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????

BarometerOpen to 

atmosphere Closed

Vacuum

1 atm 1 atm

1 atm1 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 mercurymass air density air Volume air density air Area Height airmass Hg density Hg Vol Hg density Hg Area h HgForce g mass air g mass mercury

g mass air g mass meForcePressureArea Area

== ⋅ = ⋅ ⋅= ⋅ = ⋅ ⋅

= ⋅ = ⋅

⋅ ⋅= = =

( )rcuryArea

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 barometerMass 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.

33

33

1.2 9.8 8.6 10

13.6 10 9.8

Air

Hg

P dghkgP d g h mmkg lP d

101

0.760 101

kPa

g h m al 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

185.2 10M kg= ⋅

29.8 secmg F Mg= =

2 14 25.0 10oA d mπ= ⋅ = ⋅

612.6 10od m= ⋅

182

514 2

5.2 10 9.8 sec 1.01 10 15 10

mkgF MgP Pa AtmA A m

⋅ ⋅= = = = ⋅ =

⋅

18

314 3 3

12 5.2 10 1.2 1.2

2 5 10 4.3 10M kg kg gd mA h m

⋅= = = =

⋅ ⋅ ⋅ ⋅ ⋅

A simple manometer, a device for measuring the pressure of a gas in a container.

( ) or P 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 thedecomposition of Lime stone (CaCO3) in a closed end manometer, the

height of the mercury is 341.6 mm Hg. Calculate the CO2 pressure intorr, atmospheres, and kilopascals, and bars.Plan: The pressure is in mmHg, so we use the conversion factorsto find the pressure in the other units.Solution:

PCO2 (torr) = 341.6 mm Hg x = 341.6 torr1 torr1 mm Hg

converting from mmHg to torr:

converting from torr to atm:PCO2( atm) = 341.6 torr x = 0.4495 atm1 atm

760 torrconverting from atm to kPa:

PCO2(kPa) = 0.4495 atm x = 45.54 kPa101.325 kPa1 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 nV T b b b P n

V n a a a T P

⋅ = =

= ⋅ =

= ⋅ =

Boyle’s Law : P - V inversely proportionalwhen 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 kP V k P VV 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= ⋅ = ⋅ ⋅

FPA

=

F g m= ⋅

mP g g d hA

= = ⋅ ⋅

2 2 3

3

3sec sec

1 101.3 10 10.079.8 1.026 9.8 1.026 1 10kg kgm m liter

liter liter m

P Atm Pah mg d

⋅= = = =

⋅ ⋅ ⋅ ⋅ ⋅

Boyle’s Law - A gas bubble in the ocean!

A bubble of gas is released by the submarine “Alvin” at a depth of6000 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:

16000 6000 176.534

Atmx ft ft Atm waterft

= = ⋅ = −

1 1 188.5P x Atm= + =

1 1 2 2

2

2

177.5 1 1177.5 0.18

PV PVAtm 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 TemperatureV = 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 2V2

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 K297.0 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 nV n n nV 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 inthe atmosphere, if the volume of 2.67 g of SF6 at 1.143 atm and 28.5 oCis 2.93 m3, what will be the mass of SF6 in a container whose volume is543.9 m3 at 1.143 atm and 28.5 oC?Plan: Since the temperature and pressure are the same it is a V - nproblem, 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 moln m nC m n M d M C MV 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:

V2V1

n2 = n1 x = 0.0183 mol SF6 x = 3.40 mol SF6

2.67g SF6146.07g 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 22 1

1 2 1

543.92.67 496.2.93

m m Vd m m gV 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 0.08201 273.15

PV nRTPV Atm AtmR mol KnT mole K

=⋅ −= = = −⋅

8.3141 101.3

0.0820

1 101.3 and 1 100

JR mol K JAtmAtmR 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=nRTn = 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 BalloonNeed 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

128.960.082 290

1.218

W

W W

Air

gM Air moln 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 2901.218 0.9468373Balloon W Air

PV nRT n T n TTP 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 Balloong gm V d d ll l

m g kg

Δ = ⋅ − = − = ⋅ ⋅ −

Δ = ⋅ =

Hot Air Balloon: Weight Distribution

component pounds kilograms

100,000 cu ft (2,800 m3) envelope 250 113.45-passenger basket 140 63.5double burner 50 22.73 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 723.5100,000 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?