Foundations of College Chemistry, 14 th Ed. Morris Hein and Susan Arena Air in a hot air balloon expands upon heating. Some air escapes from the top, lowering

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


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


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


Measuring Pressure


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


Pressure Dependence


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


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


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


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


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


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


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


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


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


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


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


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


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


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


Better Coffee

Artificial noses could sniff

out cancer or explosives!

Better Science

http://www.scs.illinois.edu/suslick/smell_seeing.html

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


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


Documents

Foundations of College Chemistry, 14 th Ed. Morris Hein and Susan Arena Air in a hot air balloon expands upon heating. Some air escapes from the top, lowering