Properties of Gas and Properties of Gas and VapoursVapours

GasGas

• 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 UnitsSystems of Units

• Sistem Internationale d’Unites

length 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)

• American engineering units

Length ft

Mass pound mass (lbm)

Mol lbm-mol (lbmmol)

time second (s)

temperature rankin (R)

Electrical current ampere (A)

Light intensity candela (cd)

ForceForce

• F=ma1 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

2 2m

2

f

1 kg m / s

N 1

g cm / sdyne

32.174 lb ft / s

lb

WeightWeight

• 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 VolumeMass 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

FlowratesFlowrates

• 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 CompositionChemical 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 weightMolecular 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

• Pressure is force per unit area– Unit : N/m2, dynes/cm2, dan lbf/in2

– SI Unit: N/m2 also known as Pascal (Pa)

P = Po + g/gc h

P(mm Hg) = Po (mmHg) + h (mmHg)

Pabs = Pgauge + Patm

Po (N/m2)

P (N/m2)h (m)

density

A (m2)

PressurePressurePressure = 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 fractionsMass and mole fractions

• Mass Fraction, xA

• Mole Fraction, yA

• Mass % is 100 xA, mole % is 100 yA

lb total

A lb or

g total

A g or

kg total

A kg

mass totalA mass

xm

mA

mole-lb total

A mole-lb or

mol totalA mol

or kmol total

A kmolmole total

A moleyA

Average molecular weightAverage molecular weight

• Based on mole fraction

• Based on mass fraction

component all

i2211 y iMMyMyM

component all2

2

1

11

i

i

M

x

M

x

M

x

M

ConcentrationConcentration

• 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

No. 68F (20C) is not double 50F (10C)

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 0C.• 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 scalesTemperature scales

Is 20C twice as hot as 10C?

Is 20 kg twice as heavy as 10 kg?

Temperature vs. VolumeTemperature vs. Volume

5

10

15

20

25

30

Vol

ume

(mL)

Temperature (C) 0 100

– 273

• 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.• –273C 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 + 273.62C = ? K: K=C+273 = 62 + 273 = 335 K

• Notice that kelvin is represented as K not K.

The Kelvin Temperature ScaleThe Kelvin Temperature Scale

Standard Conditions for gasesStandard 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 Ts Ps Vs ns

SI 273 K 1 atm 0.022415 m3 1 mol CSS 273 K 1 atm 22.415 L 1 mol American 492oR 1 atm 359.05 ft3 1 lb-mole

Normal Temperature & Pressure (NTP)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 ppmVapor 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

MWPmmHg

MWmmHgPC vapor

vapormequibliriu

82.5345.24760

106

NTPat 760

1010 66

mmHg

mmHgP

P

Pppm vapor

ambient

vapormequibliriu

Estimation of Vapor PressureEstimation of Vapor Pressure

)(]log[

TC

BAPvapor

Antoine Equation

Ideal GasesIdeal 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…

• 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

P = absolute pressure of the gasV = volume or volumetric flow rate of gasn = number of moles or molar flow rate of the gasR = the gas constantT = absolute temperature of gas

• Generally applicable at low pressure (< 1 atm) and temperature > 0oC.

Ideal Gas LawIdeal Gas Law

RTnV PnRT or PV

Ideal Gas LawIdeal Gas Law• The equation can also be written as

where = 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

PV RT

V

Example Example Application of Ideal Gas LawApplication 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

P

RTnVMWnm

83 HC

min

mol 65.7

K15.393 mol.K

L.bar 08314.0

min

L 250 bar1

RT

VPn

min

g337

mol

g09.44

min

mol65.7m

Working SessionWorking 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 Application of Standard Temperature and Pressure (STP)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

Standard cubic meters (SCM) m3(STP)Standard cubic feet (SCF) ft3(STP)

molelb

(STP)ft359

mol

)liters(STP2.24

mol

(STP)m0.0224

33

s

ss n

VV̂

Example – Conversion from Standard Example – Conversion from Standard Conditions (STP)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?

83 HCMWnm

min

g337

mol

g09.44

min

mol65.7m

min

mol 65.7

K15.393mol

L 4.22 bar01325.1

K15.273min

L 250 bar1

TV̂P

TVPn

ss

s

Working SessionWorking 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?

• If the input and output streams at indicated temperatures and pressures can be reasonably assumed to follow ideal gas behaviour, then

Effect of Temperature and Effect of Temperature and Pressure on Ideal GasesPressure on Ideal Gases

2

1

22

11

222111

T

T

VP

VP

nRT VP and nRT VP

Processn mol

V1, T1, P1

n mol

V2, T2, P2

Example - Standard and True Example - Standard and True Volumetric Flow RatesVolumetric 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,

h

mol1598

STP m0224.0

mol1 x

h

STPm8.35

3

3

Example 3 - Standard and True Example 3 - Standard and True Volumetric Flow RatesVolumetric Flow Rates

(ii) If the temperature and pressure of the gas are 30oC and 152 kPa, calculate the actual volumetric flow rate,

(Are there alternative methods to solve this problem?)

h

m8.26

Pa150000

K303K.mol

Pa.m314.8

h

mol1598

P

nRTV

3

3

Working SessionWorking 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 LawIdeal 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 : PV = nRT …. (1)From partial pressure: pAV = nART …. (2)

Dividing Eq. (1) by Eq. (2) : or 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

p

P

n

nyA A

A

Dalton’s Law and its applicationDalton’s Law and its application

i species offraction mole theis yi

ambientipartial PyP

i species offraction mole theis yi

1ambient

n

iipartialambienttotal PyPPP

NTPat 45.24

1010 66

MWCy

P

Pii

partial

NTPat 45.24

)/( 3

iii MW

mmgCppm

Ideal Gas Mixture - Amagat LawIdeal 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 : PV = nRT …. (1)From partial pressure: PvA = nA RT …. (2)

Dividing Eq. (1) by Eq. (2) : or vA = yA V

Thus the ideal partial volume of ideal gas add up to the total volume V :

vA + vB + vC + ... = (yA + yB + yC + ... )V = V

v

V

n

nyA A

A

Working SessionWorking Session

An ideal gas mixture contains 35% helium, 20% methane and 45% nitrogen by volume at 2.00 atm absolute and 90oC. Calculate

(a) the partial pressure of each component.(b) the mass fraction of each component.(c) the average molecular weight of the gas.(d) the density of the gas in kg/m3.

Vapor-Liquid SystemVapor-Liquid System

Multicomponent Gas-Liquid SystemsMulticomponent Gas-Liquid Systems

A, B(liquid)

yA, yB

@ T , PpA = yAP; pB = yBP

F = 2 + C - P = 2 specify 2 of T,P, yH2O

In general, From Raoult’s and Dalton’s Law

Valid• when xA ==> 1.0, • For the entire range of compositions for mixtures of similar substances

pA= yAP = xApA*(T)

xA, xB

Multicomponent Gas-Liquid SystemsMulticomponent Gas-Liquid Systems

Henry’s Law (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

pA= yAP = xAHA(T)

Air, NH3

@ T , P

water

water, NH3

Air, (less) NH3

===> (A = NH3)

End ofEnd ofProperties of Gases & VapoursProperties of Gases & Vapours