1

Chapter 5Chapter 5

The Gas LawsThe Gas Laws

2

PressurePressure Force per unit area.Force per unit area. Gas molecules fill container.Gas molecules fill container. Molecules move around and hit Molecules move around and hit

sides.sides. Collisions are the force.Collisions are the force. Container has the area.Container has the area. Measured with a barometer.Measured with a barometer.

3

BarometerBarometer The pressure of the The pressure of the

atmosphere at sea atmosphere at sea level will hold a level will hold a column of mercury column of mercury 760 mm Hg.760 mm Hg.

1 atm = 760 mm Hg1 atm = 760 mm Hg

1 atm Pressure

760 mm Hg

Vacuum

4

ManometerManometer

Gas

h

Column of Column of mercury to mercury to measure measure pressure.pressure.

h is how much h is how much lower the lower the pressure is pressure is than outside. than outside.

5

ManometerManometer h is how much h is how much

higher the gas higher the gas pressure is than pressure is than the atmosphere.the atmosphere.

h

Gas

6

Units of pressureUnits of pressure 1 atmosphere = 760 mm Hg1 atmosphere = 760 mm Hg 1 mm Hg = 1 torr1 mm Hg = 1 torr 1 atm = 101,325 Pascals = 101.325 kPa1 atm = 101,325 Pascals = 101.325 kPa Can make conversion factors from Can make conversion factors from

these.these. What is 724 mm Hg in kPa?What is 724 mm Hg in kPa? in torr?in torr? in atm?in atm?

7

The Gas LawsThe Gas Laws Boyle’s LawBoyle’s Law Pressure and volume are inversely Pressure and volume are inversely

related at constant temperature.related at constant temperature. PV= kPV= k As one goes up, the other goes As one goes up, the other goes

down.down. PP11VV11 = P = P22 V V22

GraphicallyGraphically

8

V

P (at constant T)

9

V

1/P (at constant T)

Slope = k

10

PV

P (at constant T)

CO2

O2

22.4

1 L

atm

11

ExamplesExamples 20.5 L of nitrogen at 25ºC and 742 20.5 L of nitrogen at 25ºC and 742

torr are compressed to 9.8 atm at torr are compressed to 9.8 atm at constant T. What is the new volume?constant T. What is the new volume?

30.6 mL of carbon dioxide at 740 torr 30.6 mL of carbon dioxide at 740 torr is expanded at constant temperature is expanded at constant temperature to 750 mL. What is the final pressure to 750 mL. What is the final pressure in kPa? in kPa?

12

Charle’s LawCharle’s Law Volume of a gas varies directly with Volume of a gas varies directly with

the absolute temperature at constant the absolute temperature at constant pressure.pressure.

V = kT (if T is in Kelvin)V = kT (if T is in Kelvin)

VV1 1 = V = V22

T T11 = T = T22

GraphicallyGraphically

13

V (

L)

T (ºC)

He

H2O

CH4

H2

-273.15ºC

14

ExamplesExamples What would the final volume be if 247 What would the final volume be if 247

mL of gas at 22ºC is heated to 98ºC , mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant? if the pressure is held constant?

At what temperature would 40.5 L of At what temperature would 40.5 L of gas at 23.4ºC have a volume of 81.0 gas at 23.4ºC have a volume of 81.0 L at constant pressure? L at constant pressure?

15

Avogadro's LawAvogadro's Law Avagadro’sAvagadro’s At constant temperature and At constant temperature and

pressure, the volume of gas is directly pressure, the volume of gas is directly related to the number of moles.related to the number of moles.

V = k n (n is the number of moles)V = k n (n is the number of moles)

VV1 1 = V = V22

n n11 = n = n22

16

ExampleExample A 5.20 L sample at 18.0 C and 2.00 A 5.20 L sample at 18.0 C and 2.00

atm pressure contains 0.436 moles of atm pressure contains 0.436 moles of gas. If we add an additional 1.27 gas. If we add an additional 1.27 moles of the gas at the same moles of the gas at the same temperature and pressure, what will temperature and pressure, what will the total volume occupied by the gas the total volume occupied by the gas be?be?

17

Gay- Lussac LawGay- Lussac Law At constant volume, pressure and At constant volume, pressure and

absolute temperature are directly absolute temperature are directly related.related.

P = k TP = k T

PP1 1 = P = P22

T T11 = T = T22

18

Combined Gas LawCombined Gas Law If the moles of gas remains constant, If the moles of gas remains constant,

use this formula and cancel out the use this formula and cancel out the other things that don’t change.other things that don’t change.

PP1 1 VV11 = P = P22 V V22

.. T T11 T T22

19

ExamplesExamples A deodorant can has a volume of 175 mL A deodorant can has a volume of 175 mL

and a pressure of 3.8 atm at 22ºC. What and a pressure of 3.8 atm at 22ºC. What would the pressure be if the can was would the pressure be if the can was heated to 100.ºC?heated to 100.ºC?

What volume of gas could the can release What volume of gas could the can release at 22ºC and 743 torr?at 22ºC and 743 torr?

A sample of gas has a volume of 4.18 L at A sample of gas has a volume of 4.18 L at 29 C and 732 torr. What would its volume 29 C and 732 torr. What would its volume be at 24.8 C and 756 torr?be at 24.8 C and 756 torr?

20

Kinetic Molecular TheoryKinetic Molecular Theory Theory tells why the things happen.Theory tells why the things happen. explains why ideal gases behave the explains why ideal gases behave the

way they do.way they do. Assumptions that simplify the Assumptions that simplify the

theory, but don’t work in real gases.theory, but don’t work in real gases. The particles are so small we can The particles are so small we can

ignore their volume.ignore their volume. The particles are in constant motion The particles are in constant motion

and their collisions cause pressure. and their collisions cause pressure.

21

Kinetic Molecular TheoryKinetic Molecular Theory The particles do not affect each The particles do not affect each

other, neither attracting or repelling.other, neither attracting or repelling. The average kinetic energy is The average kinetic energy is

proportional to the Kelvin proportional to the Kelvin temperature.temperature.

22

Ideal Gas LawIdeal Gas Law PV = nRTPV = nRT V = 22.41 L at 1 atm, 0ºC, n = 1 mole, V = 22.41 L at 1 atm, 0ºC, n = 1 mole,

what is R?what is R? R is the ideal gas constant.R is the ideal gas constant. R = 0.0821 L atm/ mol KR = 0.0821 L atm/ mol K Tells you about a gas is NOW.Tells you about a gas is NOW. The other laws tell you about a gas The other laws tell you about a gas

when it changes. when it changes.

23

Ideal Gas LawIdeal Gas Law An An equation of stateequation of state.. Independent of how you end up Independent of how you end up

where you are at. Does not depend where you are at. Does not depend on the path.on the path.

Given 3 you can determine the Given 3 you can determine the fourth.fourth.

An Empirical Equation - based on An Empirical Equation - based on experimental evidence.experimental evidence.

24

Ideal Gas LawIdeal Gas Law A hypothetical substance - the ideal A hypothetical substance - the ideal

gasgas Think of it as a limit.Think of it as a limit. Gases only approach ideal behavior Gases only approach ideal behavior

at at low pressure low pressure (< 1 atm) and (< 1 atm) and high high temperature.temperature.

25

ExamplesExamples A 47.3 L container containing 1.62 mol of He A 47.3 L container containing 1.62 mol of He

is heated until the pressure reaches 1.85 atm. is heated until the pressure reaches 1.85 atm. What is the temperature?What is the temperature?

Kr gas in a 18.5 L cylinder exerts a pressure of Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is the mass of Kr?8.61 atm at 24.8ºC What is the mass of Kr?

What volume will 1.18 moles of OWhat volume will 1.18 moles of O22 occupy at occupy at

STP?STP?

26

Let’s try this one!Let’s try this one! A sample containing 15.0 g of dry ice A sample containing 15.0 g of dry ice

(CO(CO22 (s)), is put into a balloon and (s)), is put into a balloon and

allowed to sublime according to the allowed to sublime according to the equation:equation:

COCO22 (s) → CO (s) → CO22(g)(g)

How big will the balloon be at 22.0°C How big will the balloon be at 22.0°C and 1.04 atm?and 1.04 atm?

27

Gas Density and Molar MassGas Density and Molar Mass D = m/VD = m/V Let Let MM stand for molar mass stand for molar mass MM = m/n = m/n n= PV/RTn= PV/RT MM = m = m

PV/RT PV/RT MM = mRT = m RT = DRT = mRT = m RT = DRT

PV PV V P V P P P

28

Examples Examples A gas at 34.0 °C and 1.75 atm has a A gas at 34.0 °C and 1.75 atm has a

density of 3.40 g/L. Calculate the density of 3.40 g/L. Calculate the molar mass of the gas.molar mass of the gas.

What is the density of ammonia at What is the density of ammonia at 23ºC and 735 torr?23ºC and 735 torr?

29

Gases and StoichiometryGases and Stoichiometry Reactions happen in molesReactions happen in moles At Standard Temperature and At Standard Temperature and

Pressure Pressure (STP, 0ºC and 1 atm): (STP, 0ºC and 1 atm):

1 mole of any gas takes up 22.42 L of 1 mole of any gas takes up 22.42 L of space.space.

If not at STP, use the ideal gas law to If not at STP, use the ideal gas law to calculate moles of reactant or calculate moles of reactant or volume of product.volume of product.

30

ExamplesExamples Mercury can be achieved by the Mercury can be achieved by the

following reactionfollowing reaction

What volume of oxygen gas can be What volume of oxygen gas can be produced from 4.10 g of mercury (II) produced from 4.10 g of mercury (II) oxide at STP?oxide at STP?

At 400.ºC and 740 torr?At 400.ºC and 740 torr?

HgO Hg(l) + O (g) heat

2

31

ExamplesExamples Using the following reactionUsing the following reaction

calculate the mass of sodium hydrogen calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L carbonate necessary to produce 2.87 L of carbon dioxide at 25ºC and 2.00 atm.of carbon dioxide at 25ºC and 2.00 atm.

If 27 L of gas are produced at 26ºC and If 27 L of gas are produced at 26ºC and 745 torr when 2.6 L of HCl are added 745 torr when 2.6 L of HCl are added what is the concentration of HCl?what is the concentration of HCl?

NaCl(aq) + CO (g) +H O(l)2 2

NaHCO (s) + HCl 3

32

ExamplesExamples Consider the following reactionConsider the following reaction

What volume of NO at 1.0 atm and What volume of NO at 1.0 atm and 1000ºC can be produced from 10.0 L 1000ºC can be produced from 10.0 L of NHof NH33 and excess O and excess O22 at the same at the same

temperture and pressure?temperture and pressure? What volume of OWhat volume of O22 measured at STP measured at STP

will be consumed when 10.0 kg NHwill be consumed when 10.0 kg NH33 is is

reacted?reacted?

4NH (g) + 5 O 4 NO(g) + 6H O(g)3 22 ( )g

33

The Same reactionThe Same reaction

What mass of HWhat mass of H22O will be produced O will be produced

from 65.0 L of Ofrom 65.0 L of O22 and 75.0 L of NH and 75.0 L of NH33

both measured at STP? both measured at STP? What volume Of NO would be What volume Of NO would be

produced?produced? What mass of NO is produced from What mass of NO is produced from

500. L of NH3 at 250.0ºC and 3.00 atm?500. L of NH3 at 250.0ºC and 3.00 atm?

4NH (g) + 5 O 4 NO(g) + 6H O(g)3 22 ( )g

34

Dalton’s LawDalton’s Law The total pressure in a container is The total pressure in a container is

the sum of the pressure each gas the sum of the pressure each gas would exert if it were alone in the would exert if it were alone in the container.container.

The total pressure is the sum of the The total pressure is the sum of the partial pressures.partial pressures.

PPTotalTotal = P = P11 + P + P22 + P + P33 + P + P44 + P + P55 ... ...

For each P = nRT/VFor each P = nRT/V

35

Dalton's LawDalton's Law PPTotalTotal = n = n11RT + nRT + n22RT + nRT + n33RT +...RT +...

V V V V V V In the same container R, T and V are the In the same container R, T and V are the

same.same.

PPTotalTotal = (n = (n11+ n+ n22 + n + n33+...)RT+...)RT

V V

PPTotalTotal = (n = (nTotalTotal)RT)RT

V V

36

The mole fractionThe mole fraction Ratio of moles of the substance to Ratio of moles of the substance to

the total moles.the total moles.

symbol is Greek letter chi symbol is Greek letter chi

= n= n11 = P= P1 1

n nTotal Total PPTotalTotal

37

ExamplesExamples A volume of 2.0L of He at 46 °C and 1.2 atm A volume of 2.0L of He at 46 °C and 1.2 atm

pressure was added to a vessel that pressure was added to a vessel that contained 4.5L of Ncontained 4.5L of N22 at STP. What is the at STP. What is the

total pressure and the partial pressure of total pressure and the partial pressure of each gas at STP after the He was added?each gas at STP after the He was added?

Calculate the # of moles of NCalculate the # of moles of N22..

Calculate the mole fractions of NCalculate the mole fractions of N22 and He, and He,

using the mole data, then the pressure data.using the mole data, then the pressure data.

38

ExamplesExamples

3.50 L

O2

1.50 L

N2

2.70 atm When these valves are opened, what is each When these valves are opened, what is each

partial pressure and the total pressure? partial pressure and the total pressure? (hint: what (hint: what happens to the volume when the valves are opened)happens to the volume when the valves are opened)

4.00 L

CH4

4.58 atm 0.752 atm

39

Vapor PressureVapor Pressure Water evaporates!Water evaporates! When that water evaporates, the When that water evaporates, the

vapor has a pressure.vapor has a pressure. Gases are often collected over water Gases are often collected over water

so the vapor. pressure of water must so the vapor. pressure of water must be subtracted from the total be subtracted from the total pressure.pressure.

It must be given.It must be given.

40

ExampleExample NN22O can be produced by the following O can be produced by the following

reactionreaction

What volume of NWhat volume of N22O collected over O collected over

water at a total pressure of 94 kPa and water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g of 22ºC can be produced from 2.6 g of NHNH44NONO33? ( the vapor pressure of ? ( the vapor pressure of

water at 22ºC is 21 torr)water at 22ºC is 21 torr)

NH NO NO (g) + 2H O4 heat

23 2( ) ( )s l

41

What MKT tells usWhat MKT tells us Applying the laws of physics, the Applying the laws of physics, the

expression expression (KE) = ½ mu(KE) = ½ mu2 2 represents represents the average KE of a gas particlethe average KE of a gas particle

Using this idea and applying it to the Using this idea and applying it to the ideal gas law (see appendix 2 for ideal gas law (see appendix 2 for derivation) we arrive at a very derivation) we arrive at a very important relationship:important relationship:

(KE)(KE)avgavg = 3/2 RT = 3/2 RT This the meaning of the Kelvin This the meaning of the Kelvin

temperature of a gas.temperature of a gas.

42

Root Mean Square VelocityRoot Mean Square Velocity

u represents the average particle u represents the average particle velocity.velocity.

____ u u 22 is the average particle velocity is the average particle velocity

squared.squared.

The root mean square velocity is The root mean square velocity is

u u 2 = 2 = uurmsrms

43

Combine these two equationsCombine these two equations (KE)(KE)avgavg = N = NAA(1/2 mu (1/2 mu 22 ) )

(KE)(KE)avgavg = 3/2 RT = 3/2 RT

Where M is the molar mass in kg/mole, Where M is the molar mass in kg/mole, and R has the units 8.3145 J/Kmol.and R has the units 8.3145 J/Kmol.

The velocity will be in m/sThe velocity will be in m/s

u = 3RT

Mrms

44

Example Example Calculate the root mean square velocity of Calculate the root mean square velocity of

Helium at 25ºCHelium at 25ºC What do we know?What do we know?– T= 25 C + 273 = 298 KT= 25 C + 273 = 298 K– R= 8.314 J/K molR= 8.314 J/K mol

What information do we need?What information do we need?– What is the mass of a mole of He in Kg?What is the mass of a mole of He in Kg?

Now plug into the formulaNow plug into the formula Since the units of J are Kg m2/s2, the resulting Since the units of J are Kg m2/s2, the resulting

units are appropriate for velocity!units are appropriate for velocity!

45

Calculate the root mean square Calculate the root mean square velocity of Carbon dioxide at 25ºC.velocity of Carbon dioxide at 25ºC.

Calculate the root mean square Calculate the root mean square velocity of chlorine at 25ºC.velocity of chlorine at 25ºC.

46

Range of velocitiesRange of velocities The average distance a molecule The average distance a molecule

travels before colliding with another travels before colliding with another is called the is called the mean free path mean free path and is and is small (near 10small (near 10-7-7))

Temperature is an average. There are Temperature is an average. There are molecules of many speeds in the molecules of many speeds in the average.average.

Shown on a graph called a velocity Shown on a graph called a velocity distributiondistribution

47

num

ber

of p

arti

cles

Molecular Velocity

273 K

48

num

ber

of p

arti

cles

Molecular Velocity

273 K

1273 K

49

num

ber

of p

arti

cles

Molecular Velocity

273 K

1273 K

1273 K

50

VelocityVelocity What happens as the temperature What happens as the temperature

increases?increases? The average velocity increases as The average velocity increases as

temperature increases.temperature increases. The spread of velocities increases as The spread of velocities increases as

temperature increases.temperature increases.

51

EffusionEffusion Passage of gas through a small hole, Passage of gas through a small hole,

into a vacuum.into a vacuum. The effusion rate measures how fast The effusion rate measures how fast

this happens.this happens. Graham’s Law the rate of effusion is Graham’s Law the rate of effusion is

inversely proportional to the square inversely proportional to the square root of the mass of its particles.root of the mass of its particles.

Rate of effusion for gas 1

Rate of effusion for gas 2

M

M

2

1

52

DerivingDeriving The rate of effusion should be The rate of effusion should be

proportional to uproportional to urmsrms

Effusion Rate 1 = uEffusion Rate 1 = urms rms 11

Effusion Rate 2 = u Effusion Rate 2 = urms rms 22

effusion rate 1

effusion rate 2

u 1

u 2

3RT

M

3RT

M2

M

Mrms

rms

1 2

1

53

DiffusionDiffusion The spreading of a gas through a room.The spreading of a gas through a room. Slow considering molecules move at 100’s of Slow considering molecules move at 100’s of

meters per second.meters per second. Collisions with other molecules slow down Collisions with other molecules slow down

diffusions.diffusions. Best estimate is Graham’s Law.Best estimate is Graham’s Law.

54

ExamplesExamples A compound effuses through a porous A compound effuses through a porous

cylinder 3.20 time faster than helium. What is cylinder 3.20 time faster than helium. What is it’s molar mass?it’s molar mass?

If 0.00251 mol of NHIf 0.00251 mol of NH33 effuse through a hole effuse through a hole

in 2.47 min, how much HCl would effuse in in 2.47 min, how much HCl would effuse in the same time?the same time?

A sample of NA sample of N22 effuses through a hole in 38 effuses through a hole in 38

seconds. what must be the molecular weight seconds. what must be the molecular weight of gas that effuses in 55 seconds under of gas that effuses in 55 seconds under identical conditions? identical conditions?

55

Real GasesReal Gases Real molecules do take up space and Real molecules do take up space and

they do interact with each other they do interact with each other (especially polar molecules).(especially polar molecules).

Need to add correction factors to the Need to add correction factors to the ideal gas law to account for these.ideal gas law to account for these.

56

Volume CorrectionVolume Correction The actual volume free to move in is less The actual volume free to move in is less

because of particle size.because of particle size. More molecules will have more effect.More molecules will have more effect. Corrected volume V’ = V - nbCorrected volume V’ = V - nb b is a constant that differs for each gas.b is a constant that differs for each gas.

P’ = P’ = nRTnRT

(V-nb) (V-nb)

57

Pressure correctionPressure correction Because the molecules are attracted Because the molecules are attracted

to each other, the pressure on the to each other, the pressure on the container will be less than idealcontainer will be less than ideal

depends on the number of molecules depends on the number of molecules per liter.per liter.

since two molecules interact, the since two molecules interact, the effect must be squared.effect must be squared.

Pobserved = P’ - a

2

( )Vn

58

AltogetherAltogether PPobsobs= nRT - a n = nRT - a n 22

V-nb VV-nb V

Called the Van der Wall’s equation if Called the Van der Wall’s equation if

rearrangedrearranged

Corrected Corrected Corrected Corrected Pressure Pressure Volume Volume

( )

P + an

V x V - nb nRTobs

2

59

Where does it come fromWhere does it come from a and b are determined by a and b are determined by

experiment.experiment. Different for each gas.Different for each gas. Bigger molecules have larger “b”.Bigger molecules have larger “b”. ““a” depends on both size and a” depends on both size and

polarity.polarity. once given, plug and chug.once given, plug and chug.

60

ExampleExample Calculate the pressure exerted by Calculate the pressure exerted by

0.5000 mol Cl0.5000 mol Cl22 in a 1.000 L container in a 1.000 L container

at 25.0ºCat 25.0ºC Using the ideal gas law.Using the ideal gas law. Van der Waal’s equationVan der Waal’s equation

– a = 6.49 atm La = 6.49 atm L22 /mol /mol22

– b = 0.0562 L/molb = 0.0562 L/mol

61

Let’s ReviewLet’s Review Chemmybear.comChemmybear.com