The Kinetic Molecular Theory

explains the forces between

molecules and the energy that

they possess. This theory has 3

basic assumptions.

Kinetic Molecular Theory

1. Matter is composed 1. Matter is composed of small particles of small particles

(molecules). (molecules).

Volume is the measure of space Volume is the measure of space in between the molecules and in between the molecules and not the space the molecules not the space the molecules

contain themselves.contain themselves.

2. The molecules are in constant motion.

   Solid - Molecules are held close to each other by their attractions of charge. They will bend and/or vibrate, but will stay in close proximity.

   Liquid - Molecules will flow or glide over one another, but stay toward the bottom of the container. Motion is a bit more random than that of a solid.

   Gas - Molecules are in continual straight line motion. The kinetic energy of the molecule is greater than the attractive force between them, thus they are much farther apart and move freely of each other.

3. When the molecules 3. When the molecules collide with each other, collide with each other, or with the walls of a or with the walls of a container, there is no container, there is no

loss of energy.loss of energy.

Properties of GasesProperties of Gases

• Gases can be compressed into smaller volumes

• ↑ P causes ↓ V

• so: ↑ P causes ↑density

Properties of GasesProperties of Gases

• Gases exert pressure on surroundings. So to confine a gas, pressure must be exerted on it. (Newton’s third law: for every action, there is an equal and opposite reaction)

Properties of GasesProperties of Gases

Gases expand without limits, therefore a gas will completely fill the container it occupies

Gas PropertiesGas PropertiesCollisions between gas particles and

between particles and the container walls are elastic collisions

Properties of GasesProperties of Gases• Gas particles are in constant, rapid,

random motion.

Have kinetic energy

Gas PropertiesGas Properties• Fluidity

Gas particles glide easily past one another since the attractive forces are minimal.

Properties of GasesProperties of Gases

• Gases diffuse into one another. Gases mix completely, ie gases are miscible.

The amount of gas can be known if the pressure, temperature and volume are known.

TEMPERATURE AND KINETIC ENERGYTEMPERATURE AND KINETIC ENERGY  

• Different substances have the same average KE when they are at the same temperature.

• The velocity of the particles depends on their atomic or molecular mass (formula weight). The greater their mass, the slower their velocity.

PressurePressurePressure – force / unit area

Units: Newtons / m2 (Pascals)Atmospheres (atm)mm HgTorr

1 atm = 14.7 psi = 1.01 x 105 Pa = 760 mm Hg = 760 Torr =101.3kPa

PressurePressure

http://www.iun.edu/~cpanhd/C101webnotes/gases/barometer.html

GAS LAWSGAS LAWSTHREE VARIABLES THAT THREE VARIABLES THAT

DETERMINE GAS PRESSURE:DETERMINE GAS PRESSURE:

• Temperature

• Amount (# of moles)

• Volume

GAS LAWSGAS LAWSTHREE VARIABLES THAT THREE VARIABLES THAT

DETERMINE GAS PRESSURE:DETERMINE GAS PRESSURE:

Temperature – the average Kinetic Energy of the gas. KE = ½ mv²

If the Temperature ↑ then the Pressure ↑

GAS LAWSGAS LAWSTHREE VARIABLES THAT THREE VARIABLES THAT

DETERMINE GAS PRESSURE:DETERMINE GAS PRESSURE:

Amount (# of moles) – the number of particles of a gas in a closed container

If the Amount ↑ then the Pressure↑

GAS LAWSGAS LAWSTHREE VARIABLES THAT THREE VARIABLES THAT

DETERMINE GAS PRESSURE:DETERMINE GAS PRESSURE:

Volume of the sealed, but flexible container of the gas

If the Volume ↑ then the Pressure ↓ If the Volume ↓ then the Pressure ↑

Gas LawsGas Laws

Boyle’s Law

↑ ↑ P causes ↓ VP causes ↓ V

PP11VV11 = P = P22VV22

PV = constantPV = constant

The PEEP in the Vacuum

STPSTPStandard Temperature and Standard Temperature and

Pressure (STP)Pressure (STP) is defined as is defined as 00ooC and 1 atm of pressure.C and 1 atm of pressure.

A balloon initially occupies 12.4 L at 1.00 atm. What will be the volume at 0.800 atm?

P1

=1.00 atm

V1

=12.4 L

P2

=0.800 atm

V2

=

?

PP11VV11 = P = P22VV22

PP11VV11 = = VV22PP22

1.00 atm * 12.4L = V1.00 atm * 12.4L = V22*0.800 atm0.800 atm

V2= 15.5L

A 5.00 L container is filled with gas at 1.00 A 5.00 L container is filled with gas at 1.00 atm. A piston then compresses the gas to atm. A piston then compresses the gas to

0.250 L. What is the new pressure? 0.250 L. What is the new pressure?

1.00 atm

5.00 L0.250L

P1 =

V1 =

P2 =

V2 =

PP11VV11 = P = P22VV22

PP11VV11 = V = V22

PP22

?

1.00 atm* 5.00 L = 0.250L *X

X = 20.0atm

• Example #3: 2.50 L of a gas was at an unknown pressure. However, at standard pressure, its volume was measured to be 8.00 L. What was the unknown pressure?

• End E 4/7

Charles LawCharles Law

↑ T causes ↑ V

As the temperature increases, the volume increases.

(in K not oC)

V/T = constant

V1/T1 = V2/T2

Charles’s LawCharles’s Law• If the amount of a gas and its pressure

remain constant, then• As the temperature of a gas increases,

the volume of the gas increases. ↑T =↑V

• As the temperature of a gas decreases, the volume of the gas decreases

↓T =↓V

(T2*V1)/T1 = V2

A 450 mL piston assembly containing gas initially at 293 K is suddenly heated to 450 K.

What will be the new volume of the piston assembly?

V1= 450 mL

T1

=293 K

V2

=?

T2

=450K

V1/T1 = V2/T2

691 mL = V2

(450K * 450 mL) =V2/293 K

• Example #2: 4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C?

• Example #3: 5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure?

KELVINSKELVINS• EVERY TEMPERATURE USED IN A

CALCULATION MUST BE IN KELVINS, NOT DEGREES CELSIUS.

Gay –Lussac’s LawGay –Lussac’s LawThe pressure of a fixed mass of gas at

a constant volume varies directly with the Kelvin Temperature

Or

Increased Temperature = Increased Pressure

Gay –Lussac’s LawGay –Lussac’s Law• If the temperature of a container is

increased, the pressure increases.

↑T =↑P• If the temperature of a container is

decreased, the pressure decreases.

↓T =↓P

P1 P2

T1 T2

Don’t forget temperature is in Kelvin

Gay –Lussac’s LawGay –Lussac’s Law

=

If the pressure of a chamber was 100 atm at 600 K, what will be the temperature of the chamber if the pressure was raised to 600 atm?

P1=

T1=

P2

=T2=

100 atm

600K

600 atm

? K

PP11 = = PP22

TT1 1 TT22

100 atm

600K

=600 atm

? K

100 atm

600K=

600 atm

? K

100 atm 600K= 600 atm *? K *

100 atm

600K=

600 atm

? K

100 atm

600K= 600 atm * ? K

? K = 360000 atm* K100 K

? K = 3600K

• Example #2: 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure?

Avogadro's LawAvogadro's Law• Gives the relationship between

volume and amount when pressure and temperature are held constant.

• V1 / n1 = V2 /n2 or V1 = V2 n1 n2

Avogadro's LawAvogadro's Law• If the amount of gas in a container

is increased, the volume increases.

↑n =↑V• If the amount of gas in a container is

decreased, the volume decreases.↓n =↓v

• 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?

THE GAS LAWSTHE GAS LAWS

Boyle’s P1V1 = P2V2

Charles’s V1 / T1 = V2 / T2

Gay-Lussac's P1 / T1 = P2 / T2

Avogadro's V1 / n1 = V2/ n2

Combined Gas LawCombined Gas LawWrite Boyle's Law:

P1V1 = P2V2

Multiply by Charles Law:

P1V12 / T1 = P2V2

2 / T2

Multiply by Gay-Lussac's Law:

P12V1

2 / T12 = P2

2V22 / T2

2

Take the square root to get the combined gas law:

P1V1 / T1 = P2V2 / T2

or

P1V1T2 = P2V2T1

• 2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP?

• P1 = P2 =

• V1= V2=

• T1= T2=

745.0 mm Hg

2.00 L

(25.0 +273)

760 mm Hg

(0+273)

X

P1V1 / T1 = P2V2 / T2

745.0mm Hg 2.00L / 298 K = 760 mm Hg *X / 273 K

V2 = 1.79 L

Ideal Gas LawIdeal Gas LawTo derive the Ideal Gas Law, we

combine all the gas lawsEach of the laws has a constant, k.

Each k is different. The product of all the constants (k) is

written as R. R depends on the units used.

R

Ideal Gas LawIdeal Gas Law• PV= nRT• P = Pressure• V = Volume (must be in Liters)• n = number of moles• T = Temperature(must be in Kelvin)R =Universal Gas Constant (Write this down)

R =0.0821 (L*atm)/(mol*K)or

R = 8.314 (L*kPa)/(mol*K)or

R = 62.36 (L*mm Hg)/(mol*K)

• How many moles of gases are contained in a can with a volume of 555 mL and a pressure of 600.0 kPa at 20oC? (R = 8.314 L*kPa)/(mol*K)

N = 0.137 mole

• Calculate the pressure exerted by 43 moles of nitrogen in a 65 L cylinder at 5oC?

(Since there are no pressure units given you can use any R value you want)

P= 15 atm

P= 1529 kPa

P= 11468 mmHg

If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature?

PV=nRT

P= 5.6 atm

V= 12 L

n = 4 mole

R= 0.0821 (L*atm)/(mol*K)

T= x

5.6 atm *12 L = 4 mol * 0.0821 L* atm *Xmol K

67.2 atm L =0.3284 x L atm/K

67.2atm L /0.3284 L atm /K = x

204.6 K

A steel tank having a volume of 5.82 L contains 8.78 g of O2 under a pressure of 1.35 atm.  What is the temperature -- in degrees Celsius -- of the O2?

PV=nRTR = 0.0821 (L*atm)/mol*K)1.35 atm *5.82L = (8.78g/32g/mol)(0.0821L atm/mol K) T7.874 atm L = T 0.0225atm L/K

T= 349.509 K = 349.509-273= 76.5oC

A sample of an unknown weighing 2.1025 grams is found to occupy 2.850 L at 22.0°C and .9737atm How many moles of the gas are present?

(0.97 atm) (2.85 L) = (n) (0.08206 L atm / mol K) (295.0 K)2.775 atm L = n *24.20 L atm/mol

2.775 atm L = n 24.20 L atm mole

0.1136 mole = n

Mole buck Question What is the molar mass of the unknown gas?