Chapter 5
Gases
Chapter 9/1
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
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
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
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
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
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
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
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
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.
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
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
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
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!
Gas Stoichiometry
15
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
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
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
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
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
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
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).
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
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.
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
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
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