ITK-233 Termodinamika Teknik Kimia I. Dicky Dermawan www.dickydermawan.net78.net firstname.lastname@example.org. 3 sks. 2 - PVT Behavior of Fluid, Equation of State. P – T Diagram P – v Diagram. T-v Diagram. Equation of State.
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2 - PVT Behavior of Fluid, Equation of State
For 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)
For liquid acetone at 20oC & 1 bar:
Β = 1.487 x 10-3oC-1 κ = 62 x x 10-6 bar-1 V= 1.287 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.
Express the volume expansivityβ 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 44.18 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.
Generally, volume expansivityβ and isothermal compressibility κ depend on T and P.
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.
For 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 = 0.125 cm3 g-1
Calculate 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)
Five 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
A 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.
For ideal gas: Z = 1, thus
The 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.
Thus H is also a funcion of temperature only
PV = constant
TV-1 = constant
TP-1 = constant
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 = ∞
Air 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:
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
An ideal gas undergoes the following sequence of mechanically reversible processes in a closed system:
Sketch 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
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:
Assuming mechanical reversibility, calculateW, Q, DU, and DH for each of the three processes.
Sketch each path in a single PV diagram
An 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.
Data: CV = 5/2 R, CP = 7/2 R
An 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:
Calculate W, Q, DUt, and DHtfor each of these processes, and sketch the paths of all processes on a single PV diagram
A 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:
Sketch 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
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 degree
Given that the vapor pressure of n-butane at 350 K & 9.4573 bar, find the molar volumes of saturated vapor and saturated liquid n-butane at these conditions as given by:
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:
and its modifications are inspired by volume expansion virialeqn and are used in the petroleum & natural gas industries for light hydrocarbons.
Reported values for the virial coefficients of isopropanol vapor at 200oC are:B = -388 cm3 mol-1 C = -26000 cm6 mol-2
Calculate V and Z for isopropanol at 200oC & 10 bar by:
Calculate 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 = -152.5 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
A 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.
A 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?
Calculate 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 = -152.5 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 = 647.1 K; PC = 220.55 bar; = 55.9 cm3mol-1
1 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 pressure
Followed by adiabatic expansion until the its pressure become 120 kPa
Finally step, gas is isothermally compressed to initial condition
From 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!