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
PV  n R T

or
PV  R T
• P & T are absolute
• R = (universal / ideal) gas constant
3
 pressure  volum e   M   1   L 
R
 2



 
 m oles   tem perature   L t   m oles  T 
Gas Constants
1.987 cal /g mol K
82.06 atm cm3 / g mol K
1.987 BTU / lb mol R
0.08206 atm L / g mol K
10.73 psia ft3 / lb mol R
21.9 in Hg ft3 / lb mol R
8.314 kPa m3 / kg mol K
0.7302 atm ft3 / lb mol R
8.314 J / g mol K
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
Partial Pressure Definition
nA
nB
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
PTV=nTRT
∑ Vi = VT (Amagat’s Law of partial volumes)
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
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
P
RT


a


V  b V (V  b)
R 2Tc2
a  0.42747
Pc
  [ 1   (1  Tr1/ 2 ) ]2
RT
b  0.08664 c
Pc
  (0.480 1.57  0.176 2 )
Plan B – Pure Real Gases
Compressibility Factor (Z)
PV = z nRT
z → fudge factor for non-ideal behavior
z → ƒ (Tr, Vr,Pr)
where:
T
Tr 
Tc
P
Pr 
Pc
V
Vr 
Vc
Compressibility Factor - Z
Generalized Compressibility Factor
Reduced Pressure - Pr
Real Gas Mixtures
Two approaches
1)
Zmean = ∑ zi yi
PV = Zmean n RT
2)
Z  f ( Pr' , Tr' )
P
P  '
Pc
'
r
Pc'   yi Pci
T
T  '
Tc
'
r
Tc'   yi Tci
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:
a) Assuming that the mixture obeys the ideal gas laws
b) 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
B
log10 ( p )  A 
T C
*
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.
ln( p
*
Toluene
3096.52
)  16.0137
(53.67  T )
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
RS 
pvapor
P*
pH 2O
RH  *
PH 2O
RS or RH 
n
vapor
n vapor at saturation
P*  vapor pressure at Tem perature T
Pvapor  partial pressure
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:
a) Percent relative saturation of the mixture.
b) Percent saturation of the mixture.
c) 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
Vapor
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
Toluene
Boiling Pt = 88.0 C
Benzene
Boiling Pt = 59.2 C
Benzene – Toluene
Temperature / Composition Plot
Vapor
Benzene – 0.55
Toluene – 0.45
Feed
P = 0.50 atm
T = 80 °C
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
yi  Ki xi
Relative Volatility
Ki
 ij 
Kj
Temperature
Vaporization “Flash” Curve
Flash curve
0.0
0.5
V/F
1.0
Bubble Point / Dew Point
*
1
*
2
Bubble Point
p x1 p x2
 yi  P  P  1.0
Total
Total
Dew Point
y1 PTotal y2 PTotal
 xi  p*  p*  1.0
1
2
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:
ln P *  A 
B
C T
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