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Gasses
• Zumdahl (6th Ed) Chapter 5 Sections 1-3
• The nature of gasses (the air we breathe)
• Scale of gasses in atmosphere.
• The ideal Gas (I.G.)
– How it behaves
- Solve Gas Problems
- Subsumes Boyle, Charles, Avogadro Laws
• Problems 5.21-24, 5.26, 5.29-32, 5.35, 5.37-38.

How big are molecules?
Water has a density (weighs) 1 gram/cc.
Boil the water and the density is 1 gram/liter.
Convert that to moles per liter:
What is the expansion factor?
What does this tell us about what a gas is?

How big are molecules?
Liquid water has a density (weighs) 1 gram/cc.
What is the volume and diameter of a water molecule?
(Assume it is a sphere).
What is the molar concentration?
Water vapor at one Atm pressure:
density is 1.2 gram/liter.
What is the number of moles per liter?
What is the average distance between vapor molecules.
Compare that to the distance between water molecules in liquid
water.
Compare the volume that N water molecules occupy in the gas
state with that in the liquid state? (i.e. What is the expansion
factor?

Why do gas molecules stay as a gas?
• Consider liquid N2 or solid CO2.
– What makes them condense?
• Temperature is the best predictor of the state of matter;
it is a direct measure of the kinetic energy inherent in
molecules.
• So the trade off is the molecular attraction to cause
matter to condense with the kinetic energy inherent in
matter (at 300K) to overcome adhesion.
• Consider the “gedanken” experiment of an isolated gas
next to a solid object in contact with a gas at Room
Temp.
• Temperature makes molecules move that
overcomes the “weaker” attractions tending to
make the solid.

Gas Molecules in a Box: Lots of Empty Space
Wbsite for the kinetic demo:
http://ccl.northwestern.edu/netlogo/models/run.cgi?GasLabGasinaBox.919.607
Trace shows the
path of one
particle, includes
collisions with other
particles.
Average distance
between particles
(at 1STP) is about
10 times particle
diameter.

Hot Air Balloon
Why does a
hot air balloon
rise? How hot
does the gas
have to be to
carry the
people and the
basket below?

How may units of Pressure are there?
• The Torr (or Torricelli)
• Mm Hg 1Torr=1mm(Hg)
• Atmospheres (Atm) 1 Atm = 760 Torr
• Pounds per square inch 1Atm = 14.7 lb/in2
• Bars SI units ~ 1.01 Atm
• The SI unit itself: the Pascal. 1 Bar=105Pa
• How are they all connected????

Barometer
Open to
atmosphere Closed
Vacuum
1 atm 1 atm
1 atm
1 atm
The mercury column exerts a
force over the cross‐sectional
area of the tube.
The pressure exerted by the
mercury column is exactly
balanced by the pressure of the
atmosphere.
P h
What is pressure; how do we
measure it?
g d= ⋅ ⋅
pressure
height accel. due
to gravity
density

A torricellian barometer Relation of Pressure, mass, force, Area
( ) ( )
( )
_ _
_ _ _ _ _
_ _ _ _ _
_ _
_ _
mass air mass mercury
mass air density air Volume air density air Area Height air
mass Hg density Hg Vol Hg density Hg Area h Hg
Force g mass air g mass mercury
g mass air g mass meForce
Pressure
Area Area
=
= ⋅ = ⋅ ⋅
= ⋅ = ⋅ ⋅
= ⋅ = ⋅
⋅ ⋅
= = =
( )rcury
Area
Hg is very dense, density is 13. 6 g/cc;
Air is sparse, and is equivalent to a column of air of a density of
1.2 g/l up to 8.6 km.
The air does not have uniform density but falls off exponentially
due to the force of gravity and its own weight.

A torricellian barometer
Mass of the Hg in the tube must equal the effective
mass of the air pushing down. .
P = 2.28 pounds / cm2 = 14.7 pounds / in2 = 760 mm Hg = 1.00 atm.
3
3
3
3
1.2 9.8 8.6 10
13.6 10 9.8
Air
Hg
P dgh
kgP d g h m
m
kg lP d
101
0.760 101
kPa
g h m a
l m
=
= ⋅ ⋅ = ⋅ ⋅ ⋅
= ⋅ ⋅ = ⋅ ⋅ ⋅ kP
=
=

Why do we have pressure from the atmosphere?
Gas in the air is held by gravity:
How thick is the earth’s atmosphere?
What is pressure?
Why don’t we “feel” 1 Atm of pressure?
Why is it hard to walk on grass in high-heeled shoes?
How deep would the liquid be if the air were to liquefy
(which it does when the temperature is below 80K).

Why do we have pressure from the atmosphere?
There is a lot of gas in the air and the gas is held by gravity. The force of gravity
brings the molecules to the earth’s surface but they do not just lay on the ground,
they have too much energy for that and they bounce around.
The mass of the atmosphere is
And the gravitational constant is
The earth is a ball with diameter
So the earths surface area is
Pressure is due to all that mass spread over the entire surface of the earth, which
generates a force due to gravity:
Half of the earth’s atmosphere is contained in the first 4.3 km. So the density of the
air is:
The density and the pressure drop exponentially as one goes away from the earth
18
5.2 10M kg= ⋅
29.8
sec
mg F Mg= =
2 14 2
5.0 10oA d mπ= ⋅ = ⋅
6
12.6 10od m= ⋅
18
2
5
14 2
5.2 10 9.8
sec 1.01 10 1
5 10
mkgF Mg
P Pa Atm
A A m
⋅ ⋅
= = = = ⋅ =
⋅
18
314 3 3
1
2 5.2 10
1.2 1.2
2 5 10 4.3 10
M kg kg gd
mA h m
⋅
= = = =
⋅ ⋅ ⋅ ⋅ ⋅

A simple manometer, a device for measuring the
pressure of a gas in a container.
( )orP hgd P gd h= Δ = ⋅Δ

How does a car tire work?
• A car tire is usually inflated to 35 psi (or 2 to 3 Atm over
atmospheric pressure).
• What fraction of the tire is filled with air? At one
atmosphere 0.1% are molecules, at 10 Atmosphere 1%
is actually air. 10 Atmospheres is 140 psi; which is
above the high pressure bike racing tires. And only 1%
of the space in the tire is taken up by the air!!!!
• Why does air in a tire work? The more air, the more
molecules hit the inside of the tire and push out.
• How well would a tire work below 80K? (Moon tires?)
• Air molecules hit the wall of the tire more often on the
inside than the outside. The molecules hitting the wall of
the tire push on the wall. They don’t loose kinetic
energy. Why?

Tire Pressure: Volume decreases, pressure goes up
This is Boyle’s law (developed in the 1650s before tires)

Problem: A chemist collects a sample of Carbon dioxide from the
decomposition of Lime stone (CaCO3) in a closed end manometer, the
height of the mercury is 341.6 mm Hg. Calculate the CO2 pressure in
torr, atmospheres, and kilopascals, and bars.
Plan: The pressure is in mmHg, so we use the conversion factors
to find the pressure in the other units.
Solution:
PCO2 (torr) = 341.6 mm Hg x = 341.6 torr
1 torr
1 mm Hg
converting from mmHg to torr:
converting from torr to atm:
PCO2( atm) = 341.6 torr x = 0.4495 atm
1 atm
760 torr
converting from atm to kPa:
PCO2(kPa) = 0.4495 atm x = 45.54 kPa
101.325 kPa
1 atm
Converting Units of Pressure

Gas Laws: an Overview
• Picture: Molecules bounce around in a box.
• Parameters needed to describe the situation:
• Number of gas molecules; Size of the box; Temperature
• The outcome is that the molecules produce pressure by
hitting the sides of the box. We can measure pressure.
• The summary relation among all of these quantities is
• Demonstrate the relations by various experiment
– Boyle’s Law
– Charles’ Law
– Avogadro’s Law
PV nRT=
( )
( )
( , )
,
,
P V k k k T n
V T b b b P n
V n a a a T P
⋅ = =
= ⋅ =
= ⋅ =

Boyle’s Law : P - V inversely proportional
when n and T are constant in a gas sample
• Pressure is inversely proportional to Volume
• Change of Conditions if n and T are constant !
• P1V1 = k P2V2 = k
• Then :
• When the volume decreases and the molecules move at
the same speed the time between wall-hits is less, so the
pressure is higher.
( ),
k k
P V k P V
V P
k k n T
⋅ = = =
=
1 1 2 2P V P V=

Plotting Boyle’s data from Table 5.1.
Syringe Demo

Boyle’s Law : Balloon
• A balloon has a volume of 0.55 L at sea level
(1.0 atm) and is allowed to rise to an altitude of 6.5
km, where the pressure is 0.40 atm. Assume that the
temperature remains constant(which obviously is not
true), what is the final volume of the balloon?
• P1 = 1.0 atm P2 = 0.40 atm
• V1 = 0.55 L V2 = ?
• V2 =

Boyle’s Law : Balloon
• A balloon has a volume of 0.55 L at sea level
(1.0 atm) and is allowed to rise to an altitude of 6.5
km, where the pressure is 0.40 atm. Assume that the
temperature remains constant(which obviously is not
true), what is the final volume of the balloon?
• P1 = 1.0 atm P2 = 0.40 atm
• V1 = 0.55 L V2 = ?
• V2 = V1 x P1/P2 = (0.55 L) x (1.0 atm / 0.40 atm)
• V2 = 1.4 L

Determine the pressure of water
• The mass of water in a column h units
high and with a foot print of area A is
related to the density:
• The Pressure is related to the Force of
the water.
• The force of the water is proportional to
the mass:
• Solve for Pressure in terms of height of
the water:
• Water density is about 1g/cc=1kg/liter
• Atmospheric pressure is 101.3kPa
• Solve for the height in meters using SI
units of pressure.
m d V d h A= ⋅ = ⋅ ⋅
F
P
A
=
F g m= ⋅
m
P g g d h
A
= = ⋅ ⋅
2 2 3
3
3
sec sec
1 101.3 10
10.07
9.8 1.026 9.8 1.026 1 10kg kgm m liter
liter liter m
P Atm Pa
h m
g d
⋅
= = = =
⋅ ⋅ ⋅ ⋅ ⋅

Boyle’s Law - A gas bubble in the ocean!
A bubble of gas is released by the submarine “Alvin” at a depth of
6000 ft in the ocean, as part of a research expedition to study under
water volcanism. Assume that the ocean is isothermal( the same
temperature through out ),a gas bubble is released that had an initial
volume of 1.00 cm3, what size will it be at the surface at a pressure of
1.00 atm? (We will assume that the density of sea water is 1.026 g/cm3,
and use the mass of Hg in a barometer for comparison!)
Initial Conditions Final Conditions
V 1 = 1.00 cm3
P 1 = ?
V 2 = ?
P 2 = 1.00 atm
Each 10 meters of sea water generates a force of 1 Atmosphere; This is a depth of
1840 meters. So 185 Atmosphere of pressure.

Ocean Bubble Calculation
For every 10 meters or 32 feet there is an additional
atmosphere of pressure.
Therefore: 32 feet water = 1 Atmosphere pressure
The total pressure on the bubble (add one for air) is
The final volume then:
1
6000 6000 176.5
34
Atm
x ft ft Atm water
ft
= = ⋅ = −
1 1 188.5P x Atm= + =
1 1 2 2
2
2
177.5 1 1
177.5 0.18
PV PV
Atm cc Atm V
V cc l
=
⋅ = ⋅
= =

Diving Questions
If the diver took a 1 liter balloon with him
to that depth, how large would the
balloon be at 40 meters?
A diver can dive with SCUBA equipment to
40 meters. A deep but doable “recreation”
depth, 132 ft. Start to get N2 narcosis
without special precautions. What is the
pressure at that depth?

Charles’ Law:
Plots of V versus T (C) for several gases.

Charles Law - V - T- proportional
• At constant pressure and for a fixed amount (# moles)
of gas:
Volume is proportional to Temperature :
V= constant x Absolute Temperature
V = T x b
• Change of conditions problems:
• Since V = T x b or V1 / T1 = V2 / T2 or:
T1
V1
T2 or
=
V2
T1 = V1 x
T 2
V2
Temperatures must be expressed in Kelvin!

Volume -- Temperature
• The box has a piston on the top that is at 1
Atm and can move up and down to keep
the pressure constant and keep the gas in
the box. The temperature drops to from
117C to 58.5C, the new volume will be as
a percentage of the original volume?
• A balloon in Antarctica in a building is at
room temperature ( 75o F ) and has a
volume of 20.0 L . What will be its volume
outside where the temperature is -70oF ?

Charles Law Problem - I
• A balloon in Antarctica in a building is at room
temperature ( 75o F ) and has a volume of 20.0
L . What will be its volume outside where the
temperature is -70oF ?
• V1 = 20.0 L V2 = ?
• T1 = 75o F T2 = -70o F
• o C = ( o F - 32 ) 5/9
• T1 = ( 75 - 32 )5/9 = 23.9o C
• K = 23.9o C + 273.15 = 297.0 K
• T2 = ( -70 - 32 ) 5/9 = - 56.7o C
• K = - 56.7o C + 273.15 = 216.4 K

Antarctic Balloon Problem - II
• V1 / T1 = V2 / T2 V2 = V1 x ( T2 / T1 )
• V2 = 20.0 L x
• V2 = 14.6 L
• The Balloon shrinks from 20 L to 15 L !!!!!!!
• Just by going outside !!!!!
216.4 K
297.0 K

Combo Problem
• Typical car tire is at 35 psi at 100F (or 40C). If
the temperature drops to -40F, what is the tire
pressure?
• First Assume the tire volume does not change
• Then redo assuming the tire volume drops by
5%.

Variations on Ideal Gas Equation
• During chemical and physical processes, any of the four
variables in the ideal gas equation may be fixed.
• Thus, PV=nRT can be rearranged for the fixed variables:
– for a fixed amount at constant temperature
• P V = nRT = constant Boyle’s Law
• P2V2=nRT=P1V1
– for a fixed amount at constant pressure
• V / T = nR / P = constant Charles’ Law
– for a fixed pressure and temperature
• V = n (RT/P) or V/n = constant Avogadro’s Law
Use I.G. E.o.S. : PV=nRT
and rearrange as needed

The volume of gas is directly proportional to the amount of gas
(in moles), when measured at the same P and T:
This is the most amazing and puzzling law. Why does the
mass of the gas molecules not matter? He, CO2 and propane
obey exactly the same gas law. Why?
For problems where P,T are fixed go from state 1 to state 2:
Concentration, C, of the gas does not change, @ fixed P and T
Avogadro’s Law - Moles and Volume
( )or where ,V n V a n a a P T∝ = ⋅ =
2 2 1 2
1 1 1 2
or C=
V a n
V n n n
V n V V
= ⋅
= =

Standard Molar Volumes; density depends on gas

One mole of gas in familiar object

Problem: Sulfur hexafluoride is a gas used to trace pollutant plumes in
the atmosphere, if the volume of 2.67 g of SF6 at 1.143 atm and 28.5 oC
is 2.93 m3, what will be the mass of SF6 in a container whose volume is
543.9 m3 at 1.143 atm and 28.5 oC?
Plan: Since the temperature and pressure are the same it is a V - n
problem, so we can use Avogadro’s Law to calculate the moles of the
gas, then use the molecular mass to calculate the mass of the gas.
Avogadro’s Law: Volume and Amount of Gas
( )6 146.07
and therefore
W
W W W
gM SF
mol
n m n
C m n M d M C M
V V V
=
= = ⋅ = = ⋅ = ⋅

Reduced Problem: 2.67 g of SF6 has volume 2.93 m3;
What is mass of SF6 in volume 543.9 m3
Solution
Alternate Solution:
V2
V1
n2 = n1 x = 0.0183 mol SF6 x = 3.40 mol SF6
2.67g SF6
146.07g SF6/mol
= 0.0183 mol SF6
543.9 m3
2.93 m3
mass SF6 = 3.40 mol SF6 x 146.07 g SF6 / mol = 496 g SF6
Calculation: Avogdaro’s Law
1 2 2
2 1
1 2 1
543.9
2.67 496.
2.93
m m V
d m m g
V V V
= = ∴ = = ⋅ =

Use Avogadro’s Principle to Obtain the Gas
Constant, R
• One mole occupies 22.4 liters at STP
• How many Joules of energy are in 1 liter-
Atmosphere (PV has units of energy!)
1 22.4
0.0820
1 273.15
PV nRT
PV Atm AtmR
mol KnT mole K
=
⋅ −= = =
−⋅
8.314
1 101.3
0.0820
1 101.3 and 1 100
J
R mol K J
AtmAtmR
mol K
Atm kPa Bar kPa
−= = =
−−
−
= =

Summary: I.G.
• The ideal gas equation combines both Boyle’s, Charles’
and Avogadro’s law into one easy-to-remember law:
PV=nRT
n = number of moles of gas in volume V
• R = Ideal gas constant
• R = 0.08206 L atm / (mol K) = 0.08206 L atm mol-1 K-1
• In SI units R = 8.314 Pa-m3/ (mol K) = 8.314 J mol-1 K-1
• An ideal gas is one for which both the volume of molecules
and forces between the molecules are so small that they
have insignificant effect on its P-V-T behavior.
Independent of substance, in the limit that n/V →0,
all gases behave ideally. Usually true below 2 atm.

Hot Air Balloon
Why does a hot air balloon rise?
Typical hot air balloons displace
100,000 cu ft (2,800 m3). The air
in the balloon is typically heated to
around 100C (max op temp is
120C). Air is 78% N2, 21% O2
(<1% Ar, other). Temperature is
17C. What is the total mass that
can be carried?

Hot Air Balloon
Need the number of moles inside the balloon at
the higher temperature. The density of air is
( )
( ) ( )1
1
0.78 28 0.21 32 0.01 40 28.96
1
28.96
0.082 290
1.218
W
W W
Air
gM Air
mol
n P Atmgd M Air M Air
mol l AtmV RT
mol
gd
l
= ⋅ + ⋅ + ⋅ =
= ⋅ = ⋅ =
−⋅
=
Use: P,V are constant, so as T goes up n goes down.
( )
1 1 2 2
1
2 2
or or
290
1.218 0.9468
373
Balloon W Air
PV nRT nT n T
TP gd M Air d
lRT T
= =
= ⋅ = = ⋅ =
The difference in mass between the mass of the air and the
balloon is what you can carry.
( ) ( ) ( )6
6
1.218 0.9468 2.8 10 1.218 0.9468
0.76 10 760
Air Balloon
g gm V d d l
l l
m g kg
Δ = ⋅ − = − = ⋅ ⋅ −
Δ = ⋅ =

Hot Air Balloon: Weight Distribution
component pounds
kilogra
ms
100,000 cu ft (2,800 m3) envelope 250 113.4
5-passenger basket 140 63.5
double burner 50 22.7
3 20-gallon (75.7-liter) fuel tanks
full of propane
3 × 135 =
405
183.7
5 passengers
5 × 150 =
750
340.2
sub total 1595 723.5
100,000 cu ft (2,800 m3) of heated
air
5922 2686.2
total
(3.76 tons)
7517
3409.7
using a density of 0.9486 kg/m³ for dry air heated to 210 °F (99 °C).

Inflated Dual Airbags
How do we store
the gas so well?
The primary
chemical reactant
is NaN3(S).
How much powder
do we need in the
air bag?

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