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 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) 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)

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

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) 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)

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 Mole Fraction, yA Mass % is 100 xA, mole % is 100 yA

Average molecular weight Based on mole fraction Based on mass fraction

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 No. 68F (20C) is not double 50F (10C) Is 20C twice as hot as 10C? Is 20 kg twice as heavy as 10 kg? No. 68F (20C) is not double 50F (10C) 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 25 mL at 22C 31.6 mL, 23.1 mL 10 15 20 25 30 Volume (mL) 25 mL at 22C 31.6 mL, 23.1 mL Y=0.0847x + 23.137 Temperature (C) 100 – 273

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 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 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

Estimation of Vapor Pressure 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 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 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

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

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 Standard cubic meters (SCM)  m3(STP) Standard cubic feet (SCF)  ft3(STP) Solution Molar quantities and absolute temperature and pressure must be used n = 1100 kg/h = 19.0 kmol/h 58.1 kg/kmol T = 633 K, P = 3.00 atm From standard equation we have V = 19.0 kmol 22.4 m3 (STP) 633 K 1.00 atm = 329 m3/h h kmol 272 K 3.00 atm

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?

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 Process n mol 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

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,

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, (Are there alternative methods to solve this problem?)

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 : 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

Dalton’s Law and its application

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 : 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

Working Session An ideal gas mixture contains 35% helium, 20% methane and 45% nitrogen by volume at 2.00 atm absolute and 90oC. Calculate the partial pressure of each component. the mass fraction of each component. the average molecular weight of the gas. the density of the gas in kg/m3.

Vapor-Liquid System

Multicomponent Gas-Liquid Systems yA, yB A, B (liquid) @ T , P pA = yAP; pB = yBP xA, xB 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)

Multicomponent Gas-Liquid Systems Air, (less) NH3 water @ T , P Air, NH3 water, NH3 pA= yAP = xAHA(T) 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 ===> (A = NH3)

End of Properties of Gases & Vapours