foundations of college chemistry, 14 th ed. morris hein and susan arena air in a hot air balloon...
Post on 20-Dec-2015
215 views
TRANSCRIPT
Foundations of College Chemistry, 14th Ed. Morris Hein and Susan Arena
Air in a hot air balloon expands upon heating. Some air escapes from the top, lowering the air density, making the balloon buoyant.
12 The Gaseous State of Matter
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
12.1 Properties of Gases
A. Measuring the Pressure of a Gas
B. Pressure Dependence: Number of Molecules and Temperature
12.2 Boyle’s Law
12.3 Charles’ Law
12.4 Avogadro’s Law
A. Mole-Mass-Volume Calculations
12.5 Combined Gas Laws
12.6 Ideal Gas Law
A. Kinetic-Molecular Theory
B. Real Gases
12.7 Dalton’s Law of Partial Pressures
12.8 Density of Gases
12.9 Gas Stoichiometry
Chapter Outline
© 2014 John Wiley & Sons, Inc. All rights reserved.
Properties of Gases
© 2014 John Wiley & Sons, Inc. All rights reserved.
Gases:
i) Have indefinite volume
ii) Have indefinite shape
Expand to fill a container
Assume the shape of a container
iii) Have low densitiesExample
dair = 1.2 g/L at 25 °C
dwater = 1.0 g/mL at 25 °C
iv) Have high velocities and kinetic energies
Volume occupied by
1 mol of H2O:
as a liquid (18 mL)
as a gas (22.4 L)
Measuring Pressure
© 2014 John Wiley & Sons, Inc. All rights reserved.
Pressure: Force per unit area
Pressure =area
force
Pressure results from gas molecule collisions
with the container walls.
Pressure depends on:
1) The number of gas molecules2) Gas temperature
3) Volume occupied by the gas
SI unit of pressure is the pascal (Pa) = 1 newton/meter2
1 atm = 760 mm Hg = 760 torr = 101.3 kPa = 1.013 bar = 14.69 psi
Unit Conversions:
× 101.3 kPa760 mm Hg
Convert 740. mm Hg to a) atm and b) kPa.
= 0.974 atm740. mm Hg
a) 1 atm = 760 mm HgUse the conversion factor:
= 98.63 kPa740. mm Hg
b)Use the conversion factor:101.3 kPa = 760 mm Hg
× 1 atm 760 mm Hg
Practicing Pressure Conversions
© 2014 John Wiley & Sons, Inc. All rights reserved.
Measuring Pressure
© 2014 John Wiley & Sons, Inc. All rights reserved.
Measuring Pressure
1) Invert a long tube of Hg over an open dish of Hg.
Use a Barometer
2) Hg will be supported (pushed up) by the
pressure of the atmosphere.3) Height of Hg column can
be used to measure
pressure.
1) On the Number of Molecules
Pressure (P ) is directly proportional to the number of gas molecules present (n ) at constant
temperature (T ) and volume (V ).
Increasing n creates more frequent collisions with the
container walls, increasing the pressure
1 mol H2
P = 1 atm0.5 mol H2
P = 0.5 atm2 mol H2
P = 2 atm
V = 22.4 LT = 25.0 °C
Pressure Dependence
© 2014 John Wiley & Sons, Inc. All rights reserved.
Pressure Dependence
© 2014 John Wiley & Sons, Inc. All rights reserved.
Pressure is directly proportional to temperature when
moles (n ) and volume (V ) are held constant.
Increasing T causes: a) more frequent and b) higher energy collisions
0.1 mol of gasin a 1L container
T = 0 °C T = 100 °C
2.24 atm 3.06 atm
2) On Temperature
The volume of a fixed quantity of gas is inversely proportional to the pressure exerted by the gas at
constant mass and temperature.
PV = constant (k ) P V 1or
P = k ×V 1
Most common form:
P1V1 = P2V2
Graph showing inverse PV relationship
Boyle’s Law
© 2014 John Wiley & Sons, Inc. All rights reserved.
730. mm Hg600. mm Hg
×
What volume will 3.5 L of a gas occupy if the pressure is
changed from 730. mm Hg to 600. mm Hg?
V1 = 3.5 L P1 = 730. mm Hg P2 = 600. mm Hg
P1V1 = P2V2
= 4.3 L 3.5 LV2 =
Knowns
Solve For V2 V2 = P2
P1V1
Calculate
Boyle’s Law Problems
© 2014 John Wiley & Sons, Inc. All rights reserved.
Kelvin Temperature ScaleDerived from the relationship between temperature
and volume of a gas.
As a gas is cooled by 1 ºC increments, the gas volume decreases in increments of 1/273.
All gases are expected to have zero volume if cooled to −273 ºC.
V -T relationship of methane(CH4) with extrapolation (-----) to absolute zero.
Temperature in Gas Law Problems
© 2014 John Wiley & Sons, Inc. All rights reserved.
This temperature (−273 ºC) is referred to as absolute zero.
Absolute zero is the temperature (0 K) when the volume of an ideal gas becomes zero.
All gas law problems use the Kelvin temperature scale!
TK = T°C + 273
Kelvin temperature
Celsius temperature
Temperature in Gas Law Problems
© 2014 John Wiley & Sons, Inc. All rights reserved.
The volume of a fixed quantity of gas is directly proportional to the absolute temperature of the
gas at constant pressure.
Most common form:
V TV = k T
or
V1 V2
T1 T2
=
VT
= k
Charles’ Law
© 2014 John Wiley & Sons, Inc. All rights reserved.
3.0 L of H2 gas at −15 ºC is allowed to warm to 27 ºC at
constant pressure. What is the gas volume at 27 ºC?
V1 = 3.0 L
T1 = −15 ºC = 258 K T2 = 27 ºC = 300. K
Knowns
Solving For V2
V2 = T
1
V1T2
Calculate
V1 V2
T1 T2
=
300. K258 K
= 3.5 L3.0 L=V2 = T1
V1T2 ×
Charles’ Law Problems
© 2014 John Wiley & Sons, Inc. All rights reserved.
Equal volumes of different gases at constant T and P
contain the same number of molecules.
1 volume unit4 molecules
1 volume unit4 molecules
2 volume units8 molecules
Avogadro’s Law
© 2014 John Wiley & Sons, Inc. All rights reserved.
Given the following gas phase reaction:
N2 + 3 H2 2 NH3
If 12.0 L of H2 gas are present, what volume of N2
gas isrequired for complete reaction? T and P are held
constant.By Avogadro’s Law, we can use the reaction
stoichiometryto predict the N2 gas needed.
Knowns
Solving For VN2
Calculate
VH2 = 12.0 L
12.0 L H2 × 1 L N2
3 L H2
= 4.00 L N2
required
Avogadro’s Law
© 2014 John Wiley & Sons, Inc. All rights reserved.
At constant T and P, how many liters of O2 are required
to make 45.6 L of H2O?a) 11.4 L
b) 45.6 L
c) 22.8 L
d) 91.2 L
Given the following gas phase reaction:
2 H2 + O2 2 H2O
45.6 L H2O × 1 L O2
2 L H2O
= 22.8 L O2
required
Sense Check: Less moles of O2 equal less L of O2!
Avogadro’s Law
© 2014 John Wiley & Sons, Inc. All rights reserved.
A combination of Boyle’s and Charles’ Laws.
Used in problems involving changes in P, T, and Vwith a constant amount of gas.
The volume of a fixed quantity of gas depends on the
temperature and pressure.It is not possible to state the volume of gas without
stating the temperature and pressure.
0.00 °C (273.15 K) and 1 atm (760 torr)
Standard Temperature and Pressure (STP):
P1V1 P2V2T
1
T2
=
Combined Gas Laws
© 2014 John Wiley & Sons, Inc. All rights reserved.
A single equation relating all properties of a gas.
where R is the universal gas constant
Constant n and T Constant P and TConstant n and P
V 1/PBoyle’s Law
V TCharles’
Law
V nAvogadro’s Law
PV = nRT
Ideal Gas Law
© 2014 John Wiley & Sons, Inc. All rights reserved.
mol . K
R is derived from conditions at STP. Calculate R.
Knowns
Solving For R
Calculate
R
PV nT
=
P = 1.00 atm V = 22.4 L T = 273 K n = 1.00 mol
R = P × Vn × T
× 22.4 L1.00 mol × 273 K
= 0.0821 L . atm1.00 atm=
PV = nRT
Units are critical in ideal gas problems!
Ideal Gas Constant
© 2014 John Wiley & Sons, Inc. All rights reserved.
A general theory developed to explain the behaviorand theory of gases, based on the motion of
particles.Assumptions of Kinetic Molecular Theory (KMT):
1) Gases consist of tiny particles.
2) The distance between particles is large when compared
to particle size. The volume occupied by a gas is
mostly empty space.3) Gas particles have no attraction for one another.
4) Gas particles move linearly in all directions, frequently
colliding with the container walls or other particles.
Kinetic Molecular Theory
© 2014 John Wiley & Sons, Inc. All rights reserved.
What the Nose Knows
Sensing low concentrations of chemicals is useful!
Dogs use smell to detect many drugs, explosives, etc.
based on trace amounts of chemical compounds in the air.
For more information, see: http://www.scs.illinois.edu/suslick/smell_seeing.html
Chemistry in Action
© 2014 John Wiley & Sons, Inc. All rights reserved.
Better Coffee
Artificial noses could sniff
out cancer or explosives!
Better Science
1) Explain atmospheric pressure and how it is measured. 2) Be able to convert between the various units of pressure.
12.1 Properties of Gases
3) Use Boyle’s Law to calculate changes in pressure or volume of a gas at constant temperature.
12.2 Boyle’s Law
4) Use Charles’ Law to calculate changes in temperature or volume of a gas at constant pressure.
12.3 Charles’ Law
Learning Objectives
© 2014 John Wiley & Sons, Inc. All rights reserved.
5) Solve problems using the relationships between moles, mass, and volume of gases.
12.4 Avogadro’s Law
6) Use the combined gas law to calculate changes in pressure, volume, or temperature of a gas sample.
12.5 Combined Gas Law
7) Use the ideal gas law to solve problems involving pressure, volume, temperature, and moles of a gas.
12.6 Ideal Gas Law
Learning Objectives
© 2014 John Wiley & Sons, Inc. All rights reserved.