1

Lecture 9 Professor Hicks

Inorganic Chemistry (CHE151)

Properties of gases

Three factors affect the properties of gases

1) P = pressure

2) T = Kelvin temperature

3) V = volume

4) n = number of moles of gas

2

Pressure

• Force divided by area

• Force visualized as all matter piled up on

top of a surface

Calculate the pressure due to a 150 pound man

if his feet have a total area of 40 square inches

pressure = 150 pounds

40 square inches = 3.75 lbs/in2

force

area

Pressure on earth

pressure

1.0 atm

Column of air above

every square inch

weighs 14.7 pounds

1.0 atm = 14.7 psi

Earth

Earth’s atmosphere

Space

3

Scuba diving space

Earth’s atmosphere

Water

pressure

1.0 atm

the 10 m column

of water above

every square inch

weighs 14.6 pounds 10

meters

Column of air above

every square inch

weighs 14.7 pounds

pressure

1.0 atm

+

= 2 atm!

Units of pressure

1) Pounds per square inch (psi)

2) Atmospheres (atm) average P on earth = 1 atm

3) Millimeters of mercury (mm Hg) aka torr

pressure on earth

pressure

1.0 atm

Earth

Earth’s atmosphere

space

column

of air

Column of

mercury

metal

760 mm

A 760 mm column Hg

creates same force as

an air column of same

area that goes from

earth up to space

Mercury metal (Hg) is

very dense (13.6 g/ml)

Column

of air

from earth

to space

1.0 atm = 760 mm Hg

4

Barometer

vacuum P = 0

liquid

mercury (Hg)

or water

atmospheric = Patm

pressure

P mercury column 2

= PHg2

P mercury column 1

= PHg1

h=PHg2 – PHg1

Patm + PHg1 = PHg2

Patm = PHg2 – PHg1

PHg2

Patm

+ PHg1

Patm = h

Average day on earth

h = 760 mm mercury

1.0 atm = 760 mm Hg

Water has a density 1/13th that of mercury

barometer using water will have h = 13 x 760 mm

h = height difference

between columns

measured with a ruler

Used to measure the

atmospheric pressure

Manometer

Liquid

mercury (Hg)

or water

Atmospheric = Patm

pressure

+ PHg1

Ps Patm

+ PHg2

Pgas + PHg1 = Patm + PHg2

P mercury 1

= PHg1

Unknown P

of sample

= Psam

P mercury 2

= PHg2

Pgas = Patm + PHg2 - PHg1

h Pgas = Patm + h

Used to measure

unknown pressures

Patm is found with

a barometer

5

Empirical gas laws (discovered by experimentation)

• Charles’ Law

• Boyles Law

= constant V

T

Applies to a sample of gas

kept at constant temperature

Applies to a sample of gas

kept at constant pressure

(T must be in Kelvins)

PV = constant

Boyles Law PV = constant Applies to a sample of gas

kept at constant temperature

The deepest part of the ocean is known as the Mariana’s Trench.

The depth is 11,000 meters! This produces a pressure of 1101 atm!

Estimate the volume of a 2.0 liter soda bottle filled with air on the earth’s

surface if it was brought to the bottom of the Mariana’s Trench without

changing its temperature.

PV = constant (Boyles Law) On earths surface, for this sample of air

the Boyles law constant is

PV = 2.0 liters x 1.0 = 2.0 lit*atm 1101 atm x V = 2.0 lit*atm

V= 2/1100 lit = 0.0018 liter

6

Empirical gas laws Charles’ Law

= constant V

T

applies to a sample of gas

kept at constant pressure

(T must be in Kelvins)

Liquid nitrogen boils at -196 C. It is commonly used to

achieve low temperatures in the laboratory. If a balloon

is filled with 1.55 liters of air at 25 C, what volume will

the balloon shrink to if it is placed in a liquid nitrogen bath?

V/T = constant (Charles law)

the Charles law constant under the lab conditions is

V/T = 1.55 liters / 298 K

= 5.20 x 10-3 lit / K

V/T = 5.20 x 10 -3

Always express temperature in

Kelvin's when using gas laws

V/77 = 5.20 x 10 -3

K= 273 + C K= 273 + 25 = 298

K= 273 + -196 = 77 K

V = 77 x 5.20 x 10 -3

V = 0.400 liters

at 77 K

Charles’ Law and Kelvin

Temperature scale V = constant*T

Volume cannot become negative

T = -273 C

Absolute zero!

gas sample 1

gas sample 2

gas sample 3

extrapolate to

V=0

Find T = -273 C

equation for a line

y = mx + b

m = Charles’ law

constant

b = 0

Basis of the Kelvin temperature scale

samples of gas taken and cooled.

Volume measured at each

temperature

0

7

e. 6666 cm water

8

5.19 A gas occupying a volume of 725 mL at a pressure

of 0.970 atm is allowed to expand at constant

temperature until its pressure reaches 0.541 atm.

What is its final volume?

5.23 A 36.4-L volume of methane gas is heated from

25°C to 88°C at constant pressure. What is the final

volume of the gas?

9

5.30 Why is the density of a gas much lower than that of

a liquid or solid under atmospheric conditions? What

units are normally used to express the density of

gases?

Ideal Gas Law PV = nRT

• R is called the universal gas constant

• R replaces all constants in the named laws

R = 8.314

Joules

mole K

= 0.0821 liter*atm

mole K

SI Units

most

used

SI unit is not used as much when actually working with gases

since SI unit V is the m3 which is too big to use in the lab

10

divide

stoichiometric

number

mass

compound

moles of

formula units,

molecules,

atoms,

or ions

number

formula units,

molecules,

atoms,

or ions

multiply

molar

mass

divide

molar

mass

multiply

molar

mass

divide

molar

mass

divide

Avogadro's

number

multiply

Avogadro's

number

divide

Avogadro's

number

multiply

Avogadro's

number

equivalents

volume

n= PV

RT V= nRT

P

A sample of nitrogen gas kept in a container of

volume 1.6 L and at a temperature of 36°C exerts

a pressure of 4.8 atm. Calculate the number of

moles of gas present.

11

Given that 5.5 moles of carbon monoxide gas are

present in a container of volume 23.4 L, what is the

pressure of the gas (in atm) if the temperature is

166°C?

.

A certain amount of gas at 37°C and at a pressure

of 0.60 atm is contained in a glass vessel. Suppose

that the vessel can withstand a pressure of 3.00

atm. How high can you raise the temperature of the

gas without bursting the vessel?

12

Molar Mass from Ideal Gas Law

• Molar Mass units grams/mole

• If you can measure a mass and a # moles

their ratio is the substance’s molar mass!

• Molar Mass = mass sample / # moles

• evaporate a liquid gas

• If you can measure P, V, T you can

calculate n!

• Condense the vapor and measure the

mass of the liquid

A quantity of gas weighing 14.20 g at 1482 torr and

44°C occupies a volume of 5.40 L. What is its molar

mass?

13

N2

H2O O2 N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

Air = N2 + O2 + H2O

Pressure each gas would

have by itself (same T and V)

PN2 PO2 PH2O

N2

H2O O2 N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

N2

H2O O2 N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

N2

H2O O2 N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

Partial

Pressure

written as P subscript

gas example partial

pressure of N2 is PN2

Daltons Law of

Partial Pressures

For a mixture of gases

Pgas (total) = Pgas1 + Pgas2 + Pgas3 …

For example air

Pair = PN2 + PO2 + PH2O … etc

+ + N2

H2OO2N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

N2

H2OO2N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

N2

N2

N2

N2 N2

N2

N2

N2

N2

N2

N2

N2 N2

N2

N2

N2

O2

O2O2

O2

O2O2

H2O

H2O

H2O

H2O

H2O

H2O

H2O

H2O=

14

Partial pressure CO2

and breathing rate

lungs release CO2

cell

cells put CO2 into blood

lungs

[CO2]

blood

cells

add

CO2

breathing

removes

CO2

CO2 levels are detected in the lower brain

Higher CO2 stimulates faster breathing

CO2 is an unwanted byproduct

of metabolism but CO2 is the gas

the brain monitors to determine

breathing rate not O2

When a cell uses more energy it releases more CO2

N2

H2OO2N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

N2

H2OO2N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

N2

N2

N2

N2 N2

N2

N2

N2

N2

N2

N2

N2 N2

N2

N2

N2

O2

O2O2

O2

O2O2

H2O

H2O

H2O

H2O

H2O

H2O

H2O

H2O

Mole Fraction ( )

• Ratio of moles of gases

• If expressed as %

aka molar percentage

N2 = moles N2

total moles gas sample N2

H2OO2N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

N2

H2OO2N2

H2O

O2

N2

H2O

O2

N2

H2O

N2

N2

N2

N2

N2

N2

N2

N2 N2

N2

N2

N2

N2

N2

N2

N2 N2

N2

N2

N2

=

mole fraction

of nitrogen

molar percentage N2 = x 100% moles N2

total moles gas sample

moles N2

total moles gas sample

Note this slide was out of order

15

Partial pressure O2

& scuba diving

Space

Earth’s atmosphere

Water

pressure

1.0 atm

90

meters!

pressure

9 atm

+ =10 atm!

if you breath air down partial

pressure of O2 is 10x higher!

PO2 = 2.1 atm!

if you breath trimix 10/70

PO2 = 1.0 atm!

(10% of 10 atm)

trimix 10/70 is a mixture of

gases used for extremely

deep dives. It has 10% O2

to reduce P02

breathing high

PO2 can be toxic

scuba divers breath

trimix to decrease PO2

O2 is 21% of air so

21% of 1.0 atm is

PO2 = 0.21 atm

K2 is the second

highest mountain

in world 28,251 ft

Patm = 0.33 atm

If you breathe air on top of K2 PO2 is reduced b/c all pressure is lower

P02 = 0.07 atm

is reduced to about 1/3 of what we have at sea level

but if you breathe pure O2 instead of air on top of K2

100% O2 means Ptotal = PO2

P02 = 0.33 atm

(mountain climbers are being almost suffocated !)

Mountain climbers breathe pure O2 to increase PO2

Life requires a certain range of PO2 not % O2

Partial pressure and

mountain climbing

16

Density of a gas

• D= m / V

• Density is intensive

• Imagine 1 mole as the sample size

• Mass of 1 mole of a gas is its Molar Mass

• Volume of 1 mole of gas depends on T

and P

• V= RT/P

• Density = Molar Mass / RT /P

A sample of air contains only nitrogen and oxygen

gases whose partial pressures are 0.60 atm and

0.10 atm, respectively. Calculate the total pressure

and the mole fractions of the gases.

17

A mixture of gases contains CH4, C2H6, and C3H8.

If the total pressure is 2.50 atm and the numbers

of moles of the gases present are 0.21 mole for

CH4, 0.36mole for C2H6, and 0.16 mole for C3H8,

calculate the partial pressures of the gases.

A 2.5-L flask at 25°C contains a mixture of three

gases, N2, He, and Ne, at partial pressures of 0.12

atm for N2, 0.18 atm for He, and 0.46 atm for Ne.

(a) Calculate the total pressure of the mixture.

(b) Calculate the volume in liters at STP occupied

by He and Ne if the N2 is removed selectively.

18

Calculate the density of argon gas at 300

K and a pressure 1.1 atm.

Kinetic Molecular Theory

• Model to explain properties

of gases

• 3 postulates of KMT

1) Gas particles small compared to distance between them

2) Average kinetic energy proportional to the Kelvin temperature

3) Collisions are all elastic

(no KE turned into heat in collisions)

Ludwig Boltzmann

19

KMT and pressure

• In Kinetic Molecular Theory the force part of

pressure is due to collisions of gas molecules.

Area

Force created by constant collisions of gas particles

P = Force

Area

increase

pressure

Boyles Law

when a gas is compressed into smaller

volume collisions with wall become

more frequent increasing the pressure

smaller volume

higher pressure

(same temp)

(PV=constant)

revisited

20

KMT and Temperature

• Kinetic energy of a gas is proportional to

Kelvin temperature

• At higher temperatures gas molecules

move more quickly

As temperature increases two effects

• More frequent collisions with walls

• More energetic collisions with walls

Charles’ Law

When cooled the gas molecules slow down

collide less often and not as hard.

This causes the volume to shrink

under the applied pressure.

smaller volume

lower temperature

(same pressure)

decrease

temperature

revisited (V/T = constant)

21

Boltzmann Speed/Energy

Distribution

Ludwig Boltzmann

• Increasing temperature

increases average speed

• Higher speeds = higher energy

- kinetic energy = ½mv2

higher

temperature

% m

ole

cu

les

with

a s

pe

ed

0 300 600 900 1200

speed (m/s)

lower

temperature

50% 50%

average

speed

• Superior teaching

increases students understanding

• Better understanding higher MCAT scores

Boltzmann Energy Distribution

like the Hicks Grade Distribution

other

classes

Hicks’

class

% s

tud

en

ts

with

a s

co

re

0 10 20 30 40 45

MCAT Score

Charles Hicks

class averages

on MCAT

medical school

MCAT requirement class

Hicks’

lesser

instructors

20%

4%

% students with

required grade

20% 4%

22

0 300 600 900 1200

speed (m/s)

% m

ole

cu

les

with

a s

pe

ed

0 300 600 900 1200

speed (m/s)

% m

ole

cu

les

with

a s

pe

ed

Boltzmann Speed/Energy

Distribution

• Higher speeds = higher energy

• Reactions have energy

requirements

higher

temperature

% m

ole

cu

les

with

a s

pe

ed

0 300 600 900 1200

speed (m/s)

lower

temperature

Every speed has a kinetic energy

KE = ½mv2

higher

temperature

lower

temperature

55%

80% 20%

45%

At the higher temperature larger

% of molecules have enough

kinetic energy to react

Ludwig BoltzmannLudwig Boltzmann

20% have

enough

energy

80% don’t

have enough

energy

55% have

enough

energy

45% don’t

have enough

energy

Reaction requires

this speed (energy)

or greater

23

two different molecules at the same temperature.

Which molecule is heavier?

molecule 1

molecule 2