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Transcript of 1 TOPIC 7: GASES Contents Properties of Gases The Simple Gas Laws The Ideal Gas Equation Gases in...
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TOPIC 7: GASESContents
•Properties of Gases
•The Simple Gas Laws
•The Ideal Gas Equation
•Gases in Chemical Reactions
•Mixture of Gases
•Kinetic-Molecular Theory of Gases
•Gas Properties Relating to the Kinetic-Molecular Theory
•Nonideal(Real) Gases
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PROPERTIES OF GASES
• Pressure : A force per unit area
Force(N)
Area(m2) P(Pa) =
Liquid Pressure P= g x h x d
g: acceleration of gravity
h: height of the liquid column
d: density
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BAROMETRIC PRESSURE
Standard Atmospheric (Barometric) Pressure
1,00 atm= 760 mmHg, 760 torr
101,325 kPa
1,01325 bar
1013,25 mbar
Atmospheric (Barometric)
pressure)
dHg = 13,5951 g/cm3 (0°C)
g = 9,80665 m/s2
Evangelista Torricelli, 1643
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Manometers
Gas pressure equal to barometric pressure
Gas pressure greater than barometric pressure
Gas pressure less than barometric pressure
Measurement of Gas Pressure using an open sided manometer
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The Simple Gas Laws
• Boyle’s Law P α 1V
PV = Constant
For a fixed amount of gas at a constant temperature, gas volume is inversely proportional to a gas pressure.
P1 V1 = P2 V2
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Charles’s LawCharles 1787 Gay-Lussac 1802
Volume (mL)
V α T V = b T
Temperature (oC)
Volume (mL)
Temperature (K)
The volume of a fixed amount of gas at a constant pressure is directly proportional to the Kelvin(absolute) temperature.
Absolute temperature scale: 273,15oC or 0 K,
T(K)= t(oC)+ 273,15
V1 V2
T1 T2
=
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Avogadro’s Law• Gay-Lussac 1808
Gases react by volumes in the ratio of small whole numbers.Avogadro 1811
Equal volumes of different gases compared at the same temperature and pressure contain equal number of molecules.
Equal numbers of molecules of different gases compared at the same temperature and the same pressure occupy equal volumes.
V α n or V = c n
Standard Conditions (0 C= 273,15 K and 1atm= 760 mm Hg)
1 mol gas = 22,414 L at STP
At a constant pressure and temperature:
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Combination of all Gas Laws: The ideal Gas Equation
• Boyle’s Law V α 1/P
• Charles’s Law V α T
• Avogadro’s Law V α n
PV = nRT
V α nTP
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Gas Constant
R = PVnT
= 0,082057 L atm mol-1 K-1
= 8.3145 m3 Pa mol-1 K-1
PV = nRT
= 8,3145 J mol-1 K-1
= 8,3145 m3 Pa mol-1 K-1
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Example: What is the volume occupied by 13,7 g Cl2(g) sample at 45 C and 745 mm Hg?
1 atm = 760 mmHg; R = 0,08206 L atm /(mol K); Cl:35,5
193,071
17,13980,0
760
174515,31815,27345
2
2 Clg
Clmolgn
mmHg
atmmmHgPT
Solution
LV
atm
KKmolatmLmolV
P
nRTV
14,5
980,0
15,318)/(08206,0193,0
Practice: What is the pressure exerted by 1,00 x 1020 molecules of N2 in a 350 ml volume of container at 175 C ?
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General Gas Equation
R = = P2V2
n2T2
P1V1
n1T1
=
PsVs
nsTs
PiVi
niTi
We often apply it in cases in which one or two of the gas properties are held constant and we can simplify the equation by eliminating these constants
In the cases of the comparison of two gases, General Gas Equation must be used. In other cases the ideal gas equation is rather relevant.
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Applications of the Ideal Gas Equations
Is the amount of gas given or asked?
Use the General Gas Equation by comparing the initial and final conditions
PiVi = PsVs
Ti Ts
Vi=Vs
Pi = Ps
Ti Ts
No
Yes If the mass of gas is constant use the Ideal Gas Equation PV=nRT
If the mass of gas is variable use the
General Gas Equation.
PiVi = PsVs
niTi nsTs
Boyle’s Law
PiVi = PsVs
Ti=Ts
Vi = Vs
Ti Ts
Pi = Ps
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Molar Mass DeterminationPropylene is an important commercial chemical. It is used in the synthesis of other organic chemicals and in plastic production. A glass vessel weighs 40,1305 g, when clean,dry and evacuated ; 138,2410 g when filled with water at 25°C (density of water δ= 0,9970 g/cm3) and 40.2959 g when filled with propylene gas at 740,3 mm Hg and 24,0°C . What is the molar mass of propylene?
Strategy:
Find out Vves , mgas ; Use the Ideal Gas Equation
Vves = mH2O / dH2O = (138,2410 g – 40,1305 g) / (0,9970 g cm-3)
mgas:
= 0,1654 g
mgas = m - mempty= (40,2959 g – 40,1305 g)
= 98,41 cm3 = 0,09841 L
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Ideal Gas Equation:
PV = nRT PV = mM
RT M = m
PVRT
M = (0,9741 atm)(0,09841 L)
(0,6145 g)(0,08206 L atm mol-1 K-1)(297,2 K)
M = 42,08 g/mol
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Gas Densities
PV = mM
RT MP
RTVm
= d =
PV = nRT ve d = mV
, n = mM
Gas densities differ from solid and liquid densities in two important ways:
1- Gas densities depend strongly on temperature and pressure; increasing as the gas pressure increases, decreasing as the temperature increases. Densities of liquids and solids also depend somewhat on temperature, but they depend far less on pressure
2- The density of a gas is directly proportional to its molar mass. No simple realtionship exists between density and molar mass for liquids and solids.
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Gases in Chemical Reactions
• Use the stoichiometric factors to relate the amount of a gas to amounts of other reactants or products.
• Use the ideal gas equation to relate the amount of gas to volume,temperature and pressure.
• Law of combining volumes can be modified with the other laws
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Law of Combining Volumes
• At times, if the reactants and/or products involved in a stoichiometric calculation are gases, we can use a particularly simple approach:
2NO(g) + O2 (g) 2NO2 (g)
2 mol NO + 1 mol O2 (g) 2 mol NO2(g)
Suppose the gases are compared at the same T and P , in this case one mol of gas occupies a particular volume 1V , 2 mol of gas 2V and 3 mol of gas 3V of liters.
2NO(g) + O2 (g) 2NO2 (g)
2 L NO(g) + 1 L O2 (g) 2 L NO2(g)
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Example: The decomposition of sodium azide, NaN3, produces N2(g). Together with the necessary devices to initiate the reaction and trap the sodium metal formed, this reaction is used in air bag safety systems. What volume of N2(g) measured at 735 mm Hg and 26°C is produced when 70,0 g NaN3 is decomposed?
2 NaN3(s) → 2 Na(l) + 3 N2(g) Calculate the mole of N2 :
Calculate the volume of N2
nN2 = 70 g N3 x
1 mol NaN3
65,01 g N3/mol N3
x3 mol N2
2 mol NaN3
= 1,62 mol N2
= 41,1 L
P
nRTV = =
(735 mm Hg)
(1,62 mol)(0,08206 L atm mol-1 K-1)(299 K)
760 mm Hg1.00 atm
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Mixtures of Gases
• Partial pressure– Dalton’s law of partial pressures:
The total pressure of a mixture of gases is the sum of partial pressures of the components of the mixture. Ptop= Pa + Pb + Pc + …
• Simple gas laws and ideal gas equation are applicable to a mixture of gases such as air.
• A simple approach to working with gas mixtures is to use ntot, total amount in moles
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Dalton’s Law of Partial Pressures
Ptot = Pa + Pb +…
Va = naRT/Ptot ve Vtot = Va + Vb+…
Va
Vtot
naRT/Ptot
ntopRT/Ptot= =
na
ntot
Pa
Ptot
naRT/Vtot
ntotRT/Vtot= =
na
ntot
na
ntot
= a (Mole
Fraction)
Remember
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Kinetic-Molecular Theory of Gases• A gas is composed of a very large number of extremely
small particles in constant, random, straight line
motion. Molecules of gas are seperated by great
distances(The molecules are treated as if they have a
mass but no volume,so called point masses.
• Molecules collide with one another and with the walls of
their container very rapidly. There are assumed to be no
forces between molecules. That is each molecule acts
independently of all others.In a collection of molecules
at constant temperature the total energy remains
constant.
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Gas Properties relating to the Kinetic Molecular Theory
Diffusion- The migration of molecules of different substances as a result of random molecular motion.
EffusionThe escape of gas molecules from their container through a tiny orifice or pin hole.
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Graham’s Law
M
RT3
A
BA
Brms
Arms
M
M
3RT/MB
3RT/M
)(u
)(u
Graham’s Law:The rates of effusion of two different gases are inversely proportional to the square roots of their molecular mass.
Graham’s Law applies only if certain conditions are met. For effusion the gas pressure must be very low, not as a jet of gas.
Rate of effusion, urms=
3 RT=NAmv2,Note that the product NAm represents the mass of 1 mol of molecules,
The molar mass M,so:
EK = 3/2RT
M
RTuv
M
RTv
RTvM
rms
3
32
3
2
1
2
2
2
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Effusion
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Example
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Real Gases• Compressibility factor: PV/nRT = 1
• PV= nRT – Ideal gas behaviour
– PV/nRT > 1 – At very high pressures
– PV/nRT < 1 – Where intermolecular forces of attraction exist
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Real Gases