Properties of Gas and
Vapours
Gas
• Atoms and molecules are constantly
moving
•Gases are formed when the energy in
the system exceeds all of the attractive
forces between molecules
•In the gas state, molecules move
quickly and are free to move in any
direction, spreading out long distances
• They have little interaction with each
other beyond occasionally bumping
into one another
•Because individual molecules are
widely separated and can move around
easily in the gas state, gases can be
compressed.
Systems of Units
• Sistem Internationale d’Unites
• American engineering units
length
Length
ft
Mass
pound mass (lbm)
Mol
lbm-mol (lbmmol)
time
second (s)
temperature rankin (R)
Electrical current ampere (A)
Light intensity candela (cd)
meter (SI)
centimeter (CGS)
mass
kilogram (SI)
gram (CGS)
Mol
gram-mol (mol)
time
second (s)
temperature kelvin (K)
Electrical current ampere (A)
Light intensity candela (cd)
Force
• F=ma
1 newton (N) = 1 kg. m/s2
1 dyne = 1 g.cm/s2
1 lbf = 32.174 lbm.ft/s2
• In changing the unit of force, a conversion
factor gc is required
gc  1
kg  m / s
N
2
 1
g  cm / s
dyne
2
 32. 174
lb m  ft / s
lb f
2
Weight
• A class of force
W = mg/gc
• Value of g/gc at 45o latitude
g = 9.8066 m/s2
g/gc = 9.8066 N/kg
gc = 1 kgm/Ns2
g = 980.66 cm/s2
g/gc = 980.66 dyne/g
gc = 1 g.cm/dyne.s2
g = 32.174 ft/s2
g/gc = 1 lbf/ lbm
gc = 32.174 lbm.ft/lbf.s2
Mass and Volume
• density ( ) is mass per volume
– kg/m3, g/cm3, and lbm/ft3
• Specific volume is volume per mass
– m3/kg, cm3/g, and ft3/lbm
– Inverse of dnsity
• Specific gravity is the ratio of density  to ref
– SG = / ref
– Common reference  is water at 4.0 oC
– ref(H2O, 4.0oC )
=
1.000 g/cm3
=
1000 kg/m3
=
62.43 lbm/ft3
Flowrates
• Mass flowrate (mass/time)
kg/s or lbm/s
• Volumetric flowrate (volume/time) m3/s or ft3 /s
• Fluid density () can be used to convert the
mass flowrate to volumetric flowrate or vice
versa.
m (kg fluid/s)
V (m3 fluid/s)
Chemical Composition
• Atomic weight – mass of an atom at the
scale given to 12C having mass of 12.
• Molecular weight (MW) –total weight of all
atoms in a molecule
– Atomic weight of oxygen (O) = 16
– Molecular weight of oxygen (O2) = 32
• Gram-mole or mole- molecular weight
– Units used - gmol, lbm-mol, kmol
Molecular weight
• How many moles in 34 kg ammonia (NH3): (MW
= 17)
34 kg NH3 1 kmol NH3 = 2 kmol NH3
17 kg NH3
• 4 lb-moles ammonia
4 lb-mole NH3 17 lbm NH3
= 68 lbmNH3
1 lb-mole NH3
• one gram-mol consists of 6.02 x 10 23
(Avogadro number) molecule
Pressure
• Pressure is force per unit area
– Unit : N/m2, dynes/cm2, dan lbf/in2
– SI Unit: N/m2 also known as Pascal (Pa)
Po (N/m2)
A (m2)
h (m)
P (N/m2)
density
P = Po +  g/gc h
P(mm Hg) = Po (mmHg) + h (mmHg)
Pabs = Pgauge + Patm
Pressure
Pressure = Force / area
P at sea level 760 mmHg
P other altitude = Psea level x 2(-altitude in ft/18000) mm Hg
Pother altitude = 760x 2(-altitude in ft/18000) mmHg
Pother altitude = 1x 2(-altitude in ft/18000) atm
For ventilation, unit of P is in of water
1 atm = 407.6 inch of water
Pabsolute = Pgauge + ambient Pressure
Mass and mole fractions
• Mass Fraction, xA
xA

kg A


total mass  total kg
mass A
or
gA
total
g
or
lb m A
total lb m




• Mole Fraction, yA
yA
mole
A
 kmol A


total mole  total kmol
or
mol A
total mol
or
lb - mole A
total
• Mass % is 100 xA, mole % is 100 yA
lb - mole



Average molecular weight
• Based on mole fraction
M  y1 M 1  y 2 M 2   
yM
i
all component
• Based on mass fraction
1
M

x1
M1

x2
M2
 

all component
xi
Mi
i
Concentration
• Mass concentration (mass of component per
volume of solution) (g/cm3, lbm/ft3 or kg/m3)
• Mole concentration is the number of moles of
component per unit volume of the solution
(mol/cm3, lb-mol/ft3 or kmol/m3)
• Molar is gram-mol of solute per liter of
solution
Temperature scales
Is 20C twice as hot as 10C?
No. 68F (20C) is not double 50F (10C)
Is 20 kg twice as heavy as 10 kg?
Yes. 44 lb (20 kg) is double 22 lb (10 kg)
What’s the difference?
• Weights (kg or lb) have a minimum value of 0.
• But the smallest temperature is not 0C.
• We saw that doubling P yields half the V.
• Yet, to investigate the effect of doubling temperature, we first have to
know what that means.
• An experiment with a fixed volume of gas in a cylinder will reveal the
relationship of V vs. T…
Temperature vs. Volume
30
25
15
10
5
– 273
0
Temperature (C)
100
Volume (mL)
20
The Kelvin Temperature Scale
• If a volume vs. temperature graph is plotted for
gases, most lines can be interpolated so that when
volume is 0 the temperature is -273 C.
• Naturally, gases don’t really reach a 0 volume, but the
spaces between molecules approach 0.
• At this point all molecular movement stops.
• –273C is known as “absolute zero” (no EK)
• Lord Kelvin suggested that a reasonable temperature
scale should start at a true zero value.
• He kept the convenient units of C, but started at
absolute zero. Thus, K = C + 273.
62C = ? K: K=C+273 = 62 + 273 = 335 K
• Notice that kelvin is represented as K not K.
Standard Conditions for gases
• Using PVT equation is easy, provided you have a
set of R constant value with different units.
• A way to avoid this is by dividing the gas law
from process condition with given chosen
reference condition
Standard Conditions for gases
System
SI
CSS
American
Ts
273 K
273 K
492oR
Ps
1 atm
1 atm
1 atm
Vs
0.022415 m3
22.415 L
359.05 ft3
ns
1 mol
1 mol
1 lb-mole
Normal Temperature & Pressure (NTP)
• Normal Condition (NTP)
Pressure = 1 atm
25 oC, gas fulfill 24.45 L considered normal
21 oC considered normal by ACGIH for ventilation,
considered a typical thermostat setting at home.
20 oC is considered normal by NIOSH
By default we consider 25 oC as normal
Vapor Pressure and ppm
• Vapor pressure is the partial pressure exerter
by airborne molecules that is in equilibrium
with its liquid .
Cequilibrium in mg/m3 can be computed using by
the following relationships
C equibliriu m 
ppm
Pvapor mmHg  MW  10
760 mmHg  24 . 45
equibliriu m

Pvapor  10
Pambient
6

6
 53 . 82  Pvapor  MW
Pvapor mmHg  10
760 mmHg
6
at
NTP
Estimation of Vapor Pressure
log[ Pvapor ]  A 
B
(C  T )
Antoine Equation
Ideal Gases
• An “ideal” gas exhibits certain theoretical
properties. Specifically, an ideal gas …
–
–
–
Obeys all of the gas laws under all conditions.
Does not condense into a liquid when cooled.
Shows perfectly straight lines when its V and T & P and T
relationships are plotted on a graph.
• In reality, there are no gases that fit this definition
perfectly. We assume that gases are ideal to
simplify our calculations.
• We have done calculations using several gas laws
(Boyle’s Law, Charles’s Law, Combined Gas Law).
There is one more to know…
Ideal Gas Law
• Equation of state relates the quantity (mass or moles) and
volume of gas with temperature and pressure
• The simplest and mostly used is ideal gas law
PV  nRT or
P V  n RT
P = absolute pressure of the gas
V = volume or volumetric flow rate of gas
n = number of moles or molar flow rate of the gas
R = the gas constant
T = absolute temperature of gas
• Generally applicable at low pressure (< 1 atm) and
temperature > 0oC.
Ideal Gas Law
• The equation can also be written as
PV  RT
where V
= V/n is the molar volume of the gas
• Any gas is presented by the above equation is known as an
ideal gas or perfect gas
• 1 mol of ideal gas at 0oC and 1 atm occupies 22.415 L,
whether the gas is argon, nitrogen, or any other single
species or mixture of gases
Example
Application of Ideal Gas Law
Propane at 120oC and 1 bar absolute passes through a flow meter
that reads 250 L/min. What is the mass flow rate of the gas?
How many ways can we calculate the mass flow rate? What additional
information is needed?
…. Using ideal gas law

  n MW
m
n 
P V
RT
  7 . 65
m
C3H8

1

n RT
V 
P
bar

 0 .08314

mol
min


 250

L.bar
mol.K





min 
L
 393 .15
 7 . 65
K
g 
g

 44 .09
  337
mol 
min


mol
min
Working Session
100 g/h of ethylene (C2H4) flows through a pipe at 120oC and 1.2
atm and 100 g/h of butene (C4H8) flows through a second pipe at the
same pressure and temperature. Which of the following quantities
differ for the two gases;
(a) the volumetric flowrate
(b) specific molar volume (L/mol)
(c) mass density (g/L)
(Assume ideal gas behaviour)
Application of Standard Temperature and
Pressure (STP)
• Reference temperature (0oC, 273K, 32oF and 492oR) and
pressure (1 atm) are commonly known as STP
• The other related values is easy to commit to memory like
the relation of
Vˆ s 
Vs
ns
3
 0.0224
m (STP)
mol
 2.24
liters(STP )
mol
Standard cubic meters (SCM)  m3(STP)
Standard cubic feet (SCF)  ft3(STP)
3
 359
ft (STP)
lb  mole
Example – Conversion from Standard
Conditions (STP)
Recall Example 1…….
Propane at 120oC and 1 bar absolute passes through a flow meter
that reads 250 L/min. What is the mass flow rate of the gas?

  n MW
m
P VT s
n 

ˆ
P V T
s
s
m  7 . 65
C3H
8

1

bar  250

1 .01325
mol
min

  273 .15 K
min 
L

bar  22 .4



  393 .15 K
mol 
g 
g

 44 .09
  337
mol 
min

L
 7 .65

mol
min
Working Session
The pressure gauge on a 20 m3 of nitrogen at 25oC reads 10 bar.
Estimate the mass of nitrogen in the tank by
(i) direct solution of the ideal gas equation of state and
(ii) conversion from standard conditions.
What does pressure reading obtained from a pressure gauge
reading indicate?
Effect of Temperature and Pressure
on Ideal Gases
n mol
n mol
Process
V1, T1, P1
V2, T2, P2
• If the input and output streams at indicated temperatures
and pressures can be reasonably assumed to follow ideal gas
behaviour, then
P1V 1  nRT

1
P 2 V 2  nRT
and
P1V 1
P 2V 2

T1
T2
2
Example - Standard and True
Volumetric Flow Rates
The volumetric flow rate of an ideal gas is given as 35.8 SCMH (i.e
m3/h at STP).
(i) Calculate the molar flow rate (mol/h),
(ii) If the temperature and pressure of the gas are 30oC and 152
kPa, calculate the actual volumetric flow rate.
_______________________________________________________
(i) Molar flow (mol/h) at STP
At STP, 1 mol of an ideal gas occupies 0.0244 m3. Thus,
35 .8
m
3
 STP 
h
x
1 mol
0 .0224 m
3
 STP 
 1598
mol
h
Example 3 - Standard and True Volumetric
Flow Rates
(ii) If the temperature and pressure of the gas are 30oC and 152
kPa, calculate the actual volumetric flow rate,
V 
nRT
P

3
mol  
m . Pa

 1598
  8 . 314
h  
mol . K

150000
Pa


  303 K



 26 . 8
(Are there alternative methods to solve this problem?)
m
h
3
Working Session
A stream of air enters a 7.50-cm ID pipe at a velocity of 60 m/s at
27oC and 1.80 bar (gauge). At a point downstream, the air flows
through a 5.00-cm ID pipe at 60oC and 1.53 bar (gauge). What is
the velocity of the gas at this point?
Ideal Gas Mixtures - Dalton Law
•
•
Suppose nA moles of substance A, nB moles of B and nC moles of C
and so on, are contained in a volume V at temperature T and total
pressure P.
The partial pressure pA of A in the mixture is defined as the
pressure exerted by nA moles of A alone occupied at the same total
volume V only for ideal gases at the same temperature T
From ideal gas law :
From partial pressure:
Dividing Eq. (1) by Eq. (2) :
PV = nRT
pAV = nART
…. (2)
pA
P

nA
n
 yA
or
…. (1)
pA = yA P
Thus, the ideal partial pressure of ideal gas add up to the total
pressure P
pA + pB + pC + ... = (yA + yB + yC + ... )P = P
Dalton’s Law and its application
Ptotal  Pambient 
P
y i is the mole fraction
n
partial


y i Pambient
i 1
of species i
Ppartial  y i  Pambient
y i is the mole fraction
of species i
Ppartial
6
P
 10  y i   10  C i
6
ppm i  C i ( mg / m )
3
24 . 45
MW i
24 . 45
MW
at NTP
at NTP
Ideal Gas Mixture - Amagat Law
•
•
Suppose nA moles of substance A, nB moles of B and nC moles of
C and so on, are contained in a volume V at temperature T and total
pressure P.
The partial volume vA of A in the mixture is defined as the volume
that would be occupied by nA moles of A alone only for ideal gases at
the total pressure P at the same temperature T of the mixture
From ideal gas law :
From partial pressure:
Dividing Eq. (1) by Eq. (2) :
PV = nRT
PvA = nA RT …. (2)
vA
V

nA
n
 yA
or
Thus the ideal partial volume of
ideal gas add up to the total volume V :
vA + vB + vC + ... = (yA + yB + yC + ... )V = V
…. (1)
vA = y A V
Working Session
An ideal gas mixture contains 35% helium, 20% methane and 45%
nitrogen by volume at 2.00 atm absolute and 90oC. Calculate
(a)
(b)
(c)
(d)
the
the
the
the
partial pressure of each component.
mass fraction of each component.
average molecular weight of the gas.
density of the gas in kg/m3.
Vapor-Liquid System
Multicomponent Gas-Liquid Systems
A, B
(liquid)
yA, yB
@T,P
pA = yAP; pB = yBP
xA, xB
F=2+C-P=2
In general,
From Raoult’s and Dalton’s Law
specify 2 of T,P, yH2O
pA= yAP = xApA*(T)
Valid
• when xA ==> 1.0,
• For the entire range of compositions for mixtures of similar substances
Multicomponent Gas-Liquid Systems
Air,
(less) NH3
water
@T,P
Air, NH3
water, NH3
Henry’s Law
pA= yAP = xAHA(T)
===> (A = NH3)
(HA(T) - Henry’s law constant for a specific solvent)
Valid when xA ==> 0.0, provided that A does not dissociate, ionize
or react in the liquid phase
==> often applied to solutions of non-condensable gases
End of
Properties of Gases & Vapours