ep103 sen lnt sep11 1a
TRANSCRIPT
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Gases – Summary of Related equations
Notes prepared byProf Dr Hikmat Al Salim
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Gases – Behavior & Properties
Mathematical Descriptions KM- Model of GasesPVTn Kinetic EnergyImpirical Gas laws Molecular speedIdeal Gas laws Real GasesMixtures Application
Behavior & Properties
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Elements that exist as gases at 25 oC and 1 atmosphere
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Four Physical Quantities for GasesPhys. Qty. Symbol SI unit Other common
unitspressure P Pascal
(Pa)atm, mm Hg, torr, psi
volume V m3 dm3, L, mL, cm3
temp. T K °C, °F
moles n mol
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Physical Characteristics of Gases
• Gases assume the volume and shape of their containers.
• Gases are the most compressible state of matter.
• Gases will mix evenly and completely when confined to the same container.
• Gases have much lower densities than liquids and solids.
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Assumptions of the Kinetic Theory of Gases
1.Gases are composed of separate, tiny particles called molecules (or atoms)
2.Gas molecules are in constant, rapid, straight line motion (which means that gas molecules have kinetic energy ( KE = ½ mv² )
3.The collisions between molecules are completely elastic (when molecules collide, there is no exchange of energy
4.The molecules of a gas have no attraction or repulsion for each other
5.Each molecule in a gas has a different velocity
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6. Gas molecules exert neither attractive nor repulsive forces on one another.
7. These particles are so small, compared with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero).
8. The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas
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Elastic vs. Inelastic Collisions
8
3
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Particles in an ideal gas…
– have no volume.– have elastic collisions. – are in constant, random,
straight-line motion.– don’t attract or repel each
other.– have an avg. KE directly
related to Kelvin temperature.
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Particles in a REAL gas…
– have their own volume– attract each other
• Gas behavior is most ideal…– at low pressures– at high temperatures– in nonpolar atoms/molecules
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Pressure - Temperature - Volume Relationship
P T V
Gay-Lussac’s P T
Charles V T
P T
V P T V
Boyle’s P 1V ___
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P 1/VP x V = constant Both T & n constant
P1 x V1 = P2 x V2 Both T & n are constant
Boyle’s Law
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• Gas properties can be modeled using Gas properties can be modeled using math. Model depends on:math. Model depends on:
V = volume of the gas (L)V = volume of the gas (L)T = temperature (K)T = temperature (K)n = amount (moles)n = amount (moles)P = pressure (atmospheres )P = pressure (atmospheres )
ALL temperatures in the entire ALL temperatures in the entire semester MUST be in Kelvin!!! No semester MUST be in Kelvin!!! No Exceptions!Exceptions!
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The Quantity-Volume Relationship: Avogadro’s Law
• Avogadro’s Law: the volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.
• Mathematically:
The Gas LawsThe Gas Laws
V=constant x n
2
2
1
1
nV
nV
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Charles and Gay-Lussac's Law
The relationship between T & V, at const n & P is called Charles and Gay-Lussac's Law .Charles did the original work, which was verified by Gay-Lussac. They observed that if the P is held constant, the volume V is equal to a constant times the T V = constant X T .
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Ideal Gas Equation
Charles’ law: V T(at constant n and P)Avogadro’s law: V n(at constant P and T)
Boyle’s law: V (at constant n and T)1P
V nTP
V = constant x = R nTP
nTP
R is the gas constant
PV = nRT
Avogadro’s law expresses (quantity & volume relationship)
We can combine the above laws to give:
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R = 8.314 J K-1 mol-1 if Pressure is in kilopascals( kPa) Volume is in litres(L) Temperature is in Kelvin(K)
R = 0.0821 L atm K-1 mol-1 if Pressure is in atmospheres (atm) Volume is in litres(L) Temperature is in Kelvin(K)
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PV = nRT
R = PVnT =
(1 atm)(22.414L)(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. This is known as the standard molar volume.
To calculate R
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A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 = V2 x T1
V1
1.54 L x 398.15 K3.20 L= = 192 K
V1/T1 = V2/T2
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A steel tank has a volume of 438 L and is filed with 0.885 kg of O2. Calculate the pressure of O2 at 21oC.
29427321 T
molegkgn 7.27
32 885.0
atmxxVnRTP 53,1
4382940821.07.27
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What is the volume (in liters) occupied by 49.8 g of HCl at STP?
PV = nRT
V = nRTP
T = 0 0C = 273.15 K
P = 1 atm
n = 49.8 g x 1 mol HCl36.45 g HCl
= 1.37 mol
V =1 atm
1.37 mol x 0.0821 x 273.15 KL•atmmol•K
V = 30.6 L
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Argon is an inert gas used in light bulbs to retard the vaporization of the filament. A certain light bulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the light bulb (in atm)?
PV = nRT n, V and R are constantnRV = P
T = constant
P1
T1
P2
T2=
P1 = 1.20 atmT1 = 291 K
P2 = ?T2 = 358 K
P2 = P1 x T2
T1
= 1.20 atm x 358 K291 K
= 1.48 atm
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Combined Gas Law EquationFOR IDEAL GASES
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Density (D) Calculations and Molar Mass
d = mV = PM
RTm is the mass of the gas in gM is the molar mass of the gasMolar mass is the weight of one mole (or 6.02 x 1023 molecules) of any chemical compounds
Molar Mass (M ) of a Gaseous SubstanceFrom above equation, we get:
dRTPM = d is the density of the gas in g/L
5.4
gingastheofmassmRTMmPV
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MOLAR MASS and DENSITY
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EXAMPLE• Determine the molar mass of an unknown
gas that has a volume of 72.5 mL at a temperature of 68°C, a pressure of 0.980 atm, and a mass of 0.207 g.
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Exercise• What is the density of carbon tetrachloride
vapor at 714 torr and 125 oC ?• The molar mass of CCl4 is 154 g/mol
LgdKKmolatmLmolgtorratmtorrd
/43.4)15.398( / 0821.0
)/154)(760/1)(714(
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Gas Mixtures and Partial Gas Mixtures and Partial PressuresPressures
etcPPPPtotal .....321
•Dalton’s Law: in a gas mixture the total pressure is given by the sum of partial pressures of each component:. i.e :
•If each gas obeys the ideal gas equation, i.e.:
VRTnP ii
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Partial Pressures and Mole Fractions
• Let ni be the number of moles of gas i exerting a partial pressure Pi, then
where i is the mole fraction (ni/nt).totalii PXP
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Consider a case in which two gases, A and B, are in a container of volume V.
PA = nARTV
PB = nBRTV
nA is the number of moles of A
nB is the number of moles of B
PT = PA + PB XA = nA
nA + nBXB =
nB
nA + nB
PA = XA PT PB = XB PT
Pi = Xi PT
5.6
X = mole fraction
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The mass percentage composition of dry air at sea level is approximately N2: 75.5; O2: 23.2; Ar: 1.3. What is the partial pressure of each component when the total pressure is 1 atm?
Solution: assume the amount of each type of molecule present in 100g of air, thus the mass of N2, O2 and Ar are 75.5g, 23.2g and 1.3g
respectively.
45.30033.0725.069.2
033.0/95.39
3.1)(
725.0/322.23)(
69.2/02.28
5.75)(
2
2
Totalmol
molmolggArn
molmolggOn
molmolggNn Mole fractions = each mole / total mole
Partial = mole fraction X total pressure
0096.010096.0)(21.0121.0)(78.0178.0)(
0096.045.3/033.0)(21.045.3/725.0)(
78.045.3/69.2)(
2
2
2
2
XArpXOPxNP
ArxOxNx
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• Compressibility of Gases• Boyle’s Law
• P proportional to collision rate with wall• Collision rate a number density• Number density a 1/V• P proportional to 1/V
• Charles’ Law• P proportional to collision rate with wall• Collision rate a average kinetic energy of
gas molecules• Average kinetic energy proportional to T• P proportional to T
Kinetic theory of gases and …
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Kinetic theory of gases and …
• Avogadro’s LawP collision rate with wallCollision rate number densityNumber density nP n
• Dalton’s Law of Partial PressuresMolecules do not attract or repel one anotherP exerted by one type of molecule is unaffected by the
presence of another gasPtotal = Pi
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Stoichiometric Relationships
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Applications of the Ideal Gas Law
• Isobaric - constant pressure• Isothermal - constant temperature• Isochoric - constant volume• Isopycnic - constant density• Isosteric - constant specific
volume
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