gases chapter 5. a gas -uniformly fills any container. -mixes completely with any other gas -exerts...
TRANSCRIPT
GASESGASES
Chapter 5
A GasA Gas
- Uniformly fills any container.
- Mixes completely with any other gas
- Exerts pressure on its surroundings.
05_47
h = 760mm Hgfor standardatmosphere
Vacuum
Simple barometer invented by Evangelista Torricelli
PressurePressure
- is equal to force/unit area
- SI units = Newton/meter2 = 1 Pascal (Pa)
- 1 standard atmosphere = 101,325 Pa
- 1 standard atmosphere = 1 atm =
760 mm Hg = 760 torr
Pressure Unit ConversionsPressure Unit ConversionsThe pressure of a tire is measured to be 28
psi. What would the pressure in atmospheres, torr, and pascals.
(28 psi)(1.000 atm/14.69 psi) = 1.9 atm
(28 psi)(1.000 atm/14.69 psi)(760.0 torr/1.000atm) = 1.4 x 103 torr
(28 psi)(1.000 atm/14.69 psi)(101,325 Pa/1.000 atm) = 1.9 x 105 Pa
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h h
Atmosphericpressure (Patm)
Atmosphericpressure (Patm)
Gaspressure (Pgas)less thanatmospheric pressure
Gaspressure (Pgas)greater thanatmospheric pressure
(Pgas) = (Patm) - h (Pgas) = (Patm) + h
(a) (b)
Simple Manometer
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Pext
Pext
Volume is decreased
Volume of a gas decreases as pressure increases at constant temperature
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P (
in H
g)
0
V(in3)
20 40 60
50
100
PP2
V
2V
(a)
00
1/P (in Hg)
0.01 0.02 0.03
20
40slope = k
(b)
V(in3 )
BOYLE’S LAW DATA
P vs V V vs 1/P
Boyle’s LawBoyle’s Law**
(Pressure)( Volume) = Constant (T = constant)
P1V1 = P2V2 (T = constant)
V 1/P (T = constant)
(*Holds precisely only at very low pressures.)
Boyle’s Law CalculationsBoyle’s Law CalculationsA 1.5-L sample of gaseous CCl2F2 has a pressure of 56
torr. If the pressure is changed to 150 torr, will the volume of the gas increase or decrease? What will the new volume be?
Decrease
P1 = 56 torr
P2 = 150 torr
V1 = 1.5 L
V2 = ?
V1P1 = V2P2
V2 = V1P1/P2
V2 = (1.5 L)(56 torr)/(150 torr)V2 = 0.56 L
Boyle’s Law CalculationsBoyle’s Law CalculationsIn an automobile engine the initial cylinder
volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The mixture is 1.00 atm, what is the final pressure?
P1 = 1.00 atm
P2 = ?
V1 = 0.725 L
V2 = 0.075 L
V1P1 = V2P2
P2 = V1P1/V2
P2 = (0.725 L)(1.00 atm)/(0.075 L)P2 = 9.7 atm
Is this answer reasonable?
A gas that A gas that strictly obeys strictly obeys Boyle’s Law is called an Boyle’s Law is called an
ideal gasideal gas..
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PV
(L
• at
m)
022.25
P (atm)
1.000.750.500.25
22.30
22.35
22.40
22.45Ne
O2
CO2
Ideal
Plot of PV vs. P for several gases at pressuresbelow 1 atm.
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V(L
)
-300
T(ºC)
-200 -100 0 100 200
1
2
3
6
4
5
300
-273.2 ºC
N2O
H2
H2O
CH4
He
Plot of V vs. T(oC) for several gases
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Pext
Pext
Energy (heat) added
Volume of a gas increases as heat is addedwhen pressure is held constant.
Charles’s LawCharles’s Law
The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin.
V = bT (P = constant)
b = a proportionality constant
Charles’s LawCharles’s Law
VT
VT
P1
1
2
2 ( constant)
Charles’s Law CalculationsCharles’s Law CalculationsConsider a gas with a a volume of 0.675 L at 35 oC and
1 atm pressure. What is the temperature (in Co) of the gas when its volume is 0.535 L at 1 atm pressure?
V1 = 0.675 L
V2 = 0.535 L
T1 = 35 oC + 273 = 308 K
T2 = ?
V1/V2 = T1/T2
T2 = T1 V2/V1
T2 = (308 K)(0.535 L)/(0.675 L)T2 = 244 K -273 T2 = - 29 oC
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Increase volume to return to originalpressure
Gas cylinder
Moles of gasincreases
Pext Pext
Pext
At constant temperature and pressure, increasing the moles of a gas increases its volume.
Avogadro’s LawAvogadro’s Law
For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).
V = an
a = proportionality constant
V = volume of the gas
n = number of moles of gas
AVOGADRO’S LAWAVOGADRO’S LAW
V1/V2 = n1/n2
AVOGADRO’S LAWAVOGADRO’S LAW
A 12.2 L sample containing 0.50 mol of oxygen gas, O2, at a pressure of 1.00 atm and a temperature of 25 oC is converted to ozone, O3, at the same temperature and pressure, what will be the volume of the ozone? 3 O2(g) ---> 2 O3(g)
(0.50 mol O2)(2 mol O3/3 mol O2) = 0.33 mol O3
V1 = 12.2 L
V2 = ?
n1 = 0.50 mol
n2 = 0.33 mol
V1/V2 = n1/n2
V2 = V1 n2/n1
V2 = (12.2 L)(0.33 mol)/(0.50 mol)V2 = 8.1 L
COMBINED GAS LAWCOMBINED GAS LAW
V1/ V2 = P2 T1/ P1 T2
COMBINED GAS LAWCOMBINED GAS LAW
V1/ V2 = P2 T1/ P1 T2
V2 = V1P1T2/P2T1
V2 = (309 K)(0.454 atm)(3.48 L)
(258 K)(0.616 atm)
V2 = 3.07 L
What will be the new volume of a gas under the following conditions?
V1 = 3.48 LV2 = ?P1 = 0.454 atmP2 = 0.616 atmT1 = - 15 oC + 273 = 258 KT2 = 36 oC + 273T2= 309 K
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Pext
Pext
Temperature is increased
Pressure exerted by a gas increases as temperature increases provided volume remains constant.
If the volume of a gas is held constant, then V1 / V2 = 1.Therefore:
P1 / P2 = T1 / T2
Ideal Gas LawIdeal Gas Law
- An equation of state for a gas.
- “state” is the condition of the gas at a given time.
PV = nRT
IDEAL GASIDEAL GAS
1. Molecules are infinitely far apart.
2. Zero attractive forces exist between the molecules.
3. Molecules are infinitely small--zero molecular volume.
What is an example of an ideal gas?
REAL GASREAL GAS
1. Molecules are relatively far apart compared to their size.
2. Very small attractive forces exist between molecules.
3. The volume of the molecule is small compared to the distance between molecules.
What is an example of a real gas?
Ideal Gas LawIdeal Gas Law
PV = nRT
R = proportionality constant
= 0.08206 L atm mol
P = pressure in atm
V = volume in liters
n = moles
T = temperature in Kelvins
Holds closely at P < 1 atm
Ideal Gas Law CalculationsIdeal Gas Law CalculationsA 1.5 mol sample of radon gas has a volume of 21.0 L at
33 oC. What is the pressure of the gas?
p = ?
V = 21.0 L
n = 1.5 mol
T = 33 oC + 273
T = 306 K
R = 0.08206 Latm/molK
pV = nRTp = nRT/Vp = (1.5mol)(0.08206Latm/molK)(306K)
(21.0L)p = 1.8 atm
Ideal Gas Law CalculationsIdeal Gas Law CalculationsA sample of hydrogen gas, H2, has a volume of 8.56 L at a
temperature of O oC and a pressure of 1.5 atm. Calculate the number of moles of hydrogen present.
p = 1.5 atm
V = 8.56 L
R = 0.08206 Latm/molK
n = ?
T = O oC + 273
T = 273K
pV = nRTn = pV/RTn = (1.5 atm)(8.56L) (0.08206 Latm/molK)(273K)n = 0.57 mol
Standard Temperature Standard Temperature and Pressureand Pressure
“STP”
P = 1 atmosphere
T = C
The molar volume of an ideal gas is 22.42 liters at STP
GAS STOICHIOMETRYGAS STOICHIOMETRY
1. Mass-Volume
2. Volume-Volume
Molar VolumeMolar Volume
pV = nRT
V = nRT/p
V = (1.00 mol)(0.08206 Latm/molK)(273K)
(1.00 atm)
V = 22.4 L
Gas StoichiometryGas StoichiometryNot at STP (Continued)Not at STP (Continued)
p = 1.00 atm
V = ?n = 1.28 x 10-1 mol
R = 0.08206 Latm/molK
T = 25 oC + 273 = 298 K
pV = nRTV = nRT/pV = (1.28 x 10-1mol)(0.08206Latm/molK)(298K)
(1.00 atm)V = 3.13 L O2
Gases at STPGases at STP
A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present?
(1.75L N2)(1.000 mol/22.4 L) = 7.81 x 10-2 mol N2
Gas StoichiometryGas Stoichiometryat STPat STP
Quicklime, CaO, is produced by heating calcium carbonate, CaCO3. Calculate the volume of CO2 produced at STP from the decomposition of 152 g of CaCO3. CaCO3(s) ---> CaO(s) + CO2(g)
(152g CaCO3)(1 mol/100.1g)(1mol CO2/1mol CaCO3) (22.4L/1mol) = 34.1L CO2
Note: This method only works when the gas is at STP!!!!!
Volume-VolumeVolume-VolumeIf 25.0 L of hydrogen reacts with an excess of
nitrogen gas, how much ammonia gas will be produced? All gases are measured at the same temperature and pressure.
2N2(g) + 3H2(g) ----> 2NH3(g)
(25.0 L H2)(2 mol NH3/3 mol H2) = 16.7 L NH3
MOLAR MASS OF A GASMOLAR MASS OF A GAS
n = m/M
n = number of moles
m = mass
M = molar mass
MOLAR MASS OF A GASMOLAR MASS OF A GAS
P = mRT/VM
or
P = DRT/M
therefore:
M = DRT/P
Molar Mass CalculationsMolar Mass Calculations
M = ? d = 1.95 g/L
T = 27 oC + 273 p = 1.50 atm
T = 300. K R = 0.08206 Latm/mol K
M = dRT/pM = (1.95g/L)(0.08206Latm/molK)(300.K)
(1.5atm)M = 32.0 g/mol
Dalton’s Law of Dalton’s Law of Partial PressuresPartial Pressures
For a mixture of gases in a container,
PTotal = P1 + P2 + P3 + . . .
Dalton’s Law of Partial Dalton’s Law of Partial Pressures CalculationsPressures Calculations
A mixture of nitrogen gas at a pressure of 1.25 atm, oxygen at 2.55 atm, and carbon dioxide at .33 atm would have what total pressure?
PTotal = P1 + P2 + P3
PTotal = 1.25 atm + 2.55 atm + .33 atm
Ptotal = 4.13 atm
Water Vapor PressureWater Vapor Pressure2KClO3(s) ----> 2KCl(s) + 3O2(g)
When a sample of potassium chlorate is decomposed and the oxygen produced collected by water displacement, the oxygen has a volume of 0.650 L at a temperature of 22 oC. The combined pressure of the oxygen and water vapor is 754 torr (water vapor pressure at 22 oC is 21 torr). How many moles of oxygen are produced?
Pox = Ptotal - PHOH
Pox = 754 torr - 21 torr
pox = 733 torr
MOLE FRACTIONMOLE FRACTION
-- the ratio of the number of moles of a given component in a mixture to the total number of moles of the mixture.
1 = n1/ ntotal
1 = V1/ Vtotal
1 = P1 / Ptotal (volume & temperature constant)
Kinetic Molecular TheoryKinetic Molecular Theory
1. Volume of individual particles is zero.
2. Collisions of particles with container walls cause pressure exerted by gas.
3. Particles exert no forces on each other.
4. Average kinetic energy Kelvin temperature of a gas.
05_58
Rela
tive
nu
mbe
r o
f O
2 m
ole
cule
sw
ith g
ive
n v
elo
city
0
Molecular velocity (m/s)
4 x 102 8 x102
Plot of relative number of oxygen molecules with a given velocity at STP (Boltzmann Distribution).
05_59
Rel
ativ
e nu
mbe
r of
N2
mol
ecul
esw
ith g
iven
vel
ocity
0Velocity (m/s)1000 2000 3000
273 K
1273 K
2273 K
Plot of relative number of nitrogen molecules witha given velocity at three different temperatures.
The Meaning of The Meaning of TemperatureTemperature
Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)
(K E )3
2av g R T
M
RTrms
3
Effusion: describes the passage of gas into an evacuated chamber.
Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.
05_60
Gas
Vacuum
Pinhole
Effusion of a gas into an evacuated chamber
Rate of effusion for gas 1Rate of effusion for gas 2
2
1
MM
Distance traveled by gas 1Distance traveled by gas 2
2
1
MM
Effusion:Effusion:
Diffusion:Diffusion:
Real GasesReal Gases
Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).
05_62
00
P(atm)
200 400 600 800
1.0
2.0
PV
nRTCO2
H2
CH4N2
Ideal
gas
1000
Plots of PV/nRT vs. P for several gases at 200 K.Note the significant deviation from ideal behavior.
Real GasesReal Gases
[ ]P a V nb nRTobs2( / ) n V
corrected pressurecorrected pressure corrected volumecorrected volume
PPidealideal VVidealideal
05_68
Con
cen
tra
tion (
pp
m)
4:0
00
Time of day
0.1
0.2
0.3
0.4
0.5
6:0
0
8:0
0
10
:00
Noo
n
2:0
0
4:0
0
6:0
0
NO
NO2
O3
Molecules of unburnedfuel (petroleum)
Otherpollutants
Concentration for some smog componentsvs. time of day
NO2(g) NO(g) + O(g)
O(g) + O2(g) O3(g)
NO(g) + 1/2 O2(g) NO2(g)__________________________
3/2 O2(g) O3(g)
What substances represent intermediates?
Which substance represents the catalyst?
Chemistry is a gas!!Chemistry is a gas!!