Agenda

• Day 80 – Gases- Intro - 11. 1 & 11.7 &

12.2

• Lesson: PPT

• Handouts: 1.PPT Handout, 2. ISU - Kinetic

Molecular Theory and Real vs. Ideal Gases

Sections ( W1)

Macroscopic vs. Microscopic

Representation

States of Matter

Solids have greater attractions, that’s why they stay together (whereas gases disperse). Liquids have forces that are strong enough to keep their molecules together (thus, they are incompressible with a definite volume). However, these forces are not strong enough to fix molecules in place (shape is not fixed) Gases have forces so weak that, at room temperature, molecules move freely (shape & volume change). The large spaces between molecules makes gases compressible.

Kinetic Energy

Temperature affects state.

Types of Motion:

- vibrational (small motion from side to side),

- rotational (spinning),

- translational (straight line motion).

Solids: vibrational,

Liquids: all,

Gases: mostly translational, also rotational (and

vibrational).

Kinetic energy is the energy of motion.

Temperature = average kinetic energy

The average speed will increase with higher temperature.

How do solids and liquids differ from gases?

• Gases – molecules/particles in chaotic motion whose average kinetic energy is greater than the attractive forces between molecules

• Liquids – attractive intermolecular forces between

molecules is comparable to kinetic energies of molecules: these molecules are held in close proximity, but still move in a rather chaotic motion.

• Solids – intermolecular forces are sufficiently strong

relative to kinetic energy. Molecules are virtually locked in place. Often this is very orderly (crystalline structures).

IntErmolecular Forces – Review

Intermolecular Forces (IM) - These are weak forces

that exist between molecules and are responsible

for the fact that the substance can exist as a liquid

(and/or solid). Molecules of gases have little to no

intermolecular forces (which is why gases have no

definite shape or volume). Dipole-dipole forces,

London dispersion, and hydrogen bonding are

intermolecular forces.

IntrAmolecular Forces – Review

IntrAmolecular Forces are the bonds or forces

within the atoms of a molecule; the two bonds

connecting two hydrogen atoms to an atom of

oxygen in water are intrAmolecular forces. The

force holding water molecules together so they

don’t evaporate are intErmolecular forces and are

About 1/6 to 1/10th as strong as single covalent

bonds. Covalent, ionic, and metallic bonding are

intramolecular forces. Metallic bonding is the

strongest of the three.

Types of (intermolecular) forces (aka van derWaals forces):

London Dispersion Forces (aka induced dipole- induced dipole IM forces) - weakest of the IM forces

(1) found in pure non-polar liquids

(2) found in a solution of 2 different non-polar liquids

London dispersion forces are found holding any type

of matter together, but these are the only ones found

holding nonpolar molecules together!) Dispersion

forces result from the constant movement of all

particles. When molecules come close to one another,

their electron clouds repel each other and a temporary

dipole is formed. A dipole is like a magnet; it is any

object that has a negative charge on one end and a

positive charge on the other end. These dipoles are

attracted to each other. Example of I2 molecules being

held together by dispersion forces:

Dipole – Dipole IM Forces (stronger then dispersion forces) –

(1) found in pure polar liquids. Ex: between two molecules of methanol

(2) found in solutions of 2 different polar liquids. Ex: between water and ammonia

Dipole-dipole IM forces occurs when one polar

molecule is attracted to another polar molecule.

Remember that polar molecules are those that are

unequally sharing electrons, because of this all polar

molecules are dipoles! Dipole –dipole forces occur

when two or more polar molecules are together, the

molecules orient themselves so that the oppositely

charged regions line up with each other. (Dipole-dipole

forces are just like London dispersion forces, but much

stronger because the dipole is constant and not

induced). Example of HCl molecules being held

together by dipole-dipole forces:

Hydrogen Bonding (strongest of all)

(1) occurs in all water solutions along with other forces

(2) are NEVER found as the only IM force. Dipole-dipole will always be one of the other IM forces here.

Hydrogen bonds are a special case of very strong

dipole-dipole interactions. They are not really chemical

bonds in the formal case. Strong hydrogen bonding

occurs among polar covalent molecules containing H

and one of the three small, highly electronegative

elements – F, O, or N. Hydrogen bonding is exactly the

same as dipole-dipole forces except that a hydrogen

atom from one molecule (which is in a polar bond) is

attracted to a fluorine, oxygen or nitrogen atom (FON)

of another molecule. Example of H2O molecules being

held together by hydrogen bonds:

Practice: Draw the dot structure and shape of the

following molecules and then determine its polarity

and which type of intermolecular force would hold two

or more of the following molecules together.

1. Br2

2. HBr

3. CO2

4. H2S

5. HF

Characteristics/Descriptions of Liquids

Volatile – when 2 liquids are compared, the one

that evaporates more readily is said to be volatile.

Boiling Point – when the vapor pressure of the liquid

equals the atmospheric pressure the liquid boils.

Water boils at 100.C. Water has a relatively high

boiling point. Draw water’s dot structure and

determine its molar mass:

What accounts for the difference in boiling points

between water and H2S, with a boiling point of 60.0C?

Boiling Point and KE

Only the fastest moving molecules are able to overcome the attractive forces of their neighbors and leave the surface of the liquid. The slower molecules are left behind. Thus the average kinetic energy (i.e. temperature) decreases.

Vapor Pressure – the pressure exerted by a gas above a liquid in a closed container. Vapor pressure increases with increasing temperature. Because the rate of evaporation increases with increasing temperature, vapor pressures of liquids always increases as temperature increases. Vapor pressure decreases with increased IM forces. Water has low vapor pressure. It is good that water has a low vapor pressure, otherwise what would happen to the rivers & lakes? An easy way to remember vapor pressure is to ask yourself how fast the liquid evaporates, the faster the rate of evaporation, the higher the vapor pressure. Example: water and rubbing alcohol

Other facts about boiling points for all liquids:

a) normal boiling point is the boiling point of a

substance at 1 atm of pressure

b) boiling point is unique for each liquid; it does not

depend on volume or surface area

c) boiling point increases with increased molecular

mass if IM forces are the same

d) boiling point increases with increased IM forces

Heat Capacity (or specific heat) – the amount of heat

needed to change the temperature of 1.0 g of a

substance by 1.0°C. Because water has such a high heat

capacity, it is cooler at the beach on a warm day. The

water absorbs the heat and lowers the air

temperature.

SUMMARY WHEN PREDICTING PROPERTIES – Remember water has a:

(1) high boiling point

(2) high surface tension

(3) high heat capacity

(4) low vapor pressure because of dipole-dipole

and hydrogen bonding

Next, look at the type of intramolecular forces

(metallic, ionic, covalent). If all molecules being

compared are covalent, then look at any

intermolecular forces present. From the strongest to

the weakest are:

Metallic

Ionic

Covalent

a. Hydrogen bonding with dipole-dipole

b. dipole-dipole

c. London dispersion

Intramolecular forces

Intremolecular forces

Let’s look at some of the Nature of Gases:

1. Expansion – gases do NOT have a definite shape or volume.

2. Fluidity – gas particles glide past one another, called fluid just like a liquid.

3. Compressibility – can be compressed because gases take up mostly empty space.

4. Diffusion – gases spread out and mix without stirring and without a current. Gases mix completely unless they react with each other.

Kinetic Molecular Theory of Gases

1. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.

Kinetic Molecular Theory of Gases

2. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy

3. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.

4. Gas molecules exert neither attractive nor repulsive forces on one another.

5. Each gas molecule “behaves” as if it were alone in container (due to #3 and #4)

Demo 1: Diffusion of KMnO4

Observations: KMnO4 (potassium permanganate) diffuses faster in hot water.

Explanation: KMnO4 dissolves in water. As the KMnO4 dissolves, it collides with H2O molecules and spreads out. Because molecules in hot water are moving faster, the solid dissolves faster and diffuses faster.

Cold water

Hot water

Demo 2: thermal expansion of a liquid

Observations: coloured water moves slowly up the tube

Explanation: heat increases the kinetic energy of liquid particles. The particles move faster (greater vibrational, rotational, and translational energy). This greater movement increases the distance between molecules. Thus, the volume expands.

Glass tube

Rubber stopper

Coloured water in florence

flask

Hot plate

Demo 3: Heating Mercury

Observations: beads are propelled upwards when mercury is heated.

Explanation: heat increases the kinetic energy, so that the particles move faster. The fastest moving mercury molecules (that boil off) transfer their energy to the beads, causing them to jump.

Would this work with water?

Air

Glass tube

Beads

Heat source

Mercury

Will a gas diffuse faster or slower in a vacuum?

Gas particles travel in a straight line until they hit another particle. A vacuum is devoid of molecules, so particles of a gas placed in a vacuum will not bump into anything until they hit the side of the container. Thus, a gas will diffuse faster in a vacuum.

Pressure – KMT Viewpoint

• Origin of Pressure – Gas molecules hitting container walls

– Temp, KE, # collisions, P

– Volume, # collisions, P

Pressure – Macroscopic Viewpoint

• Temp, KE, Force, P

• Volume, Area, P

Pressure = Force Area

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Atmospheric Pressure

Barometer

Ideal vs. Real Gases

• Ideal gases always obey the kinetic theory. (Closest to ideal would be the noble gases.)

• Real gases vary from the kinetic theory at various temperatures and pressures.

• The greatest deviation from ideal gas behavior occurs at:

high pressure

higher density of gas molecules

– Molecules are closer together so:

» finite volume of gas molecules more important

» attraction between molecules more important

• This can also occur when the temperature is low – the gas will change state form gas to liquid

Real Gases

In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

Curves for 1 mol of gas

Real Gases

Even the same gas (e.g. nitrogen) will show wildly different behavior under high pressure at different temperatures.

Volume – refers to the space matter (gas) occupies. Measured in liters (L).

Pressure – the number of times particles collide with each other and the walls of the container (force exerted on a given area). Measured in atmospheres (atm).

1atm = 760 millimeters Hg ( Barometers use Hg)

1atm = 760 torr (Named after Torricelli for the invention of the

barometer)

1atm = 101.3 kPa – kilopascals

Memorize Table 1. P. 542

STP: standard temperature and pressure (0°C, 101.3 kPa or 1atm).

SATP: standard ambient temperature and pressure (25°C, 100 kPa).

Definitions

How Much Pressure? Q - A balloon is filled with pure oxygen. What is the

pressure of the oxygen in the balloon?

A - Atmospheric pressure. If it was not, then the balloon would expand or shrink.

Q - A windbag is blown up with exhaled air. What is the pressure of oxygen in the bag?

A – Around 16 - 21% of atmospheric pressure (O2 is 16% of exhaled air, 21% of atmosphere)

Q - A solid container is filled with pure oxygen. What is the pressure in the container?

A - It could be anything. The container is solid and therefore cannot shrink or expand.

Sample Problems

Convert 4.40 atm to mmHg.

Convert 212.4kPa to mmHg.

Temperature – average kinetic energy -

as temperate increases gas particles move faster,

as temperature decreases gas particles move slower.

Measured in Kelvin (K). K = 273 + C

Number of Moles – tells you how much of a certain

gas you have

1 mole = number of grams of the compound or

element (molar mass)

Definitions

Demo 4: Kill da wabbit Question: what will happen when the air pressure surrounding a balloon is decreased? Why?

Explanation: A balloon normally stays the same size because the pressure inside the balloon is equal to the pressure outside the balloon (rate of collisions with the wall of the balloon is equal). Reducing the pressure on the outside eliminates opposing collisions, allowing the balloon to expand.

Agenda

• Day 81 – Gas Laws

• Lesson: PPT,

• Handouts: 1.PPT Handout 2. ISU- The Gas

Laws ( W2-3)

How do all of pressure, temperature, volume, and

amount of a gas relate to each other? Rules for

solving gas law problems:

1st write down what is given and what is unknown,

2nd identify the gas law you want to use, and

3rd rearrange the formula to solve for the unknown

and then solve the problem.

(If temperature is involved, it MUST be converted

to Kelvin! K = 273 + C)

Gas Laws

Boyle’s Law - Pressure and Volume (when temperature remains constant) V1 = initial or old volume

V1P1 = V2P2 V2 = final or new volume

P1 = initial or old pressure

P2 = final or new pressure

Inverse Relationship (As pressure increases, volume

decreases and as pressure decreases, volume

increases.)

Effect of Pressure on Volume Boyle’s Law

5

1

3

1 atm

1

3

2 atm

5

1

3

5 atm

5

Which picture represents what the gas will look like when the pressure is doubled?

(Assume constant n, T)

P a 1/V

P x V = constant

P1 x V1 = P2 x V2

5.3

Boyle’s Law

Constant temperature Constant amount of gas

Kinetic Molecular theory of gases and …

• Boyle’s Law

P a collision rate with wall

Collision rate Increases with decreased volume

P a 1/V

Increase P, decrease volume

An increase in pressure decreases the volume of air in the lungs.

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL 154 mL

= = 4460 mmHg

Sample Problems What is the new pressure when 80.0mL of gas at

500.mmHg is moved to a 100.mL container?

A gas at 800.torr of pressure has a volume of 5.00L.

What volume does this gas occupy at 1.00X103torr

of pressure?

Charles’ Law -Volume and Temperature (when pressure is constant)

V1/T1 = V2/T2 V1 = initial or old volume

V2 = final or new volume

T1 = initial or old temperature

T2 = final or new temperature

Direct Relationship (As temperature increases,

volume increases and as temperature decreases,

volume decreases.)

As T increases, V Increases

V a T V = kT V/T = k

V1/T1 = V2/T2

Constant pressure Constant amount of gas

The volume of a gas increases with and increase in temperature.

If we place a balloon in

liquid nitrogen it

shrinks:

How Volume Varies With Temperature

So, gases shrink if cooled.

Conversely, if we heat a gas it

expands (as in a hot air balloon).

Let’s take a closer look at temperature before we try to find the exact relationship of V vs. T.

Which picture represents what the gas will look like when the temperature is increased?

(Assume constant n, P)

Kinetic theory of gases and …

Charles’ Law

-Average kinetic energy a T

-Increase T, Gas Molecules hit walls with greater

Force, this Increases the Pressure

BUT since pressure must remain constant, and only

volume can change

-Volume Increase to reduce Pressure

-Increase Temperature, Increase Volume

No. 68F (20C) is not double 50F (10C)

Yes. 44 lb (20 kg) is double 22 lb (10 kg)

What’s the difference?

• Weights (kg or lb) have a minimum value of 0.

• But the smallest temperature is not 0C.

• We saw that doubling P yields half the V.

• Yet, to investigate the effect of doubling temperature, we first have to know what that means.

• An experiment with a fixed volume of gas in a cylinder will reveal the relationship of V vs. T…

Temperature scales Is 20C twice as hot as 10C?

Is 20 kg twice as heavy as 10 kg?

Temperature vs. Volume Graph

5

10

15

20

25

30

Vo

lum

e (

mL

)

Temperature (C) 0 100 – 273

25 mL at 22C

31.6 mL, 23.1 mL

Y=0.0847x + 23.137

Determination of Absolute Zero

• If a volume vs. temperature graph is plotted for gases, most lines can be interpolated so that when volume is 0 the temperature is -273 C.

• Naturally, gases don’t really reach a 0 volume, but the spaces between molecules approach 0.

• At this point all molecular movement stops. • –273C is known as “absolute zero” (no EK) • Lord Kelvin suggested that a reasonable temperature scale

should start at a true zero value. • He kept the convenient units of C, but started at absolute

zero. Thus, K = C + 273. 62C = ? K: K=C+273 = 62 + 273 = 335 K • Notice that kelvin is represented as K not K.

The Kelvin Temperature Scale

What is the approximate temperature for absolute zero in degrees Celsius and kelvin?

Calculate the missing temperatures

0C = _______ K 100C = _______ K

100 K = _______ C – 30C = _______ K

300 K = _______ C 403 K = _______ C

25C = _______ K 0 K = _______ C

Kelvin Practice

273 373

– 173 243

27 130

298 – 273

Absolute zero is – 273C or 0 K

A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K 3.20 L

= = 192 K

V1/T1 = V2/T2

2. A sample of gas occupies 3.5 L at 300 K. What volume

will it occupy at 200 K?

3. If a 1 L balloon is heated from 22°C to 100°C, what

will its new volume be?

V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K Using Charles’ law: V1/T1 = V2/T2 3.5 L / 300 K = V2 / 200 K V2 = (3.5 L/300 K) x (200 K) = 2.3 L

V1 = 1 L, T1 = 22°C = 295 K V2 = ?, T2 = 100 °C = 373 K V1/T1 = V2/T2, 1 L / 295 K = V2 / 373 K V2 = (1 L/295 K) x (373 K) = 1.26 L

Gay-Lussac’s Law - Pressure and Temperature (when volume is constant) P1/T1 = P2/T2 P1 = initial or old pressure

P2 = final or new pressure

T1 = initial or old temperature

T2 = final or new temperature

Direct Relationship - As the temperature of

the gas increases the pressure of the gas

Increases.

Sample Problems

The gas in an aerosol can is at 3atm of pressure at

298K. What would the gas pressure in the can be at

325K?

At 120.C the pressure of a sample of nitrogen gas is

769torr. What will the pressure be at 205C?

The Combined

Gas Law

Combining the gas laws • So far we have seen three gas laws:

Jacques Charles Robert Boyle P1V1 =

P2V2 V1

T1

=

V2

T2 These are all subsets of a more

encompassing law: the combined gas law

P1

T1

= P2

T2

P1V1 P2V2

T1 T2 =

Joseph Louis Gay-Lussac

Combined Gas Law Equations

P1 =

P2T1V2

T2V1

V1 =

P2T1V2

T2P1

T2 =

P2T1V2

P1V1

T1 =

P1T2V1

P2V2

P2 =

P1T2V1

T1V2

V2 =

P1T2V1

P2T1

Sample Problems A helium filled balloon has a volume of 50.0mL at 298K and 1.08atm. What volume will it have at 0.855atm and 203K? Given 700.mL of oxygen at 7.00C and 7.90atm of pressure, what volume does is occupy at 27.0C and 4.90atm of pressure?

Note: any unit for pressure will work, provided the same units are used throughout.

The only unit that MUST be used is K for temperature.

V1 = 50.0 ml, P1 = 101 kPa

V2 = 12.5 mL, P2 = ? T1 = T2

P1V1

T1

=

P2V2

T2

(101 kPa)(50.0 mL)

(T1)

=

(P2)(12.5 mL)

(T2)

(101 kPa)(50.0 mL)(T2)

(T1)(12.5 mL) = (P2)

= 404 kPa

Notice that T cancels out if T1 = T2

V1 = 0.10 L, T1 = 298 K

V2 = ?, T2 = 463 P1 = P2

P1V1

T1

=

P2V2

T2

(P1)(0.10 L)

(298 K)

=

(P2)(V2)

(463)

(P1)(0.10 L)(463 K)

(P2)(298 K) = (V2)

= 0.16 L

Notice that P cancels out if P1 = P2

P1 = 150 kPa, T1 = 308 K

P2 = 250 kPa, T2 = ? V1 = V2

P1V1

T1

=

P2V2

T2

(150 kPa)(V1)

(308 K)

=

(250 kPa)(V2)

(T2)

(250 kPa)(V2)(308 K)

(150 kPa)(V1) = (T2)

= 513 K

= 240 °C Notice that V cancels out if V1 = V2

P1 = 100 kPa, V1 = 5.00 L, T1 = 293 K

P2 = 90 kPa, V2 = ?, T2 = 308 K

P1V1

T1

=

P2V2

T2

(100 kPa)(5.00 L)

(293 K)

=

(90 kPa)(V2)

(308 K)

(100 kPa)(5.00 L)(308 K)

(90 kPa)(293 K) = (V2)

= 5.84 L

P1 = 800 kPa, V1 = 1.0 L, T1 = 303 K

P2 = 100 kPa, V2 = ?, T2 = 298 K

P1V1

T1

=

P2V2

T2

(800 kPa)(1.0 L)

(303 K)

=

(100 kPa)(V2)

(298 K)

(800 kPa)(1.0 L)(298 K)

(100 kPa)(303 K) = (V2)

= 7.9 L

P1 = 6.5 atm, V1 = 2.0 mL, T1 = 283 K

P2 = 0.95 atm, V2 = ?, T2 = 297 K

P1V1

T1

=

P2V2

T2

(6.5 atm)(2.0 mL)

(283 K)

=

(0.95 atm)(V2)

(297 K)

(6.5 atm)(2.0 mL)(297 K)

(0.95 atm)(283 K) = (V2)

= 14 mL

33. The amount of gas (i.e. number of moles of gas) does not change.

The Ideal Gas Law

PV = nRT

Agenda

• Day 82 – Ideal Gas Law

• Read: Section 12.2 -Make your own notes

on this section

• Lesson: PPT,

• Handouts: 1.PPT Handout 2. ISU- The Ideal

Gas Law

( W4)

Ideal Gas Equation

Charles’ law: V a T (at constant n and P)

Avogadro’s law: V a n (at constant P and T)

Boyle’s law: V a (at constant n and T) 1 P

V a nT

P

V = constant x = R nT

P

nT

P R is the gas constant

PV = nRT

R = 0.082057 L • atm / (mol • K)

Ideal Gases An “ideal” gas exhibits certain theoretical properties.

Specifically, an ideal gas …

• Obeys all of the gas laws under all conditions.

• Does not condense into a liquid when cooled.

• Shows perfectly straight lines when its V and T & P and T relationships are plotted on a graph.

In reality, there are no gases that fit this definition perfectly. We assume that gases are ideal to simplify our calculations.

We have done calculations using several gas laws (Boyle’s Law, Charles’s Law, Combined Gas Law). There is one more to know…

Sample problems How many moles of H2 is in a 3.1 L sample of H2 measured at 300 kPa and 20°C?

PV = nRT

(300 kPa)(3.1 L) = n (8.31 kPa•L/K•mol)(293 K)

(8.31 kPa•L/K•mol)(293 K)

(300 kPa)(3.1 L) = n = 0.38 mol

How many grams of O2 are in a 315 mL container that has a pressure of 12.0 atm at 25°C?

P = 300 kPa, V = 3.10 L, T = 293 K

PV = nRT

(8.31 kPa•L/K•mol)(298 K)

(1215.9 kPa)(0.315 L) = n = 0.1547 mol

P= 1215.9 kPa, V= 0.315 L, T= 298 K

0.1547 mol x 32.00 g/mol = 4.95 g

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV = nRT n, V and R are constant

nR V

= P T

= constant

P1

T1

P2

T2 =

P1 = 1.20 atm T1 = 291 K

P2 = ? T2 = 358 K

P2 = P1 x T2

T1 = 1.20 atm x 358 K

291 K = 1.48 atm

Solving for Density and /or Molar Mass of a gas using the Ideal Gas Law Density (units are g/L) Use the Ideal Gas Law to find moles (n), convert n to grams OR use the Ideal Gas Law to find the volume. Divide m (in grams) by the volume.

Molar Mass (units are g/mol) If density is given, use the density of the gas to determine the molar mass (use 1 L at the volume and solve for n). If a mass is given, use the Ideal Gas Law to solve for n and then find the molar mass.

Ideal Gas Law Questions 1. How many moles of CO2(g) is in a 5.6 L sample of CO2

measured at STP?

2. a) Calculate the volume of 4.50 mol of SO2(g) measured at STP. b) What volume would this occupy at 25°C and 150 kPa? (solve this 2 ways)

3. How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C?

4. At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate its molar mass.

5. 98 mL of an unknown gas weighs 0.087 g at SATP. Calculate the molar mass of the gas. Can you determine the identity of this unknown gas?

Sample Problems What is the density of a sample of ammonia gas, NH3, if the pressure is 0.928 atm and the temperature is 63.0C?

What is the density of argon gas at a pressure of 551 torr and a temperature of 25.0C?

Sample Problems The density of a gas was found to be 2.00g/L at 1.50atm and 27.0C. What is the molar mass of the gas?

What is the molar mass of a gas if 0.427g of the gas occupies a volume of 125mL at 20.0C and 0.980atm?

Types of Problems

• Make Substitution into

PV = nRT

22

22

11

11

Tn

VP

Tn

VP

Given initial conditions, determine final conditions; Cancel out what is constant

molegMolarMass

gmassnmoles

/,

,)(

Volume

massDensity

Density (d) Calculations

d = m V

= PM RT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRT

P M = d is the density of the gas in g/L

P=101.325 kPa, V=5.6 L, T=273 K PV = nRT

(101.3 kPa)(5.6 L) = n (8.31 kPa•L/K•mol)(273 K)

1. Moles of CO2 is in a 5.6 L at STP?

(8.31 kPa•L/K•mol)(273 K)

(101.325 kPa)(5.6 L) = n = 0.25 mol

2. a) Volume of 4.50 mol of SO2 at STP.

P= 101.3 kPa, n= 4.50 mol, T= 273 K PV=nRT

(101.3 kPa)(V)=(4.5 mol)(8.31 kPa•L/K•mol)(273 K)

(101.3 kPa)

(4.50 mol)(8.31 kPa•L/K•mol)(273 K) V = = 100.8 L

2. b) Volume at 25°C and 150 kPa (two ways)?

Given: P = 150 kPa, n = 4.50 mol, T = 298 K

(150 kPa)

(4.50 mol)(8.31 kPa•L/K•mol)(298 K) V = = 74.3 L

From a): P = 101.3 kPa, V = 100.8 L, T = 273 K

Now P = 150 kPa, V = ?, T = 298 K

P1V1

T1

=

P2V2

T2

(101.3 kPa)(100 L)

(273 K)

=

(150 kPa)(V2)

(298 K)

(101.3 kPa)(100.8 L)(298 K)

(273 K)(150 kPa) = (V2)

= 74.3 L

3. How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C?

PV = nRT

(8.31 kPa•L/K•mol)(303 K)

(1000 kPa)(10.0 L) = n = 3.97 mol

P= 1000 kPa, V= 10.0 L, T= 303 K

3.97 mol x 70.9 g/mol = 282 g

4. At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate molar mass. PV = nRT

(8.31 kPa•L/K•mol)(423 K)

(100 kPa)(1.00 L) = n = 0.02845 mol

P= 100 kPa, V= 1.00 L, T= 423 K

g/mol = 2.506 g / 0.02845 mol = 88.1 g/mol

5. 98 mL of an unknown gas weighs 0.081 g at SATP. Calculate the molar mass.

PV = nRT

(8.31 kPa•L/K•mol)(298 K)

(100 kPa)(0.098 L) = n = 0.00396 mol

P= 100 kPa, V= 0.098 L, T= 298 K

g/mol = 0.081 g / 0.00396 mol = 20.47 g/mol

It’s probably neon

(neon has a molar mass of 20.18 g/mol)

Agenda

• Day 83 – Avogadro’s Law of Combining

Volumes and Partial Pressures

• Read Section 12.1 & 12.4 - Make your

own study notes on this section.

• Lesson: PPT,

• Handouts: 1.PPT Handout 2. ISU- ( W4)

Avogadro’s Law

V a number of moles (n)

V = constant x n

V1/n1 = V2/n2

Constant temperature Constant pressure

Kinetic theory of gases and …

Avogadro’s law states that equal volumes of gases at the same temperature and pressure contain equal number of particles

• Avogadro’s Law

More moles of gas, more collisions with walls of container

More collisions, higher pressure

BUT since pressure must remain constant and only volume can change

Volume increases to decrease pressure to original value

Which picture represents what the gas will look like when the moles of gas is doubled?

(Assume constant P, T)

Avogadro’s Law

The pressure of each gas in a mixture is called the partial pressure of that gas. Daltons Law of Partial Pressure states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.

Daltons Law of Partial Pressures

• So far: pure gases

• Many gases are actually mixtures of two or more gases:

– air: O2, N2 , H2O, etc

• How do mixtures of gases behave?

Gas Mixtures--Partial Pressure

Gas Mixtures--Partial Pressure

P= 8 kPa

N2 (g)

P= 6 kPa

O2 (g)

P= 9 kPa

CO2 (g)

Gas Mixtures--Partial Pressure

• Ptotal = PO2 + PN2 + PCO2

• So for this example:

Ptotal = 6 kPa + 8 kPa + 9 kPa

= 23 kPa

Dalton’s Law of Partial Pressures

V and T are

constant

P1 P2 Ptotal = P1 + P2

Kinetic theory of gases and … • Dalton’s Law of Partial Pressures

Molecules do not attract or repel one another

P exerted by one type of molecule is unaffected by the presence of another gas

Ptotal = SPi

In other words, at constant T and V,

Ptotal depends only on the total number of moles of gas present Ptotal is independent of the type (or types) of gases present.

This works according to the KMT because at the same temperature molecules of different gases have the same Kinetic Energy.

Partial Pressure-Mole Fraction

• When describing a mixture of gases, it is useful to know the relative amount of each type of gas.

• Mole fraction (X): a dimensionless number that expresses the ratio of the number of moles of one component compared to the total number of moles in a mixture.

Mole Fraction • If a gas mixture contains 5.0 mol O2(g), 3.0 mol

H2O(g), and 12.0 mol N2(g), what is the mole fraction of oxygen?

XO=

• On the exam, you must be able to calculate the mole fraction of each component of a gas mixture.

nO2

nt

= 5.0 mol

20.0 mol = 0.25

Partial Pressure

• The partial pressure of a gas in a mixture can be found:

PA = XA Ptotal

where PA = partial pressure of gas A

XA = mole fraction of gas A

Ptotal = total pressure of mixture

Partial Pressure Calculation A mixture of gases contains 0.51 mol N2, 0.28 mol H2, and 0.52 mol NH3. If the total pressure of the mixture is 2.35 atm, what is the partial pressure of H2?

PH2 = XH2

Ptotal

XH2=

0.28 mol

0.28 mol + 0.51 mol + 0.52 mol = 0.21

PH2 = 0.21 x 2.35 = 0.50 atm

Sample Problems A mixture of gases has the following partial pressure for the component gases at 20.0C in a volume of 2.00L: oxygen 180.torr, nitrogen 320.torr, and hydrogen 246torr. Calculate the pressure of the mixture.

What is the final pressure of a 1.50L mixture of gases produced from 1.50L of neon at 0.3947atm, 800.mL of nitrogen at 150.mmHg and 1.2oL of oxygen at 25.3kPa? Assume constant temperature. (Hint use Boyle’s law.)

In the lab

• Chemical reaction producing gas

eg: NH4NO2 (s) N2(g) + H2O (l)

Determine number of moles (amount) of gas collected?

When a gas is collected over water; you always have a mixture of that gas and

water.

Partial Pressures

• When one collects a gas over water, there is water vapor mixed in with the gas. Ptotal = Pgas + PH2O

• To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

• Table1 P. 596 shows water vapor pressure (T dep)

Vapour Pressure Defined • Vapour pressure is the pressure exerted by a vapour. E.g.

the H2O(g) in a sealed container.

• Yet, molecules both leave and join the surface, so vapour pressure also pushes molecules up.

Eventually the air above the water is filled with vapour pushing down. As temperature , more molecules fill the air, and vapour pressure .

• To measure vapour pressure we can heat a sample of liquid on top of a column of Hg and see the pressure it exerts at different °C.

Measuring Vapour Pressure

• When the vapour pressure is equal to the atmospheric pressure (Patm), the push out is enough to overcome Patm and boiling occurs.

Vap

our

pres

sure

Temperature

Vapour pressure for H2O

°C kPa °C kPa

10 1.23 50 12.33

20 2.34 75 38.54

30 4.17 100

See pg. 464 for more

101.3

• Thus, water will boil at a temperature below 100 °C if the atmospheric pressure is reduced.

Collecting gases over water • Many times gases are collected over H2O • Often we want to know the volume of dry gas at STP

(useful for stoichiometry). For this we must make 3 corrections: 1. The level of water inside and outside the tube must be

level (so pressure inside is equal to the pressure outside).

2. The water vapour pressure must be subtracted from the total pressure (to get the pressure of the dry gas).

3. Finally, values are converted to STP using the combined gas law.

Sample calculation A gas was collected over 21°C H2O. After equalizing water levels, the volume was 325 mL. Give the volume of dry gas at STP (Patm=102.9 kPa).

Step 1: Determine vapour pressure (pg. 596) At 21°C vapour pressure is 2.49 kPa

Step 2: Calculate the pressure of dry gas Pgas = Patm - PH2O = 102.9 - 2.49 = 100.41 kPa

Step 3: List all of the data T1 = 294 K, V1 = 325 mL, P1 = 100.41 kPa

Step 4: Convert to STP

(P1)(V1)(T2)

(P2)(T1) V2=

(100.4 kPa)(325 mL)(273 K)

(101.325 kPa)(294 K) =

= 299 mL

Assignment 1. 37.8 mL of O2 is collected by the downward

displacement of water at 24°C and an atmospheric pressure of 102.4 kPa. What is the volume of dry oxygen measured at STP?

2. 236 mL of H2 is collected over water at 22°C and at an atmospheric pressure of 99.8 kPa. What is the volume of dry H2 at STP?

3. If H2 is collected over water at 22°C and an atmospheric pressure of 100.8 kPa, what is the partial pressure of the H2 when the water level inside the gas bottle is equal to the water level outside the bottle?

1)

V1 = 37.8 mL, P1 = 99.42 kPa, T1 = 297 K

V2 = ?, P2 = 101.3 kPa, T2 = 273 K

P1V1

T1

=

P2V2

T2

(99.42 kPa)(37.8 mL)

(297 K) =

(101.3 kPa)(V2)

(273 K)

(99.42 kPa)(37.8 mL)(273 K)

(297 K)(101.3 kPa) = (V2) = 34.1 mL

Vapor pressure at 24C = 2.98 kPa

Pgas = Patm - Pvapor

= 102.4 kPa - 2.98 kPa

= 99.42 kPa = P1

2)

V1 = 236 mL, P1 = 97.16 kPa, T1 = 295 K

V2 = ?, P2 = 101.3 kPa, T2 = 273 K

P1V1

T1

=

P2V2

T2

(97.16 kPa)(236 mL)

(295 K) =

(101.3 kPa)(V2)

(273 K)

(97.16 kPa)(236 mL)(273 K)

(295 K)(101.3 kPa) = (V2) = 209 mL

Vapor pressure at 22C = 2.64 kPa

Pgas = Patm - Pvapor

= 99.8 kPa - 2.64 kPa

= 97.16 kPa = P1

Answers 3 - Total pressure = PH2 + PH2O

100.8 kPa = PH2 + 2.64 kPa

100.8 kPa - 2.64 kPa = PH2 = 98.16 kPa

Gas

Stoichiometry

Agenda

• Day 84 – Stoichiometry and Gases

• Look over Tutorial 1 on P. 598 and Tutorial 2 on P. 600.

• Lesson: PPT,

• Handouts: 1.PPT Handout 2. ISU- Stoichiometry and gases

( W4)

Molar Volume of Gases

Recall that 1 mole of a compound contains 6.022 X 1023 molecules of that compound – it doesn’t matter what the compound is. One mole of any gas, at STP, will occupy the same volume as one mole of any other gas at the same temperature and pressure, despite any mass differences. The volume occupied by one mole of a gas at STP is known as the standard molar volume of a gas. It has been found to be 22.4liters. We can use this as a new conversion factor 1mol of gas/22.4L of same gas. (Avogadro’s Law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules).

Sample Problems

What volume, in L, is occupied by 32.0 grams of oxygen gas at STP?

If all the reactants and products are gases in a chemical reaction and the pressure and temperature remain constant throughout the problem:

• In stoichiometry problems, the coefficients of a balanced chemical equation will also represent the volumes of gases found at a fixed temperature and pressure

• When solving such problems gas volumes are equivalent to their molar ratios

Gas Stoichiometry • We have looked at stoichiometry: 1) using masses

& molar masses, & 2) concentrations. • We can use stoichiometry for gas reactions. • As before, we need to consider mole ratios when

examining reactions quantitatively.

• At times you will be able to use 22.4 L/mol at STP and 24.8 L/mol at SATP as shortcuts.

grams (x) moles (x) moles (y) grams (y)

molar mass of y mole ratio from balanced equation

molar mass of x

P, V, T (x)

P, V, T (y)

PV = nRT

Sample problem 1 CH4 burns in O2, producing CO2 and H2O(g). A 1.22 L CH4 cylinder, at 15°C, registers a pressure of 328 kPa.

a) What volume of O2 at SATP will be required to react completely with all of the CH4?

First: CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

PV = nRT

(8.31 kPa•L/K•mol)(288 K)

(328 kPa)(1.22 L) = n = 0.167 mol

P = 328 kPa, V = 1.22 L, T = 288 K

# mol O2= 0.167 mol CH4 2 mol O2

1 mol CH4 x = 0.334 mol

PV = nRT

(100 kPa)

(0.334 mol)(8.31 kPa•L/K•mol)(298 K) =V = 8.28 L

P= 100 kPa, n= 0.334 mol, T= 298 K

or # L = 0.334 mol x 24.8 L/mol = 8.28 L

Sample problem 1 continued CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

b) How many grams of H2O(g) are produced?

c) What volume of CO2 (at STP) is produced if only 2.15 g of the CH4 was burned?

# g H2O= 0.167 mol CH4 2 mol H2O

1 mol CH4 x = 6.02 g

H2O

18.02 g H2O

1 mol H2O x

# mol CO2= 2.15 g CH4 1 mol CH4

16.05 g CH4 x = 0.134

mol CO2

1 mol CO2

1 mol CH4 x

PV = nRT P = 101.3 kPa, n = 0.134 mol, T = 273 K

(101.3 KPa)

(0.134 mol)(8.31 kPa•L/K•mol)(273 K) = V = 3.00 L CO2

or # L = 0.134 mol x 22.4 L/mol = 3.00 L

Sample problem 2 Ammonia (NH3) gas can be synthesized from nitrogen gas + hydrogen gas. What volume of ammonia at 450 kPa and 80°C can be obtained from the complete reaction of 7.5 kg hydrogen?

# mol NH3= 7500 g H2 1 mol H2

2.02 g H2 x = 2475 mol

2 mol NH3

3 mol H2 x

PV = nRT P = 450 kPa, n = 2475 mol, T = 353 K

(450 KPa)

(2475 mol)(8.31)(353 K) = V = 16 135 L NH3

First we need a balanced equation: N2(g) + 3H2(g) 2NH3(g)

Sample problem 3 Hydrogen gas (and NaOH) is produced when sodium metal is added to water. What mass of Na is needed to produce 20.0 L of H2 at STP?

First we need a balanced equation: 2Na(s) + 2H2O(l) H2(g) + 2NaOH(aq)

# g Na= 0.893 mol H2 = 41.1 g Na

2 mol Na

1 mol H2 x

22.99 g Na

1 mol Na x

PV = nRT

(8.31 kPa•L/K•mol)(273 K)

(101.3 kPa)(20.0 L) = n = 0.893 mol H2

P= 101.3 kPa, V= 20.0 L, T= 273 K

or # mol = 20.0 L x 1 mol / 22.4 L = 0.893 mol

Assignment 1. What volume of oxygen at STP is needed to

completely burn 15 g of methanol (CH3OH) in a fondue burner? (CO2 + H2O are products)

2. When sodium chloride is heated to 800°C it can be electrolytically decomposed into Na metal & chlorine (Cl2) gas. What volume of chlorine gas is produced (at 800°C and 100 kPa) if 105 g of Na is also produced?

3. What mass of propane (C3H8) can be burned using 100 L of air at SATP? Note: 1) air is 20% O2, so 100 L of air holds 20 L O2, 2) CO2 and H2O are the products of this reaction.

4. A 5.0 L tank holds 13 atm of propane (C3H8) at 10°C. What volume of O2 at 10°C & 103 kPa will be required to react with all of the propane?

5. Nitroglycerin explodes according to: 4 C3H5(NO3)3(l) 12 CO2(g) + 6 N2(g) + 10 H2O(g) +

O2(g)

a) Calculate the volume, at STP, of each product formed by the reaction of 100 g of C3H5(NO3)3.

b) 200 g of C3H5(NO3)3 is ignited (and completely decomposes) in an otherwise empty 50 L gas cylinder. What will the pressure in the cylinder be if the temperature stabilizes at 220°C?

Answers 1. 3O2(g) + 2CH3OH(l) 2CO2(g) + 4H2O(g)

# L O2=

15 g CH3OH 1 mol CH3OH

32.05 g CH3OH x

= 15.7 L O2

3 mol O2

2 mol CH3OH x

22.4 L O2

1 mol O2

x

2. 2NaCl(l) 2Na(l) + Cl2(g)

# mol Cl2= 105 g Na 1 mol Na

22.99 g Na x

1 mol Cl2

2 mol Na x

PV = nRT P = 100 kPa, n = 2.284 mol, T = 1073 K

(100 KPa)

(2.284 mol)(8.31)(1073 K) = V = 204 L Cl2

= 2.284 mol Cl2

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

# g C3H8=

20 L O2 1 mol O2

24.8 L O2

x

= 7.1 g C3H8

1 mol C3H8

5 mol O2

x 44.11 g C3H8

1 mol C3H8

x

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

PV = nRT

# mol O2= 2.8 mol C3H8 5 mol O2

1 mol C3H8

x = 14 mol O2

(8.31)(283 K)

(1317 kPa)(5.0 L) n = = 2.8 mol C3H8

PV = nRT P = 103 kPa, n = 14 mol, T = 283 K

(103 KPa)

(14 mol)(8.31)(283 K) = V = 320 L O2

5. # mol C3H5(NO3)3=

100 g C3H5(NO3)3 1 mol C3H5(NO3)3

227.11 g C3H5(NO3)3

x = 0.4403 mol

# L CO2= 0.4403 mol C3H5(NO3)3

12 mol CO2

4 mol C3H5(NO3)3

x 22.4 L 1 mol

x = 29.6 L CO2

# L N2= 0.4403 mol C3H5(NO3)3

6 mol N2

4 mol C3H5(NO3)3

x 22.4 L 1 mol

x = 14.8 L N2

# L H2O= 0.4403 mol C3H5(NO3)3

10 mol H2O

4 mol C3H5(NO3)3

x 22.4 L 1 mol

x = 24.7 L H2O

# L O2= 0.4403 mol C3H5(NO3)3

1 mol O2

4 mol C3H5(NO3)3

x 22.4 L 1 mol

x = 2.47 L O2

5. # mol C3H5(NO3)3=

200 g C3H5(NO3)3 1 mol C3H5(NO3)3

227.11 g C3H5(NO3)3

x = 0.8806 mol

# mol all gases=

0.8806 mol C3H5(NO3)3 29 mol gases

4 mol C3H5(NO3)3

x = 6.385 mol all gases

PV = nRT V = 50 L, n = 6.385 mol, T = 493 K

(50 L)

(6.385 mol)(8.31)(493 K) = P = 523 kPa