we use many that were designed with the knowledge of the

A. Properties of Gases

eg)

technologies properties of

gases SCUBA equipment, hot air balloons,jackhammers

3.1 Gases and the Kinetic Molecular Theory

Chapter 3

gases have several distinct macroscopic (visible)

compressible ie)

gases expandie)

gases have low resistance to flow (viscosity) …allows them to escape quickly through small openings

properties:

gases are

pressure = volume

as temperature increases

temperature = volume (not confined)temperature =pressure (confined)

gases havelow densities

gases mix evenly and completely, they all aremiscible

gases have they of the container they are in

no shape or volume,fill the shape

we need to describe how gases behave on the molecular level

B. Kinetic Molecular Theory

models

the says that all particles are

kinetic molecular theory in motion at all times

an is defined by the following characteristics:

ideal gas (which is hypothetical)

1.the gas molecules are in where they until they

constant random motion move in a straight line collide

2. the gas molecules are “point masses”

3. the only interaction between molecules of the gas and container are collisions where is

with a particle or wall of the container

(they have but no ) mass volume

elastic collisions…kinetic energy

conserved

do not have these perfect characteristics however their behaviour is real gases not that far offof ideal gases

Assignment: p. 101 #1-9

A. Atmospheric Pressure

although gas molecules have , the Earth’s keeps them near the

very little mass gravitational

pull the surface of the planet

force per unit area

pressure is exerted in all directions to the same extent

3.2 Gases and Pressure

pressure =

which creates our atmosphere

atmospheric pressure is the that a column of air exerts on a on the

air is as altitude pressure is exerted

less compressed increases less

forceparticular area

Earth’s surface

B. Measuring Pressure

Pascal and Perier used to prove that atmospheric pressure

the work of Pascal, Perier and Torricelli all led to the development of the mercury barometer

Hg()decreases with altitude

there are several different units used to measure pressure: millimetres of mercury (mmHg)

the

the

the

Pascal (Pa)

kilopascal (kPa)

atmosphere (atm)

the bar

you will be using the standard unit of in gas law calculations and therefore you must be able to convert mmHg and atm to kPa

kPa

memorize the following standard atmospheric pressures:

760 mmHg =1 atm = 101.325 kPa

to convert other units of pressure to kPa, set up aratio

another conversion:1 bar = 100 kPa

Example 1

Convert 650 mmHg to kPa.

101.325 kPa = x

760 mmHg 650 mmHg x = 86.6… kPa

 

Example 2

Convert 2.5 atm to kPa.

101.325 kPa = x

1 atm 2.5 atmx = 253.3… kPa

 

Try These:

Convert the following pressures to kPa (unrounded):

1.     4.0 atm

2.     855 mmHg

3.     0.625 atm

4.     150 mmHg  

405.3 kPa

113.9…kPa

63.3…kPa

19.9…kPa

C. Boyle’s Law

Irish scientist studied the relationship between the and of gases at

Robert Boyle pressure volume

constant temperatures

pressure on the is caused by the

walls of a container collisions of the gas molecules with the walls

as you of a contained gas, there is

reduce the volume less room

more collisions = higher pressure

for the gas particles so theycollide more

http://michele.usc.edu/java/gas/gassim.html

Volume vs. Pressure for a Gas

Volume (L)

Pressure (kPa)

Boyle’s Law states that the volume of a gas varies inversely with the pressure at a constant temperature and mass

eg) lungs – to inhale, we the volume of our chest cavity which the pressure which makes the air move in

increasedecreases

breath-hold diving – all air containing spaces in body as pressure with depth…this doesn’t happen with SCUBA gear

shrink increaseseg)

where:

P1V1 = P2V2

P1, P2 = pressures in kPa V1, V2 = volumes in L

Example 1A balloon is filled with 30.0 L of helium gas at 100 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (assume constant temperature)

P1 = 100 kPaP2 = 25.0 kPaV1 = 30.0 L

P1V1 = P2V2

(100 kPa)(30.0 L) = (25.0 kPa) V2

V2 = 120 L

Example 2The pressure on 2.50 L of anesthetic gas is 100 kPa. If 6.25 L of gas is the required volume, what pressure must it be under assuming constant temperature? P1 = 100 kPaV1 = 2.50 LV2 = 6.25 L

P1V1 = P2V2

(100 kPa)(2.50 L) = P2 (6.25 L)

P2 = 40.0 kPa

when of a gas are graphed, the plot is

it was also noticed that when these linear plots were down to all the lines at one point

volume vs. temperature linear

A. Volume vs. Temperature3.3 Gases and Pressure

(as long as amount of gas and pressure were constant)

extrapolated zero volume,converged

Volume vs. Temperature for a Gas

Volume (L)

Temperature (C)

-273.15 C

the temperature when the volume of a gas is is

, in 1848, suggested that this is the lowest possible temperature or

he established a new temperature scale which is called the scale in his honour

273.15C

Lord Kelvin

absolute zero

Kelvin

zero

is used for temperature in

is used for temperature in

t C

T K

= = -273.15C absolute zero zero Kelvin

to go from C to K…you add 273.15

eg) 0C = 25C = -30C = -273.15C =

273.15 K298.15 K243.15 K

0 K

B. Charles’ Law (and

Joseph Louis Gay-Lussac) noticed that there was a relationship between the and of a gas

Jacques Charles

temperature volume

as temperature , so does the of the gas molecules

as the molecules move , they exert pressure

the volume of the gas will until it reaches

increases kinetic energy

faster higher

expand under this pressure atmospheric

pressure

where:

V1 = V2 T1 T2

T1, T2 = temperatures in KV1, V2 = volumes in L

Charles’ Law states that the volume of a gas varies directly with the temperature at a constant pressure and mass

Example 1A balloon was inflated at 27C and has a volume of 4.0 L. If it is heated to 57C, what is the new volume? (assume constant pressure)

T1 = 27 C = 300.15 KV1 = 4.0 L T2 = 57 C = 330.15 K

V1 = V2

T1 T2

(4.0 L) = V2

(300.15 K) 330.15 KV2 = 4.39… LV2 = 4.4 L

Example 2A sample of gas occupies 6.8 L at 110C. What will the final temperature be in C when the volume is decreased to 5.6 L?

T1 = 110 C = 383.15 KV1 = 6.8 L V2 = 5.6 L

V1 = V2

T1 T2 (6.8 L) = 5.6 L(383.15 K) T2

T2 = 315.5… K – 273.15T2 = 42C

now we’ll combine Boyle’s Law and Charles’ Law

P1V1 = P2V2

T1 T2

where: V1, V2 = volumes in L T1, T2 = temperatures in KP1, P2 = pressures in kPa

A. Combined Gas Law Calculations4.1 Gases and the Kinetic Molecular Theory

Chapter 4

= standard temperature and pressure

= standard ambient temperature and pressure

= 273.15 K (0 C) and 101.325 kPa

= 298.15 K (25C) and 100.000 kPa

STP

SATP

Example 1A weather balloon is filled with H2(g) at 20C and 100 kPa. It has a volume of 7.50 L. It rises to an altitude where the air temperature is -36C and the pressure is 28 kPa. What is the new volume of the balloon? T1 = 20 C = 293.15 KV1 = 7.50 L P1 = 100 kPaP2 = 28 kPa T2 = -36 C = 237.15 K

P1V1 = P2V2

T1 T2

(100 kPa)(7.50 L) = (28 kPa)V2

(293.15 K) (237.15 K)V2 = 21.6…L V2 = 22 L

Example 2A large syringe was filled with 50.0 mL of ammonia gas at STP. If the gas was compressed to 25.0 mL with a pressure of 210 kPa, what was the final temperature in C?

T1 = 0 C = 273.15 KV1 = 0.0500 L P1 = 101.325 kPaP2 = 210 kPa V2 = 0.0250 L

P1V1 = P2V2

T1 T2

(101.325 kPa)(0.0500 L) = (210 kPa)(0.0250 L) (273.15 K) T2T2 = 283.0…K – 273.15

T2 = 9.91C

B. Combining Volumes of Gases Gay-Lussac analyzed

that involved

he studied the of the gaseous reactants and products and concluded that the gases combine in

chemical reactionsgases

volumes

very simple proportions

the states that, when gases react, the of the gaseous reactants and products, measured at constant temperature and pressure, are always in

eg) 1 N2(g) + 3 H2(g) 2 NH3(g)

1 : 3 : 2 is the volume ratio For every one mole of nitrogen gas and three moles of hydrogen gas, 2 moles of ammonia gas is produced

Law of Combining Volumesvolumes

whole number ratios

Formula: v1 = v2

n1 = n2Balancing coefficients

ExampleWhat volume of nitrogen is used up if 100 mL of ammonia is formed in a composition reaction? (Compare ratios of volume:coefficients for what values you know, and what values you are solving for)

N2(g) + 3H2(g) 2NH3(g)

x mL = 100 mL

1 = 2

x mL = 50.0 mL

x mL 100 mL

Formula: v1 = v2

n1 = n2

this law combines all four variables

where:

PV = nRT

V = volume in L T = temperature in K P = pressure in kPan = number of moles in mol R = universal gas constant = 8.314 kPaL/molK

(P, V, n and T)

A. Ideal Gas Law Calculations 4.2 Ideal Gas Law

intoone equation:

Example 1What is the volume of 10.8 mol of oxygen gas at 100.00 kPa and 15.5C?

n = 10.8 molT = 15.5 C = 288.65 KP = 100.00 kPaR = 8.314 kPaL/molK

PV = nRT(100.00kPa)V = (10.8mol)(8.314 kPaL/molK)(288.65K)

V = 259.1…L = 259 L

Example 2A rigid steel vessel with a volume of 20.0 L is filled with nitrogen gas to a pressure of 20 000 kPa at 27.0C. What is the number of moles of nitrogen?

T = 27 C = 300.15 KV = 20.0 L P = 20 000 kPaR = 8.314 kPaL/molK

PV = nRT(20 000kPa)(20.0L) = n(8.314 kPaL/molK)(300.15K)

n = 160.2…mol = 160 mol

Example 3What is the pressure exerted by 15.5 g of methane, CH4(g), if it occupies a volume of 10.0 L at 25C?

T = 25 C = 298.15 KV = 10.0 L R = 8.314 kPaL/molKm = 15.5 gM = 16.05 g/mol

PV = nRTP(10.0L) = (0.965…mol)(8.314 kPaL/molK)(298.15K)

P = 239.3…kPa = 2.4 102 kPa

n = m M = 15.5 g 16.05 g/mol

= 0.965…mol

OR PV = mRT M

Example 4What is the mass of hydrogen gas contained in a 4.5 L weather balloon at 25C and 102.0 kPa?

T = 25 C = 248.15 KV = 4.5 L P = 102.0 kPaR = 8.314 kPaL/molKM = 2.02 g/mol PV = mRT

M(102.0 kPa)(4.5L) =m(8.314 kPaL/molK)(248.15K) 2.02 g/mol (102.0)(4.5)= m(1021.346….)

(102.0)(4.5) = m

(1021.346..)= 0.449…g= 0.45 g

B. Dalton’s Law of Partial Pressures Dalton’s law of partial pressures states that

that in a mixture of gases that do the is the

Ptotal = P1 + P2 + P3

not react chemically,total pressure sum of the

partial pressures of each individual gas

ExampleTwo gases are pumped into a 32.0 L reaction vessel at 25.0 C one after another. 6.20 mol of O2(g) is pumped in first, then 8.30 mol of H2(g) is pumped in. What would the pressure gauge reading be after each gas is pumped in? O2(g)

n = 6.20 molT = 25.0 C = 298.15 KV = 32.0 L R = 8.314 kPaL/molK

PV = nRT

P(32.0L) = (6.20 mol)(8.314 kPaL/molK)(298.15K) P = 480.2…kPa

1st reading = 480 kPa

O2(g) + H2(g)

n =6.20 mol + 8.30 mol = 14.5 molT = 25.0 C = 298.15 KV = 32.0 L R = 8.314 kPaL/molK

PV = nRTP(32.0L) = (14.5 mol)(8.314 kPaL/molK)(298.15K) P = 1123.2…

kPa 2nd reading = 1.12 x 103 kPa

Or you can calculate the pressure of oxygen, then the pressure of hydrogen separately and add the pressure’s together to get the same answer. This follows Dalton’s Law of Partial Pressures. Give it a try:

C. Ideal Gases and Real Gases for ideal gases we assume that there are no

between the molecules of the gas

in real gases, however, there are between the molecules

we don’t have to worry about considering this in our calculations because at standard P and T conditions, the molecules are and are they much with each other

intermolecular attractions

attractions

very far apart moving very quickly don’t interact

real gases behave like ideal gases at

real gases deviate from ideal gas behaviour at

high temperaturesand low pressures

very low temperatures(moving very slowly)very high pressures

(molecules close together)and

Review assignment p. 156 #1-25 (omit 12, 21, 24)