Chemistry-Thermodynamics Study Notes Whole semester (54 pages), Study notes of Thermodynamics

Download Chemistry-Thermodynamics Study Notes Whole semester (54 pages) and more Study notes Thermodynamics in PDF only on Docsity! Introduction and Basic Concepts | 01 INTRODUCTION Thermodynamics is the study of (1) all energy in all its form, (2) conversion of energy from one form to another, (3) transfer of energy, and (4) its effect on the properties. LAWS OF THERMODYNAMICS Zeroth Law of Thermodynamics states that if a body A is in equilibrium with body B, then they have the same temperature. First Law of Thermodynamics states the conservation of energy and introduces the concept of internal energy. Second Law of Thermodynamics dictates the limits on the conversion of heat into work and provides the yard stick to measure the performance of various processes. It also tells whether a particular process is feasible or not and specifies the direction in which a process will proceed. Consequently, it also introduces the concept of entropy. Third Law of Thermodynamics defines the absolute zero of entropy Three measures of amount of size are used are the mass (m), number of moles (n) and total volume (Vt). The number of moles can be calculated by dividing the mass with its molecular or atomic weight (M) which has a unit of amu or g/mol. We must note that the mole used in thermodynamics and other subjects refers to the gram- mole of Chemical Engineering and Industrial Process Calculations. Total volume, representing the size of a system, is a defined quantity given ass the product of three lengths. It has the following relations to the specific volume or molar volume. 𝑉𝑡 ≡ 𝑚𝑉𝑚 𝑉𝑡 ≡ 𝑛𝑉𝑠 Volume along with density are independent of the size of a system and are examples of intensive thermodynamic variables. Measures of Amount of Size Introduction and Basic Concepts | 02 Force The SI unit for force is Newton (N), which is derived from Newton’s second law which states that the acceleration of any particular mass is directly resultant to the resultant force and inversely proportional to its mass. To translate these qualities mathematically a constant gc is introduced. 𝐹 = 𝑚𝑎 𝑔𝑐 Wherein gc = 1 (kg-m/s2)/N = 32.2 (lbm-ft/s2)/lbf Temperature It is defined as degree or intensity of heat present in a substance or object expressed accordingly to a comparative scale. There are four temperature scales present with the most common being in Celsius (OC) and the SI unit being in Kelvin (K). TEMPERATURE CONVERSION ℃ = 5 9 (℉ − 32) ℉ = 9 5 (℃+ 32) 𝐾 = ℃+ 273.15 Fahrenheit to Celsius Celsius to Fahrenheit Celsius to Kelvin R = ℉+ 459.67 Fahrenheit to Rankine Pressure The pressure (P) exerted of a fluid on a surface is the normal force exerted by the fluid per unit area of the surface. 𝑃 = 𝐹 𝐴 with a unit of 𝑁 𝑚2 or 𝑘𝑔 𝑚∗𝑠2 or Pa Introduction and Basic Concepts | 05 Gen. Eq. (WORK) 𝑑𝑊 = −𝑃𝑜𝑝𝑝𝑜𝑠𝑖𝑛𝑔𝑑𝑉 We can consider this equation for both reversible and irreversible processes. A reversible process where the direction can be reversed at any point by an infinitesimal change in external conditions while an irreversible process cannot. While work for irreversible process follow the equation above, the reversible process of gases can follow: 𝑑𝑊 = −𝑃𝑔𝑎𝑠𝑑𝑉 Heat is the energy driven by differences in temperature. Similar to work, it is energy in transit denoted by symbol (Q). Heat and Work, by convention, is positive if it is absorbed done on the system. It is negative if it is released or done by the system. The mathematical statement for heat states that it is directly proportional to the change in temperature having a constant known as heat capacity. 𝑄 = 𝑛𝐶∆𝑇 𝑜𝑟 𝑄 = 𝑚𝐶∆𝑇 When integrated, this equation yields the work of a finite process. By convention, work is regarded positive when the displacement is in the same direction as the applied force and negative if it goes the opposite direction. The work which accompanies a change in the volume of a fluid can be expressed as: HEAT> System and Surroundings System is the region of space being studied while surroundings are the rest of the universe Volumetric Properties of Pure Substances | 06 VOLUMETRIC PROPERTIES OF PURE SUBSTANCES PVT Relations of Pure Substances In single-phase regions, the relationship of pressure, volume and temperature can be described analytically as f(P,V,T) = 0 or also known as the PVT Equation of state. An equation of state may be solved for any one of the three quantities P,V or T as a function of the other two. We can set these functions as: 𝑑𝑃 = ቤ 𝜕𝑃 𝜕𝑉 𝑇 𝑑𝑉 + ቤ 𝜕𝑃 𝜕𝑇 𝑉 𝑑𝑇 𝑑𝑇 = ቤ 𝜕𝑇 𝜕𝑉 𝑃 𝑑𝑉 + ቤ 𝜕𝑇 𝜕𝑃 𝑉 𝑑𝑃 𝑑𝑉 = ቤ 𝜕𝑉 𝜕𝑃 𝑇 𝑑𝑃 + ቤ 𝜕𝑉 𝜕𝑇 𝑃 𝑑𝑇 The partial derivates in the last equation have definite physical meanings, and are related to two properties namely, volume expansivity and isothermal compressibility, which are commonly tabulated for liquids. 𝑉𝑜𝑙𝑢𝑚𝑒 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑣𝑖𝑡𝑦 𝛽 ≡ 1 𝑉 ቤ 𝜕𝑉 𝜕𝑇 𝑃 𝐼𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝜅 ≡ − 1 𝑉 ቤ 𝜕𝑉 𝜕𝑃 𝑇 When substituting these properties in the equation, it would result to: 𝑑𝑉 𝑉 = 𝛽𝑑𝑇 − 𝜅𝑑𝑃 Volumetric Properties of Pure Substances | 07 Ideal Gas Laws And evaluating the equation, we can calculate changes in properties for liquids as: For Adiabatic Processes Where: 𝑙𝑛 𝑉2 𝑉1 = 𝛽 𝑇2 − 𝑇1 − 𝜅 𝑃2 − 𝑃1 For small changes in T and P, little error is introduced if they are assumed constant. GAS LAW FORMULAS 𝑃1𝑉1 = 𝑃2𝑉2 𝑉1 𝑇1 = 𝑉2 𝑇2 Boyle’s Law (Isothermal) Charles’ Law (Isobaric) 𝑃1 𝑇1 = 𝑃2 𝑇2 Gay-Lussac’s Law (Isochoric) 𝑃𝑉 = 𝑛𝑅𝑇 Ideal Gas Law 𝑇1𝑉 𝛾−1 1 = 𝑇2𝑉 𝛾−1 2 𝑇1𝑃 1−𝛾 𝛾 1 = 𝑇2𝑃 1−𝛾 𝛾 2 𝑃1𝑉 𝛾 1 = 𝑃2𝑉 𝛾 2 𝐶𝑃 = 𝐶𝑉 + 𝑅 𝛾 ≡ 𝐶𝑃 𝐶𝑉 𝐶𝑃 = 5 2 𝑅 𝐶𝑃 = 7 2 𝑅 Monatomic Ideal Gas Diatomic Ideal Gas First Law of Thermodynamics | 10 Internal Energy (U) is the energy of a substance due to the configuration and activity of its molecules, atoms and subatomic units. It is the result of the chaotic motion of gas molecules, rotational motion of molecules or group of atoms which are free to move within a molecule, internal vibration of atoms within the molecule, and motion of electrons within the atom. Internal Energy Enthalpy (H) is a thermodynamic property which describes a mathematical expression of: Enthalpy 𝐻 ≡ 𝑈 + 𝑃𝑉 𝑑𝐻 = 𝑑𝑈 + 𝑑(𝑃𝑉) Any body has a capacity to contain heat mathematically described as: Heat Capacity 𝐶 ≡ 𝑑 Τ𝑄 𝑑 𝑇 At constant volume, it refers to the energy required to raise the temperature of a unit quantity of a substance by one degree as the volume is maintained constant. 𝐶 = ( 𝜕𝑈 𝜕𝑇 )𝑉 𝑄 = 𝑛∆𝑈 = 𝑛𝐶𝑉∆𝑇 At constant pressure, it refers to the energy required to raise the temperature of a unit quantity of a substance by one degree as the pressure is maintained constant. 𝐶 = ( 𝜕𝐻 𝜕𝑇 )𝑃 𝑄 = 𝑛∆𝐻 = 𝑛𝐶𝑃∆𝑇 For solid and liquids, Cp = Cv while for ideal gases, Cp = Cv + R First Law of Thermodynamics | 11 Energy Balances Closed Systems Following the law of conservation energy wherein energy can neither be created nor destroyed during a process, only be transformed. Thus, the energy closed in the system and the energy entering the system is equal to the energy stored in a system and energy exiting the system. The energy balance will therefore become: Gen. Eq. (ENERGY BALANCES) 𝑑𝑈 + 𝑑𝐾𝐸 + 𝑑𝑃𝐸 ≡ 𝑑𝑄 + 𝑑𝑊 For closed systems, they undergo processes that cause no changes in external potential energy and external kinetic energy. For such process, there is a flow of energy, but no flow of mass across the boundary of the system thus the energy balance equation becomes: ∆𝑈 = 𝑄 +𝑊 Isolated Systems For isolated systems, there is no exchange of energy between the system and surroundings and only experiences internal thermodynamic equilibrium. Thus: ∆𝑈 = 0 ∆𝐾𝐸 = 0 ∆𝑃𝐸 = 0 𝑄 = 0 𝑊 = 0 Open Systems The laws of mass and energy conservation apply to all processes, to open as well as to closed systems. There are processes in which open systems includes closed systems as special cases. An open system is where both mass and energy have an exchange with both the system and the surrounding. Usually, these types of systems are considered in steady-state. Steady state conditions are characterized by: 1. There is no accumulation or depletion of mass and energy within the system over the time considered. 2. The mass flow rate is the same at all points along the path of flow of fluid 3. The properties of the fluid may vary from point to point of the system but the properties at a given point are constant with time First Law of Thermodynamics | 12 Wherein the following nomenclature are: ሶ𝑚 = mass flow rate (kg/s) q = volumetric flow rate (m3/s) u = velocity (m/s) A = Cross-sectional area perpendicular to the flow (m2) ρ= density (kg/m3) V = Specific Volume (m3/kg) G = Mass velocity (kg/s-m2) P = Pressure T = Temperature U = Internal Energy (kJ/kg) h = Elevation from a reference plane Q = Heat (kJ/kg) Ws = Shaft Work (kJ/kg) H = Enthalpy (kJ/kg) KE = Kinetic Energy (kJ/Kg) PE = Potential Energy (kJ/kg) Since it follows the first law of thermodynamics, the derived energy balance equation will still be ∆𝑈 + ∆𝐾𝐸 + ∆𝑃𝐸 = 𝑄 +𝑊 The work considered in this type of system consists of two types of work considerably: 1. Flow Work or Energy from Pressure – The entering fluid carries flow energy (P1V1) as a result of being pushed into the system by the fluid immediately behind it. Similarly, the fluid on passing the second point does work (P2V2) by pushing the fluid just ahead of it. Therefore, a net flow work done by the system would be P2V2 – P1V1 or delta PV. Second Law of Thermodynamics | 15 SECOND LAW OF THERMODYNAMICS The Second Law of Thermodynamic States that: 1. The spontaneous flow of heat is unidirectional from higher temperature to the lower temperature 2. All naturally occurring processes always tend to change spontaneously in a direction that will lead to equilibrium 3. Not all of the heat absorbed by a system can be converted into work without leaving permanent changes Heat Engine An engine is a device or machine that produces work from heat in a cyclic process. Essential to all heat engine cycles are absorption of heat from a hot reservoir at a high temperature, production of work and rejection of heat to a cold reservoir at a lower temperature. Gen. Eq. (HEAT ENGINE) 𝑄𝐻 = 𝑄𝐶 + 𝑊 THERMAL EFFICIENCY η = 𝑊 𝑄𝐻 = 𝑄𝐻 − 𝑄𝐶 𝑄𝐻 = 1 − 𝑄𝐶 𝑄𝐻 Second Law of Thermodynamics | 16 A Carnot Engine is an ideal engine whose theory states that for two given heat reservoirs, no engine can have a higher thermal efficiency than that of a Carnot engine. To consider a Carnot Engine Cycle: 1. A system is initially in thermal equilibrium at Tc with a cold reservoir undergoes a reversible adiabatic compression that causes its temperature to rise that of a hot reservoir, TH (Path 4 – 1). 2. The system maintains its contact with the hot reservoir at TH and undergoes a reversible isothermal expansion during which heat QH is absorbed from the hot reservoir (Path 1 – 2). 3. The system undergoes a reversible adiabatic expansion opposite direction of step 1 that brings its temperature back to that of the cold reservoir at Tc (Path 2 – 3). 4. The systems maintains contact with the reservoir at Tc and undergoes a reversible isothermal compression that returns to the initial state with rejection of heat QC to the cold reservoir (Path 3 – 4). For any Carnot Engine, the ratio between the heat at the reservoirs is equal to the ratio of their respective temperatures. Carnot Engine 𝑄𝐻 𝑇𝐻 = 𝑄𝐶 𝑇𝐶 Using this theorem, we can calculate the maxim efficiency an engine can give being. MAXIMUM THERMAL EFFICIENCY η𝑚𝑎𝑥 = 𝑊 𝑇𝐻 = 𝑇𝐻 − 𝑇𝐶 𝑇𝐻 = 1 − 𝑇𝐶 𝑇𝐻 Second Law of Thermodynamics | 17 Refrigerators A refrigerator is also an application of the second law. It absorbs the heat at a low temperature and through additional work (such as electricity) rejects the heat at a higher temperature. The system is referred to as the refrigerant. While the flow diagrams indicate a reverse flow of direction, the equation for the heat engine remains the same as other heat engines. 𝑄𝐻 𝑇𝐻 = 𝑄𝐶 𝑇𝐶 Entropy Entropy (S) is a state function which allows us to predict the natural direction of a process. It is regarded as a measure of the order and disorder of particles in a system. The larger the space, the more possible arrangements particles can exist within a system. Recall the relation derived in the Carnot Engine, and remove the absolute value signs we get the respective equation. 𝑄𝐻 𝑇𝐻 = −𝑄𝐶 𝑇𝐶 Putting them on the same side, we have an equation that describes entropy. Similar to Internal Energy and Enthalpy, within a system, the change will be zero. 𝑄𝐻 𝑇𝐻 + 𝑄𝐶 𝑇𝐶 = 0 = ∆𝑆 Second Law of Thermodynamics | 20 ሶ𝑆𝑇 + ሶ𝑆𝐺 = 𝑑𝑆𝐶𝑉 𝑑𝑡 Entropy can be transported across the control surface by means of heat flow. If heat flows at the rate dotQj across a portion of the control surface at a temperature T csj, the resulting rate of transport is dotQj/ Tcsj. The sum would be: 𝛴 ሶ𝑄𝑗 𝑇𝑐𝑠,𝑗 Entropy can also be transported by flowing streams. Each stream entering or leaving the control volume carries with it, entropy for which the transport rate is dotms. The net rate of transport into control value would then be: ሶ𝑚𝑆𝑖𝑛 - ሶ𝑚𝑆𝑜𝑢𝑡 The entropy balance can be then written as: T in the equation is the temperature of the surroundings. If dotSG is equal to zero, then it is a reversible process. If it is greater than zero, then the process is irreversible Entropy Balance (OPEN) ෍ ሶ𝑄𝑗 𝑇𝑐𝑠,𝑗 − 𝛥( ሶ𝑚𝑆)𝑓𝑠 + ሶ𝑆𝐺 = 𝑑𝑆𝐶𝑉 𝑑𝑡 For a steady-state flow process, the rate of change of entropy is zero since control volume has constant mass and entropy. The equation becomes ሶ𝑆𝐺 = 𝛥( ሶ𝑚𝑆)𝑓𝑠 − 𝛴 ሶ𝑄𝑗 𝑇ℴ,𝑗 ≥ 0 Second Law of Thermodynamics | 21 Calculation of Ideal Work Note than in a completely reversible process in a steady state flow, dotSG is equal to zero. By rearranging terms, we can derive an equation in terms of heat. ሶ𝑆𝐺 = 𝛥( ሶ𝑚𝑆)𝑓𝑠 − 𝛴 ሶ𝑄𝑗 𝑇ℴ,𝑗 = 0 ሶ𝑄𝑗 = 𝑇ℴ𝛥( ሶ𝑚𝑆)𝑓𝑠 We can then replace this value of Q in the general open system balance equation so that Entropy would become a variable of the equation. We also know that for any reversible process, the amount of work produced is the maximum, so what we can calculate is the ideal work. 𝛥[ ሶ𝑚(𝐻 + 𝐾𝐸 + 𝑃𝐸)]𝑓𝑠 = 𝑇𝜎𝛥( ሶ𝑚𝑆)𝑓𝑠 + ሶ𝑊𝑠𝑟𝑒𝑣 ሶ𝑊𝑠𝑟𝑒𝑣 = ሶ𝑊𝑠−𝑖𝑑𝑒𝑎𝑙 = 𝛥[ ሶ𝑚(𝐻 + 𝐾𝐸 + 𝑃𝐸)]𝑓𝑠 − 𝑇𝜎𝛥( ሶ𝑚𝑆)𝑓𝑠 Recall the thermodynamic efficiency of a system. We calculate for the efficiency differently is it is either required or produced. 𝜂 𝑤𝑜𝑟𝑘 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = ሶ𝑊𝑖𝑑𝑒𝑎𝑙 ሶ𝑊𝑆 𝜂 𝑤𝑜𝑟𝑘 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 = ሶ𝑊𝑆 ሶ𝑊𝑖𝑑𝑒𝑎𝑙 Lost work is work wasted as the result of irreversibility in a process. Lost work can be found in either work-producing or work-requiring processes. 𝑊𝐿𝑜𝑠𝑡 = 𝑊𝑖𝑑𝑒𝑎𝑙 −𝑊𝑎𝑐𝑡𝑢𝑎𝑙 𝑊𝐿𝑜𝑠𝑡 = 𝑊𝑎𝑐𝑡𝑢𝑎𝑙 −𝑊𝑖𝑑𝑒𝑎𝑙 Work-Producing Work-Requiring Lost Work Second Law of Thermodynamics | 22 For actual work, or an irreversible process, entropy cannot be equated to zero ሶ𝑆𝐺 = 𝛥( ሶ𝑚𝑆)𝑓𝑠 − 𝛴 ሶ𝑄𝑗 𝑇ℴ,𝑗 ሶ𝑄 = 𝑇ℴ𝛥( ሶ𝑚𝑆)𝑓𝑠 − 𝑇ℴ ሶ𝑆𝐺 We can then substitute this value for Q in the open system balance equation for an irreversible process using the general equation. For both irreversible and reversible open system balances, both contain changes in enthalpy, kinetic and potential energy multiplied to the mass rate. We can use this to substitute one equation to the other. 𝛥[ ሶ𝑚(𝐻 + 𝐾𝐸 + 𝑃𝐸)]𝑓𝑠 =𝑇ℴ𝛥( ሶ𝑚𝑆)𝑓𝑠 − 𝑇ℴ ሶ𝑆𝐺 + ሶ𝑊𝑠−𝑎𝑐𝑡𝑢𝑎𝑙 ሶ𝑊𝑠−𝑖𝑑𝑒𝑎𝑙 + 𝑇𝜎𝛥( ሶ𝑚𝑆)𝑓𝑠 = 𝛥[ ሶ𝑚(𝐻 + 𝐾𝐸 + 𝑃𝐸)]𝑓𝑠 ሶ𝑊𝑠−𝑖𝑑𝑒𝑎𝑙 + 𝑇𝜎𝛥( ሶ𝑚𝑆)𝑓𝑠 = 𝑇ℴ𝛥( ሶ𝑚𝑆)𝑓𝑠 − 𝑇ℴ ሶ𝑆𝐺 + ሶ𝑊𝑠−𝑎𝑐𝑡𝑢𝑎𝑙 We now have two terms that can cancel each other out. We also know that in a work requiring process, the work lost is equal to the actual minus ideal. Therefore, we have an equation of work lost in terms of entropy. 𝑇ℴ ሶ𝑆𝐺 = ሶ𝑊𝑠−𝑎𝑐𝑡𝑢𝑎𝑙 − ሶ𝑊𝑠−𝑖𝑑𝑒𝑎𝑙 ሶ𝑊𝑙𝑜𝑠𝑡 = 𝑇ℴ ሶ𝑆𝐺 The greater the irreversibility of a process, the greater the rate of entropy generation and the greater the amount of energy that becomes unavailable for work. Thus, every irreversibility carries with it a price. Production of Power from Heat | 25 The Rankine cycle addresses the problems of the Carnot cycle by superheating the steam. Rankine Cycle 3 - 4 (Condenser) - A constant pressure and constant temperature process produces saturated liquid 4 - 1 (Pump) 1 - A (Boiler) - Reversible, adiabatic, isentropic pumping of saturated liquid to the pressure of the boiler producing subcooled liquid A - B (Boiler) - Heating of subcooled liquid to its saturation temperature (Sensible heat) - Vaporization at constant temperature and pressure (Latent Heat) B- 2 (Boiler) 2 – 3(Turbine) - Superheating of steam way above saturation temperature (Sensible Heat) - Reversible, adiabatic, isentropic expansion of vapor from boiler pressure to condenser pressure. Work for Pumps 𝑊𝑠 = ∆𝐻 = 𝑉∆𝑃 Only applicable at constant Entropy Refrigeration and Liquefaction | 26 REFRIGERATION AND LIQUEFACTION Heat absorbed at a low temperature is continuously rejected to the surroundings at a higher temperature, or essentially, a reversed heat-engine cycle. Vapor-Compression Refrigeration Cycle The path taken by the system is the reverse of an engine, and usually thermodynamic graphs can be used to get the properties of the system at different points. Taking the energy balance equations at the four different components: 𝑂𝐸𝐵 𝑖𝑛 𝑡ℎ𝑒 𝐸𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑜𝑟: 𝑂𝐸𝐵 𝑖𝑛 𝑡ℎ𝑒 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟: 𝑑𝐻 + 𝑑𝐾𝐸 + 𝑑𝑃𝐸 = 𝑄 + 𝑊𝑠 𝑑𝐻 + 𝑑𝐾𝐸 + 𝑑𝑃𝐸 = 𝑄 + 𝑊𝑠 𝑑𝐻 + (0) + (0) = 𝑄 + (0) 𝑑𝐻 + (0) + (0) = (0) + 𝑊𝑠 𝑄𝐶 = 𝐻2 −𝐻1 𝑊𝑠 = 𝐻3 − 𝐻2 𝑂𝐸𝐵 𝑖𝑛 𝑡ℎ𝑒 𝐶𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟: 𝑂𝐸𝐵 𝑖𝑛 𝑡ℎ𝑒 𝑇ℎ𝑟𝑜𝑡𝑡𝑙𝑒 𝑉𝑎𝑙𝑣𝑒: 𝑑𝐻 + 𝑑𝐾𝐸 + 𝑑𝑃𝐸 = 𝑄 + 𝑊𝑠 𝑑𝐻 + 𝑑𝐾𝐸 + 𝑑𝑃𝐸 = 𝑄 + 𝑊𝑠 𝑑𝐻 + (0) + (0) = 𝑄 + (0) 𝑑𝐻 + (0) + (0) = (0) + (0) 𝑄𝐻 = 𝐻4 −𝐻3 𝐻1 − 𝐻4 = 0 Usually H2 is a saturated vapor, and H4 is a saturated liquid Refrigeration and Liquefaction | 27 The measure of effectiveness of a refrigerator is its coefficient of performance which is the heat absorbed at the lower temperature divided by the net work Absorption-Refrigeration Unit Coefficient of Performance 𝜔 = 𝑄𝐶 𝑊 For Carnot Refrigerator 𝜔 = 𝑇𝐶 𝑇𝐻 − 𝑇𝐶 Other Types of Refrigeration Units Two-stage Cascade Refrigeration Liquefaction Liquefaction is the process of changing a gas (natural state) to liquid. Condensation on the other hand changes a vapor (natural state is liquid) back to its liquid state. Liquefaction involves three main processes 1. Heat exchange/Cooling (Constant P) (A to 1) 2. Expansion process from which work is obtained (A to 2) 3. Throttling process (A to 3) (A to B ; B to A’ ; A’ to 3’) Vapor Liquid Equilibrium Introduction | 30 In more general context, consider a volatile substance i contained in a gas-liquid system in equilibrium at temperature T and pressure P. Raoult’s Law states that the partial pressure of substance i in the gas phase is equal to the product of mole fraction of i (xi) in the liquid phase and the vapor pressure of pure liquid𝑷𝒊 ∘ at temperature T. The mathematical statement of Raoult’s Law is: 𝑷𝒊 = 𝒙𝒊𝑷𝒊 ∘ - a relation of partial pressure of one component in the vapor phase to the mole fraction of the same component in the liquid phase. Consider a binary mixture with components A and B: Note: x - mole fraction in the liquid phase y = mole fraction in the vapor phase xA + xB = 1 yA + yB = 1 Dalton’s Law of Partial Pressures: 𝑃 = 𝑃𝐴 + 𝑃𝐵 Raoult’s Law with Dalton’s Law of Partial Pressures: 𝑃 = 𝑥𝐴𝑃𝐴 ∘ + 𝑥𝐵𝑃𝐵∘ Vapor is an ideal gas: 𝑦𝐴 = 𝑃𝐴 𝑃 𝑦𝐵 = 𝑃𝐵 𝑃 Bubble Point Temperature - The temperature when the liquid starts to vaporize Dew Point Temperature - The temperature when the vapor starts to condense Mathematical Statement of Raoult’s Law Henry’s Law Henry’s law is applied when a gas solute is dissolved in the liquid phase and the critical temperature of the gas is lower than the temperature of application. Henry’s law states that the partial pressure of the species in the vapor phase is directly proportional to its liquid-phase mole fraction. 𝑦𝑖𝑃 = 𝑃𝑖 = 𝑥𝑖ℋ𝑖 where ℋ𝑖 = 𝐻𝑒𝑛𝑟𝑦′𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Vapor Liquid Equilibrium Introduction | 31 At low pressure, where gases can be considered ideal, it may be necessary Modified Raoult’s law results when 𝜸𝒊, an activity coefficient is inserted into Raoult’s law (solution (l) is non-ideal/ real soln): 𝑷𝒊 = 𝒙𝒊𝛾𝑖 𝑷𝒊 𝒔𝒂𝒕𝒏 𝑷 = σ𝒊𝒙𝒊 𝜸𝒊𝑷𝒊 𝒔𝒂𝒕𝒏 𝒚𝒊 = 𝒙𝒊𝛾𝑖 𝑷𝒊 𝒔𝒂𝒕𝒏 𝑷 σ𝒊𝒚𝒊 = 𝟏 The activity coefficient, 𝛾𝑖, is introduced into Raoult’s law to account for liquid-phase non- idealities. Activity coefficients are obtained from experimental data and using empirical equations. Some methods of estimating activity coefficients are shown in Appendix H (Smith, 8th Edition). VLE By Modified Raoult’s Law Azeotropes An azeotrope is a point for which the dew point curve and the bubble point curve coincide, and maintains its constant boiling point and composition throughout distillation Positive azeotrope - boiling point of azeotrope is either less than the boiling points of its components Negative azeotrope - boiling point of azeotrope is greater than the boiling point of any of its components Azeotropic composition: x1 = y1 and x2 = y2 Solution: A quantity called relative volatility is represented by 𝛼. It is a measure of separability of volatile components in a mixture For a binary system (alpha is relative volatility): 𝛼12 = 𝑦1/𝑥1 𝑦2/𝑥2 For azeotrope: x1 = y1 and x2 = y2, 𝛼12 = 1 (meaning: they will vaporize together) 𝑃1 = 𝑥1𝛾1𝑃1 𝑠𝑎𝑡 𝑦1 = 𝑥1𝛾1𝑃1 𝑠𝑎𝑡 𝑃 𝑦1 𝑥1 = 𝛾1𝑃1 𝑠𝑎𝑡 𝑃 𝑦2 𝑥2 = 𝛾2𝑃2 𝑠𝑎𝑡 𝑃 𝛼12 = 𝛾1𝑃1 𝑠𝑎𝑡 𝛾2𝑃2 𝑠𝑎𝑡 = 1.0 𝜸𝟏 𝜸𝟐 = 𝑷𝟐 𝒔𝒂𝒕 𝑷𝟏 𝒔𝒂𝒕 Solutions Thermodynamics Theory | 32 SOLUTIONS THERMODYNAMICS THEORY FUNDAMENTAL PROPERTY RELATIONS Homogenous Phases of Constant Composition Recall the First of Thermodynamics in a closed system with (n) number of moles: 𝑑 𝑛𝑈 = 𝑑𝑄𝑟𝑒𝑣 + 𝑑𝑊𝑟𝑒𝑣 Recall the Second Law of Thermodynamics: 𝑑𝑄𝑟𝑒𝑣 = 𝑇𝑑(𝑛𝑆) Recall the PV-Work Equation: 𝑑𝑊𝑟𝑒𝑣 = −𝑃𝑑(𝑛𝑉) Combining the three equation results to: 𝒅 𝒏𝑼 = 𝑻𝒅 𝒏𝑺 − 𝑷𝒅(𝒏𝑽) Additional Thermodynamic Properties by Definition: 1. Enthalpy 𝐻 ≡ 𝑈 + 𝑃𝑉 𝑑 𝑛𝐻 = 𝑑(𝑛𝑈) + [𝑃 𝑑 𝑛𝑉 + 𝑛𝑉 𝑑𝑃] 2. Helmholtz Energy 𝐺 ≡ 𝐻 − 𝑇𝑆 𝑑 𝑛𝐺 = 𝑑 𝑛𝐻 − [𝑇 𝑑 𝑛𝑆 + 𝑛𝑆 𝑑𝑇] 3. Gibbs Energy 𝐴 ≡ 𝑈 − 𝑇𝑆 𝑑 𝑛𝐴 = 𝑑 𝑛𝑈 − [𝑇 𝑑 𝑛𝑆 + 𝑛𝑆 𝑑𝑇] Helmholtz Energy (A) - Energy available to do non-PV work in a thermodynamic closed system at constant volume and temperature. Gibbs Energy (G) - the energy available to do non-PV work in a thermodynamic closed system at constant pressure and temperature. Gibbs energy relates the tendency of a physical or chemical system to simultaneously lower its energy H and increase its disorder S in a spontaneous natural process. Solution Thermodynamics is the application of thermodynamics in gas mixtures and liquid solutions. Solutions Thermodynamics Theory | 35 For the vapor () phase: 𝑑(𝑛𝐺)𝛼= (𝑛𝑉)𝛼𝑑𝑃 − (𝑛𝑆)𝛼𝑑𝑇 +෍ 𝑖 𝜇𝑖 𝛼 𝑑𝑛𝑖 𝛼 For the liquid () phase: 𝑑(𝑛𝐺)𝛽= (𝑛𝑉)𝛽𝑑𝑃 − (𝑛𝑆)𝛽𝑑𝑇 +෍ 𝑖 𝜇𝑖 𝛽 𝑑𝑛𝑖 𝛽 For the whole two-phase system, considered to be a closed system recall that: 𝑑 𝑛𝐺 = 𝑛𝑉 𝑑𝑃 − 𝑛𝑆 𝑑 𝑇 Total-System Property: 𝑛𝐺 = (𝑛𝐺)𝛼+(𝑛𝐺)𝛽 Change in Total-System Property = Sum of Changes in Phase Properties 𝑑 𝑛𝐺 = 𝑑(𝑛𝐺)𝛼+𝑑(𝑛𝐺)𝛽 𝑑 𝑛𝐺 = 𝑛𝑉 𝑑𝑃 − 𝑛𝑆 𝑑 𝑇 + σ𝑖 𝜇𝑖 𝛼 𝑑𝑛𝑖 𝛼 +σ𝑖 𝜇𝑖 𝛽 𝑑𝑛𝑖 𝛽 At equilibrium, the equation becomes σ𝑖 𝜇𝑖 𝛼 𝑑𝑛𝑖 𝛼 + σ𝑖 𝜇𝑖 𝛽 𝑑𝑛𝑖 𝛽 = 0 Note that the number of moles in the liquid phase and in the vapor phase are a result of mass transfer between the two phases. Since it is a closed system in equilibrium, no new moles are added, and law of mass conservation applies. Therefore, any addition of moles in the vapor phase has an equivalent decrease in the liquid phase. Mathematically: 𝑑𝑛𝑖 𝛼 = −𝑑𝑛𝑖 𝛽 Substituting the value in the previous equation σ𝑖 𝜇𝑖 𝛼 𝑑𝑛𝑖 𝛼 + σ𝑖 𝜇𝑖 𝛽 (−𝑑𝑛𝑖 𝛼) = 0 Rearranging we get the term ෍ 𝑖 (𝜇𝑖 𝛼 − 𝜇𝑖 𝛽 )𝑑𝑛𝑖 𝛼 = 0 For the statement to hold true then 𝜇𝑖 𝛼 = 𝜇𝑖 𝛽 Solutions Thermodynamics Theory | 36 The chemical potential of a pure component is an intensive property 𝜇𝑖,𝑝𝑢𝑟𝑒 = 𝐺𝑖 𝑖𝑛 𝐽/𝑚𝑜𝑙 The chemical potential of a species in a mixture is a partial molar Gibbs energy 𝜇𝑖 = ҧ𝐺𝑖 = 𝜕 𝑛𝐺 𝜕𝑛𝑖 𝑃,𝑇,𝑛𝑗 𝑖𝑛 𝐽/𝑚𝑜𝑙 For any generic property of the solution, we denote it as 𝑀 𝑎𝑛𝑑 ഥ𝑀𝑖 which can be unit-mass or unit-mole based. ഥ𝑀𝑖 refers to any partial molar property of species i which is defined as ഥ𝑀𝑖 ≡ 𝜕 𝑛𝑀 𝜕𝑛𝑖 𝑃,𝑇,𝑛𝑗 ഥ𝑀𝑖 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑀 𝑜𝑓 𝑚𝑖𝑥𝑡𝑢𝑟𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝 𝑖 Solution properties 𝑀, 𝑉, 𝑈, 𝐻, 𝑆, 𝐺 Partial properties ഥ𝑀𝑖 ത𝑉𝑖, ഥ𝑈𝑖 , ഥ𝐻𝑖 , ҧ𝑆𝑖 , ҧ𝐺𝑖 Pure-species properties 𝑀𝑖 , 𝑉𝑖, 𝑈𝑖 , 𝐻𝑖 , 𝑆𝑖 , 𝐺𝑖 PARTIAL PROPERTIES Equations Relating Molar and Partial Molar Properties Similar to how 𝒏𝑮 = 𝑮(𝑷, 𝑻, 𝒏𝟏, 𝒏𝟐, 𝒏𝟑, … , 𝒏𝒊) can be differentiated, we can do the same for any property M. Eqn (1): 𝒏𝑴 = 𝒇(𝑷, 𝑻, 𝒏𝟏, 𝒏𝟐, … , 𝒏𝒊) Eqn (2): 𝑑 𝑛𝑀 = 𝜕 𝑛𝑀 𝜕𝑃 𝑇,𝑛 𝑑𝑃 + 𝜕 𝑛𝑀 𝜕𝑇 𝑃,𝑛 𝑑𝑇 + σ𝑖 𝜕 𝑛𝑀 𝜕𝑛𝑖 𝑃,𝑇,𝑛𝑗 𝑑𝑛𝑖 By definition of ഥ𝑀𝑖, we can replace it in the equation: Eqn (3): 𝑑 𝑛𝑀 = 𝑛 𝜕 𝑀 𝜕𝑃 𝑇,𝑥 𝑑𝑃 + 𝑛 𝜕 𝑀 𝜕𝑇 𝑃,𝑥 𝑑𝑇 + σ𝑖 ഥ𝑀𝑖 𝑑𝑛𝑖 The equation is in simpler form, where subscript x denotes differentiation at constant composition. Because mole fraction xi is ni/n, the equation becomes: Eqn (4a): 𝑑𝑀 = 𝜕𝑀 𝜕𝑃 𝑇,𝑥 𝑑𝑃 + 𝜕𝑀 𝜕𝑇 𝑃,𝑥 𝑑𝑇 + σ𝑖 ഥ𝑀𝑖 𝑑𝑥𝑖 Eqn (4b): 𝑀 = σ𝑖 𝑥𝑖 ഥ𝑀𝑖 (defines the summability) Eqn (5): 𝑛𝑀 = σ𝑖 𝑛𝑖 ഥ𝑀𝑖 (alternative expression) For example ,we use to property of Volume so we write 𝑉 instead of 𝑀 and 𝑉𝑖 instead of ഥ𝑀𝑖. Therefore, if we look at the partial molar volume in terms of the summability: 𝑽 = 𝒙𝟏ഥ𝑽𝟏 + 𝒙𝟐ഥ𝑽𝟐 + … + 𝒙𝒊ഥ𝑽𝒊 + … + 𝒙𝑵ഥ𝑽𝑵 Solutions Thermodynamics Theory | 37 Gibbs/Duhem Equation 𝝏𝑴 𝝏𝑷 𝑻,𝒙 𝒅𝑷 + 𝝏𝑴 𝝏𝑻 𝑷,𝒙 𝒅𝑻 −෍ 𝒊 𝒙𝒊𝒅ഥ𝑴𝒊 = 𝟎 at constant temperature and pressure, Gibbs/Duhem Eqn is reduced to: ෍ 𝒊 𝒙𝒊𝒅ഥ𝑴𝒊 = 𝟎 Partial Properties of Binary Solutions For any binary solution we have components with mole fractions 𝒙𝟏and 𝒙𝟐, where 𝒙𝟐can also be expressed as (𝟏 − 𝒙𝟏) From the summability equation 𝑀 = σ𝑖 𝑥𝑖 ഥ𝑀𝑖, Partial Properties for binary solutions are written as 𝑴 = 𝒙𝟏 ഥ𝑴𝟏 + 𝒙𝟐 ഥ𝑴𝟐 (A) Differentiating the equation (using product rule) 𝒅𝑴 = 𝒙𝟏𝒅ഥ𝑴𝟏 + ഥ𝑴𝟏𝒅𝒙𝟏 + 𝒙𝟐𝒅ഥ𝑴𝟐 + ഥ𝑴𝟐𝒅𝒙𝟐 (B) At constant temperature and pressure, the Gibbs/Duhem Equation becomes: 𝒙𝟏𝒅ഥ𝑴𝟏 + 𝒙𝟐𝒅ഥ𝑴𝟐 = 𝟎 (C) Since 𝑥1 + 𝑥2 = 1 it follows that 𝑑𝑥1 = −𝑑𝑥2. Eliminating 𝑑𝑥2 in equation (B) and combining the result with equation (C) give: 𝒅𝑴 𝒅𝒙𝟏 = ഥ𝑴𝟏 − ഥ𝑴𝟐 or 𝒅𝑴 𝒅𝒙𝟐 = ഥ𝑴𝟐 − ഥ𝑴𝟏 (D) Eliminating 𝒙𝟏 or 𝒙𝟐 in equation A: 𝑀 = ഥ𝑀1 − 𝑥2( ഥ𝑀1 − ഥ𝑀2) or 𝑀 = 𝑥1( ഥ𝑀1 − ഥ𝑀2) + ഥ𝑀2 and combining with equation (D) result to: ഥ𝑴𝟏 = 𝑴+ 𝒙𝟐 𝒅𝑴 𝒅𝒙𝟏 (E1) or ഥ𝑴𝟐 = 𝑴− 𝒙𝟏 𝒅𝑴 𝒅𝒙𝟏 (E2) Gibbs/Duhem equation (C) may be written in derivatives form: At constant T and P: 𝒙𝟏 𝒅ഥ𝑴𝟏 𝒅𝒙𝟏 + 𝒙𝟐 𝒅ഥ𝑴𝟐 𝑑𝑥1 = 𝟎 or 𝒅ഥ𝑴𝟏 𝒅𝒙𝟏 = − 𝒙𝟐 𝒙𝟏 𝒅ഥ𝑴𝟐 𝑑𝑥1 (F) lim 𝑥1→1 𝒅ഥ𝑴𝟏 𝒅𝒙𝟏 = 0 (Provided lim 𝑥1→1 𝒅ഥ𝑴𝟐 𝒅𝒙𝟏 is finite) (G) or: lim 𝑥2→1 𝒅ഥ𝑴𝟐 𝒅𝒙𝟏 = 0 (Provided lim 𝑥2→1 𝒅ഥ𝑴𝟏 𝒅𝒙𝟏 is finite) Solutions Thermodynamics Theory | 40 Partial Pressure (pi): The partial pressure of species i in an ideal gas mixture is the pressure that species i would exert if it alone occupied the molar volume of the mixture. In a mixture of gases, mole ratio = pressure ratio: 𝑝𝑖 𝑃 = 𝑛𝑖 𝑛 = 𝑦𝑖 𝑝𝑖 = 𝑦𝑖𝑃 = 𝑦𝑖 𝑅𝑇 𝑉𝑖𝑔 𝑦𝑖 = 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 𝑖 Partial Molar Enthalpy: ഥ𝑯𝒊 𝒊𝒈 𝑻, 𝑷 = 𝑯𝒊 𝒊𝒈 𝑻, 𝒑𝒊 = 𝑯𝒊 𝒊𝒈 𝑻, 𝑷 Partial Molar Internal energy: ഥ𝑼𝒊 𝒊𝒈 𝑻, 𝑷 = 𝑼𝒊 𝒊𝒈 𝑻, 𝒑𝒊 = 𝑼𝒊 𝒊𝒈 𝑻,𝑷 Partial Molar Enthalpy and Internal Energy are independent in changes in Pressure Recall that the Change in Entropy for Ideal Gases is: 𝑑𝑆𝑖𝑔 = 𝐶𝑝𝑖𝑔 𝑑𝑇 𝑇 − 𝑅 𝑑𝑃 𝑃 Integration from 𝑝𝑖 𝑡𝑜 𝑃 and at const T gives: 𝑆𝑖 𝑖𝑔 𝑇, 𝑃 − 𝑆𝑖 𝑖𝑔 𝑇, 𝑝𝑖 = −𝑅 𝑙𝑛 𝑃 𝑝𝑖 = −𝑅 𝑙𝑛 𝑃 𝑦𝑖𝑃 = 𝑅𝑙𝑛𝑦𝑖 𝑆𝑖 𝑖𝑔 𝑇, 𝑝𝑖 = 𝑆𝑖 𝑖𝑔 𝑇, 𝑃 − 𝑅𝑙𝑛𝑦𝑖 𝑆𝑖 𝑖𝑔 𝑇, 𝑝𝑖 = ҧ𝑆𝑖 𝑖𝑔 (𝑇, 𝑃) Partial Molar Entropy: ത𝑺𝒊 𝒊𝒈 𝑻,𝑷 = 𝑺𝒊 𝒊𝒈 𝑻, 𝑷 − 𝑹𝒍𝒏𝒚𝒊 Partial molar Gibbs Free Energy For the Gibbs energy of an ideal gas mixture, 𝐺𝑖𝑔 = 𝐻𝑖𝑔 − 𝑇𝑆𝑖𝑔; the parallel relation for partial properties is ҧ𝐺𝑖 𝑖𝑔 = ഥ𝐻𝑖 𝑖𝑔 − 𝑇 ҧ𝑆𝑖 𝑖𝑔 ҧ𝐺𝑖 𝑖𝑔 = ഥ𝐻𝑖 𝑖𝑔 − 𝑇 ҧ𝑆𝑖 𝑖𝑔 is combined with molar enthalpy and molar entropy relations: ҧ𝐺𝑖 𝑖𝑔 = 𝐻𝑖 𝑖𝑔 − 𝑇(𝑆𝑖 𝑖𝑔 − 𝑅𝑙𝑛𝑦𝑖) ҧ𝐺𝑖 𝑖𝑔 = 𝐻𝑖 𝑖𝑔 − 𝑇𝑆𝑖 𝑖𝑔 + 𝑅𝑇𝑙𝑛𝑦𝑖 𝝁𝒊 𝒊𝒈 ≡ ഥ𝑮𝒊 𝒊𝒈 = 𝑮𝒊𝒈 + 𝑹𝑻𝒍𝒏𝒚𝒊 Solutions Thermodynamics Theory | 41 The Summability Relation: 𝑴 =෍ 𝒊 𝒙𝒊 ഥ𝑴𝒊 𝑯𝒊𝒈 =σ𝒊𝒚𝒊𝑯𝒊 𝒊𝒈 𝑺𝒊𝒈 =෍ 𝒊 𝒚𝒊 𝑺𝒊 𝒊𝒈 − 𝑹෍ 𝒊 𝒚𝒊 𝒍𝒏 𝒚𝒊 𝑮𝒊𝒈 = σ𝒊𝒚𝒊 𝑮𝒊 𝒊𝒈 + 𝑹𝑻σ𝒊𝒚𝒊 𝒍𝒏 𝒚𝒊 Molar Change of Mixing: 𝚫𝑴 = 𝑴−σ𝒊𝑴𝒊 𝚫𝑯𝒊𝒈 = 𝚫𝑼𝒊𝒈 = 𝚫𝑽𝒊𝒈 = 𝟎 𝚫𝑺𝒊𝒈 = −𝑹σ𝒊𝒚𝒊 𝒍𝒏 𝒚𝒊 𝚫𝑮𝒊𝒈 = 𝑹𝑻σ𝒊𝒚𝒊 𝒍𝒏 𝒚𝒊 Solutions Thermodynamics Applications | 42 SOLUTIONS THERMODYNAMICS APPLICATIONS FUGACITY AND ACTIVITY CONCEPTS When the equation ∆𝐺 = 𝑛𝑅𝑇𝑙𝑛 𝑃2 𝑃1 (ideal gas and at constant temperature) is applied to real gases, particularly at higher pressures, it is found that the change in Gibbs free energy is not reproduced by this simple relation. In cases of non-ideal behavior, V is no longer given by 𝑛𝑅𝑇 𝑃 but by some more complicated function of the pressure Fugacity and Fugacity Coefficient Fugacity – a quantitative measure of escaping tendency of a substance from a particular state (a kind of pressure). Consider the property relation: 𝑑𝐺 = 𝑉𝑑𝑃 − 𝑆𝑑𝑇 𝑑𝐺 = 𝑅𝑇 𝑃 𝑑𝑃 − 𝑆𝑑𝑇 Integration at constant T gives: 𝐺 = 𝑅𝑇𝑙𝑛𝑃 + Γ(𝑇) where Γ(𝑇)= integration constant dependent only on the temperature and the nature of the substance For a pure species i in the ideal gas state: 𝐺𝑖 𝑖𝑔 = 𝑅𝑇𝑙𝑛𝑃 + Γ𝑖(𝑇) For a real fluid species i: 𝑮𝒊 = 𝑹𝑻𝒍𝒏𝒇𝒊 + 𝚪𝒊(𝑻) where 𝑓𝑖 is the fugacity Subtracting the two equation results to 𝐺𝑖 − 𝐺𝑖 𝑖𝑔 = 𝑅𝑇𝑙𝑛 𝑓𝑖 𝑃 Where: 𝐺𝑖 − 𝐺𝑖 𝑖𝑔 = 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 𝐺𝑖𝑏𝑏𝑠 𝑒𝑛𝑒𝑟𝑔𝑦, 𝐺𝑖 𝑅 𝑓𝑖 𝑃 ≡ ∅𝑖 = 𝑓𝑢𝑔𝑎𝑐𝑖𝑡𝑦 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡, 𝑝𝑢𝑟𝑒 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 𝑖, 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑙𝑒𝑠𝑠 Residual Property – is the difference of property of the real substance from the ideal gas at same temperature and pressure 𝑀𝑅 ≡ 𝑀 −𝑀𝑖𝑔 Solutions Thermodynamics Applications | 45 (1) Calculate 1 and 2 using the modified Raoult’s Law 𝜸𝒊 = 𝒚𝒊𝑷 𝒙𝒊𝑷𝒊 𝒔𝒂𝒕𝒏 (2) Calculate the value of 𝑮 𝑬 𝑹𝑻 and 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 𝑮𝑬 𝑹𝑻 = 𝒙𝟏𝒍𝒏𝜸𝟏 + 𝒙𝟐𝒍𝒏𝜸𝟐 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 = 𝟏 𝒙𝟏𝒙𝟐 (𝒙𝟏𝒍𝒏𝜸𝟏 + 𝒙𝟐𝒍𝒏𝜸𝟐) (3) Plot 𝒍𝒏𝜸𝟏, 𝒍𝒏𝜸𝟐, 𝑮𝑬 𝑹𝑻 , and 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 versus 𝒙𝟏 LIQUID PHASE PROPERTIES FROM VLE DATA Determination of i from Experimental VLE Data Reduction and Margules Equation Margules Equation: 𝑮𝑬 𝑹𝑻 = (𝑨𝟐𝟏𝒙𝟏 + 𝑨𝟏𝟐𝒙𝟐)𝒙𝟏𝒙𝟐 Multiplying the equation by n and converting all mole fractions to moles, the right side 𝑥1 is replaced by 𝑛1 𝑛1+𝑛2 and 𝑥2 by 𝑛2 𝑛1+𝑛2 . With 𝑛 = 𝑛1 + 𝑛2 , this gives: 𝒏𝑮𝑬 𝑹𝑻 = (𝑨𝟐𝟏𝒏𝟏 + 𝑨𝟏𝟐𝒏𝟐) 𝒏𝟏𝒏𝟐 (𝒏𝟏+𝒏𝟐)𝟐 Activity Coefficients from Margules Correlation The Margules equation was developed by the data reduction method. Of the data sets of points shown in the graph of liquid-phase properties and their correlation, the 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 vs 𝒙𝟏 closely conform to a linear equation. Consider that the linear relation is given by the equation: 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 = 𝑨𝟐𝟏𝒙𝟏 + 𝑨𝟏𝟐𝒙𝟐 (Margules Equation) Alternative equation: 𝑮 𝑬 𝑹𝑻 = (𝑨𝟐𝟏𝒙𝟏 + 𝑨𝟏𝟐𝒙𝟐)𝒙𝟏𝒙𝟐 where A21 and A12 are constants (Margules parameters). The linear correlation of Margules Equation: 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 = 𝑨𝟐𝟏𝒙𝟏 + 𝑨𝟏𝟐𝒙𝟐 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 = 𝑨𝟐𝟏𝒙𝟏 + 𝑨𝟏𝟐(𝟏 − 𝒙𝟏) 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 = 𝑨𝟐𝟏𝒙𝟏 + 𝑨𝟏𝟐 − 𝑨𝟏𝟐𝒙𝟏 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 = 𝑨𝟐𝟏 − 𝑨𝟏𝟐 𝒙𝟏 + 𝑨𝟏𝟐 𝑦 = 𝑚𝑥 + 𝑏 𝑦 = 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 𝑚 = 𝑨𝟐𝟏 − 𝑨𝟏𝟐 𝑏 = 𝑨𝟏𝟐 𝑙𝑛𝛾1 = 𝑥2 2[𝐴12 + 2 𝐴21 − 𝐴12 𝑥1] 𝑙𝑛𝛾2 = 𝑥1 2[𝐴21 + 2 𝐴12 − 𝐴21 𝑥2] Solutions Thermodynamics Applications | 46 Differentiating with respect to 𝒏𝟏: 𝝏(𝒏𝑮𝑬/𝑹𝑻) 𝝏𝒏𝟏 𝑷,𝑻,𝒏𝟐 = 𝒏𝟐 (𝑨𝟐𝟏𝒏𝟏 + 𝑨𝟏𝟐𝒏𝟐) 𝟏 (𝒏𝟏+𝒏𝟐)𝟐 − 𝟐𝒏𝟏 (𝒏𝟏+𝒏𝟐)𝟑 + 𝒏𝟏𝒏𝟐 (𝒏𝟏+𝒏𝟐)𝟐 𝑙𝑛𝛾1 = 𝝏(𝒏𝑮𝑬/𝑹𝑻) 𝝏𝒏𝟏 𝑷,𝑻,𝒏𝟐 = 𝒏𝟐 (𝑨𝟐𝟏𝒏𝟏 + 𝑨𝟏𝟐𝒏𝟐) 𝟏 (𝒏𝟏+𝒏𝟐) 𝟐 − 𝟐𝒏𝟏 (𝒏𝟏+𝒏𝟐) 𝟑 + 𝒏𝟏𝒏𝟐 (𝒏𝟏+𝒏𝟐) 𝟐 Reconversion of the 𝑛𝑖 to 𝑥𝑖 (𝑛1 = 𝑛𝑥1; 𝑛2 = 𝑛𝑥2) gives: 𝑙𝑛𝛾1 = 𝒙𝟐 (𝑨𝟐𝟏𝒙𝟏 + 𝑨𝟏𝟐𝒏𝒙𝟐)(𝟏 − 𝟐𝒙𝟏) + 𝑨𝟐𝟏𝒙𝟏 Further reduction, noting that 𝑥2 = 1 − 𝑥1, leads to: 𝒍𝒏𝜸𝟏 = 𝒙𝟐 𝟐[𝑨𝟏𝟐 + 𝟐 𝑨𝟐𝟏 − 𝑨𝟏𝟐 𝒙𝟏] And similarly: 𝒍𝒏𝜸𝟐 = 𝒙𝟏 𝟐[𝑨𝟐𝟏 + 𝟐 𝑨𝟏𝟐 − 𝑨𝟐𝟏 𝒙𝟐] For limiting conditions of infinite dilution, they become 𝒍𝒏𝜸𝟏 ∞ = 𝑨𝟏𝟐(𝒙𝟏 = 𝟎) and 𝒍𝒏𝜸𝟐 ∞ = 𝑨𝟐𝟏(𝒙𝟐 = 𝟎) Thermodynamic Consistency If the experimental data are inconsistent with the Gibbs/Duhem equation, then the data are incorrect due to systematic error in the data. Because correlating equations for GE/RT impose consistency on derived activity coefficient, no such correlation exists that can precisely reproduce P-x1-y1 data that are inconsistent. Our purpose now is to develop a simple test for consistency with respect to the Gibbs/Duhem equation of a P-x1-y1 data set. Consider the summability and Gibbs/Duhem equations for a binary system 𝑮𝑬 𝑹𝑻 = 𝒙𝟏𝒍𝒏𝜸𝟏 + 𝒙𝟐𝒍𝒏𝜸𝟐 𝒙𝟏 𝒅𝒍𝒏𝜸𝟏 + 𝒙𝟐𝒅𝒍𝒏𝜸𝟐 = 𝟎 𝒅 𝑮𝑬/𝑹𝑻 𝒅𝒙𝟏 = 𝒍𝒏𝜸𝟏 − 𝒍𝒏𝜸𝟐 + 𝒙𝟏 𝒅𝒍𝒏𝜸𝟏 + 𝒙𝟐𝒅𝒍𝒏𝜸𝟐 (correlation) 𝒅 𝑮𝑬/𝑹𝑻 ∗ 𝒅𝒙𝟏 = 𝒍𝒏𝜸𝟏 ∗ − 𝒍𝒏𝜸𝟐 ∗ + 𝒙𝟏 𝒅𝒍𝒏𝜸𝟏 ∗ + 𝒙𝟐𝒅𝒍𝒏𝜸𝟐 ∗ (experimental) 𝒅 𝑮𝑬/𝑹𝑻 𝒅𝒙𝟏 − 𝒅 𝑮𝑬/𝑹𝑻 ∗ 𝒅𝒙𝟏 = 𝒍𝒏 𝜸𝟏 𝜸𝟐 − 𝒍𝒏 𝜸𝟏 ∗ 𝜸𝟐∗ − 𝒙𝟏 𝒅𝒍𝒏𝜸𝟏 ∗ 𝒅𝒙𝟏 + 𝒙𝟐 𝒅𝒍𝒏𝜸𝟐 ∗ 𝒅𝒙𝟏 The differences between like terms are residuals, which may be represented by a  notation. The preceding equation then becomes: 𝒅𝜹 𝑮𝑬/𝑹𝑻 𝒅𝒙𝟏 = 𝜹𝒍𝒏 𝜸𝟏 𝜸𝟐 − 𝒙𝟏 𝒅𝒍𝒏𝜸𝟏 ∗ 𝒅𝒙𝟏 + 𝒙𝟐 𝒅𝒍𝒏𝜸𝟐 ∗ 𝒅𝒙𝟏 Solutions Thermodynamics Applications | 47 If a data set is reduced so as to make the residuals in 𝑮𝑬/𝑹𝑻 scatter about zero, then the derivative 𝒅𝜹 𝑮𝑬/𝑹𝑻 𝒅𝒙𝟏 is effectively zero, reducing the preceding equation to: 𝜹𝒍𝒏 𝜸𝟏 𝜸𝟐 = 𝒙𝟏 𝒅𝒍𝒏𝜸𝟏 ∗ 𝒅𝒙𝟏 + 𝒙𝟐 𝒅𝒍𝒏𝜸𝟐 ∗ 𝒅𝒙𝟏 where 𝜹𝒍𝒏 𝜸𝟏 𝜸𝟐 is a direct measure of deviation from the Gibbs/Duhem Equation Average values of the residuals < 0.03 high degree of consistency < 0.10 acceptable > 0.10 not acceptable, data contain significant error Van Laar Equation 𝒙𝟏𝒙𝟐 𝑮𝑬/𝑹𝑻 = 𝒙𝟏 𝑨𝟐𝟏 ′ + 𝒙𝟐 𝑨𝟏𝟐 ′ = 𝑨𝟏𝟐 ′ 𝒙𝟏 + 𝑨𝟐𝟏 ′ 𝒙𝟐 𝑨𝟏𝟐 ′ 𝑨𝟐𝟏 ′ 𝑮𝑬 𝒙𝟏𝒙𝟐𝑹𝑻 = 𝑨𝟏𝟐 ′ 𝑨𝟐𝟏 ′ 𝑨𝟏𝟐 ′ 𝒙𝟏 + 𝑨𝟐𝟏 ′ 𝒙𝟐 The activity coefficients implied by this equation are: 𝑙𝑛𝛾1 = 𝑨𝟏𝟐 ′ 𝟏 + 𝑨𝟏𝟐 ′ 𝒙𝟏 𝑨𝟐𝟏 ′ 𝒙𝟐 −𝟐 𝑙𝑛𝛾2 = 𝑨𝟐𝟏 ′ 𝟏 + 𝑨𝟐𝟏 ′ 𝒙𝟐 𝑨𝟏𝟐 ′ 𝒙𝟏 −𝟐 When 𝑥1 = 0, 𝒍𝒏𝜸𝟏∞ = 𝑨𝟏𝟐 ′ ; when 𝑥2 = 0, 𝒍𝒏𝜸𝟐∞ = 𝑨𝟐𝟏 ′ Replacing x2 by 1-x1, the van Laar Equation becomes 𝒙𝟏𝒙𝟐 𝑮𝑬/𝑹𝑻 = 𝒙𝟏 𝑨𝟐𝟏 ′ + 𝒙𝟐 𝑨𝟏𝟐 ′ = 𝟏 𝑨𝟐𝟏 ′ − 𝟏 𝑨𝟏𝟐 ′ 𝒙𝟏 + 𝟏 𝑨𝟏𝟐 ′ 𝒎 = 𝟏 𝑨𝟐𝟏 ′ − 𝟏 𝑨𝟏𝟐 ′ 𝒃 = 𝟏 𝑨𝟏𝟐 ′ 𝒎+ 𝒃 = 𝟏 𝑨𝟐𝟏 ′ 𝑨𝟏𝟐 ′ = 𝟏 𝒃 𝑨𝟐𝟏 ′ = 𝟏 𝒎+𝒃 Solutions Thermodynamics Applications | 50 (10)Calculate for the Excess Gibb’s Energy 𝐺 𝐸 𝑅𝑇 = 𝑥1𝑙𝑛𝛾1 + 𝑥2𝑙𝑛𝛾2 using the new calculated values (11)Calculate for the Residual Gibb’s Energy by subtracting the calculated from the experimental and get the average values Margules Van Laar x1 Calculated lnγ1 Calculated lnγ2 Calculated γ1 Calculated γ2 Calculated lnγ1 Calculated lnγ2 Calculated γ1 Calculated γ2 0.0000 0.3844 0.0000 1.4687 1.0000 0.4109 1.5082 0.0895 0.2908 0.0043 1.3374 1.0043 0.2912 0.0054 1.3380 1.0054 0.1981 0.1993 0.0195 1.2205 1.0197 0.1896 0.0221 1.2087 1.0224 0.3193 0.1225 0.0461 1.1303 1.0472 0.1142 0.0480 1.1210 1.0492 0.4232 0.0749 0.0740 1.0778 1.0768 0.0712 0.0733 1.0738 1.0760 0.5119 0.0457 0.0996 1.0468 1.1047 0.0455 0.0957 1.0466 1.1004 0.6096 0.0236 0.1275 1.0239 1.1360 0.0259 0.1206 1.0262 1.1282 0.7135 0.0095 0.1548 1.0096 1.1674 0.0124 0.1468 1.0125 1.1581 0.7934 0.0037 0.1725 1.0037 1.1883 0.0059 0.1665 1.0059 1.1812 0.9102 0.0003 0.1906 1.0003 1.2100 0.0010 0.1944 1.0010 1.2145 1.0000 0.0000 0.1963 1.0000 1.2169 0.2149 1.2398 Margules Van Laar x1 Calculated GE/RT Residual GE/RT Calculated GE/RT Residual GE/RT 0.0000 0.00000 0.00000 0.00000 0.0000 0.0895 0.02995 -0.00177 0.03096 -0.0008 0.1981 0.05514 0.00081 0.05529 0.0010 0.3193 0.07049 0.00259 0.06918 0.0013 0.4232 0.07440 0.00198 0.07238 0.0000 0.5119 0.07199 0.00118 0.07000 -0.0008 0.6096 0.06419 0.00071 0.06286 -0.0006 0.7135 0.05114 0.00036 0.05089 0.0001 0.7934 0.03855 0.00011 0.03908 0.0007 0.9102 0.01743 -0.00112 0.01835 -0.0002 1.0000 0.00000 0.00000 0.00000 0.0000 Average 0.00044 Average 0.00005 (12) Get the residual 𝑙𝑛𝛾1 and 𝑙𝑛𝛾2 by subtracting them from the experimental. (13)To obtain 𝛿𝑙𝑛 𝛾1 𝛾2 , subtract the residual 𝑙𝑛𝛾1 with the 𝑙𝑛𝛾2 residual. Get the average values Solutions Thermodynamics Applications | 51 (14) To get the new calculated pressures, Use Dalton’s Law of Partial Pressures together with modified Raoult’s Law, 𝑃 = 𝑥1𝛾1𝑃1 𝑠𝑎𝑡 + 𝑥2𝛾2𝑃2 𝑠𝑎𝑡 (15) 𝑦1 can be computed by 𝑦1 = 𝑥1𝑦1𝑃1 𝑠𝑎𝑡 𝑃 Margules Van Laar x1 Residual lnγ1 Residual lnγ2 𝜹𝒍𝒏 𝜸𝟏 𝜸𝟐 Residual lnγ1 Residual lnγ2 𝜹𝒍𝒏 𝜸𝟏 𝜸𝟐 0.0000 0.3844 0.0000 0.3844 0.4109 0.0000 0.4109 0.0895 0.0252 -0.0044 0.0296 0.0256 -0.0034 0.0290 0.1981 0.0268 -0.0056 0.0324 0.0171 -0.0030 0.0201 0.3193 0.0147 -0.0031 0.0177 0.0064 -0.0011 0.0076 0.4232 0.0060 -0.0010 0.0070 0.0023 -0.0017 0.0040 0.5119 0.0028 -0.0005 0.0033 0.0026 -0.0044 0.0070 0.6096 0.0008 0.0005 0.0003 0.0031 -0.0064 0.0095 0.7135 -0.0009 0.0036 -0.0045 0.0019 -0.0044 0.0064 0.7934 0.0004 -0.0010 0.0014 0.0026 -0.0070 0.0096 0.9102 0.0034 -0.0465 0.0499 0.0040 -0.0428 0.0468 1.0000 0.0000 0.1963 -0.1963 0.0000 0.2149 -0.2149 Average 0.02957 Average 0.03054 Margules Van Laar x1 Calculated P Calculated y1 Calculated P Calculated y1 0.0000 12.3000 0.0000 12.3000 0.0000 0.0895 15.5676 0.2775 15.5815 0.2774 0.1981 18.7839 0.4645 18.7258 0.4615 0.3193 21.7925 0.5977 21.7025 0.5952 0.4232 24.1012 0.6830 24.0340 0.6824 0.5119 25.9704 0.7446 25.9411 0.7453 0.6096 27.9817 0.8050 27.9945 0.8065 0.7135 30.1105 0.8634 30.1522 0.8646 0.7934 31.7586 0.9049 31.8049 0.9056 0.9102 34.1967 0.9609 34.2231 0.9608 1.0000 36.0900 1.0000 36.0900 1.0000 Solutions Thermodynamics Applications | 52 From here we and the graph of thermodynamic consistency, we can see that the two models, Margules and van Laar do not deviate much from the experimental values. PROPERTY CHANGES OF MIXING For any ideal solution, the summability relation of partial properties is expressed as: 𝑀𝑖𝑑 = ෍ 𝑖 𝑥𝑖 ഥ𝑀 𝑖𝑑 Applying the different properties G,S,V and H result to: 𝐺𝑖𝑑 = σ𝑖 𝑥𝑖 𝐺𝑖 − 𝑅𝑇σ𝑖 𝑥𝑖 ln 𝑥𝑖 𝑆𝑖𝑑 = σ𝑖 𝑥𝑖 𝑆𝑖 + 𝑅σ𝑖 𝑥𝑖 ln 𝑥𝑖 𝑉𝑖𝑑 = σ𝑖 𝑥𝑖 𝑉𝑖 𝐻𝑖𝑑 = σ𝑖 𝑥𝑖 𝐻𝑖 Recall that by definition, the expression of the excess property for any ideal solution is expressed as: 𝑀𝐸 ≡ 𝑀 − 𝑀𝑖𝑑 Using this relation with the summability of ideal solutions, the excess properties can be written as 𝐺𝐸 = 𝐺 − σ𝑖 𝑥𝑖 𝐺𝑖 − 𝑅𝑇σ𝑖 𝑥𝑖 ln 𝑥𝑖 𝑆𝐸 = 𝑆 − σ𝑖 𝑥𝑖 𝑆𝑖 + 𝑅σ𝑖 𝑥𝑖 ln 𝑥𝑖 𝑉𝐸 = 𝑉 − σ𝑖 𝑥𝑖 𝑉𝑖 𝐻𝐸 = 𝐻 − σ𝑖 𝑥𝑖 𝐻𝑖