properties of gases

Kinetic Molecular TheoryPostulates of the Kinetic Molecular Theory of Gases

1. Gases consist of particles (atoms or molecules) in constant, straight-line motion.

2. Gas particles do not attract or repel each other (no interactions). Particles collide with each other and surfaces elastically. Collisions with walls of container define pressure (P = F/A).

3. Gas particles are small, compared with the distances betweenthem. Hence, the volume (size) of the individual particles can be assumedto be negligible (zero).

4. The average kinetic energy of the gas particles is directly proportionalto the Kelvin temperature of the gas

Properties of GasesGases expand to fill any container.

– random motion, no attraction

Gases are fluids (like liquids).– particles flow easily

Gases have very low densities.– lots of empty space; particles spaced far apart

Gases are easily compressible.– empty space reduced to smaller volume

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Collisions of Gas ParticlesPressure = collisions on container walls

• In a smaller container - particles have less room to move.

• Particles hit the sides of the container more often.

• This causes an increase in pressure.

• As volume decreases: pressure increases.

Changing the Size of the Container

Pressure = Force/Area

KEY UNITS AT SEA LEVEL (alsoKEY UNITS AT SEA LEVEL (alsoknown as standard pressure)known as standard pressure)

101.325 kPa (kilopascal)

1 atm

760 mm Hg

760 torr

14.7 psi


2m

NkPa

Sea level

Barometers

Mount Everest

Sea level On top of Mount EverestSea level

Temperature

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

32FC 95 K = ºC + 273

Always use temperature in Kelvin when working with gases. Std temperature = 273 K


STP

Standard Temperature & PressureStandard Temperature & Pressure

0°C 273 K

1 atm 101.325 kPa- OR -

STP


Boyle’s Law

1 atm

4 Liters

• As the pressure on a gas increases2 atm

2 Liters

• As the pressure on a gas increases -

the volume decreases

• Pressure and volume are inversely related

Boyle’s Law Illustrated

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 404

The pressure and volume of a gas are inversely related

•at constant mass & temp

Boyle’s Law

PV = kCourtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Volume(mL)

Pressure(torr)

P.V(mL.torr)

10.0

20.0

30.0

40.0

760.0

379.6

253.2

191.0

7.60 x 103

7.59 x 103

7.60 x 103

7.64 x 103

P1 x V1 = P2 x V2

Boyle’s Law exampleBoyle’s Law example

A quantity of gas under a pressure of 106.6 kPa has a volumeof 380 cm3. What is the volume of the gas at standard pressure, if the temperature is held constant?

P1 x V1 = P2 x V2

(106.6 kPa) x (380 cm3) = (101.3 kPa) x (V2)

V2 = 400 cm3

Charles’s Law

Timberlake, Chemistry 7th Edition, page 259

• If you start with 1 liter of gas at 1 atm pressure and 300 K

• and heat it to 600 K one of 2 things happens

300 K

• Either the volume will increase to 2 liters at 1 atm

300 K600 K

300 K 600 K

Or the pressure will increase to 2 atm.

The volume and absolute temperature (K) of a gas are directly related

–at constant mass & pressure

kT

V

Charles’ Law


Volume

(mL)

Temperature

(K)

V / T

(mL / K)

40.0

44.0

47.7

51.3

273.2

298.2

323.2

348.2

0.146

0.148

0.148

0.147

V1 / T1 = V2 / T2

The pressure and absolute temperature (K) of a gas are directly related – at constant mass & volume

Gay-Lussac’s Law


Temperature (K)

Pressure(torr)

P/T(torr/K)

248 691.6 2.79

273 760.0 2.78

298 828.4 2.78

373 1,041.2 2.79

kT

P P1 / T1 = P2 / T2

= kPVPTVT

PVT

Combined Gas Law

P1V1

T1

=P2V2

T2

P1V1T2 = P2V2T1Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

A quantity of gas has a volume of 400 cm3 at STP. What volume will it occupy at 35oC and 83.3 kPa?

P1 = 101.325 kPaT1 = 273 KV1 = 400 cm3

P2 = 83.3 kPaT2 = 35oC + 273 = 308 KV2 = ? cm3

(101.325 kPa) x (400 cm3) = (83.3 kPa) x (V2)

273 K 308 K

V2 = 548.9 cm3

1 1 2 2

1 2

PV PV

T T

The Combined Gas Law

The Combined Gas Law

When measured at STP, a quantity of gas has a volume of 500 cm3. What volume will it occupy at 20oC and 93.3 kPa?

P1 = 101.325 kPaT1 = 273 KV1 = 500 cm3

P2 = 93.3 kPaT2 = 20oC + 273 = 293 KV2 = ? cm3

(101.325 kPa) x (500 cm3) = (93.3 kPa) x (V2)

273 K 293 K

V2 = 582.8 cm3

1 1 2 2

1 2

PV PV

T T

Molar Volume (Avogadro)

Timberlake, Chemistry 7th Edition, page 268

1 mol of all gases @ STP have a volume of 22.4 L

Avogadro’s Law V1/n1 = V2/n2

Ideal Gas Law

PV = nRTPV = nRTPV = nRTPV = nRTBrings together all gas properties.

P = pressureV = volume (must be in liters)n = molesR = universal gas constant (0.082 or 8.314)T = temperature (must be in Kelvin)

Can be derived from experiment and theory.

Ideal Gas Law

What is the pressure of 0.18 mol of a gas in a 1.2 L flask at 298 K?

P = ? atmn = 0.18 molT = 298 KV = 1.2 LR = .082 (L x atm)/(mol x K)

P x (1.2 L) = (0.18 mol) x (.082) x (298 K)

P = 3.7 atm

PV = nRTPV = nRT

Gas Density

D = (MM)P/RTD = (MM)P/RTD = (MM)P/RTD = (MM)P/RT

Larger particles are more dense. Gases are more dense athigher pressures and lower temperatures

D = densityP = pressureMM = molar massR = universal gas constantT = temperature (must be in Kelvin)

Can be derived from experiment and theory.

Gas Problems

1. The density of an unknown gas is 0.010g/ml. What is the molar mass of this gas measured at -11.00C and 3.25 atm? Use proper sig figs.

2. What is the volume of 3.35 mol of gas which has a measured temperature of 47.00C and a pressure of 185 kPa? Use proper sig figs.

Gas Problems

1. The density of an unknown gas is 0.010g/ml. What is the molar mass of this gas measured at -11.00C and 3.25 atm? Use proper sig figs.

Molar mass = ? g/mol D = 0.010 g/mlT = 262 KP = 3.25 atmR = .082 (L atm)/(mol K)

g/mol = (0.010g/ml) x (.082atm L/mol K) x (262 K) x (1/3.25 atm) x (1000ml/1 L)

P = 66 g/mol

Gas Problems

2. What is the volume of 3.35 mol of gas which has a measured temperature of 47.00C and a pressure of 185 kPa? Use proper sig figs.

PV = nRT

V = ? L n = 3.35 molT = 320 KP = 185 kPaR = 8.314 (L kPa)/(mol K)

(185 kPa) x (V) = (3.35 mol) x (8.314 L kPa/mol K) x (320 K)

V = 48.2 L

Dalton’s Law

The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + ...


In a gaseous mixture, a gas’s partial pressure is the one the gas would exert if it were by itself in the container.

The mole ratio in a mixture of gases determines each gas’s partial pressure.

Gas Mixtures and Dalton’s

Law

Gas Collected Over Water

When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.

GIVEN:

PH2 = ?

Ptotal = 94.4 kPa

PH2O = 2.6 kPa

WORK:

Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.6 kPa

PH2 = 91.8 kPa

Dalton’s Law Hydrogen gas is collected over water at 22°C. Find the

pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

Look up water-vapor pressure on p.10 for 22°C.

The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.


41.7 kPa

Dalton’s Law

The total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4kPa. What is the partial pressure of each gas.

PHe =

PNe =

3 mol He7 mol gas

(97.4 kPa) =

55.7 kPa4 mol Ne7 mol gas

(97.4 kPa) =

?

?

Dalton’s Law

Suppose you are given four containers – three filled with noble gases.The first 1 L container is filled with argon and exerts a pressure of 2 atm.The second 3 liter container is filled with krypton and has a pressure of 380 mm Hg. The third 0.5 L container is filled with xenon and has a pressure of 607.8 kPa. If all these gases were transferred into an empty2 L container…what would be the pressure in the “new” container?

PAr = 2 atm Pxe 607.8 kPa

PKr = 380 mm HgPtotal = ?

V = 1 liter V = 3 liters V = 0.5 liter V = 2 liters

…just add them up

PAr = 2 atm Pxe 607.8 kPa

PKr = 380 mm HgPtotal = ?

V = 1 liter V = 3 liters V = 0.5 liter V = 2 liters

Dalton’s Law of Partial Pressures

“Total Pressure = Sum of the Partial Pressures”PT = PAr + PKr + PXe + …

PT = 1 atm + 0.75 atm + 1.5 atmPT = 3.25 atm

P1 x V1 = P2 x V2 P1 x V1 = P2 x V2

(0.5 atm) (3L) = (X atm) (2L) (6 atm) (0.5 L) = (X atm) (2L)

PKr = 0.75 atm Pxe = 1.5 atm

Partial Pressure

A gas is collected over water at 649 torr and 26.00C. If its volume when collected is 2.99 L, what is its volume at STP? Use proper sig figs.

P1V1/T1 = P2V2/T2 PT = PG + Pw

V2 = ? L V1 = 2.99 LT1 = 299 KT2 = 273 KPT = 649 torrP1 = 86.5 kPa – 3.4 kPa = 83.1 kPa P2 = 101.325 kPa

(83.1 x 2.99) / 299 = (101.325 x V2) / 273

V2 = 2.24 L

Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloricacid. Pressure =107.3 kPa; temperature = 88oC.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

Gas Stoichiometry

Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloricacid. Pressure =107.3 kPa; temperature = 88oC.

Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

38.2 g excess V = ? L H2

P = 107.3 kPaT = 88oC (361 K)

Gas Stoichiometry

22

2 H mol 0.584 Zn mol 1

H mol 1

Zn g 65.4Zn mol 1

Zn g 38.2 H mol X

L 16.3 3.107

(361) (8.314) mol 0.584

PT R n

V T R n V P

At STP, we’d use 22.4 L per mol, but we aren’t at STP.

Pressure and Balloons

A

B = pressure exerted ON balloonA = pressure exerted BY balloon

BWhen balloon is being filled:

PA > PB

When balloon is filled and tied:PA = PB

When balloon deflates:PA < PB

When the balloons are untied,will the large balloon (A) inflatethe small balloon (B); will they end up the same size or will the small balloon inflate the large balloon?

Why?

Balloon Riddle

A

B

C

properties of gases

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