The Kinetic Theory, Pressure

& Gas Laws

The Kinetic Theory, Pressure

& Gas Laws

GASGAS

The four quantities needed to describe a gas:

1) number of particles

2) temperature

3) pressure

4) volume

The Kinetic TheoryThe Kinetic TheoryThe Kinetic Theory explains the effects of

temperature & pressure on matter.• Idea that all particles of matter are constantly in motion

4 Assumptions: of gases for IDEAL GASES4 Assumptions: of gases for IDEAL GASES

1) All gases are composed of small particles and have no volume

2) These small particles move in continual, random, and rapid motion with no attraction or repulsion forces

3) All collisions are perfectly elastic

4) The average kinetic energy of the particles is directly proportional to its Kelvin temperature

Real vs. Ideal GasesAn ideal gas is one that follows the gas laws at all

conditions of pressure and temperature and follows all the assumptions of the KMT

An ideal gas does not really exist!!!!!

Real gases can be liquefied and sometimes solidifiedThe behavior of real gases under many conditions is

similar to ideal gases. Each gas is different.Real gases behave like ideal gases at most conditions

except for extremely high pressures and low temperatures

Kinetic Theory con’tKinetic Theory con’t

The physical behavior of a gas depends on its volume, temperature, and pressure.

TemperatureTemperature: Kelvin: Kelvin

the average kinetic energy of the particles in an object

ie. 25oC, O2 molecules 443 m/s; 1700 km/h; 1057 mi/h

Absolute ZeroAbsolute Zero::

temperature at which all molecular motion should cease; lowest possible temperature-273.15oC or 0K

Kinetic Theory con’t

Temperature is not the total amount of thermal energy a substance has absorbed.

Question 1:

How is the average kinetic energy of water molecules affected when hot water from a kettle is poured into cups at the same temperature as the water?

IT IS UNAFFECTEDIT IS UNAFFECTED

Kinetic Theory con’tQuestion 2:

By what factor does the average kinetic energy of the molecules of a gas in an aerosol container increase when the temperature is raised from 300K (27oC) to 900K?

The average kinetic energy triplesThe average kinetic energy triples

PressurePressurePressurePressure::

the force per unit area 1 object exerts on another

SI Unit – Pascal (Pa)

Gas PressureGas Pressure::

the pressure exerted by a the pressure exerted by a gasgas

it is created by it is created by collisionscollisions of gas particles with an of gas particles with an objectobject

What happens to the pressure when you add more gas particles without changing its volume or temperature? Why?

What happens to the pressure when you decrease the volume of the container for a fixed mass of a gas at a constant temperature? Why?

What must happen to a gas’s temperature in order for its pressure to decrease? Why?

PressurePressurePascalPascal: (Pa): (Pa)

• the SI unit for pressure

• equivalent to 1N/m2

Millimeter of MercuryMillimeter of Mercury: (mmHg): (mmHg)

pressure needed to support a column of mercury 1 mm high

Pressure con’t

AtmosphereAtmosphere: (atm): (atm)

the pressure required to support 760 mm of mercury in a mercury barometer

TorrTorr: (torr): (torr)

named after Evangelista Torricelli who invented the barometer

BarometerBarometer::

a closed-arm manometer used to measure pressure

Pressure con’t

1 atm = 760 mmHg 1 atm = 760 mmHg

1 atm = 101.325 kPa1 atm = 101.325 kPa

760 mmHg = 101.325 kPa760 mmHg = 101.325 kPa

1 torr = 1 mmHg1 torr = 1 mmHg

1 atm = 760 torr1 atm = 760 torr

1 atm = 760 mmHg = 760 torr = 101.325 kPa

Question 3: Convert the following:

a. 4.328 atm to kPa b. 328 kPa to mmHg

c. 3290 Pa to atm

4.328 atm x

= 438.5 kPa

328 kPa x

= 2460 mmHg

3290 Pa x

= 0.0325 atm

101.325 kPa 1.0 atm

760 mmHg 101.325 kPa

1.0 atm 101325 Pa

manometer: a devise used to measure pressure of a gas

Two Types of Manometers1) closed-arm:

used to measure the actual or “absolute” gas pressure

*known as a barometer

2) open-arm manometer:one arm of the manometer is open to the

atmospheric air to measure the gas pressure of a confined gas

Avogadro’s HypothesisAvogadro’s Hypothesis

Equal volumes of gases at the same temperature and pressure contain equal numbers of particles

States of MatterStates of Matter

1)1) GASES:GASES:

are independent of one another moving in straight lines until they collide with something that affects its direction and possibly its speed

have no definite shape or volume

gases assume the shape of the container

2) Liquids:2) Liquids:

have a definite volume but will take the shape of their container

particles are attracted to one another but have enough energy to slide past each other

no bond is formed between the particles

3) Solids:3) Solids:particles possess relatively fixed positions and vibrate around that fixed point

attractive forces hold the particles extremely close but the particles of the solid are still traveling in straight paths between colliding with its exceedingly close neighboring particles

the physical state of a substance at STP depends on the attractive forces verses the energy of the particles

Ionic compounds:tend to be solids with strong electric charges

Molecular compoundsare attracted by van der Waals forces

high molecular mass compounds tend to be solids

nonpolar molecules of low molecular mass tend to be gases

*the greater the mass and polarity, the more likely the compound will be a solid or liquid

4) Plasma:4) Plasma:

matter at temperatures greater than 5000oC causing a state where the matter is composed of electrons and positive ions

magnetohydrodynamics

VaporizationVaporizationVaporizationVaporization::

the conversion of a liquid to a gas or vapor below its boiling point (bp)

EvaporationEvaporation::

the vaporization of an uncontained liquid

Vapor PressureVapor Pressure::

pressure created by a vapor in equilibrium with its liquid

Vaporization con’t

Boiling PointBoiling Point: (bp): (bp)

temperature at which the vapor pressure of the liquid is equal to the external pressure

Normal Boiling PointNormal Boiling Point::

boiling point for a substance at 1 atm

* Boiling is a cooling process

Vaporization con’t

evaporation

Liquid Vapor (gas) condensation

Heat of Fusion:

the additional amount of energy required to cause a phase change from a solid to liquid once a substance reaches its melting/freezing point

Heat of Vaporization:

the additional amount of energy required to cause a phase change from a gas to liquid once a substance reaches its boiling/condensation point

Phase Changes

Phase Diagram

Triple Point:

temperature and pressure where all three phases are at equilibrium

Critical Point

at the critical temperature and critical pressure of the substance, beyond this point the liquid and gas phases become indistinguishable

JOHN DALTON

DALTON’S LAW OF PARTIAL DALTON’S LAW OF PARTIAL PRESSURESPRESSURES

At a constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the gases

PARTIAL PRESSURE:

the pressure(contribution) each gas in a mixture makes to the total pressure of the mixture

PPtotaltotal = P = P11 + P + P22 + P + P33 … …

Example:

Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other elements. What is the partial pressure of O2 at 1.00 atm if PN2 = 593.4 mmHg, PCO2 = 0.3 mmHg and Ptrace = 7.1 mmHg?

PPtotaltotal = P = PO2O2 + P + PCO2CO2 + P + PN2N2 + P + Ptracetrace

PPO2O2 = P = Ptotaltotal - (P - (PCO2CO2 + P + PN2N2 + P + Ptrtr))

= 760mmHg - (593.4 mmHg + 0.3 mmHg + 7.1 mmHg)= 760mmHg - (593.4 mmHg + 0.3 mmHg + 7.1 mmHg)

= = 159 mmHg159 mmHg

Your Turn:

Determine the total pressure of a mixture of gases if the partial pressures of the gases are PO2 = 242.5 mmHg, PHe

= 27.3 kPa, and PNe = 0.021 atm.

PT = PO2 + PHe + PNe

Look for unit agreement!!!!

We will use mmHg as the common unit for our answer

= 27.3 kPa X 760 mmHg = 205 mmHg

101.325 kPa

= 0.021 atm X 760 mmHg = 16 mmHg

1 atm

PT = 242.5 mmHg + 205 mmHg + 16 mmHg

464 mmHg

Robert BoyleRobert Boyle

When the number of particles and temperature are constant, pressure and volume are inversely proportional. (opposite)

PP11 V V11 = P = P22 V V22

PP11 = =

initial pressure

VV11 = =

initial volume

PP22 = =

final pressure

VV22 = =

final volume

Example:A balloon is filled with 30.0 L of helium gas at 1.00 atm.

What is the volume when the balloon rises to an altitude where the pressure is only 185.3 mmHg? Assume temperature remains constant.

30.0 L = V1

1.00 atm = P1

185.3 mmHg = P2

? L = V2

Look for unit agreement

1.00 atm = 760 mmHg

P1V1 = P2V2

Divide both sides by P2

V2 =

V2 = (760 mmHg)(30.0 L) = 185.3 mmHg

= 123 L

P1V1

P2

Jacques CharlesJacques Charles

Charles’ LawCharles’ Lawstates that the volume of a fixed mass of gas is directly

proportional to its Kelvin temperature if the pressure is kept constant

V1 = V2 T1 T2

*temperature must always be in Kelvin

A 225 cm3 volume of gas is collected at 58.0oC. What volume, in liters, would this sample of gas occupy at standard temperature?

V1 = 225 cm3

V2 = ?

T1 = 58.0oC

T2 = 0.0oC

Charles’ Law

: 273.15 + 58.0oC = 331.2K

: 273.15 + 0.0oC = 273.2K

V2 = V1T2

T1

= (225 cm3)(273.2 K)

331.2 K

= 186 cm3 · 1 L =

1000 cm3

= 0.186 L

Your Turn:

A balloon, inflated in an air-conditioned room at 27.0oC, has a volume of 4.0 L. It is heated to a temperature of 57.3oC. What is the new volume of the balloon if the pressure remains constant?

T1 = 27.0oC

V1 = 4.0 L

T2 = 57.3oC

V2 = ????

Convert all temperatures to KelvinConvert all temperatures to Kelvin

V1 = V2 T1 T2

V2 = V1 T2 T1

V2 = (4.0 L)(330.5K) 300.2 K

= 4.4 L

Joseph Louis Gay-Lussac

Gay-Lussac’s LawGay-Lussac’s Law

the pressure of a given mass of gas is directly proportional to the Kelvin temperature if the volume is held constant

P1 = P2

T1 T2

An acetylene gas cylinder has a pressure of 24350 mmHg at a temperature of 19.49oC. What would the internal pressure be if the temperature was increased to 100.0oC? Assume no change the volume of the cylinder.

P1 = 24350 mmHg

T1 = 19.49oC + 273.15 = 292.64 K

P2 = X

T2 = 100.00oC + 273.15 = 373.15K

P1 = P2

T1 T2

P2 = P2T1

T2 X = (24350 mmHg)(373.15 K)

294.64 K

24350 mmHg = X 292.64 K 373.15K

X = 31050 mmHgX = 31050 mmHg

Your Turn:

A gas has a pressure of 50.0 atm at 540 K. What will the temperature, in Celsius, be if the pressure is increased to 8330 kPa?

P1 = 50.0 atm

T1 = 540 K

P2 = 8330 kPa

T2 = X

Unit agreement must be kept. Choose atm or kPa

50.0 atm x 101.325 kPa = 1.00 atm

= 5070 kPa

5070 kPa = 8330 kPa 540 K X

890 K

oC = 890 – 273 = 617oC

Combined Gas LawCombined Gas LawCombines Boyle’s, Charles’ and Gay-Lussac’s laws

together

P1V1 = P2V2 T1 T2

all three laws can be derived from the combined gas law by removing the variable with a constant value

The volume of a gas measured at 75.6 kPa pressure and 60.0oC is to be corrected to correspond to the volume it would occupy at STP. The measured volume of the gas is 10.0 cm3.

P1 = 75.6 kPa

V1 = 10.0 cm3

T1 = 60.0oC = 333.2 K

P2 = 101.325 kPa

T2 = 0.0oC = 273.2 K

V2 = ????? cm3

P1V1 = P2V2 T1 T2

V2 = P1V1T2 P2 T1

V2 = (75.6 kPa)(273.2 K)(10.0 cm3) (101.325 kPa)(333.2 K)

= 6.12 cm3

Your Turn:A cylinder of compressed oxygen gas has a volume of

30.0 L and 100.0 atm pressure at 300.0 K. The cylinder is cooled until the pressure is 5.00 atm. What is the new temperature, in Celsius, of the gas in the cylinder?

V1 = 30.0 L

P1 = 100.0 atm

T1 = 27.0oC = 300.2 K

V2 = 30.0 L

P2 = 5.00 atm

T2 = ???? K

TT22 = = PP22TT11

P P11

TT22 = = (5.00 atm)(300.2 K)(5.00 atm)(300.2 K)

100.0 atm 100.0 atm

= 15.0 K= 15.0 K

15.0 K = 273.15 + 15.0 K = 273.15 + ooCC

ooC = C = - - 258.2 258.2

IDEAL GAS LAW

AVOGADRO’S HYPOTHESIS:

gases at the same temperature, gases at the same temperature, pressure and volume contain the same pressure and volume contain the same

number of particlesnumber of particles

IDEAL GAS LAW IDEAL GAS LAW

PV = nRT PerVNeRT

P = PressureV = Volume in liters or dmliters or dm33

n = molesR = ideal gas constant

determined by the unit of pressureT = temperature in Kelvin

R = 0.0821 atm·L

mol·K

= 8.31 kPakPa·L

mol·K

= 62.4 mmHgmmHg·L

mol·K

Ideal Gas LawA rigid steel cylinder with a volume of 20.0 L is

filled with nitrogen gas to a final pressure of 200.0 atm at 27.0oC. How many moles of nitrogen gas does the cylinder contain?

V = 20.0 LP = 200.0 atmT = 27.0oC + 273.2 = 300.2Kn = ? moln = ? molCAN ONLY BE PV=nRTR=?R = 0.0821 atm·L/K·mol

PV = nRTPV = nRT

n = n = PVPV

RTRT

= = (200.0 atm)(20.0 L)(200.0 atm)(20.0 L)

(0.0821 (0.0821 atmatm· L· L)(300.2 K))(300.2 K)

mol· Kmol· K

all units cancel except molall units cancel except mol

= = 162 mol162 mol

Your turn:

Determine the pressure that 453.38 g of oxygen gas would exert if it is put into a container with a volume of 25.0 mL and a temperature of 17.4oC.

m = 453.38 g O2

V = 25.0 mL

T = 17.4oC

P = ?

Must be PV=nRT

LOOK FOR UNIT AGREEMENTLOOK FOR UNIT AGREEMENT

n = 453.38 g O2 X 1 mol O2 = 14.2 mol

31.9988 g O2

V = 25.0 mL = 0.0250 L

T = 17.4oC + 273.2 = 290.6 K

P = ? atm

R = 0.0821 L atm/K mol

P = nRT

V

P = (14.2 mol)(0.0821 L atm/K mol)(290.6K)

0.0250 L

P = 13600 atmP = 13600 atm

Graham’s Law of EffusionDiffusion:

the tendency of atoms, ions or molecules to move toward areas of lower concentration until there is a uniform composition

Effusion:

occurs as a gas escapes through a tiny hole in a container

Graham’s Law

the rate of effusion of a gas in inversely proportional to the square root of its molar mass

RateA = molar mass B RateB molar mass A

Example

Which gas effuses faster, carbon dioxide or neon? By how much faster?

Remember, lighterlighter moves faster

CO2 = 44.0098 g/mol

Ne = 20.179 g/mol

Ne is faster because its molar mass is smaller

RateNe = 44.0098 g/mol RateCO2 20.179 g/mol

= 1.477

neon gas is 1.477 times faster than carbon dioxide

Example 2Determine the molar mass for a gas that is 0.6372

times as fast as oxygen gas.

Rateunknown = Molar mass oxygen Rateoxygen molar mass

unknown

RateRateunknown =unknown = 0.6372 0.6372

RateRateoxygenoxygen

0.6372 = 0.6372 = 31.9988 g/mol31.9988 g/mol XX

0.4060 = 0.4060 = 31.9988 g/mol31.9988 g/mol XX

X = X = 78.81 g/mol78.81 g/mol

2222

14.83 g zinc reacts with an excess of hydrochloric acid, HCl, in the production of hydrogen gas. Calculate the pressure, in kPa, of the gas collected if it has a volume of 2.150 L at a temperature of 25.0oC. Assume 100% yield.

Zn + HCl → ZnCl2 + H2

mass Zn = 14.83 g

PH2 = ? kPa

VH2 = 2.150 L

TH2 = 25.0oC + 273.2 = 298.2K

All means Gas Law Stoichiometry

22

Zn + 2HCl → ZnCl2 + H2

14.83 g Zn X 1mol Zn X 1 mol H2 =

65.39 g Zn 1 mol Zn

= 0.2268 mol H2

P = nRT

V

P = (0.2268 mol)(8.31 kPa·L/K·mol)(298.2 K) 2.150 L

P = 261.4 kPaP = 261.4 kPa

Acetylene, C2H2, is a gas used in cutting metal. If 1.70 x 103 g of acetylene gas undergoes a complete combustion reaction, calculate the mass of oxygen gas that would be required to react with all the acetylene gas. Second, calculate the pressure of the oxygen gas if the container the acetylene is in has a volume of 55.00 L and a temperature of 84.56oC.

C2H2 + O2 → CO2 + H2O2222 4455

1.70 x 103 g C2H2 X1 mol C2H2 X

26.0379 g C2H2

5 mol O2 =

2 mol C2H2

163 mol O2 X

1 mol O2

31.998 g O2 =

5220 g O5220 g O22

n = 163 mol

V = 55.00 L

T = 84.56 + 273.15 = 357.71 K

P = (163 mol)(0.0821 L· atm/ K·mol)(357.71 K)

55.00 L

87.0 atm87.0 atm

Calculate the mass of nitrogen gas required to react hydrogen gas with a pressure of 356.33 kPa, a temperature of 288.3 K and a volume of 377.4 mL in the production of ammonia.

NN22 + H + H22 → NH→ NH33

nnH2H2 = PV = PV

RTRT

(356.33 kPa)(0.3774 L)

(8.31 kPa·L)(288.3 K)

K·mol

= 0.05613 mol H0.05613 mol H22

33 22

= 0.05613 mol H2 X X 1 mol N2 X

3 mol H2

28.0134 g N2 =

1 mol N2

= 0.5241 g N= 0.5241 g N2 2