Ideal Gases

Ideal Gases All properties = ƒ(T, p) “compressible” Describe ƒ(T, p) with equations of state Consider gas in a balloon P V = nRT • at constant p & T ..... V ~ n ........ n = number of moles • at constant T ..... V ~ 1/p ..... Boyle's Law • at constant p ..... V ~ T ..... Charles' Law

• thus, V ~ nT/p • Equations of State .... change ~ to = ..... Equation • the simplest equation of state is the Ideal Gas Law or P & T are absolute R = (universal / ideal) gas constant

Gas Constants 1.987 cal /g mol K 1.987 BTU / lb mol R 10.73 psia ft3 / lb mol R 8.314 kPa m3 / kg mol K 8.314 J / g mol K 82.06 atm cm3 / g mol K 0.08206 atm L / g mol K 21.9 in Hg ft3 / lb mol R 0.7302 atm ft3 / lb mol R 1545.3 ft lbf / lb mol R

Standard Conditions System T P V / n SI 273.15 K 101.325 kPa 22.415 m3 / kg mol Universal Scientific 0.0 C 760 mm Hg 22.415 liters / g mol Natural Gas Industry 60 F 14.696 psia 379.5 ft3 / lb mol American Engineering 32 F 1 atm 359.05 ft3 / lbmol

Ideal Gas Mixtures PTV=nTRT Partial Pressure Definition nA+nB = nT (mole balance) Therefore: PA + PB = PT at constant T & V ∑ Pi = PT (Dalton’s Law of partial pressure) Likewise for constant P & T ∑ Vi = VT (Amagat’s Law of partial volumes) nA nB PTV=nTRT

Example An old way of producing hydrogen gas in the laboratory was by the reaction of sulfuric acid with zinc metal H2 SO4(l) + Zn(s) → ZnSO4(s) + H2(g) How many grams of sulfuric acid solution (98%) must act on an excess of zinc to produce 12.0 m3/hr of hydrogen at standard conditions. Assume all the acid used completely reacts.

10 Minute Problem R SO2 100 % SO2 P Air SO2 + ½ O2 → SO3 Converter In a sulfuric acid plant, sulfur is burned in the presence of excess oxygen to produce sulfur dioxide which in turn is further reacted in the next step with oxygen in a converter to produce sulfur trioxide. In the plant SO2 along with 10% excess air is fed into the process which operates at 1500°C and 1 atm. The per pass conversion of SO2 is 75% and overall conversion is 100%. If 106 m3/hr of SO2 at 1100°C and 1 atm is fed to the process as fresh feed, calculate the: SO2 + ½ O2 → SO3 (a) flow rate of the product stream P in m3/hr at 1500°C and 1 atm and its composition in mole percent; (b) flow rate of the recycle stream R in m3/hr at 1500°C and 1 atm. R SO2 100 % SO2 Converter P Air Separator

Extra Practice Problems Handout Problem Set: III-1 – III-37

Real Gases

“Real” Gas Behavior There is no such thing as an ideal gas There are conditions under which a gas can be adequately described by the ideal gas law (IGL) Conditions such that: • molecules are far enough apart • molecules are small enough • neglect inter-molecular forces and molecular volumes • under different conditions, the same gas may not be adequately described by IGL — need a more realistic law — derived from “ideal” behavior

Plan A – Pure Real Gases Equation of State Relates, P,V,T Soave-Redlich-Kwong (SRK) Equation of State

Plan B – Pure Real Gases Compressibility Factor (Z) PV = z nRT z → fudge factor for non-ideal behavior z → ƒ (Tr, Vr,Pr) where:

Compressibility Factor - Z Generalized Compressibility Factor Reduced Pressure - Pr

Real Gas Mixtures Two approaches 1) Zmean = ∑ zi yi PV = Zmean n RT 2) pseudo-critical properties

Example Seven pounds of N2 at 120°F are stored in a cylinder having a volume of 0.75 ft3. Calculate the pressure in atmospheres in the cylinder. (a) assuming N2 to be an ideal gas (b) using the compressibility factor method (c) using the SRK equation of state

Example Figure

10 Minute Problem A gas analyzes 60% methane and 40% ethylene by volume. It is desired to store 12.3 kg of this gas mixture in a cylinder having a capacity of 5.14 X 10-2 m3 at a maximum temperature of 45°C. Calculate the pressure inside the cylinder by: Assuming that the mixture obeys the ideal gas laws Using the compressibility (Z) factor determined by the pseudo critical point method.

Extra Practice Problems Handout Problem Set: III-38 – III-46

Single Component Two-Phase Systems

Single-Component Phase Equilibria It is possible to represent phase behavior on a 2-D plot (a phase diagram)

Pure Component (Water) Phase Diagram

Water Phase Diagram Application Steam Production in a Coal Fired Boiler

Vapor Pressure Estimation Antoine Equation Where: A,B,C constants p* - vapor pressure (mm Hg) T – temperature (°C)

Steam Table

Example Calculate the vapor pressure of benzene at 50°C using the Antoine Equation. Also estimate the normal boiling point of benzene (the vapor pressure at 1 atm), and compare it with the experimental value of 353.26 (taken from a handbook).

Two Phase Gas – Liquid Systems

Saturation Consider chamber of dry air at 75°C & 760 mm Hg Air is typically considered as one component • Introduce liquid water to the bottom of the chamber • Water begins to evaporate — molecules leave liquid and become vapor • Mole fraction of water in the gas, yw, increases • Thus, partial pressure, pw = yw * p increases (assumption: H2O + air is an ideal gas = good assumption) • Eventually the system reaches equilibrium that is, molecules evaporate at same rate as they condense and there is no change in the composition of either phase

At equilibrium, the gas phase is saturated with water that is, the gas holds all the water it can at this T & p the H2O in the gas phase is referred to as saturated vapor. • What is the relationship between composition of the phases and T & p ? • specifying T & P fixes the composition of both phases • specifying the composition of both phases sets T & P • if a gas at T & p contains a saturated vapor of mole fraction yv, and this vapor is the only condensable species, then the partial pressure of the vapor in the gas equals the pure component vapor pressure at the system temperature. pv = yvp = pv* (T) • for pv* (T) of water as saturated steam, see steam tables

Steam Table Examples Air & liquid water in equilibrium in closed vessel • at 180°F & 14.696 psia: pv* (180°F) = 7.511 psia • yv = 7.511/14.696 = 0.511 moles H2O/mole of mix • at 212°F & 14.696 psia: pv* (212°F) = 14.696 psia • yv = 1 moles H2O / mole of mix • at 212°F & 14.696 psia: specific volume of the liquid = 0.016719 ft3 / lbm specific volume of the vapor = 26.799 ft3 / lbm

Example Dry air at 25°C is saturated with toluene under a total pressure of 760 mm Hg abs. Is there adequate air for complete combustion of all the toluene? If so, determine the percent excess air present for combustion. Where: T → K P → mm Hg

Partial Saturation • partial saturation occurs when you heat a saturated vapor • partial saturation occurs when contact between the liquid and gas is too short to establish equilibrium pv = yvp < p*v (T) • a vapor present in a gas at less than its saturation amount is referred to as superheated and is not at equilibrium • if a gas with a single superheated vapor is cooled at constant pressure, the temperature at which the vapor becomes saturated is referred to as the dew point • at the dew point, pv = yv p = pv* (Tdp) • the difference between the system temperature and the dew point is referred to as the degrees of superheat

Relative Saturation / Relative Humidity Saturated steam: The space above the liquid can hold no more liquid vapor at the given T and p, the liquid and vapor are in equilibrium. Superheated steam: The vapor space is heated beyond the saturation condition (that is, the space is not saturated with water vapor) the liquid and vapor are not in equilibrium.

Example A gas mixture contains 0.0083 g mol of water vapor per g mol of dry CH4 at a temperature of 27°C and a total pressure of 200 kPa. Calculate the: Percent relative saturation of the mixture. Percent saturation of the mixture. Temperature to which the mixture must be heated at 200 kPa in order that the relative saturation will be 0.20.

10 Minute Problem Air saturated with water vapor at 80°F and 745.0 mm Hg abs. is passed through an air compressor and then stored in a tank at 25.0 psig and 100°F. What percentage of the water originally in the air was removed during the processing?

Vapor-Liquid Equilibrium A,B Feed A,B,C,D Liquid B,C,D

Vapor-Liquid Equilibrium V–L Systems have More Than One Condensable Component More than one component can be transferred from one phase to another Each phase may or may not contain some of each component Specifying T & p fixes the composition of each phase Specifying T (or p) and the composition of either phase, fixes p (or T) and the other composition

Benzene – Toluene Temperature / Composition Plot Boiling Pt = 88.0 C Benzene Boiling Pt = 59.2 C

Benzene – Toluene Temperature / Composition Plot Vapor Benzene – 0.55 Toluene – 0.45 P = 0.50 atm T = 80 °C Feed Liquid Benzene – 0.10 Toluene – 0.90

Henry’s Law For dilute solutions of any compound A (ie, xA < 5.0 mole percent) pA = yA ptotal = xA HA where HA is the Henry's Law constant HA values found in many books

Raoult’s Law For ideal solutions of any compound A pA = yA ptotal = xA pA* Raoult’s Law Distribution coefficient, equilibrium constant, “K” factor - KA KA = yA / xA = pA* / ptotal

“K” Factor Chart

Relative Volatility

Vaporization “Flash” Curve Temperature Flash curve 0.0 0.5 1.0 V/F

Bubble Point / Dew Point

Example Suppose that a liquid mixture of 75.0 mol % benzene in toluene is vaporized, What is the composition of the first vapor formed if the total pressure is 0.50 atm ? Vapor Pressure (Antoine Equation) Data: Where: p* is in mm Hg, T is in K Component A B C benzene 15.9008 2788.51 -52.36 toluene 16.0137 3096.52 -53.67