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Pages 422-440
Notes One Unit Five
Characteristics of GasesPressure of fluidsStandard Temperature and PressureConverting PressuresGas Laws
Pressure Versus Molecular Collision
• Pressure is caused by molecular collision
• A molecule colliding creates a force.• Catching a ball creates a force.• P=F/A
• pp 427
Pressure viewed as created in a fluid
• Created by the weight
• The deeper you go, the more weight .
pp 427
Standard Pressure, Temperature and Volume1 atm = 14.7 psi1 atm = 29.92 in Hg1 atm = 760 mm Hg1 atm = 760 torr 1 atm = 101,325 Pa 1 atm = 101.325 kPa 1 atm = 1.01325 bar 273K or 0oC K=oC+273
22.4 Liter/mole for any gas at STP
pp 427
Charles’ and Boyle’s Demo
P1=1.0atmV1=1.0LT1= 273K
P2=2.0L
T2= 546KV2=
1.0atm
Pressure is constant.Moles are constant.
V1
T1
= V2
T2
P1=1.0atmV1=1.0LT1= 273K
P2= 2.0atmV2= 0.50LT2= 273K
P1 x V1 P2 x V2=
Temperature is constant.Moles are constant.
pp 433-4401.0L273K
= 2.0L546K
1.0a x 1.0L 2.0a x0.50L=
Combined gas Law Problem One• A gas occupies 2.0 m3 at 121.2 K, exerting a pressure
of 100.0 kPa. What volume will the gas occupy at 410.0 K if the pressure is increased to 220.0 kPa?
(2.0m3 )(100.0KPa)(121.2K )
(220.0KPa )(V2 )(410.0K )
=
(2.0m3 )(100.0KPa) (410.0K )(121.2K )(220.0KPa )
= V2
V2= 3.1M3 pp 433-440
Assign variables and calculate V2.
Combined gas Law Problem Two• A 10.0 gram sample of ethane(C2H6) gas is at STP. If
the volume is changed to 26.0 liters, what is the new Kelvin temperature of the gas?
10.0g 30.0g/mol
=22.4L V1 V1= 7.46L
(101.325KPa) (7.46L)(273K)
= (101.325KPa)(26.0L)(T2)
(273K)(101.325KPa)(26.0L)(101.325KPa) (7.46L)
= T2 T2= 951Kpp 433-440
CH
2x6x
12.0 =1.0 =
24.0 6.030.0g/m
E # Mass
10.0 gram
1) Calculate Formula mass.
2) Calculate V1.
3) Assign variables and calculate T2.
Notes Two Unit Five
Grahams’ Law Calculation
Review Mass-Mass Calculation
Mass-Volume Calculation @STP
Volume-Mass Calculation @STP
Pages 441-450
Graham’s Law• Describes how speed compares between gas
molecules with different masses.
• Two different gases:
• 1)Same Temperature
• 2)Different Masses
• Kinetic energy ½ M1V12= ½ M2V2
2
pp 442
V12
½ M2V22= ½ M1V1
2
Graham’s Equation
pp 442
M2V22 M1V1
2
M1 M1
M2V22
M1
V12
= V2
2 V22
= ÷ by M1
÷ by V22
Square root of both sides
Grahams’ Law Problem One• At a high temperature molecules of chlorine gas travel 15.90cm. What is the mass of vaporized
metal (gas) under the same conditions, if the metal travels 8.97cm?
M2= 223g/m
Cl2x35.5 = 71.0g/mE # Mass
pp 442
(15.90cm)
8.97cm= = 14.9 2
Grahams’ Law Problem Two• At a certain temperature molecules of chlorine gas travel at 0.450 km/s. What is the speed of sulfur dioxide gas
under the same conditions?
V2= 0.474Km/s pp 442
Cl2x35.5 = 71.0g/mE # Mass
SO
1x2x
32.1 =16.0 =
32.1 32.064.1g/m
E # Mass
V2=
Review Mass-Mass Calculation
1) grams H2 to moles H2
5.00g÷
2) moles H2 to moles O2
2.5m H2x
3) moles O2 to grams O2
1.3mO2x
2.0gH2 /m= 2.5m H2
32.0g/m = 42gO2
(1mO2)(2mH2)
=1.3mO2
2 1 2
H 2 x 1.0 = 2.0g/mE # Mass
O 2x16.0 = 32.0g/mE # Mass
How many grams of oxygen will react with 5.00 grams of hydrogen to make water?
H2(g) + O2(g) H2O(l)
425.0g
X32.0g/m___m
___g
2.5 ___m1.3
÷2.0g/m
Mass-Volume Calculation @STP
1) grams H2 to moles H2
5.00g÷
2) moles H2 to moles O2
2.5m H2x
3) moles O2 to liters O2
1.3mO2x
2.0gH2 /m= 2.5m H2
22.4L/m = 29LO2
(1mO2)(2mH2)
=1.3mO2
2 1 2
H 2 x 1.0 = 2.0g/mE # Mass
O 2x16.0 = 32.0g/mE # Mass
How many liters of oxygen will react with 5.00 grams of hydrogen to make water?
H2(g) + O2(g) H2O(l)
295.0g
X22.4g/m___m
___L
2.5 ___m1.3
÷2.0g/mpp 449
Volume-Mass Calculation @STP
1) Liters H2 to moles H2
56.0L÷
2) moles H2 to moles O2
2.50m H2x
3) moles O2 to grams O2
1.25mO2x
22.4LH2 /m= 2.50m H2
32.0g/m = 40.0gO2
(1mO2)(2mH2)
=1.25mO2
2 1 2
H 2 x 1.0 = 2.0g/mE # Mass
O 2x16.0 = 32.0g/mE # Mass
How many grams of oxygen will react with 56.0 liters of hydrogen to make water?
H2(g) + O2(g) H2O(l)
40.056.0L
X32.0g/m____m
____g
2.50 ____m1.25
÷22.4L/mpp 449
Notes Three Unit Five
• Kinetic theory of gases
• Molar volume @ Non-STP Conditions
• R is Universal Gas Constant
Pages 452-459
THE KINETIC THEORY OF GASES• Large number of particles • Elastic collisions For a collision KEBefore=KEAfter
pp 426
THE KINETIC THEORY OF GASES• Large number of particles • Elastic collisions • No external forces
pp 433-440
THE KINETIC THEORY OF GASES• Large number of particles • Elastic collisions • No external forces • Separated by large distances
1.6x1011 times diameter
pp 426
THE KINETIC THEORY OF GASES• Large number of particles • Elastic collisions • No external forces • Separated by large distances• No forces between particles
pp 426
Finding volumes @ Non-STP Conditions
pp 446
PV=nRTWhat is…P?V?n?
T?
moles
R? Universal Gas Constant
P Vn T
R=(101.325kpa)(22.4L)
(1 m)(273.15K)R=
Ideal Gas Equation
R =(8.314 )Kpa- Lm- K
Finding Volume at Non-STPOxygen is made reacting 15.00g water at 209.0Kpa and 20.0oC.
18.0g/m= 0.833m
22.4L/m = 9.33LO2
( 1mO2 )(2m H2O) =0.417mO2
a) What volume of oxygen would be made at STP?1) grams H2O to moles H2O
2) moles H2O to moles O2
3) moles O2 to liters O2
15.00g÷
0.833m x
0.417mO2x
2H2O(l) 2H2(l) + 1O2(l)
15.00g
X22.4L/m______m
____L
0.833 0.417
÷18.0g/m
_____m
9.33
@STP
E # Mass
O 16.0 = 16.01 x18.0g/m
H 2 x 1.0 = 2.0
Finding Volume at Non-STP
PV=nRTV= (0.417m)(8.314 L•Kpa•m-1•K-1)(293.2K)
(209.0KPa)
V= 4.86L
n= 0.417mO2
n x g/m= g0.417m x 32.0g/m=13.3gO2
pp 450
c) How many moles O2 gas are produced?
d) How many grams O2 gas are produced?
b) What is the volume of gas at reaction conditions?
O 2x16.0 = 32.0g/mE # Mass
P Vn TR=
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