GASES & THEIR GASES & THEIR PHYSICAL PROPERTIESPHYSICAL PROPERTIES

Chemistry in your Future While scuba-diving with some friends, you surface to find that one of the party is barely conscious and appears to be in pain. The group had been exploring a reef 150 feet below the surface and didn’t notice when this person surfaced much earlier than the rest. Three of you immediately force this diver back down 100 feet and slowly bring him up in stages of 50 feet, staying at each depth for about 30 minutes. At the end of the process, he is fully recovered. The certified divers in the group knew how to handle the emergency, but you are the only one who studied chemistry and can explain what occurred. It has to do with the material in Chapter 5.

The Physical Properties of Gases

Air BagsAir Bags• Impact trips a sensor that activates the reaction:

• The gaseous product of the reaction occupies 450 times more space than does the solid reactant.

Pressure DefinedPressure Defined

• Gas molecules are in constant motion, colliding with each other and with the walls of their container.

• The sum of ALL these collisions is called pressure.

A Swarm of ParticlesA Swarm of Particles

• The air around us is a mixture of gases, tasteless, odorless and invisible

• We can feel its effects, and we depend on it every moment of our existence.

• It’s a swarm of molecules not unlike a swarm of gnats

Pressure• Pressure is directly proportional to the number

of gas molecules in the air; pressure can change.

Pressure and WeatherPressure and Weather• High atmospheric

pressure redirects storms (sign of fair weather).

• Low atmospheric pressure tends to draw storms in (sign of rainy weather).

• Changes in pressure are responsible for wind.

PHYSICAL PROPERTIESPHYSICAL PROPERTIES

OF OF

GASESGASES

Some Important Industrial Gases

Name - Formula Origin and use

Methane (CH4) Natural deposits; domestic fuel

Ammonia (NH3) From N2 + H2 ; fertilizers, explosives

Chlorine (Cl2) Electrolysis of seawater; bleaching and disinfecting

Oxygen (O2) Liquified air; steelmaking

Ethylene (C2H4) High-temperature decomposition of natural gas; plastics

The Three States of Matter

Br2 can exist as a gas (A), liquid (B), or solid (C) if p & T are the right values.

Important Characteristics of Gases

1) Gases are highly compressible An external force compresses the gas sample and decreases its volume, removing the external force allows the gas volume to increase.2) Gases are thermally expandable When a gas sample is heated, its volume increases, and when it is cooled its volume decreases.3) Gases have low viscosity Gases flow much easier than liquids or solids.4) Most Gases have low densities Gas densities are on the order of grams per liter whereas liquids and solids are grams per cubic cm, 1000 times greater.5) Gases are infinitely miscible

Gases mix in any proportion such as in air, a mixture of many gases.

• Helium He 4.0• Neon Ne 20.2• Argon Ar 39.9

• Hydrogen H2 2.0

• Nitrogen N2 28.0

• Nitrogen Monoxide NO 30.0

• Oxygen O2 32.0

• Hydrogen Chloride HCL 36.5

• Ozone O3 48.0

• Ammonia NH3 17.0

• Methane CH4 16.0

Substances That Are Gases under Substances That Are Gases under Normal ConditionsNormal ConditionsSubstance Formula MM(g/mol)

Pressure of the AtmospherePressure of the Atmosphere

• Called “atmospheric pressure,” or the force exerted upon us by the atmosphere above us.

• A measure of the weight of the atmosphere pressing down upon us.

• Measured using a barometer - A device that can weigh the atmosphere above us.

Pressure = Force Area

Effect of Atmospheric Pressure on Effect of Atmospheric Pressure on Objects at the Earth’s SurfaceObjects at the Earth’s Surface

In left diagram, p inside can = p outside can. In right, air is removed (pinside < poutside) and can collapses.

…measures atmospheric pressure

A Mercury Barometer

The Mystery of the Suction Pump

When you drink through a straw, you create a lower p above liquid, and the atmosphere pushes the liquid up.

Differential manometers. (a) Both columns are at the same height because both sides are exposed to the atmosphere; (b) the stopcock on the left is closed, and the stopcock on the right is open to the atmosphere. The difference in heights is a direct measure of the difference in pressure between the flask on the left and the atmospheric pressure.

Manometers are used to measure gas pressure in an expt.

Common Units of Pressure

Unit Atmospheric Pressure Scientific Field

pascal (Pa); 1.01325 x 105 Pa SI unit; physics, kilopascal(kPa) 101.325 kPa chemistry

atmosphere (atm) 1 atm* Chemistry millimeters of mercury 760 mmHg* Chemistry, medicine, ( mm Hg ) biology

torr 760 torr* Chemistry

pounds per square inch 14.7 lb/in2 Engineering ( psi or lb/in2 )bar 1.01325 bar Meteorology, chemistry, physics

Gas Property RelationshipsGas Property Relationships

• Fundamental properties of gases:– Pressure– Amount (measured in moles, represented by n)– Volume (usually expressed in liters)– Temperature (expressed in Kelvins)

• If one of these properties is changed, the others will also change.

The Gas LawsThe Gas Laws

Boyle’s Law : P - V Relationship

• Pressure is inversely proportional to volume• P = or V = or PV=k• Change of conditions problems

if n (mol of gas) and T (in Kelvin) are constant

• P1V1 = k P2V2 = k’

k = k’• Then :

PP11VV11 = P = P22VV22

kV

kP

Pressure increaseswith depth & decreases with Height

Ascending to quickly increases volume of gas in lungs leading to ailment called bends.

Experimental results for Boyle’s Law

Charles Law - V - T- Relationship

• Temperature is directly related to volume• T proportional to Volume : T = kV

• Change of conditions problem:

• Since T/V = k or T1 / V1 = T2 / V2 or:

T1

V1

T2=V2

T1 = V1 xT2

V2

• NOTE:NOTE: Temperatures must be expressed in Kelvin to avoid negative values.

Experimental Results of Charles’ Law

Avogadro’s Law - Amount and Volume

The amount of gas (moles) is directly proportional to the volume of the gas.

n V or n = kV

For a change of conditions problem we have the initial conditions,and the final conditions, and we must have the units the same.

n1 = initial moles of gas V1 = initial volume of gasn2 = final moles of gas V2 = final volume of gas

n1

V1

n2

V2

= or: n1 = n2 x V1

V2

An Experiment to Study the Relationshipbetween the Volume and the Amount of a Gas

Avogadro’s law

The volume of a gas is directly proportional to the amt. (mol) of gas

The volume of a gas confined in an insulated cylinder will remain the same when its Kelvin temperature is doubled if, at the same time, the pressure also is doubled.

Gay-Lussac’s Law - Temperature and Pressure

Change of Conditions, with No Change in the Amount of Gas

• = constant Therefore for a change

of conditions :

• T1 T2

P x V

T

P1 x V1=

P2 x V2

CalculationsCalculations

Change of Conditions: Problem - I

• A gas sample in the laboratory has a volume of 45.9 L at 25 oC and a pressure of 743 mm Hg. If the temperature is increased to 155 oC by pumping (compressing) the gas to a new volume of 3.10 ml what is the pressure?

Change of Conditions: Problem - I

• P1= 743 mm Hg x 1 atm/ 760 mm Hg = 0.978 atm

• P2 = ?

• V1 = 45.9 L V2 = 3.10 ml = 0.00310 L

• T1 = 25 oC + 273 = 298 K

• T2 = 155 oC + 273 = 428 K

Change of Conditions: Problem I continued

•

•

•

•

•

=T1 T2

P1 x V1 P2 x V2

( 0.978 atm) ( 45.9 L) P2 (0.00310 L)

( 298 K) ( 428 K)=

P2 =( 428 K) ( 0.978 atm) ( 45.9 L)

= 21000 atm( 298 K) ( 0.00310 L)

Change of Conditions: Problem II

• A weather balloon is released at the surface of the earth. If the volume was 100 m3 at the surface ( temp = 25 oC, P = 1 atm ) what will its volume be at its peak altitude of 90,000 ft where the temperature is - 90 oC and the pressure is 15 mm Hg ?

Change of Conditions: Problem II

• Initial Conditions Final Conditions

• V1 = 100 m3 V2 = ?

• T1 = 25 oC + 273.15 T2 = -90 oC + 273.15

• = 298 K = 183 K

• P1 = 1.0 atm P2 = 15 mm Hg

760 mm Hg/ atm

P2 = 0.0198 atm

Change of Conditions: Problem II continued

• P1 x V1 P2 x V2

V2 =

• V2 = =

• V2 = 3117.2282 m3 = 3,100 m3 or 30 times the volume !!!

T1 T2

= P1V1T2

T1P2

( 1.0 atm) ( 100 m3) ( 183 K)( 298 K) ( 0.0197 atm)

Change of Conditions: Problem III

• How many liters of CO2 are formed at 1.00 atm and 900 oC if 5.00 L of propane at 10.0 atm, and 25 oC is burned in excess air?

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O(g)

• 25 oC + 273 = 298 K• 900 oC + 273 = 1173 K

Change of Conditions: Problem III continued

• V1 = 5.00 L V2 = ?• P1 = 10.0 atm P2 = 1.00 atm• T1 = 298K T2 = 1173 K

• P1V1/T1 = P2V2/T2 V2 = V1P1T2/ P2T1

• V2 = = 197 L • VCO2 = (197 L C3H8) x (3 L CO2 / 1 L C3H8) =

VCO2 = 591 L CO2

( 5.00 L) (10.00 atm) (1173 K)

( 1.00 atm) ( 298 K)

Avogadro’s Law: Volume and Amount of Gas

Problem IV:

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?

Avogadro’s Law: Volume and Amount of Gas

Plan: Since the temperature and pressure are the same it is a V – n problem, so we can use Avogadro’s Law to calculate themoles of the gas, then use the molar mass to calculate the massof the gas.

Solution: Molar mass SF6 = 146.07 g/mol

V2

V1

n2 = n1 x = 0.0183 mol SF6 x = 3.39 mol SF6

2.67g SF6

146.07g SF6/mol= 0.0183 mol SF6

543.9 m3

2.93 m3

mass SF6 = 3.39 mol SF6 x 146.07 g SF6 / mol = 496 g SF6

Boyle’s Law : Balloon

Problem V:• 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?

Boyle’s Law : 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

Charles Law Problem

Problem VI:• A sample of carbon monoxide, a poisonous

gas, occupies 3.20 L at 125 oC. Calculate the temperature (in oC) at which the gas will occupy 1.54 L if the pressure remains constant.

Charles Law Problem

• V1 = 3.20 L T1 = 125oC = 398 K• V2 = 1.54 L T2 = ?

• T2 = T1 x ( V2 / V1) T2 = 398 K x • = 192 K

• T2 = 192 K oC = K - 273.15 = 192 - 273 oC = -81.2oC

1.54 L3.20 L

Ideal Gases• An ideal gas --a hypothetical gas whose

properties (P, V, and T) are completely described by the ideal gas equation (or ideal gas law)

• The ideal gas equation is:

PV= nRT R = Ideal, Universal, or Molar gas constant R = 0.08206 L atm mol-1 K-1

Boyle’s, Charles’ & Avogadro’s Laws are all combined into a single statement called the Ideal Gas lawIdeal Gas law.

Evaluation of the Ideal Gas Constant, R

Ideal gas Equation PV = nRT R = PVnT

at Standard Temperature and Pressure, the molar volume = 22.4 L P = 1.00 atm (by definition) T = 0 oC = 273.15 K (by definition) n = 1.00 moles (by definition)

R = = 0.08206 (1.00 atm) ( 22.414 L)( 1.00 mole) ( 273.15 K)

L atm mol K

or to three significant figures R = 0.0821 L atm mol K

Gas Law CalculationsGas Law Calculations

Gas Law: Solving for Pressure

Problem 1: (ydi)(ydi) Calculate the pressure in a container whose volume is 87.5 L and it is filled with 5.038kg of xenon at a temperature of 18.8 oC.

Gas Law: Solving for PressurePlan: Convert all information into the units required, andsubstitute into the ideal gas equation ( PV= nRT ).

Solution:

nXe = = 38.37014471 mol Xe 5038 g Xe131.3 g Xe / mol

T = 18.8 oC + 273.15 K = 291.95 K

PV = nRT P = nRT V

P = = 10.5108 atm = 10.5 atm

(38.37 mol )(0.0821 L atm)(291.95 K)87.5 L (mol K)

Ideal Gas Calculation - Nitrogen

Problem 2: (ydi)(ydi)

Calculate the pressure in a container holding 375 g of nitrogen gas. The volume of the container is 0.150 m3 and the temperature is 36.0 oC.

Ideal Gas Calculation - Nitrogen

• n = 375 g N2 (28.0 g N2 / mol) = 13.4 mol N2

• V = 0.150 m3 x 1000 L / m3 = 150 L• T = 36.0 oC + 273.15 = 309.2 K• PV = nRT P = nRT/V

• P = •

• P = 2.26 atm

( 13.4 mol) ( 0.08206 L atm/mol K) ( 309.2 K)

150 L

The Atmosphere: What is in it?

• The Earth’s carbon atoms travel a constant cycle through the atmosphere, from air to plants and animals and back to the air again.

• Nitrogen– Tasteless, colorless, nonflammable, relatively inert

– Nitrogen compounds are a limiting factor in plant growth.

• Oxygen– Oxygen reacts with glucose during respiration.

• Carbon dioxide– Central to plant growth through photosynthesis

• Argon, neon, and helium– Chemically unreactive

The Atmosphere: A Layered Structure

The Carbon CycleThe Carbon Cycle

Four Sections of the Atmosphere• Troposphere

– All Earth-bound life exists and all weather phenomena occur here.

– Ozone exists here as a pollutant.

• Stratosphere– Contains UV absorbing ozone.– Ozone is a natural and necessary component of this

section.

• Mesosphere and Ionosphere– “Falling stars” and the aurora borealis occur here.