chapter 5 gases chapter 9/1. 1 atm = 760 mmhg t(k) = t(°c) + 273, 1 mol so 2 = 64.07 g calculate...
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![Page 1: Chapter 5 Gases Chapter 9/1. 1 atm = 760 mmHg T(K) = t(°C) + 273, 1 mol SO 2 = 64.07 g Calculate the Volume Occupied by 637 g of SO 2 (MM 64.07) at 6.08](https://reader038.vdocuments.us/reader038/viewer/2022102907/56649d835503460f94a69399/html5/thumbnails/1.jpg)
Chapter 5
Gases
Chapter 9/1
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1 atm = 760 mmHg
T(K) = t(°C) + 273, 1 mol SO2 = 64.07 gKmolLatm 0.0821
nRT, RPV
Calculate the Volume Occupied by 637 g of SO2 (MM 64.07)
at 6.08 x 103 mmHg and –23 °C.mSO2 = 637 g, P = 6.08 x 103 mmHg, t = −23 °C,
V, L
Solution:
Solution Map:
Relationships:
Given:
Find:
P
nRT V
P, n, T, R V
L 55.2
atm 0.80
K 0520.0821mol 249.9 Kmol
Latm
P
TRnV
atm 0.08mmHg 760
atm 1mmHg 106.08 3
K 025
273C 23- (K)
T
T
g n
g 64.07
mol 1
22
2 SO mol 249.9 g 64.07
SO mol 1SO g 637
2
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m = 9.988g, n = 0.250 mol, P = 1.0197 atm, T = 300. K
density, g/L
L 5530.6atm 9711.0
K 0030.082mol .2500Kmol
Latm
P
TRnV
Solution:
Calculate the Density of a Gas at 775 torr and 27 °C if 0.250 moles Weighs 9.988 g
m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C,
density, g/L
Solution Map:
Relationships:
Given:
Find:
P
TRn V
V, m d
atm 9710.1 torr760
atm 1 torr775
K 030
273C 27 (K)
T
T
P, n, T, R V
V
m d
g/L 1.65
L 553.06
g .9889
V
md
1 atm = 760 mmHg, T(K) = t(°C) + 273
KmolLatm 0.0821
nRT, RPV
V
m d
3
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4
Molar Mass of a Gas
• One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law.
moles
gramsin massMassMolar
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m = 0.311g, V = 0.225 L, P = 1.1658 atm, T = 328 K
Molar Mass, g/mol
mol 105447.9K 2830.0821
L .2250atm 5861.1 3
Kmol
Latm
TR
VPn atm 5861.1
mmHg 760
atm 1mmHg 886
Calculate the Molar Mass of a Gas with Mass 0.311 g that Has a Volume of 0.225 L at 55 °C and 886 mmHg.
m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,
Molar Mass, g/mol
Solution:
Solution Map:
Relationships:
Given:
Find:
TR
VP n
n, m MM
K 328
273C55 (K)
T
T
P, V, T, R n
n
m MM
g/mol 31.9 mol 109.7454
g 311.03-
n
mMM
1 atm = 760 mmHg, T(K) = t(°C) + 273
Kmol
Latm 0.0821 nRT, RPV
n
m MM
5
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6
• Write a solution map:
P,V,T,R n
When using the ideal gas equation, the units of V must be L; and the units of P must be atm, or you will have to convert.
The units of T must be kelvin, K.
Information:Given: V = 0.225 L, P = 886 mmHg,
t = 55 °C, m = 0.311 gFind: molar mass, (g/mol)Equation:
PV = nRT; MM = mass/moles
Example:A sample of a gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Molar Mass
mass
PV = nRT
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m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K,
molar mass, g/mol
What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C?m=12.0 g, V= 197 L, P=380 torr, t=127°C,
molar mass, g/mol
Solution:
Solution Map:
Relationships:
Given:
Find:
TR
VPn
n, m MM
mol 0.3K 0040.0821
L 971atm 0.50
Kmol
Latm
TR
VPn atm 50.0
torr760
atm 1 torr083
K 004
273C127 (K)
T
T
P, V, T, R n
n
mMM
g/mol 4.0 mol .03
g 2.01
n
mMM
1 atm = 760 torr, T(K) = t(°C) + 273
Kmol
Latm 0.0821 nRT, RPV
n
mMM
7
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8
Partial Pressure• Each gas in the mixture exerts a pressure
independent of the other gases in the mixture.
• The pressure of a component gas in a mixture is called a partial pressure.
• The sum of the partial pressures of all the gases in a mixture equals the total pressure. Dalton’s law of partial pressures. Ptotal = Pgas A + Pgas B + Pgas C +...
atm 1.00 atm 0.01 atm 0.21 atm 0.78 Ar2O2Nair PPPP
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A Mixture of He, Ne, and Ar Has a Total Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg,
Determine the Partial Pressure of Ar in the Mixture.
PHe= 341 mmHg, PNe= 112 mmHg, Ptot = 558 mmHg
PAr, mmHg
Solution:
Solution Map:
Relationships:
Given:
Find:
Ptot, PHe, PNe PAr
Ptot = Pa + Pb + etc.
PAr = Ptot – (PHe + PNe)
mmHg 051
mmHg 112341558Ar
P
9
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10
Finding Partial Pressure• To find the partial pressure of a
gas, multiply the total pressure of the mixture by the fractional composition of the gas.
• For example, in a gas mixture that is 80.0% He and 20.0% Ne that has a total pressure of 1.0 atm, the partial pressure of He would be:
PHe = (0.800)(1.0 atm) = 0.80 atm Fractional composition =
percentage divided by 100.
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The Partial Pressure of Each Gas in a Mixture, or the Total Pressure of a Mixture, Can Be
Calculated Using the Ideal Gas Law
BAtotal
total
BAtotal
BB
AA
P PV
x R x Tn P
n n n
V
x R x Tn P
V
x R x Tn P
same theare mixture in the
everything of volumeand re temperatuthe
mixture ain B andA gasesfor
1111
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atm 8959.0
L 8.7
K 9850.0821mol 0.17Kmol
Latm
XeXe
V
TRnP
XeNetotal ,KmolLatm 0.0821 PPPnRT, RPV
Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe.
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
Solution:
Solution Map:
Relationships:
Given:
Find:
V
TRnP
Xe
Xe
nXe, V, T, R PXe
atm 2.9
atm 8950.9 atm 9.3XetotalNe
PPP
Ptot, PXe PNe
XetotalNe PPP
12
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Deep Sea Divers and Partial Pressure• It is also possible to have too much O2, a condition called
oxygen toxicity.PO2 > 1.4 atm.Oxygen toxicity can lead to muscle spasms, tunnel vision, and
convulsions.
• It is also possible to have too much N2, a condition called nitrogen narcosis.Also known as rapture of the deep.
• When diving deep, the pressure of the air that divers breathe increases, so the partial pressure of the oxygen increases.At a depth of 55 m, the partial pressure of O2 is 1.4 atm.Divers that go below 50 m use a mixture of He and O2 called heliox that
contains a lower percentage of O2 than air.
13
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14
Partial Pressure vs. Total Pressure
At a depth of 30 m, the total pressure of air in the diverslungs, and the partial pressure of all the gases in the air,
are quadrupled!
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Gas Stoichiometry
15
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16
• Write a solution map:
When using the ideal gas equation, the units of V must be L; and the units of P must be atm, or you will have to convert.
The units of T must be kelvin, K.
Information:Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 KFind: VO2, LEquation: PV = nRTConversion Factors:
1 mole KClO3 = 122.5 g 2 mole KClO3 3 moles O2
Example:How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
)()()( gss 23 O 3 KCl 2 KClO 2
nO2 VO2nKClO3g KClO3
3
3
KClO g 5122
KClO mol 1
. 3
2
KClO mol 2
O mol 3
P,T,R
PV=nRT
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17
• Apply the solution map:Find moles of O2 made.
Information:Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 KFind: VO2, LEquation: PV = nRTConversion Factors:
1 mole KClO3 = 122.5 g 2 mole KClO3 3 moles O2
Solution Map: g → mol KClO3
→ mol O2 → L O2
23
2
3
33 O mol 603
KClO mol 2
O mol 3
KClO g 122.5
KClO mol 1KClO g 294 .
Example:How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
)()()( gss 23 O 3 KCl 2 KClO 2
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18
• Apply the solution map:Convert the units.
Information:Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K, nO2 = 3.60 molesFind: VO2, LEquation: PV = nRTConversion Factors:
1 mole KClO3 = 122.5 g 2 mole KClO3 3 moles O2
Solution Map: g → mol KClO3
→ mol O2 → L O2
atm 42399.0mmHg 760
atm 1mmHg 557 P
Example:How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
)()()( gss 23 O 3 KCl 2 KClO 2
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19
• Apply the solution map:
L 90.7
atm 42399.0
K 305 0.0821mol 60.3 Kmol
Latm
V
VP
TRn
TRnVP
Information:Given: 294 g KClO3
PO2 = 0.99342 mmHg, TO2 = 305 K, nO2 = 3.60 molesFind: VO2, LEquation: PV = nRTConversion Factors:
1 mole KClO3 = 122.5 g 2 mole KClO3 3 moles O2
Solution Map: g → mol KClO3
→ mol O2 → L O2
Example:How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
)()()( gss 23 O 3 KCl 2 KClO 2
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mHgO = 10.0g, P=0.750 atm, T=313 K
VO2, L
nO2 = 0.023085 mol, P = 0.750 atm, T = 313 K
VO2, L
L 791.0
atm 0.750
K3130.08206mol 850023.0Kmol
Latm
P
TRnV
Practice—What Volume of O2 at 0.750 atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO?
2 HgO(s) 2 Hg(l) + O2(g), Continued
Solution:
Solution Map:
Relationships:
Given:
Find:
P
TRnV
2
2
O mol 850023.0
HgO mol 2
O mol 1
g 216.59
HgO mol 1HgO g 0.01
P, n, T, R V
1 atm = 760 mmHg, HgO = 216.59 g/mol2 mol HgO : 1 mol O2 Kmol
Latm 0.0821 nRT, RPV
g HgO mol HgO mol O2
HgO mol 2
O mol 1 2
g 216.59
HgO mol 1
20
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21
Calculate the Volume Occupied by 1.00 Mole of an Ideal Gas at STP.
• 1 mole of any gas at STP will occupy 22.4 L.• This volume is called the molar volume and can
be used as a conversion factor.As long as you work at STP.
1 mol 22.4 L
(1.00 atm) x V = (1.00 moles)(0.0821 )(273 K)L∙atmmol∙K
V = 22.4 L
P x V = n x R x T
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22
Molar Volume
There is so muchempty spacebetween moleculesin the gas state thatthe volume of thegas is not effectedby the size of themolecules (underideal conditions).
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Example —How Many Grams of H2O Form When 1.24 L H2 Reacts Completely with O2 at STP?
O2(g) + 2 H2(g) → 2 H2O(g)VH2 = 1.24 L, P = 1.00 atm, T = 273 K
massH2O, g
Solution:
Solution Map:
Relationships:
Given:
Find:
OH mol 1
g 02.18
2
OH g 998.0
OH mol 1
OH g 8.021
H mol 2
OH mol 2
H L 22.4
H mol 1H L .241
2
2
2
2
2
2
22
H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP2 mol H2O : 2 mol H2
OH mol 2
H mol 2
2
2
L 22.4
H mol 1 2
g H2OL H2 mol H2 mol H2O
2323
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24
• Write a solution map:
mol H2O g H2Omol H2L H2
2
2
H L 2.42
H mol 1
2
2
H mol 2
OH mol 2
Information:Given: 1.24 L H2
Find: g H2OConversion Factors:
1 mol H2O = 18.02 g2 mol H2O 2 mol H2
1 mol H2 22.4 L
Example:How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
)()( ggg OH 2 O )(H 2 222
OH mol 1
OH g 0218
2
2.
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25
• Apply the solution map:
Information:Given: 1.24 L H2
Find: g H2OConversion Factors:
1 mol H2O = 18.02 g2 mol H2O 2 mol H2
1 mol H2 22.4 LSolution Map:
L → mol H2 → mol H2O → g H2O
OH g 9880OH mol 1
OH g 18.02
H mol 2
OH mol 2
H L 22.4
H mol 1H L 1.24 2
2
2
2
2
2
22 .
Example:How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
)()( ggg OH 2 O )(H 2 222
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Practice—What Volume of O2 at STP is Generated by the Thermolysis of 10.0 g of HgO?
2 HgO(s) 2 Hg(l) + O2(g), ContinuedmHgO = 10.0 g, P = 1.00 atm, T = 273 K
VO2, L
Solution:
Solution Map:
Relationships:
Given:
Find:
g 59.216
HgO mol 1
2
2
22
O L 517.0
O mol 1
O L 2.42
HgO mol 2
O mol 1
g 216.59
HgO mol 1HgO g 0.01
HgO = 216.59 g/mol, 1 mol = 22.4 L at STP2 mol HgO : 1 mol O2
HgO mol 2
O mol 1 2
2O mol 1
L 22.4
L O2g HgO mol HgO mol O2
2626