its neighbouring molecules by means of collision. This process repeats itself again and again from neighbour to neighbour and eventually is transferred to the other end of the metal. The rate of transfer of heat in a metal rod by means of conduction is proportional to the cross- sectional area (A) of the rod and to the temperature difference (T2 – T1) between the rods and inversely proportional to the length (L) of the rod. Q T - T = κA 2 1 (18) Δt L Q is amount of heat transferred from one end of the rod to the other end in a time interval ∆t Q (rate of heat flow = ). The unit of measurement of rate of flow of heat (power) is J/s (Watt). Δt κ is a material constant called thermal conductivity of the material. The unit of measurement for thermal conductivity is W/m K. Radiation is a transfer of thermal energy by electromagnetic waves. Any hot object emits electromagnetic waves. Electromagnetic waves involve a wide range of waves including light waves. A small subset of these waves called infrared wave causes sensation of heat. It is a common experience that a light bulb also causes sensation of heat in addition to sensation of vision. This is because the electromagnetic wave emitted by the light bulb contains infrared waves. The Stefan-Boltzmann Law Joseph Stefan (1835-1893) found experimentally, and Ludwig Boltzmann (1844-1906) found theoretically, that everybody at an absolute temperature T radiates energy that is proportional to the fourth power of the absolute temperature. The result, which is called the Stefan- Boltzmann law is given by Pe = σeATe4 (19) Pe is the rate of emission of electromagnetic energy by an object of surface area A at a temperature Te in Kelvin. Rate of emission of energy is equal to energy emitted per unit time Pe = ΔQ Δt . The unit of measurement of emission rate is Watt. e is a constant called emissivity that depends on the properties of the surface. Its value is between 0 (for no emission) 10 and 1 (for perfect emission). It is unit less. σ is a universal constant called Stefan-Boltzmann constant. Its value is 5.67 ×10-8 W m 2 K 4. Radiation from a Blackbody The amount of radiation depends on the radiating surface. Polished surfaces are usually poor radiators, while blackened surfaces are usually good radiators. Good radiators of heat are also good absorbers of radiation, while poor radiators are also poor absorbers. A body that absorbs all the radiation incident upon it is called a blackbody. The name blackbody is really a misnomer, since the sun acts as a blackbody and it is certainly not black. A blackbody is a perfect absorber and a perfect emitter. The substance lampblack, a finely powdered black soot, makes a very good approximation to a blackbody. A box, whose insides are lined with a black material like lampblack, can act as a blackbody. If a tiny hole is made in the side of the box and then a light wave is made to enter the box through the hole, the light wave will be absorbed and re-emitted from the walls of the box, over and over. Such a device is called a cavity resonator. For a blackbody, the emissivity e in equation 19 is equal to 1. The rate of heat absorbed or emitted from a blackbody is Pe = σATe4 (20) Example 3: Energy radiated from the sun. If the surface temperature of the sun is approximately 5800 K, how much thermal energy is radiated from the sun per unit time? Assume that the sun can be treated as a blackbody. Solution We can find the energy radiated from the sun per unit time from equation 20. The radius of the sun is about 6.96 × 108 m. Its area is therefore A = 4 r 2 = 4π 6.96 ×108 2 = 6.09 ×1018 m 2 The heat radiated from the sun is therefore Q Pe = = σAT 4 t = 5.67 ×10-8 6.96 ×1018 5800 = 3.91×1026 J s 11 IDEAL GAS PROPERTIES A perfect gas or an ideal gas is defined as a gas having no intermolecular forces. A gas which follows the gas laws at all ranges of pressures and temperatures can be considered as an ideal gas, but no such gas exists in nature. An ideal gas is a gas in which for “n” moles having pressure “P”, volume “V” and temperature “T” there is a relation known as an equation of state, given by PV = nRT (21) m But n = M m PV = RT (22) M where m = total mass of the gas (grams) M = molar mass of the gas (grams per mole) R = Ideal or universal gas constant = 8.314 J mol-1 K-1 m = total mass of the gas (grams) m = total mass of the gas (grams) Also, m = Nma and M = N A m a (23) where N = number of atoms N A = Avogadro's number m a = mass of one atom Substitute (23) into (22) Nma PV = RT N a ma NRT PV = NKT Na R where K = = 1.38 ×10-23 JK -1 NA K = Boltzmann’s constant Hence, PV = NKT (24) HEAT CAPACITIES OF AN IDEAL GAS Let’s start with looking at Figures below, which shows two vessels A and B, each containing 1 mol of the same type of ideal gas at a temperature T and a volume V. The only difference between the two vessels is that the piston at the top of A is fixed, whereas the one at the top of 12 B is free to move against a constant external pressure P. We now consider what happens when the temperature of the gas in each vessel is slowly increased to T + dT with the addition of heat. Figure 2: Two vessels are identical except that the piston at the top of A is fixed, whereas that atop B is free to move against a constant external pressure P. Since the piston of vessel, A is fixed, the volume of the enclosed gas does not change. Consequently, the gas does no work, and we have from the first law dU = dQ - dW = dQ since dW = 0 dQ = CV dT C V = molar heat capacity at constant volume of the gas. Since dU = dQ for this process, dU A = CV dT (25) We obtained this equation assuming the volume of the gas was fixed. However, internal energy is a state function that depends on only the temperature of an ideal gas. Therefore, dU = CV dT gives the change in internal energy of an ideal gas for any process involving a temperature change dT. When the gas in vessel B is heated, it expands against the movable piston and does work dW = PdV. In this case, the heat is added at constant pressure, dQ = CP dT (26) CP = molar heat capacity at constant pressure of a gas. Furthermore, since ideal gas expands against a constant pressure, 13 d PV = d RT PdV + VdP = RdT PdV = RdT (27) Substitute (26) and (27) into the equation of 1st law of thermodynamics dU = dQ PdV dU = CP dT - RdT dU B = CP - R dT (28) We found dU for both an isochoric and an isobaric process. Because internal energy of an ideal gas depends on only the temperature, dU must be the same for both processes. That is equation (25) is equal to equation (28). dU A = dU B CV dT = CP R dT CP = CV R (29) SOME PROCESSES AND THE FIRST LAW 1. Cycle A system which performs a cycle returns all its variable to their values at the beginning. The net change in internal energy is zero. Applying first law, dQ = dU + dW dQ = dW dQ = PdV 2. Adiabatic process A process whereby the quantity of heat is constant i.e. dQ = 0 Applying first law, dQ = dU + dW 0 = dU + dW dU = dW dU = PdV 3. Isothermal process A process whereby the temperature T = constant i.e. dT = 0 14 Applying first law, dQ = dU + dW But since T is constant, then u is also constant dQ = dW dQ = PdV 4. Isochoric process A process whereby the volume of the work system remains constant i.e. dV = 0 Applying first law, dQ = dU + dW dQ = dU + PdV dQ = dU 5. Isobaric process In this process, the pressure should remain constant i.e. dP = 0 and the process is to be slow as that of isothermal process. Applying the first law, dQ = dU + dW dQ = dU + PdV ADIABATIC EXPANSION OF AN IDEAL GAS The word adiathermal means without flow of heat. A system bounded by adiathermal walls is said to be thermally isolated. Any work done on such a system produces an adiathermal change. We define a change to be adiabatic if it is both adiathermal and reversible. In an adiabatic expansion, there is no flow of heat and we have dQ = 0 The first law becomes dU = dW dU = PdV For an ideal gas, dU = CV dT , we find for 1 mole of an ideal gas, CV dT PdV PdV dT (30) CV Also for one mole of an ideal gas, 15 d PV d RT PdV VdP RdT PdV VdP dT = (31) R Equate equation (30) and (31) PdV VdP PdV R CV CV PdV CV VdP = RPdV CV VdP CV R PdV = 0 (32) Divide equation (32) by PV dP dV CV + CV + R =0 (C P = C V + R) P V dP dV CV + CP =0 divide bothsides by C V P V dP CP dV + =0 P CV V CP Βut, = ratio of molar heat capacities CV dP dV + =0 P V dP dV P + V =0 InP + InV = Constant (33) From, In A x = xInA (34) InAB = InA + InB, Using (34), equation (33) becomes InP + In V γ Constant PV γ Constant (35) nRT Substituting for P in equation (35) from the general gas equation P = , V TV γ 1 = constant (36) 16 1 Example 4: How much is required to compress 5 mol of air at 20 °C and 1 atm to of the 10 original volume by (a) An isothermal process and an adiabatic process? (b) What are the final pressures for the two cases? = 1.403 and CV = 20.68 Solution V2 1 (a) n = 5, T1 = 273 + 20 = 292K, P1 = 1 atm, = V1 10 For Isothermal process, the work done on the gas is W V2 W nRTIn V1 1 W = - 5 8.314 293 In 2.80 10 4 J 10 For adiabatic process, work done W is W nCV T2 - T1 But T2 is unknown, T1V1γ 1 = T2 V2 γ 1 γ 1 V 293 0.1 1.403 1 T2 = T1 1 741K V2 W nCV T2 - T1 5 20.68 741 293 4.63 104 J (b) For an isothermal process, P1V1 = P2 V2 V P2 = P1 1 = 110 = 10 atm V2 For an adiabatic process, P1V1 = P2 V2 V P2 = P1 1 110 25.3 atm. 1.403 V2 Example 5: Gas in a cylinder initially at a temperature of 290K and pressure 1.03 × 105 Pa is compressed to 1/8 of its original volume (a) Calculate the difference Pf Pi if done (i) Isothermally (ii) adiabatically 17 (b) What is the final temperature? (take γ = 1.5) Solution (a) (i) Pf Vf = Pi Vi Pi Vi 1 Pf = but Vf = Vi Vf 8 Pi Vi Pf = 8Pi Vi 8 8 1.03 105 Pa Hence the difference is Pf Pi 8Pi Pi 7Pi 7 1.03 105 Pa (ii) PV γ = constant Pf Vf γ = PV i i γ γ V Pf = Pi i Vf γ V Pi 8 γ Pf = Pi i Vi 8 Pf Pi 8 Pi Pi 17.6 105 Pa (b) PV γ = constant Pf Vf γ = PV i i γ Also, Tf Vf γ -1 TV i i γ -1 γ -1 V Tf Ti i Vf 290 8 1.4 1 Example 6: One mole of an ideal gas goes through the thermodynamic cycle shown in the figure below. If PA = 2.00 × 104 Pa, PD = 1.00 × 104 Pa, VA = 0.250 m3, and VB = 0.500 m3, find the work done along the path (a) AB, (b) BC, (c) CD, (d) DA, and (e) ABCDA. 18 P A B PA PD D C VA VB V Solution (a) The work done along path AB, is WAB = PA VB - VA = 2 ×104 0.50 0.25 5.00 ×103 Nm J (b) The work done along path BC, is WBC = P VB - VB 0 (c) The work done along path CD, is WCD = PD VA - VB = 1.0 ×104 0.25 0.50 = 2.50 ×103 J Note that the work done in compressing the gas is negative. (d) The work done along path DA, is WDA = P VA - VA 0 (e) The work done along path ABCDA, is Wtotal = WAB WBC WCD WDA = 5.0 ×103 + 0 - 2.5 ×103 + 0 = 2.5 ×103 J HEAT ENGINES AND THE SECOND LAW OF THERMODYNAMICS A heat engine absorbs heat from a source at high temperature, performs some amount of mechanical work, and then rejects some heat at a lower temperature. This process can be 19 visualized with the schematic diagram for an ideal heat engine, and is shown in Figure 3. The engine is represented by the circle in the diagram. The engine absorbs the quantity of heat QH from a hot temperature reservoir, at a temperature TH. Some of this absorbed heat energy is converted to work, which is shown as the pipe coming out of the engine at the right. The rest of the original absorbed heat energy is dumped as exhaust heat QC into the low-temperature reservoir at a temperature TC. Figure 3: A schematic diagram for an ideal heat engine In general, a heat engine carries some working substance through a cyclic process during which: (1) Energy is transferred by heat from a source at a high temperature, (2) Work is done by the engine, and (3) Energy is expelled by the engine by heat to a source at lower temperature. Because the engine operates in a cycle, U = 0 , and as we have already seen, the net work done is equal to the net heat absorbed by the engine, that is, W=Q But the net heat absorbed is equal to the difference between the total heat absorbed QH at the hot reservoir, and the heat rejected QC at the cold reservoir, that is, 20 Q = Q H QC Thus, the work done by the engine is equal to the difference between the heat absorbed from the hot reservoir and the heat rejected to the cold reservoir W = QH QC The efficiency (𝛈) of an engine can be defined in terms of what we get out of the system compared to what we put into the system. Heat, Qin, is put into the engine, and work, W, is performed by the engine, hence the efficiency of an engine can be defined as Work out W = = Heat in Qin W QH - QC = = QH QH QC =1 QH Thus, to make any heat engine as efficient as possible it is desirable to make QH as large as possible and QC as small as possible. It would be most desirable to have QC = 0, then the engine would be 100 % efficient. Note that this would not be a violation of the first law of thermodynamics. Refrigerators and Heat Pumps Heat engines can operate in reverse. In this case, energy is injected into the engine, modelled as work W in Active Figure 4, resulting in energy being extracted from the cold reservoir and transferred to the hot reservoir. The system now operates as a heat pump, a common example being a refrigerator. Energy QC is extracted from the interior of the refrigerator and delivered as energy QH to the warmer air in the kitchen. The work is done in the compressor unit of the refrigerator. A household air conditioner is another example of a heat pump. Some homes are both heated and cooled by heat pumps. The most efficient refrigerator or air conditioner is one that removes the greatest amount of energy from the cold reservoir in exchange for the least amount of work. 21 Figure 4: Schematic diagram of a heat pump The total heat energy exhausted to the high-temperature reservoir QH is the sum of the work done on the engine plus the heat QC extracted from the low-temperature reservoir. Thus, QH W QC The coefficient of performance (equivalent of an efficiency) for a refrigerator or an air conditioner is the magnitude of the energy extracted from the cold reservoir, Q C , divided by the work W performed by the device: Heat removed COP cooling mode = Work done QC COP = W QC COP = Q H QC The Carnot Cycle As we saw, it is desirable to get the maximum possible efficiency from a heat engine. Sadi Carnot (1796-1832) showed that the maximum efficiency of any heat engine must follow a cycle consisting of the isothermal and adiabatic paths shown in the PV diagram in Figure 5, 22 and now called the Carnot cycle. The cycle begins at point A. Let us now consider each path individually. Figure 5: A P-V diagram for a Carnot cycle. Path AB: An ideal gas is first compressed isothermally along the path AB. Since AB is an isotherm, ΔT = 0 and hence ΔU = 0. The first law therefore says that Q = W along path AB. That is, the work WAB done on the gas is equal to the heat removed from the gas QC, at the low temperature, TC. Path BC: Path BC is an adiabatic compression and hence Q = 0 along this path. The first law therefore becomes ΔU = W. That is, the work WBC done on the gas during the compression is equal to the increase in the internal energy of the gas as the temperature increases from TC to TH. Path CD: Path CD is an isothermal expansion. Hence, ΔT = 0 and ΔU = 0. Therefore, the first law becomes W = Q. That is, the heat added to the gas QH at the high temperature TH is equal to the work WCD done by the expanding gas. Path DA: Path DA is an adiabatic expansion, hence Q = 0 along this path. The first law becomes ΔU = W. Thus, the energy necessary for the work WDA done by the expanding gas comes from the decrease in the internal energy of the gas. The gas decreases in temperature from TH to TC. The net effect of the Carnot cycle is that heat QH is absorbed at a high temperature TH, mechanical work W is done by the engine, and waste heat QC is exhausted to the low temperature reservoir at a temperature TC. The net work done by the Carnot engine is W = QH QC 23 The efficiency is thus given as QC =1 QH Lord Kelvin proposed that the ratio of the heat rejected to the heat absorbed could serve as a temperature scale. Kelvin then showed that for a Carnot engine QC TC = QH TH where TC and TH are the Kelvin or absolute temperatures of the gas. Therefore, we can express the efficiency of a Carnot engine as TC =1 TH Carnot’s theorem states that “No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs”. Example 7: A heat engine takes 900 J from the hot reservoir and gives off 700 J of energy to the cold reservoir per cycle. The hot reservoir is at a temperature of 350 °C and the cold reservoir is at a temperature of 20 °C. (a) Calculate the efficiency of the engine. (b) Calculate the maximum possible efficiency that can be obtained from this engine. Solution (a) QC = 700 J, QH = 900 J, η =? QC 1 QH 700 η =1 = 22.2 O O 900 (b) TC = (20 + 273) °K = 293 °K, TH = (350 + 293) °K = 623 °K, C =? TC C 1 TH 293 ηC = 1 = 53 O O 623 Reversible process: A process in which it is possible to ret urn both the system and surroundings to their original states. Irreversible process: A process in which it is impossible to ret urn bot h the system and surroundings to their original states. 24 The Second Law of Thermodynamics The second law of thermodynamics can be stated in terms of the heat engine and entropy. The Kelvin-Planck Statement of the Second Law “No process is possible whose sole result is the complete conversion of heat into work.” The Clausius Statement of the Second Law of Thermodynamics No process is possible whose sole result is the transfer of heat from a cooler to a hotter body. The Clausius statement of the second law of thermodynamics can best be described by the refrigerator. Second law of thermodynamics based on entropy The entropy of an isolated system can never decrease with time. Entropy Entropy (S) is a state function used as a measure of the order or disorder of a system. It is defined in such way that it increases with disorder. Since it is a state function, its change depends only on the entropies of the final and initial states of the system. S = Sf Si Change in entropy is equal to the amount of heat supplied or lost per unit temperature (in °K). Q S = T where ∆S > 0 for irreversible processes (the system is approaching equilibrium) ∆S < 0 for reversible processes (the system is said to be in equilibrium). Classwork 2 Derive a formula for the work done by any gas (ideal or not) which expands (a) Isobarically (b) Isothermally Assignment 2 (a) A liquid of specific heat 3.66 J g-1 K-1 is heated from 27 °C to 77 °C. If 1g of the liquid was used, what is the entropy change of this liquid. (b) (i) Show that any flow of heat between two heat reservoirs at TH and TC where TH > TC must be from hotter to cooler body (ii) If TH = 553 K, TC = 278 K and heat transferred = 100 KJ find the change in entropy. 25 (c) If a Carnot cycle is run backward we have a refrigerator. A quantity of heat QC is taken at the lower temperature TC and a quantity of heat is given out at the higher temperature TH show that T TC (i) W= H QC TC TC (ii) K= TH TC (iii) = 1 η η KINETIC THEORY OF GASES A gas is composed of many molecules in random motion. The large-scale characteristics of gases should be explainable in terms of the motion of these molecules. The analysis of a gas at this microscopic level (the molecular level) is called the kinetic theory of gases. Assumptions of the kinetic theory: 1. A gas is composed of a very large number of molecules that are in random motion. 2. The molecules behave as rigid spheres. 3. Pressure is due to collision between the molecules and the walls of the container. 4. The volume of the individual molecules is very small compared to the total volume of the gas. 5. The collisions of the molecules with the walls and other molecules are elastic and hence there is no energy lost during a collision. 6. The forces between molecules are negligible except during a collision. Hence, there is no potential energy associated with any molecule. 7. The temperature of the gas is proportional to the average kinetic energy of the molecules. 8. Finally, we assume that the molecules obey Newton’s laws of motion. Let us consider one of the very many molecules contained in the box shown in figure 6. For simplicity we assume that the box is a cube of length L. The gas molecule has a mass m and is moving at a velocity v which can be resolve into v x , v y , and v z. 26 9. Figure 6: The kinetic theory of a gas. For the moment we only consider the motion in the x-direction. The molecule has an initial momentum pi mv x (37) To the x-axis and after striking the wall of the box, it rebounds with the same velocity and its final momentum is pf mv x (38) The change in momentum ∆p is given by Δp = pf pi 2mv x (39) The time taken for the molecule to move across the cube to the oppose face and come back is given by distance travelled 2L Time t = = (40) velocity vx Hence the change in momentum per second from (39) and (40) is given by change in momentum mv 2x (41) time L Also change in momentum of a body per second is equal to the force acting on the body mv 2x Fx = (42) L Force Fx mv 2x mv 2x Pressure Px = = = (43) Area L× L2 V 27 The total pressure P on x due to impacts by all N molecules is given by P= V m 2 v1x v 2x 2 v3x 2 ............. v 2Nx (44) If v 2x represents the mean value of the squares of all the velocity components in the x- direction, 2 v1x v 2x 2 v3x 2 ............. v 2Nx v 2 x N N v2x v1x 2 v2x 2 v3x 2 ............. v 2Nx (45) Substitute (45) into (44) Nm v2x P= (46) V From the figure, v2 v2x v2y vz2. This relation will also hold for the mean squares values: v2 v2x v2y vz2 (47) But since N is large and the molecules move randomly, it follows that the mean values v2x , v2y , and vz2 are equal. Equation (47) can be written as v2 v2x v2x v2x 3 v2x v2 v 2 x (48) 3 Using (48) in (46) Nm v2 P= (49) 3V where V = volume of the cube = volume of the gas Nm = M 1 M P = v2 3 V 1 P = v2 3 3P v2 = 2 But v 2 = v rms = mean square speed 28 3P v 2rms = (50) ρ The square root of (50) is called root mean square speed given by 3P vrms = (51) ρ TEMPERATURE AND THE KINETIC THEORY From equation (49) Nm v2 P= 3V 1 PV = N A mv 2rms 3 2 1 PV = NA mv2rms (52) 3 2 From the ideal gas law for one mole of a gas, n = 1 PV = RT (53) Equating (52) and (53) 2 1 NA mv2rms = RT 3 2 1 3 R mv2rms = T 2 2 NA 1 mv 2rms = kinetic energy of translational motion per molecule 2 R = K B = Boltzmann's constant = 1.38 ×10-23 JK -1 NA 1 3 mv 2rms = K BT (54a) 2 2 1 3 N A mv 2rms = RT (54b) 2 2 Equation (54a and 54b) shows that the average kinetic energy (K.E) of translation of a molecule is proportional to the absolute temperature of the gas. From (54a), 3K BT v rms = (55a) m 29 3RT v rms = (55b) M Equation (55a and 55b) give the expression for the root mean square speed of a gas molecule. From equation (52), we can deduct some important features of the pressure from the kinetic theory of gases as: 1. Pressure which is a large-scale quantity is related to atomic quantities (i.e. average kinetic energy, average value of the square of the molecular speed, number of molecules) 2. Pressure is proportional to the number of molecules per unit volume 3. Pressure is proportional to the average translational kinetic energy of the molecules. Example 8: A cylinder contains a mixture of helium (He) and argon (Ar) gas in equilibrium at 150 °C. I. What is the average kinetic energy of each type of molecule? II. What is the rms speed of each type of gas molecule? (The molar mass of He is 4 g/mol and that of Ar is 40 g/mol, KB = 1.38 × 10-23 and R = 8.31 J mol-1 K-1). Solution 1 3 I. mv 2rms = K BT 2 2 1 3 mv 2rms = 1.38 10 23 423 2 2 ∴ average kinetic energy per molecule = 8.76 ×10-21 J 3RT II. For He: v rms = M 3 × 8.31× 423 v rms = 4 ×10-3 = 1624 m s 3RT For Ar: v rms = M 3 × 8.31× 423 v rms = 40 ×10-3 = 513 m s 30 Class work 3 1. What is the average K.E. of the oxygen and nitrogen molecules in a room at a temperature of 20 °C? 2. Find the rms speed of an oxygen and nitrogen molecule at 20 °C Assignment 3 1. Calculate the root mean square speed of the molecule of hydrogen at 273K. If the density of hydrogen at S.T.P is 9.0 × 102 kg/m3 and pressure is 1.01 × 105 N/m2 2. Calculate the root mean square of hydrogen molecules at 100 °C. 3. The pressure of a gas in a 100-mL container is 200 kPa and the average translational K.E. of each gas is 6.0 × 10-21 J. I. How many moles are in the container? II. Find the number of gas particles in the container The Maxwell distribution of velocities The Maxwell distribution of velocities describe the thermal motions of distribution of classical particles. It states that the distribution of speed is given by no of atoms with speed between u and u + du f u du = total no of atoms N For a one-dimensional distribution of speed 1 m 2 mu 2 f u = exp 2πK BT 2K BT For a two-dimensional distribution of speed m mu 2 f u = exp 2πK BT 2K BT For a three-dimensional distribution of speed 3 m 2 mu 2 f u = exp 2πK BT 2K BT Thermal velocity 1 mv 2th = K B T 2 2K B T v 2th = m 31 Hence, in terms of one-dimensional distribution 1 1 2 u2 f u = 2 exp 2 πv th v th Figure 7: Maxwell-Boltzmann velocity distribution. We can use this to find the mean square speed, we can also use it to find the velocity v f u u du 2 v 2 = v2 = f u du f(u) = distribution of particles moving in a speed no of atoms with speed between u and u + du But f u du = total no of atoms N N f u du = no of atoms with speed between u and u + du N f u u 2du v 2 = u 2 f u du N Also, average mean speed is N f u u du v = u f u du N Note: the type of distribution that is applicable to gases is Maxwellian distribution. Hence the Maxwellian distribution for one-dimension is 32 1 m 2 mu 2 f u = exp 2πK BT 2K BT 1 m 2 = v th 2K BT 1 1 2 u2 f u = 2 exp 2 v th π v th 1 u2 f u = v 2th π 2 exp 2 v th v = u f u du v = 2 u f u du 0 1 u2 v π = 2 u 2 th 2 exp 2 du 0 v th 1 u2 = 2 v 2th π 2 0 u exp 2 v th du Standard integral: In α = u n e-αu du 2 0 n 0 1 2 3 4 5 1 π 1 1 π 1 3 π 1 In α 2 α 2 4 α3 2α 2 8 α5 α3 1 Let α = v2th 1 π u exp αu du 2 2 v = 2 α 0 1 n = 1, I n α 2 2K BT ∴ v = mπ 33 Also, 1 π exp αu 2 du 2 u 2 2 v = 2 α 0 K BT v2 = m For three-dimensional gas, v2 = v2x + v2y + v 2z 1 v = v2x + v2y + v2z 2 Also, the distribution is 3 m 2 mu 2 f u = exp 2πK BT 2K BT For volume, V 2 = V 2f V dV 0 But volume in velocity space id dV = 4πV2dV dV = 4πV2dV m 4 V2 V 2 = 4π V exp 2 dV 2πK BT 0 Vth 3 n = 4, I n α 8 5 3K B T V2 = m 3K BT V2 V2 = m 1 2 3K T 2 Vrms = V = B m Also, V (mean velocity) = V f V dV 0 1 8K T 2 V= B mπ 34 1 3K T 2 Vrms = B m Most probable velocity Figure 8: number of molecules against Vmp, Vavg and Vrms 3 m 2 2 V2 f V dV = 4π V exp 2 dV 2πK BT Vth 3 m 2 2 V2 f V = V exp 2 4π 2πK BT Vth 3 m 2 2 V2 = 4π V exp 2 2πK BT Vth 3 f m 2 V2 2 2V V2 = 4π 2Vexp 2 + V 2 exp 2 V 2πK BT Vth Vth Vth 3 m 2 V2 2V3 V2 0 = 4π 2Vexp 2 exp 2 2πK BT Vth Vth Vth 2 V 2 2Vmp V2 3 2Vmp exp 2 = 2 exp 2 Vth Vth Vth 35 3 2Vmp 2Vmp = Vth2 Vth2 = Vmp 2 1 2K T 2 Vmp = Vth = B m Example 9: 1. Calculate the temperature at which the rms speed is equal to the escape speed from the surface of the earth for molecular hydrogen 2. It is found that the most probable of molecules in a gas at equilibrium temperature T2 is the T2 same as the rms speed when its equilibrium temperature T1. Find. T1 Solution 3K BT 1. v rms = m 1 Gmm H But escape velocity can be found from m H Ve2 = 2 R 2Gm Ve2 = R v rms = v e 1 1 3K BT 2 2Gm 2 = m R 2GmmH 2 6.67 1011 6.0 1020 2 103 T= 3K BR 3 1.38 1023 6.4 106 1 2K T 2 2. Vmp = B 2 m 1 3K T 2 Vrms = B 1 m 2K BT2 3K BT1 Hence, = m m T2 3 = T1 2 36 Assignment 4: (a) Sketch the graph of the Maxwell-Boltzmann velocity distribution function: 3 m 2 mv 2 g v = 4πv 2 exp , and outline some of the key features of the 2πK BT 2K BT distribution. (b) Use the Maxwell-Boltzmann velocity distribution function g(v) in (a) to find: I. The average speed, v II. The rms speed, Vrms III. The most probable speed, Vmp IV. The ratio of Vmp to Vrms Thermodynamic potential The equilibrium conditions of a system are governed by the thermodynamic potential functions. These potential functions tell us how the state of the system will vary, given specific constraints. They include: Internal energy (U), Enthalpy (H), Helmholtz energy (f), Gibb’s function (G) and chemical potential (μ). The differential forms of the potentials are exact because we are now dealing with the state of the system. Internal energy (U): This is the total internal energy of a system and can be the sum of the kinetic and potential energies of all the constituent’s parts of the system. U = K.E + P.E (56) n =1 n =1 Using this definition of internal energy and the 2nd law of thermodynamics, we can combine the two together to give us one of the central equations of thermodynamics TdS = dU + PdV dU = TdS PdV (57) Enthalpy (H): This is a state function and is defined by H = U + PV We are more interested in the change of enthalpy dH , which is a measure of the heat of system changes state. dH = dU + PdV + VdP But dQ = dU + PdV 37 ∴ dH = dQ + VdP (58) For isobaric process dH dQ = = CP dT P dT P Since dQ = CP dT dH = C P dH = C P dT dT f f dH = C dT i i P f Hf Hi = CPdT i If a system is undergoing a reversible process, dQ = TdS From equation (58) dH = TdS + VdP At constant pressure and entropy dH =T (59a) dS P dH =V (59b) dP S Helmholtz free energy (f): The Helmholtz free energy of a system is the maximum amount of work obtainable in which there is no change in temperature. This is a state function and is defined as F = U TS The change of Helmholtz free energy is given by dF = dU TdS SdT But dU = TdS PdV ∴ dF = PdV SdT (60) For a reversible isothermal process, dF = PdV f f dF = P dV i i 38 f Ff Fi = P dV (61) i ∴ The change of Helmholtz function during a reversible isothermal process is equal to the work done on the system. Again, for reversible isothermal and isochoric processes, dF = 0, F = constant Gibb’s free energy: The Gibbs free energy of a system is the maximum amount of work obtainable in which there is no change in volume. G = H TS The change of Gibbs free energy is given by dG = dH TdS SdT But dH = TdS VdP ∴ dG = VdP SdT (62) For isobaric and isothermal processes, dG = 0, G = constant So, processes such as vaporization, sublimation, melting takes place isothermally and isobarically and can be conceived as occurring reversibly. Chemical potential: This is important when the quantity of matter is not fixed (e.g. we are dealing with a changing number of atoms within a system). When this happens, we should modify our thermodynamic relations to take account of this: dU = TdS PdV dN (63a) dF = PdV SdT dN (63b) dG = VdP SdT dN (63c) This means that there are several ways of writing the chemical potential μ, U F G = (64) N S,V N V,T N T,P We can also show that the chemical potential can be written as G (65) N “The chemical potential is the Gibbs free energy per particle provided only one type of particle is present”. 39 The state functions in terms of each other We can write infinitesimal state functions for the internal energy U, the enthalpy H, the Helmholtz free energy F, and the Gibbs free energy G: dU = TdS PdV dH = TdS VdP dF = SdT PdV dG = SdT VdP By inspection of these equations, there are natural variables which govern each of the state functions. For instance, from the formula for the Helmholtz free energy we can assume its natural variables are temperature and volume and therefore we can write F F S and P T V V T And from the Gibbs free energy, assuming its natural variables are temperature and pressure, we have G G S and V T P P T This means that if we know one of the thermodynamic potential in terms of its natural variables then we can calculate the other state functions from it. Suppose we know the Gibbs free energy in terms of its natural variables T and P, then we can write U = G PV + TS G G = G P T P T T P H = G + TS G = G T T P F = G PV G = G P P T Assignment 5: The internal energy U of a certain hydrostatic system is given by: U = AP 2 V where the constant A has the units of (pressure)-1. 40 dP (a) Find the slope, of the adiabatic path in the PV plane in terms of A, P, and V. Assume dV that one also knows the thermal expansion coefficient α and the isothermal compressibility 1 V 1 V K T. KT V P T and V T P dP (b) Find the slope of an isothermal path in the PV plane. dV (c) Find the constant volume heat capacity C V in terms of the known quantities P V T Hint : 1 T V P T V P Differential relationship: The Maxwell relations Theorem: If there is a relation between x, y, and z, then z (x, y) can be written as z z dz dx dy x y y x z z M and N = x y y x M 2z y x xy N 2z x y xy M N y x x y dz Mdx Ndy This is known as the condition for an exact differential. Maxwell relation from U As stated earlier, the equation of state for dU is dU = TdS PdV And U is a function of S and V, therefore we could rewrite this as U U dU = dS dV S V V S U U ⟹ T= and P S V V S 41 P T (66) S V V S Equation (66) is the first Maxwell relation Maxwell relation from H The equation of state for dH is dH = TdS + VdP This suggests that H is a function of S and P and could therefore be written as H H dH = dS dP S P P S H H ⟹ T= and V S P P S V T (67) S P P S Equation (67) is the second Maxwell relation Maxwell relation from F The equation of state for dF is dF = SdT PdV This suggests that F is a function of T and V and could therefore be written as F F dF = dT dV T V V T F F ⟹ S = and P T V V T P S (68) T V V T Equation (68) is the third Maxwell relation Maxwell relation from G The equation of state for dG is dG = VdP SdT This suggests that G is a function of P and T and could therefore be written as G G dG = dP dT P T T P G G ⟹ V= and S P T T P 42 V S (69) T P P T Equation (69) is the fourth Maxwell relation C C Example 10: Find the expressions for P and V in terms of P, V, and T P T V T Solution By the definitions of C V and CP , Q S CP T (I) T P T P Q S CV T (II) T V T V Now, CP S T P T P T P T S T P T P T S T (III) T P T P Using a Maxwell relation CP V T P T T T P P CP 2V T 2 (IV) P T T P Similarly, CV S T V T V T V T S T V T V T S T (V) T V T V Using a Maxwell relation 43 CV P T V T T T V V C V 2P T 2 (VI) V T T V Both the expressions in equation (IV) and (VI) are zero for a perfect gas Use of the Maxwell relations 1. Maxwell equations provide relation between measurable and unmeasurable quantities. 2. The relations can be used to solve problems in thermodynamics. 3. The Maxwell relations allow you to relate changes in one set of thermodynamic variables to other variables. Application of Maxwell relations to simple systems: Stefan-Boltzmann law The Stefan-Boltzmann law, also known as Stefan's law, states that the total energy radiated per unit surface area of a black body in unit time (known variously as the black-body irradiance, energy flux density, radiant flux, or the emissive power), j*, is directly proportional to the fourth power of the black body's thermodynamic temperature T (also called absolute temperature): J* = eσT 4 The irradiance j* has dimensions of power density (energy per time per square distance), and the SI units of measure are joules per second per square meter, or equivalently, watts per square meter. The SI unit for absolute temperature T is the kelvin. ε is the emissivity of the blackbody; if it is a perfect blackbody, ε = 1. The constant of proportionality σ, called the Stefan-Boltzmann constant or Stefan's constant, is non-fundamental in the sense that it derives from other known constants of nature. The value of the constant is 2π 5 K B4 σ= 2 3 = 5.67 ×10-8 Js -1m -2 K -4 15c h Thermodynamic derivation of the Stefan-Boltzmann law The fact that the energy density of the box containing radiation is proportional to T 4 can be derived using thermodynamics. The perfect gas theory tells us that the pressure of a gas can be expressed in terms of the mean velocity c 44 1 P = ρc 2 3 It follows from classical electrodynamics that the radiation pressure P is related to the internal energy density: u P= 3 From the first law of thermodynamics, dQ = dU PdV dQ = Amonut of radiation per unit time per unit area U = Total energy of the radiation of enclosure P = Pressure of radiation V = Volume of the enclosure 1 dQ = d uV udV 3 U Where energy density u is described as the total energy stored U per unit volume V u = V 1 dQ = udV + Vdu udV 3 4 = udV + Vdu 3 From the second law of thermodynamics, dQ = TdS 1 dS = dQ T 41 1 = udV + Vdu 3T T S S But, dS du + dV u V V u S V S 4 u and u T V 3 T 2S 2S uV Vu S S V u u V 45 V 4u V T u 3T 1 4 1 u dT because T is a function of u T 3 T T 2 du 1 41 4 u dT T 3 T 3 T2 du 11 4 u dT 3 T 3 T2 du u dT 1 4 T du du dT 4 u T Inu = 4InT + C u = CT 4 (70) Equation (70) implies that u α T 4 THIRD LAW OF THERMODYNAMICS The third law can be stated in various ways as: 1. Nernst: Near absolute zero, all reactions in a system in internal equilibrium take place with no change in entropy. 2. Planck: The entropy of all systems in internal equilibrium is the same at absolute zero, and may be taken to be zero. 3. Simon: The contribution to the entropy of a system by each aspect of the system which is in internal thermodynamic equilibrium tends to zero as T 0. Consequences of the third law 1. Heat capacities tends to zero as T 0 S S C T 0 T InT Because as T 0 , InT and S 0 , hence C 0. 2. Thermal expansion stops 0 as T Since S 0 , we have for example that S 0 P T As T 0 , but by a Maxwell relation, this implies that 46 1 V 0 V T P And hence the isobaric expansivity βP 0 3. No gases remain ideal T 0 As T 0 , both CP and CV tends to zero. 4. Curie’s law breaks down Curie’s law states that the susceptibility χ is proportional to 1 T and hence χ as T 0. 5. Unattainability of absolute zero (T = 0): It is impossible to cool to T = 0 in a finite number of steps. Real gases Most of the real gases obey Boyle’s and Charle’s law at low pressures and temperatures. But the actual behavior of real gases at elevated pressures, at low temperatures and small volume deviates considerably. For example, if you get a real gas cold enough it will liquefy and this is something that the ideal gas equation does not predict or describe. This section deals with how this additional element of real behavior can be modelled, by introducing various extensions to the ideal gas model, including those introduced by Van der Waals, dieterici equation and viral expansion. The Van der Waals gas The most commonly used model of the real gas behavior is Van der Waals gas. This is the simplest real gas model which includes (i) Intermolecular interactions (gas molecules weakly attract one another) (ii) The non-zero size of molecules (gas molecules don’t have freedom to move around in all the volume of the container, because some of the volume is occupied by the other gas molecules). The equation of state for a Van der Waals gas is a P + 2 Vm b = RT (71) Vm In this equation, the constant “a” parameterizes the strength of the intermolecular interactions, while the constant “b” accounts for the volume excluded owing to the finite size of molecules and a V 2 is the force of attraction. 47 If “a” and “b” are both set to zero, we recover the equation of state for an ideal gas. Unique feature Van der Waals equation qualitatively accounts for the heating effect observed at ordinary temperatures. Limitations The values of a and b (which are assumed to be constant) are found to vary with temperature. Thus, the results obtained from the equation are incorrect when the variation of a and b is large with respect to temperature. The equation is not accurate enough in the critical region. Example 11: Find the temperature TC , pressure PC and volume VC at the critical point of a Van der Waals gas, and calculate the ratio PC VC RTC. Solution a P + 2 Vm b = RT (I) Vm RT a P 2 (II) Vb V where Vm = V The point of inflexion can be found by using P RT 2a 3 0 (III) V T V b V 2 2P 2RT 6a 2 4 0 (IV) V T V b 3 V From (III), 2a V b 2 RT (V) V3 From (IV), 3a V b 3 RT (VI) V4 Equating (V) and (VI), we have 3 V b 2 (VII) V 48 This implies that at V = VC, where VC is the critical volume given by VC = 3b (VIII) Substituting (VIII) into (V), 8a RT = (IX) 27b Hence, T = TC where TC is the critical temperature given by 8a TC = (X) 27Rb Substituting the expressions for VC and TC back into the equation of state for a Van der Waals gas gives the critical pressure PC as a PC = (XI) 27b 2 PC VC 3 = = 0.375 (XII) RTC 8 Independent of both “a” and “b”. At the critical point, P 0 V T Change in internal energy when a gas obeys Van der Waals equation We know that dU = TdS PdV P dU = C V dT T P dV T V a Van der Waals equation is P + 2 V b = RT V RT a P 2 Vb V P R T V V b R RT a Therefore, dU = C V dT T 2 dV Vb Vb V a dU = CV dT 2 dV V 2 2 2 dV dU = CV dT a 1 1 1 V2 49 1 1 U 2 U1 = CV T2 T1 a (72) V1 V2 Change in entropy when a gas obeys Van der Waals equation TdS = dU PdV P TdS = CV dT T dV T V dT P dS = CV dV T T V a Van der Waals equation is P + 2 V b = RT V RT a P 2 Vb V P R T V V b dT R Therefore, dS = CV dV T Vb 2 2 2 dT dV dS = CV 1 1 T R 1 Vb T V b S2 S1 = CV In 2 RIn 2 (73) T1 V1 b The Dieterici equation This modification suggested in 1899 by C. Dieterici attempts to account for the fact that molecules at the wall of a containing vessel have higher potential energy than molecules in the bulk gas. The following equation was proposed. a P Vm b = RT exp 2 (74) RTV The constant “a” is a parameter which controls the strength of attractive interactions. The critical point can be identified for this model by evaluating 2 P P 2 0 V T V T And after little algebra, it yields a a TC = , PC = 2 2 , VC = 2b 4Rb 4e b 50 PC VC 2 = 2 = 0.271 RTC e Viral expansion Another method to model real gases is to take the ideal gas equation and modify it using a power series in 1 Vm (where Vm is the molar volume). This leads to the following viral expansion PVm B C = 1 2 (75) RT Vm Vm Where B, C, etc are called viral coefficients. Phase changes A change of phase of a substance occurs when the system changes from one distinct state to another. This phase change can be caused by many different factors e.g. temperature changes can cause a phase change between solid and liquid, applied magnetic fields can cause a phase change between superconductor and normal conductor. First order phase change A first order phase change of a substance is characterised by a change in the specific volume between the two phases, accompanied by a latent heat. Some typical examples of first order phase changes are The transition of a solid melting into a liquid The transition of a liquid boiling into a gas The change from superconductor to normal conductor, provided the change occurs in an applied magnetic field. Second order phase change Second order phase change of a substance is characterised by no change in the specific volume between the two phases and no accompanying latent heat. Some typical examples of a second order phase changes are: The transition from ferro-magnet to paramagnet at the Curie temperature. The transition from superconductor to normal conductor, provided there is no applied magnetic field. The change from normal liquid 4 He to superfluid 4 He below 2.2 K. 51 The Clausius-Clapeyron equation for first order phase changes When a phase change occurs, we are mostly interested in how it affects the Gibbs free energy. At thermodynamic equilibrium, the Gibbs function is at a minimum so at the transition line on the PT diagram the specific Gibbs energy is the same for both sides. The line of coexistence of the two phases is determined by the equation μ1 P, T μ 2 P, T If we move along this phase boundary, we must also have μ1 P + dP, T + dT μ 2 P + dP, T + dT So that when we change P to P + dP and T to T + dT we must have dμ1 dμ 2 G Using, dG SdT + VdP and μ N S1dT + V1dP S2dT + V2dP Where S1 and S2 are the entropy per particle in phase 1 and 2, while V1 and V2 are the volume per particle in phase 1 and 2. Rearranging this equation gives dP S2 S1 dT V2 V1 If we define the latent heat per particle as L = TΔS , then we have dP L (76) dT T V2 V1 This equation is known as Clausius-Clapeyron equation. It shows that the gradient of the phase boundary of the P-T plane is purely determined by the latent heat, the temperature at the phase boundary and the difference in volume between the two phases. Example 12: Derive an equation for the phase boundary of the liquid and gas under the assumptions that the latent heat L is temperature independent, that the vapour can be treated as an ideal gas, and that Vvapour V Vliquid. Solution dP L dT T V2 V1 If Vvapour V Vliquid and that PV = RT for one mole, then the Clausius-Clapeyron equation becomes 52 dP LP dT RT 2 Rearranging we have dP LdT P RT 2 dP L dT P R T2 L InP constant RT Hence the equation of the phase boundary is L P P0 exp RT L Where the exponential looks like a Boltzmann factor e βl with l , since R N A K B NA The Ehrenfest equation for second order phase changes In a second order phase change there is no latent heat, and so no entropy change and no volume change. Let us consider the first Ehrenfest equation for second order phase change whereby there is no change in entropy (entropy is constant). Si = Sf dSi = dSf If we consider the entropy as a function of temperature and pressure S = S T, P , then its differential is S S dS dT + dP T P P T 1 S S T dT + dP T T P P T S S V But T = CP and Maxwell relation T P P T T P V CP dT dP 1 dS T T P 1 V CP dT V dP 1 T V T P 53 1 C P dT VβdP T 1 V where β Isobaric volume expansivity V T P From dSi = dSf 1 1 CP i dT Vβi dP = CP f dT Vβf dP T T dT CP f CP i V β f βi dP T dP CP f CP i (77) dT TV β f βi This is the first Ehrenfest equation for second order phase change, the second Ehrenfest equation can be derived by considering the continuity of the volume (volume is constant) of the two phases. Vi = Vf dVi = dVf If we consider the volume as a function of temperature and pressure V = V T, P , then its differential is V V dV dT + dP T P P T 1 V 1 V V dT + V dP V T P V P T Where, 1 V β = Isobaric volume expansivity V T P 1 V K T = Isothermal compressibility V P T dV VβdT VK T dP From dVi = dVf Vβi dT - V K T i dP = Vβ f dT - V K T f dP V K T f K T i dP = V βf βi dT 54 dP = β f βi (78) dT K T f K T i Example 13: Evaluate the temperature dependence of the latent heat along the phase boundary in a liquid gas transition and hence deduce the equation of the phase boundary including this temperature dependence Solution P y ar nd ou (dP/dT)dT eb as Ph dT T From the figure above, the gradient in the temperature can be written as d P = + dT T P T P T S = SV SL L = T Where V and L refer to vapour and liquid respectively, d L ΔS dP ΔS = + dT T T P dT P T CPV - CPL SV SL dP = T P T P T dT CPV CPL dP = VV VL T T dT Using VV VL and PV = RT, we have d L CPV CPL R LP = dT T T P RT 2 Expanding 55 d L 1 dL L = dT T T dT T 2 dL = CPV CPL dT L = L0 CPV CPL T Thus, the latent heat contains a linear temperature dependence. The negative slope is because CPV CPL. Substituting the value of L in the Clausius-Clapeyron equation yields the equation of the phase boundary L C CPL InT P = P0 exp 0 PV RT R Assignment 6: Find the entropy of one mole of an ideal gas. 56
