Behavior of GASES

Gases are made up of atoms and molecules just like all other compounds, but because they are in the form of a gas we can learn a great deal more about these molecules and compounds. It might seem a bit confusing because we can’t see most gases, but we know they exist.

Elements that exist as gases at 250C and 1 atmosphere

•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.

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

Nature of Gases cont.

•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.

Collisions of Gas ParticlesCollisions of Gas Particles

•The word KINETIC refers to motion

•Kinetic energy= energy an object has because of its motion

Collisions of Gas ParticlesCollisions of Gas Particles

II. Kinetic Molecular Theory II. Kinetic Molecular Theory of Gasesof Gases

Particles of matter (any type) are in constant motion! Because we know this we have a few assumptions that we make about gases, called the Molecular Theory of Gases:

1. Particles of a gas are in constant, straight-line motion, until they collide.

They move independently from each other

Kinetic theory:

Molecular Motion

Kinetic theory:2. Gases consist of a

large number of tiny particles (molecules or atoms) ; these particles are very far apart, therefore gas is mostly empty space.

•There are no forces of attraction or repulsion between particles of gases.

Kinetic theory:

•3. Collisions between particles of a gas and the container wall are elastic. Which means there is no loss of energy.

•Total Kinetic energy remains constant.

Kinetic theory:4. The average kinetic

energy of gas particles depends on the temperature of the gas. (It is directly proportional)

KE=1/2 mv2 (m=mass in kg and

v=velocity is m/sec) Calories (cal) ..Joules (j)

measure Enegy 1 cal= 4.18 J 1Cal = 1000 cal

III. Volume, Pressure, Temperature, Number of Moles (Descriptions of Gases)

1. Volume – refers to the space matter (gas) occupies.

•Measured in liters (L).

1.00 dm3 = 1.00L = 1000 cm3 = 1000mL

Volume (V)

Pressure (Pressure (PP))•2. Pressure(P) – the number of times particles collide with each other and the walls of the container (force exerted on a given area).

•A vacuum is empty space= It has no pressure

Pressure Simulation

Atmospheric PressureAtmospheric PressureAtmospheric PressureAtmospheric Pressure

The gases in the air are exerting a pressure called atmospheric pressure

Atmospheric pressure is a result of the fact that air has mass and is colliding with everything under the sun with a force.

Atmospheric PressureAtmospheric PressureAtmospheric PressureAtmospheric Pressure

Atmospheric pressure varies with altitudealtitude

•The lower the altitude, the longer and heavier the column of air above an area of the earth.

Check the back of a box of cake mix for the difference in baking times based on the atmospheric pressure in your region.

Atmospheric PressureAtmospheric PressureAtmospheric PressureAtmospheric Pressure

Knowing atmospheric pressure is how forecasters predict the weather.

Low pressure or dropping pressure = a change of weather from fair to rain.

High pressure = clear skies & sun.

Atmospheric PressureAtmospheric PressureAtmospheric PressureAtmospheric Pressure

•Pressure is measured with a device called a barometer. (They operate on the change of pressure due to the weather)

Barometer

• At 1 atm (one atmospheric pressure) a column of mercury 760 mm high.

Dish of Mercury

Column of Mercury

1 atm Pressure

Barometer

• At 1 atm a column of mercury 760 mm high.

• A 2nd unit of pressure is mm Hg (mercury) 1 atm = 760 mm Hg

• A 3rd unit & the SI unit is the Pascal (Pa) 1 atm = 101.3 kPa

760 mm1 atm Pressure

1atm = 760 millimeters Hg (Barometers use Hg)

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

1atm = 101.3 kPa – kilopascals•1 atm = 760 mm Hg = 101.3 kPa

Measured in atmospheres (atm).

Practice: Convert 4.40 atm to mmHg.

Convert 212.4kPa to mmHg.

•3. Temperature (T) – as temperature increases gas particles move faster, as temperature decreases gas particles move slower.

• measured with a thermometer in Celsius. • calculations involving gases are made after

converting the Celsius to Kelvin temperature.

Measured in Kelvin (K).

Kelvin = 273 + C Celsius = K - 273

Temperature (T)Temperature (T)

Practice: Convert 32.0°C to K.

Convert 400. K to °C.

Amount (n)4. 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)

6.02 x 106.02 x 102323 molecules per mole of molecules per mole of the gas.the gas.

STP – “standard temperature and pressure”.

( Measured at Sea Level)

• Standard Temperature

0C= 273 K

• Standard Pressure 1.00atm= 760 torr = 760 mmHg = 101.325 kPa

V. Gas Laws - How do all of pressure, temperature, volume, and amount of a gas relate to each other?

Combined GAS Law (Initial) (Final)

Peas x Vegetables P1 x V1 = P2 x V2

Table T1 T2

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 4th solve the problem.

(If temperature is involved, it MUST be converted to Kelvin! K = 273 + C)

A. 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.)PP11 x V x V

11 = = PP2 2 x V x V22

TT1 1 T T22

#2 from Boyles Law Problem Sheet A sample of carbon dioxide occupies a volume of 3.50 liters at 125 kPa pressure. What pressure would the gas exert if the volume was decreased to 2.00 liters?

219 kPa = P219 kPa = P22

* P1 x V1 = P2 x V2

P1 = 125 kPa

P2 = X

V1 = 3.50 L

V2 = 2.00 L

125 kPa x 3.50 L = P2 X 2.00L

2.00L 2.00L

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

V1 = V2 T1 = initial or old temperature

T1 T2 T2 = final or new temperature

Direct Relationship (As temperature increases, volume increases and as temperature decreases, volume decreases.)

• #2 From Charles Law Problem Sheet• Oxygen gas is at a temperature of 40O C when it

occupies a volume of 2.3 liters. To what temperature should it be raised to occupy a volume of 6.5 liters?

T2 = 880K

V1 = V2

T1 T2

12.3 L x T2 = 6.5L X 313K

2.3L 2.3L

PP1 1 ==

PP2 2 = =

V1 = .3 L

V2 = 6.5 L

T1 = 40oC + 273 = 3l3 K

T2 = X

2.3L = 6.5L 3l3 K T2

Boyle’s Law ReviewBoyle’s Law Review

1621-16911621-1691

PP11VV11 = = PP22VV22

PP11VV11 = = PP22VV22

Boyle’s LawBoyle’s Law BOYLE's Law in Action

How does Pressure and Volume of gases relate

graphically?

How does Pressure and Volume of gases relate

graphically?

Volu

me

Volu

me

PressurePressure

PV = kPV = k

Temperature, # of particlesremain constant

Temperature, # of particlesremain constant

Boyle’s Mathematical Law:Boyle’s Mathematical Law:

since PV equals a constantsince PV equals a constant

PP11VV11 = = PP22VV22

PP11VV11 = = PP22VV22

Ex: A gas has a volume of 3.0 L Ex: A gas has a volume of 3.0 L at 2 atm. What will its volume at 2 atm. What will its volume

be at 4 atm?be at 4 atm?

If we have a given amount of a gas at If we have a given amount of a gas at a starting pressure and volume, what a starting pressure and volume, what would happen to the pressure if we would happen to the pressure if we

changed the volume? changed the volume? Or to the volume if we changed the Or to the volume if we changed the

pressure? pressure?

Boyle’s Mathematical Law:Boyle’s Mathematical Law:

1)1)List the variables or clues given:List the variables or clues given:

2)2)determine which law is being determine which law is being represented:represented:

PP11 = 2 atm = 2 atm VV11 = 3.0 L = 3.0 L

PP22 = 4 atm = 4 atm VV22 = ? = ?

P1V1 = V2 P2

3) Plug in the variables & calculate:3) Plug in the variables & calculate:

(2 atm)(2 atm)(3.0 L) =(3.0 L) =(4 atm)(4 atm)(V(V22))

PP11VV11 = = PP22VV22

PP11VV11 = = PP22VV22

Charles’s Law ReviewCharles’s Law Review

1746-18231746-1823

==VV11 VV22

TT11 TT22====

VV11 VV22

TT11 TT22

Any relation to Bernet??

Charles’ LawCharles’ LawCHARLES' Law in Action

Temp

How does Temperature and How does Temperature and Volume of gases relate Volume of gases relate

graphically?graphically?

How does Temperature and How does Temperature and Volume of gases relate Volume of gases relate

graphically?graphically?V

olu

me

V/T = k

Pressure, # of particlesremain constant

Pressure, # of particlesremain constant

Charles’s Mathematical Law:Charles’s Mathematical Law:

since V/T = ksince V/T = k

Ex: A gas has a volume of 3.0 L Ex: A gas has a volume of 3.0 L at 400K. What is its volume at at 400K. What is its volume at

500K?500K?

==VV11 V V22

TT11 T T22

If we have a given amount of a gas at If we have a given amount of a gas at a starting volume and temperature, a starting volume and temperature, what would happen to the volume if what would happen to the volume if

we changed the temperature? we changed the temperature? Or to the temperature if we changed Or to the temperature if we changed

the volume? the volume?

Charles’s Mathematical Law:Charles’s Mathematical Law:

2)2)determine which determine which law is being law is being represented:represented:

TT11 = 400K = 400K VV11 = 3.0 L = 3.0 L

TT22 = 500K = 500K VV22 = ? = ?

1)1)List the variables or clues given:List the variables or clues given:

VV11VV11

TT11TT11

VV22VV22

TT22TT22

====

3) Plug in the 3) Plug in the variables & variables & calculate:calculate:

3.0L3.0L

400K400K 500K500K

X LX L==

C. Gay-Lussac’s Law - Pressure and Temperature (when volume is constant)

• P1 = P2

T1 T2

• Direct RelationshipDirect Relationship (As temperature increases, pressure increases and as temperature decreases, pressure decreases.)

P1 x T2 = P2 x T1

#2 From Gay-Lussac’s Problem Sheet•A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at standard temperature? (STP =Remember O oC or 273 K) (Change 50.0 °C to Kelvin)

P1 = 0.370 atm

P2 = 0.820 atm

V1 =

V2 =

T1 = 40 oC + 273 = 323 K

•T2 = O oC + 273 = 273 K

P1 = P2

T1 T2

•0.370 atm = P2

323 K 273 K

P2 = 0.820 atm

D. Combined Gas Law - Pressure, Temperature, and Volume (None of the variables are constant)

P1 x V1 x T2 = P2 x V2 x T1

Combined GAS Law (Initial) (Final)

Peas x Vegetables P1 x V1 = P2 x V2

Table T1 T2

• Ex: Find the final volume of 25.0 ml of a Gas at STP. If the conditions change to 14 oC and 740 mmHg

P1 = 760 mmHg

P2 = 740 mmHg

V1 = 25.0mL

V2 = ?

T1 = 0 oC + 273 = 273 K

T2 = 14 oC + 273 = 287 K

P1 x V1 = P2 x V2

T1 T2

(760 mmHg) (25.0 ml) = (740 mmHg) (V2)

273 K 287 K

(287K) (69.6 mmHg . ml) = (740 mmHg) (V2) (287K)

K 287 K

19974 mmHg.mL = (740 mmHg) (V2)

740 mmHg 740 mmHg 27ml = V2

STP

Temperature = 273KPressure =760 mmHg

Gay-Lussac’s LawGay-Lussac’s Law

1778-1778-18501850

PP11 PP22TT11 T T22

==

Gay-Lussac’s LawGay-Lussac’s Law

Temp

Pre

ssu

re

How does Pressure and Temperature of gases relate

graphically?

How does Pressure and Temperature of gases relate

graphically?

P/T = k

Volume, # of particlesremain constant

Volume, # of particlesremain constant

since P/T = ksince P/T = k

PP11 PP22TT11 T T22

==

Ex: A gas has a pressure of Ex: A gas has a pressure of 3.0atm at 400K. What is 3.0atm at 400K. What is

its pressure at 500K?its pressure at 500K?

If we have a given amount of a gas at If we have a given amount of a gas at a starting temperature and pressure, a starting temperature and pressure, what would happen to the pressure if what would happen to the pressure if

we changed the temperature? we changed the temperature? Or to the temp. if we changed the Or to the temp. if we changed the

pressure? pressure?

Gay-Lussac’s Mathematical Law:Gay-Lussac’s Mathematical Law:

Gay-Lussac’s Mathematical Law:Gay-Lussac’s Mathematical Law:

2)2)determine which determine which law is being law is being represented:represented:

TT11 = 400K = 400K PP11 = 3.0 atm = 3.0 atm

TT22 = 500K = 500K PP22 = ? = ?

1)1)List the variables or clues given:List the variables or clues given:

P1P1

TT11

PP22

TT22

==

3) Plug in the 3) Plug in the variables & variables & calculate:calculate:

3.0atm

400K400K 500K500K

X LX L==

LAWLAWRELAT-RELAT-IONSHIPIONSHIP

LAWLAWCON-CON-

STANTSSTANTS

Boyle’sBoyle’s PP V V PP11VV1 1 = P= P22VV22 T, nT, n

Charles’Charles’ VV T T VV11/T/T11 = V = V22/T/T22 P, nP, n

Gay-Gay-Lussac’sLussac’s

PP T T PP11/T/T11 = P = P22/T/T22 V, nV, n

Summary of the Named Gas-Laws:Summary of the Named Gas-Laws:

III. 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.

E. Ideal Gas Law (PV = nRT) – to use this law, all units must be as follows:

• P = pressure in atm• V = volume in liters• n = number of moles• T = temperature in Kelvin• R = (0.0821L) (1atm)

(1mol) (1K)• R is the ideal gas constant (page 342 in book

describes where this constant came from.)

• 1f. How many moles of CH4 gas are there in 85.0L at STP?

• 2f. What volume will be occupies by 1.50grams of nitrogen monoxide gas at 348K and pressure of 300.mmHg?

• 3f. A volume of 11.2L of a gas at STP has how many moles?

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.

F Daltons Law of Partial Pressures

PT = P1 + P2 + P3 + ……. PT = total pressure

P# = the partial pressures of the individual gases

• 1e. 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.

• 2e. 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.)

Daltons Law applied to Gases Collected by Water Displacement – Figure 10-15 page 324

Patm or PT= Pgas + PH2O

Patm or PT= barometric pressure or total pressurePgas = pressure of the gas collectedPH2O = vapor pressure of water at specific temperature (Found on page 899 of you textbook.)

• 3e. Oxygen gas from the decomposition reaction of potassium chlorate was collected by water displacement at a pressure of 731torr and a temperature of 20.0C. What was the partial pressure of the oxygen gas collected?

• 4e. Solid magnesium and hydrochloric acid react producing hydrogen gas that was collected over water at a pressure of 759mmHg and measured 19.0mL. The temperature of the solution at which the gas was collected was 25.0C. What would be the pressure of the dry hydrogen gas? What would be the volume of the dry hydrogen gas at STP?

G. Solving for Density and /or Molar Mass of a gas using the Ideal Gas Law

1. 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 n (in grams) by the volume.

• 1g. What is the density of a sample of ammonia gas, NH3, if the pressure is 0.928 atm and the temperature is 63.0C?

• 2g. What is the density of argon gas at a pressure of 551 torr and a temperature of 25.0C?

2. 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.

• 3g. 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?

• 4g. 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?

H. 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.

• 1h. What volume, in L, is occupied by 32.0 grams of oxygen gas at STP?

I. Stoichiometry of Gases

Just like mole ratios can be written from an equation so can a volume ratio-same concept!

• 2CO(g) + O2 (g) 2CO2 (g)

• 1i. Using the above equation, what volume of oxygen gas is needed to react completely with 0.626L of carbon monoxide to form carbon dioxide?

• 2i. How many grams of solid calcium carbonate must be decomposed to produce 5.00L of carbon dioxide gas at STP?

• 3i. How many liters of hydrogen gas at 35.0C and 0.980atm are needed to produce 8.75L of gaseous water according to the following equation?

• WO3(s) + 3H2(g) W(s) + 3H2O(g)

J. Graham’s Law IV. Effusion and Diffusion

• Graham’s Law states that the rates of diffusion/effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container. (onions on page 352)

• Rate of diffusion/effusion of A = √(MB/MA) Rate of diffusion/effusion of B

MA or B = molar mass of that compound Gas A is the lighter, faster gasRate of diffusion/effusion is the same as the velocity (or speed) of the gas.

After the rates of diffusion/effusion for two gases are determined, the gas with the lower molar mass will be the one diffusing/effusing fastest.

• 1j. Compare the rates of effusion for hydrogen and oxygen at the same temperature and

• pressure. (Which one effuses faster and how much faster is it effusing?)

• 2j. A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.

Graham’s Law and Time

Graham’s Law and Time – the time it takes a gas to effuse is directly proportional to its molar mass.

tA = MA t = time

tB MB

• 3j. A sample of an unknown gas flows through the wall of a pours cup in 39.9 minutes. An equal volume of helium (under same temperature and pressure) flows through in 9.75 minutes. What is the molar mass of the unknown gas?

Pressure and Volume (Boyle’s Law) Gas Demonstrations

• Bell Jar – Shaving Cream

As pressure decreases the volume of the gas increases.

• Bell Jar – Balloon

• Bell Jar – Peeps

• Cartesian Diver