Chapter 09 Slide 1 Gases: Their Properties & Behavior Gases: Their Properties & Behavior 9
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Chapter 09 Slide 1
Gases: Their Properties & Behavior
Gases: Their Properties & Behavior
9
Chapter 09 Slide 2
Gas Pressure 01Gas Pressure 01
Chapter 09 Slide 3
Gas Pressure 02Gas Pressure 02
• Units of pressure: atmosphere (atm)Pa (N/m2, 101,325 Pa = 1 atm)Torr (760 Torr = 1 atm)bar (1.01325 bar = 1 atm)mm Hg (760 mm Hg = 1 atm)lb/in2 (14.696 lb/in2 = 1 atm)in Hg (29.921 in Hg = 1 atm)
Chapter 09 Slide 4
• Pressure–Volume Law (Boyle’s Law):
Boyle’s Law 01Boyle’s Law 01
Chapter 09 Slide 5
• Pressure–Volume Law (Boyle’s Law):
Boyle’s Law 01Boyle’s Law 01
Chapter 09 Slide 6
Boyle’s Law 02Boyle’s Law 02
• Pressure–Volume Law (Boyle’s Law):
• The volume of a fixed amount of gas maintained at constant temperature is inversely proportional to the gas pressure.
Pressure1Volume ∝
V1P1 = k1
Chapter 09 Slide 7
A sample of chlorine gas occupies a volume of 946 mLat a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P1 x V1 = P2 x V2
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 = P1 x V1
V2
726 mmHg x 946 mL154 mL= = 4460 mmHg
5.3
Chapter 09 Slide 8As T increases V increases
Chapter 09 Slide 9
Charles’ Law 01Charles’ Law 01
• Temperature–Volume Law (Charles’ Law):
T(K)=Temperature in KelvinT (K) = t (0C) + 273.15
Chapter 09 Slide 10
Charles’ Law 01Charles’ Law 01
• Temperature–Volume Law (Charles’ Law):
• The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin temperature of the gas.
V ∝ T
V1
T1=k1
Chapter 09 Slide 11
Variation of gas volume with temperatureat constant pressure.
5.3
V α T
T (K) = t (0C) + 273.15
Temperature must bein KelvinV = constant x T
V1/T1 = V2/T2
P1<P2<P3<P4
Chapter 09 Slide 12
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
Chapter 09 Slide 13
Avogadro’s Law 01Avogadro’s Law 01
• The Volume–Amount Law (Avogadro’s Law):
Chapter 09 Slide 14
Avogadro’s Law 01Avogadro’s Law 01
• The Volume–Amount Law (Avogadro’s Law):
• At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present.
nV ∝
111 k
nV
=
Chapter 09 Slide 15
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?
4NH3 + 5O2 4NO + 6H2O
1 mole NH3 1 mole NO
At constant T and P
1 volume NH3 1 volume NO
Chapter 09 Slide 16
Ideal Gas EquationIdeal Gas Equation
Boyle’s law: V α (at constant n and T)1
Charles’ law: V α T (at constant n and P)
Avogadro’s law: V α n (at constant P and T)
P
V αnTP
V = constant x = R nTP
nTP
R is the gas constant
PV = nRT
Chapter 09 Slide 17
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
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.
Chapter 09 Slide 18
The Ideal Gas Law 01The Ideal Gas Law 01
• Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro.
• R =The gas constant
• R = 0.08206 L·atm·K–1·mol–1
TRnVP ⋅⋅=⋅
Chapter 09 Slide 19
What is the volume (in liters) occupied by 49.8 g of HClat 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
5.4
Chapter 09 Slide 20
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulbcontaining 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 lightbulb (in atm)?
PV = nRT n, V and R are constantnRV = P
T
P1T1
= constant
P2T2
=
P1 = 1.20 atmT1 = 291 K
P2 = ?T2 = 358 K
P2 = P1 x T2T1
= 1.20 atm x 358 K291 K
= 1.48 atm
Chapter 09 Slide 21
USING GAS DENSITYUSING GAS DENSITY
The density of air at The density of air at 15 15 ooCC and and 1.00 1.00 atmatm is is 1.23 1.23 g/Lg/L. What is the molar mass of air?. What is the molar mass of air?
1. Calc. moles of air.1. Calc. moles of air.V = 1.00 L, m=1.23 g, V = 1.00 L, m=1.23 g, P = 1.00 P = 1.00 atmatm,,T = (273+15 )= 288 KT = (273+15 )= 288 Kn = PV/RT = 0.0423 mol,n = PV/RT = 0.0423 mol, m=1.23 gm=1.23 g
2.2. Calc. molar massCalc. molar massmass/mol = 1.23 g/0.0423 mol = 29.1 g/molmass/mol = 1.23 g/0.0423 mol = 29.1 g/mol
Chapter 09 Slide 22
Flash Animation - Click to ContinueFlash Animation Flash Animation -- Click to ContinueClick to Continue
Dalton’s Law of Partial Pressures 01Dalton’s Law of Partial Pressures 01
Chapter 09 Slide 23
Dalton’s Law of Partial Pressures
V and Tare
constant
P1 P2 Ptotal = P1 + P2
Chapter 09 Slide 24
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
Chapter 09 Slide 25
PT = PA + PB
nARTV
nBRTV
+ = (nA + nB)RTV
PT =
PA = nART
PA = XA PT
V PB = nBRT
PB = XB PT
V
PA / PT = nA
nA + nBXA =
nAnA + nB
PA / PT = XA
XB = nB
nA + nB
Pi = Xi PT
Chapter 09 Slide 26
A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
Xpropane = 0.116
8.24 + 0.421 + 0.116
PT = 1.37 atm
= 0.0132
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
Chapter 09 Slide 27
Gas Stoichiometry 01Gas Stoichiometry 01
• In gas stoichiometry, for a constant temperature and pressure, volume is proportional to moles.
• Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10):
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l)
Chapter 09 Slide 28
Gas Stoichiometry
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
g C6H12O6 mol C6H12O6 mol CO2 V CO2
5.60 g C6H12O61 mol C6H12O6
180 g C6H12O6x
6 mol CO2
1 mol C6H12O6x = 0.187 mol CO2
V = nRTP
0.187 mol x 0.0821 x 310.15 KL•atmmol•K
1.00 atm= = 4.76 L
Chapter 09 Slide 29
Chapter 09 Slide 30
Kinetic Molecular Theory 01Kinetic Molecular Theory 01
• This theory presents physical properties of gases in terms of the motion of individual molecules.
• Average Kinetic Energy ∝ Kelvin Temperature• Gas molecules are points separated by a great distance• Particle volume is negligible compared to gas volume• Gas molecules are in rapid random motion• Gas collisions are perfectly elastic• Gas molecules experience no attraction or repulsion
Chapter 09 Slide 31
Kinetic-Molecular TheoryKinetic-Molecular Theory
Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.
Chapter 09 Slide 32
• Average Kinetic Energy (KE) is given by:
KE = 1
2mu2
u2 = u2∑N
Kinetic Molecular Theory 03Kinetic Molecular Theory 03
Chapter 09 Slide 33
• Average Kinetic Energy (KE) is given by:
KE = 1
2mu2
u2 = u2∑N
Kinetic Molecular Theory 03Kinetic Molecular Theory 03
Chapter 09 Slide 34
Kinetic Molecular Theory 05Kinetic Molecular Theory 05
• Maxwell speed distribution curves.
Chapter 09 Slide 35
Kinetic Molecular Theory 04Kinetic Molecular Theory 04
• The Root–Mean–Square Speed: is a measure of the average molecular speed.
• Calculate the root–mean–square speeds of helium atoms and nitrogen molecules in m/s at 25°C.
• R=8.314 J/K.mol 1J = 1Kg.m2/s2
Taking square root of both sides gives the equation M
RTurms3
=MRTu 32 =
H243.1 m/s
Chapter 09 Slide 36
Graham’s Law 02Graham’s Law 02
• Effusion is when gas molecules escape , through a tiny hole into a vacuum.
Chapter 09 Slide 37
Graham’s Law 03Graham’s Law 03
• Graham’s Law: Rate of effusion is proportional to its rms speed, urms.
• For two gases at same temperature and pressure:
Rate α urms = 3RT
M
Rate1Rate2
=M2
M1= M2
M1
Chapter 09 Slide 38
A Problem to Consider
• How much faster would H2 gas effuse through an opening than methane, CH4?
)(HM)(CHM
CH of RateH of Rate
2m
4m
4
2 =
8.2g/mol 2.0g/mol 16.0
CH of RateH of Rate
4
2 ==
So hydrogen effuses 2.8 times faster than CH4
Chapter 09 Slide 39
Behavior of Real Gases 01Behavior of Real Gases 01
• Deviations result from assumptions about ideal gases.
1. Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another.
2. Volume of the molecules is negligibly small compared with that of the container.
Chapter 09 Slide 40
Behavior of Real Gases 02Behavior of Real Gases 02
• At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures.
As a result, the pressure of real gases willbe smaller than the ideal value
Chapter 09 Slide 41
Behavior of Real Gases 03Behavior of Real Gases 03
• The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value.
Chapter 09 Slide 42
Behavior of Real Gases 04Behavior of Real Gases 04
• Test of ideal gasbehavior.
Chapter 09 Slide 43
Behavior of Real Gases 05Behavior of Real Gases 05
• Test of ideal gasbehavior.
Chapter 09 Slide 44
Behavior of Real Gases 05Behavior of Real Gases 05
• Corrections for non-ideality require van der Waalsequation.
( ) nRTbnVVnaP – 2
2
=⋅⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+
IntermolecularAttractions
ExcludedVolume
Chapter 09 Slide 45
A Problem to ConsiderA Problem to Consider
• If sulfur dioxide were an “ideal” gas, the pressure at 0 oCexerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real”pressure.
Use the following values for SO2
a = 6.865 L2.atm/mol2
b = 0.05679 L/mol
Chapter 09 Slide 46
A Problem to ConsiderA Problem to Consider
• First, let’s rearrange the van der Waals equation to solve for pressure.
2
2
Van -
nb-VnRT P =
R= 0.0821 L. atm/mol. KT = 273.2 KV = 22.41 L
a = 6.865 L2.atm/mol2
b = 0.05679 L/mol
Chapter 09 Slide 47
A Problem to ConsiderA Problem to Consider
• The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the “ideal” pressure.
2
2
Van -
nb-VnRT P =
L/mol) 79mol)(0.056 (1.000 - L 22.41)K2.273)(06mol)(0.082 (1.000
P KmolatmL
⋅⋅
= 2mol
atmL2
L) 41.22() (6.865mol) (1.000
- 2
2⋅
atm0.989P =
Chapter 09 Slide 48
H2O (s) H2O (l) ΔH = 6.01 kJ
• The physical states of all reactants and products must be specified in thermochemical equations.
Thermochemical Equations
H2O (l) H2O (g) ΔH = 44.0 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) ΔH = -3013 kJ
266 g P4
1 mol P4
123.9 g P4x 3013 kJ
1 mol P4x = 6470 kJ
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