4Equation of StateFor the regions of the diagram where a single phase exist: This means an equation of state exist relating P, V, T. An EOS may be solved for any one of the three quantities as a function of the other two, viz.: V = V(T,P) Volume expansivity: Isothermal compressibility:
5ExampleFor liquid acetone at 20oC & 1 bar: Β = x 10-3 oC-1 κ = 62 x x 10-6 bar-1 V= cm3 g-1 For aceton, find: a. The value of b. The pressure generated by heating at constant V from 20oC & 1 bar to 30oC c. The change in volume for a change from 20oC & 1 bar to 0oC & 10 bar.
6Problem 3.1Express the volume expansivity β and isothermal compressibility κ as functions of density ρ and its partial derivatives. The isothermal compressibility coefficient () of water at 50oC and 1 bar is x 10-6 bar-1. To what pressure must water be compressed at 50oC to change its density by 1%? Assume that is independent of P.
7Problem 3.2 & 3.3Generally, volume expansivity β and isothermal compressibility κ depend on T and P. Prove that The Tait equation for liquids is written for an isotherm as: where V is specific or molar volume, Vo is the hypothetical molar or specific volume at P = 0 and A & B are positive constant. Find an expression for the isothermal compressibility consistent with this equation.
8Problem 3.4For liquid water the isothermal compressibility is given by: where c & b are functions of temperature only. If 1 kg of water is compressed isothermally & reversibly from 1 bar to 500 bar at 60oC, how much work is required? At 60oC, b=2700 bar and c = cm3 g-1
9Problem 3.5Calculate the reversible work done in compressing 1 ft3 of mercury at a constant temperature of 32oF from 1 atm to 3000 atm. The isothermal compressibility of mercury at 32oF is: κ/atm-1 = 3.9 x 10-6 – 0.1 x 10-9 P (atm)
10Problem 3.6Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which the temperature change from 0oC to 20oC. Determine ΔVt, W, Q, and ΔUt The properties for liquid carbon tetrachloride at 1 bar & ooC may be assumed independent of temperature: β = 1.2 x 10-3 K-1 Cp = 0.84 kJ kg-1 K-1 ρ = 1590 kg m-3
11Problem 3.7A substance for which κ is a constant undergo a mechanically reversible process from initial state (P1,V1) to final state (P2,V2), where V is molar volume.Starting with the definition of κ, show that the path of the process is described byDetermine an exact expression which gives the isothermal work done on 1 mol of this constant-κ substance
14Virial EOS & Ideal GasVirial expansion: Alternative form: Compressibility factor For ideal gas: Z = 1, thus
15The Ideal GasThe internal energy of a real gas is a function of pressure and temperature. This pressure dependency is the result of forces between the molecules. In an ideal gas, such forces does not exist. No energy would be required to alter the average intermolecular distance, and therefore no energy would be required to bring about volume & pressure changes in an ideal gas at constant temperature. In other word, the internal energy of an ideal gas is a function of temperature only.
16Implied Property Relations for an Ideal Gas Thus H is also a funcion of temperature only
20Adiabatic Process Closed System, Ideal Gas, Mechanically Reversible
21Polytropic Process Closed System, Ideal Gas, Mechanically Reversible Polytropic process can be considered as general form of process. Isobaric process : δ = 0 Isothermal process : δ = 1 Adiabatic process : δ = γ Isochoric process : δ = ∞
22Example 3.2Air is compressed from an initial condition of 1 bar & 25oC to a final state of 5 bar & 25oC by three different mechanically reversible processes in a closed system:Heating at constant volume followed by cooling at constant pressureIsothermal compressionAdiabatic compression followed by cooling at constant volume.Assume air to be an ideal gas with the constant heat capacities: CV = 5/2 R, CP = 7/2 R.Sketch the process in a PV diagram & calculate the work required, heat transferred, and the changes in internal energy & entalphy of the air for each processes
23Example 3.3An ideal gas undergoes the following sequence of mechanically reversible processes in a closed system:From an initial state of 70oC & 1 bar, it is compressed adiabatically to 150oCIt is then cooled from 150oC to 70oC at constant pressureFinally, it is expanded isothermally to its original stateSketch the process in a PV diagram & calculate W, Q, DU, and DH for each of the three processes and for the entire cycle.Take CV = 3/2 R and CP = 5/2R
24Problem 3.8 Constant volume Contant temperature Adiabatically One mole of an ideal gas with CV = 5/2 R, CP = 7/2 R expands from P1 = 8 bar & T1 = 600 K to P2 = 1 bar by each of the following path:Constant volumeContant temperatureAdiabaticallyAssuming mechanical reversibility, calculate W, Q, DU, and DH for each of the three processes.Sketch each path in a single PV diagram
25Problem 3.9An ideal gas initially at 600 K & 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 1-2 pressure decreases isothermally to 3 bar; in step 2-3 pressure decreases at constant volume to 2 bar; in step 3-4 volume decreases at constant pressure; and in step 4-1 the gas returns adiabatically to its initial step.Sketch the cycle on a PV diagramDetermine (where unknown) both T & P for states 1, 2, 3, and 4Calculate W, Q, DU, and DH for each step of the cycleData: CV = 5/2 R, CP = 7/2 R
26Problem 3.10An ideal gas, CV = 3/2 R & CP = 5/2 R is changed from P1 = 1 bar & Vt1 = 12 m3 to P2 = 12 bar & Vt2 = 1 m3 by the following mechanically reversible processes:Isothermal compressionAdiabatic compression followed by cooling at constant pressureAdiabatic compression followed by cooling at constant volumeHeating at constant volume followed by cooling at constant pressureCooling at constant pressure followed by heating at constant volumeCalculate W, Q, DUt, and DHt for each of these processes, and sketch the paths of all processes on a single PV diagram
27Example 3.4A 400 gram mass of nitrogen at 27oC is held in a vertical cylinder by a frictionless piston. The weight of the piston makes the pressure of the nitrogen 0.35 bar higher than that of the surroundings atmosphere, which is at 1 bar & 27oC. Thus the nitrogen is initially at a pressure of 1.35 bar, and is in mechanical & thermal equilibrium with its surroundings. Consider the following sequence of process:The apparatus is immersed in an ice/water bath and is allowed to come to equilibriumA variable force is slowly applied to the piston so that the nitrogen is compressed reversibly at the constant temperature of 0oC until the gas volume reaches one-half the value at the end of step a. At this point the piston is held in place by latches.
28Example 3.4 (cont’)The apparatus is removed from ice/water bath and comes to thermal equilibrium with the surrounding atmosphere at 27oCThe latches are removed, and the apparatus is allowed to return to complete equilibrium with its surroundingsSketch the entire cycle on a PV diagram, and calculate Q, W, DU & DH for the nitrogen for each step of the cycle. Nitrogen may be considered an ideal gas for which CV = 5/2 R and CP = 7/2 R
30Cubic Equation of State: van der Waals Equation (1873)
31Theorem of Corresponding State; Accentric Factor All fluids, when compared at the same reduced temperature & reduced pressure, have approximately the same compressibility factor, and all deviate from ideal gas behavior to about the sam degreeAccentric Factor(Pitzer)
37Cubic Equation of State: Redlich/Kwong Equation (1949)
38Cubic Equation of State: Soave/Redlich/Kwong Equation (1972)
39Cubic Equation of State: Peng - Robinson Equation (1976)
40Example 3.8Given that the vapor pressure of n-butane at 350 K & bar, find the molar volumes of saturated vapor and saturated liquid n-butane at these conditions as given by:Van der WaalsRedlich/KwongSoave/Redlich/KwongPeng-Robinson
42Application of The Virial Equations At low to moderate pressure, it is common to use truncated virial equation: At pressure above the range of applicability of the above eqn, the appropriate form is: Benedict/Webb/Rubin equation: and its modifications are inspired by volume expansion virial eqn and are used in the petroleum & natural gas industries for light hydrocarbons.
43Example 3.7Reported values for the virial coefficients of isopropanol vapor at 200oC are: B = -388 cm3 mol C = cm6 mol-2Calculate V and Z for isopropanol at 200oC & 10 bar by:The ideal gas equationTwo term truncated pressure expansion virial equationThree term truncated volume expansion virial equation
44Pitzer Correlation for the Second Virial Coefficient
45Generalized Correlations 4 Gases: Pitzer type: Lee -Kessler TrZ0Pr
46Generalized Correlations 4 Gases: Pitzer type: Lee -Kessler TrZ0Pr
47Generalized Correlations 4 Gases: Pitzer type: Lee -Kessler TrZ1Pr
48Generalized Correlations 4 Gases: Pitzer type: Lee -Kessler TrZ1Pr
49Real Gas, EOSCalculate Z and V for steam at 250oC and 1,800 kPa by the following: a. Ideal gas equation b. Truncated virial equation with the following experimental values of virial coefficient: B = cm3 mol-1 C = -5,800 cm6 mol-2 c. Pitzer correlation for virial equation d. Pitzer-type correlation of Lee – Kessler e. Steam table f. van der Waals equation g. Redlich/Kwong equation h. Soave/Redlich/Kwong equation i. Peng-Robinson Equation
52Correlation for Liquids: Lydersen, Greenkorn & Hougen
53Problem 3.45A 30 m3 tank contains 14 m3 of liquid n-butane in equilibrium with its vapor at 25oC. Estimate the total mass of n-butane in the tank. The vapor pressure if n-butane at the given temperature is 2.43 bar.
55Ideal GasA cylinder pressure vessel having an inside diameter of 50 cm and height of 1.25 cm. Amount of the ammonia gas is confined in that a pressure vessel. Measured pressure and temperature of gas are 30 bars and 200oC. What are the specific volume and amount of the ammonia gas in pressure vessel (in SI unit), assuming ammonia is an ideal gas?
56Real Gas, EOSCalculate Z and for steam at 250oC and 1,800 kPa by the following: The ideal gas equation The truncated virial equation with the following experimental values of virial coefficient: B = cm3 mol-1 C = -5,800 cm6mol-2 The van der Waals equation The Redlich/Kwong equation The steam table Data: critical point of steam, TC = K; PC = bar; = 55.9 cm3mol-1
571st law after diagram1 kmole of ideal gas (Cv = 3R/2) undergoes a three-step mechanically reversible cycle in a closed system.Gas in initial state at 500 kPa and 27 oC is heated at constant pressureFollowed by adiabatic expansion until the its pressure become 120 kPaFinally step, gas is isothermally compressed to initial conditionFrom that process description:Illustrate the cycle process at P-V diagram in T parameter!b. Calculate H, U, Q and W for each step of process and total!