Chapter 5Chapter 5

GasesGases

Properties of a gasProperties of a gas

- Uniformly fills any containerUniformly fills any container..- Mixes completely with any other gasMixes completely with any other gas- Exerts pressure on its surroundingsExerts pressure on its surroundings..- CompressibleCompressible- Gas pressure varies with altitudes and stormsGas pressure varies with altitudes and storms

5.1 Pressure

Measuring atmospheric pressureMeasuring atmospheric pressureTorricellian barometerTorricellian barometer

Torricelli (1608-1647) studied the problem using Torricelli (1608-1647) studied the problem using mercury rather than Hmercury rather than H22O.O.

Mercury is denser than water, so the column wasn’t Mercury is denser than water, so the column wasn’t quite so high.quite so high.

Gas PressureGas Pressure

Liquid Pressure = Liquid Pressure = g ·h ·d

P (Pa) =

Area (m2)

Force (N)

Pascal (SI units)

Pascal and Torricelli

Blaise Pascal(1623-1662)

Evangelista Torricelli(1608-1647)

BarometerBarometer

760760 mmHgmmHg

atmospheric atmospheric

pressurepressure

P = d·g·h

d - densityg - acc. of gravity

h

atmospheric atmospheric

pressurepressure

Units of PressureUnits of Pressure

One atmosphere (1 atm) One atmosphere (1 atm)

Is the average pressure of the atmosphere at Is the average pressure of the atmosphere at

sea level sea level

Is the standard atmospheric pressureIs the standard atmospheric pressure

Standard Atmospheric Pressure:1 atm = 76 cm Hg = 760 mm Hg = 760 Torr =

101,325 PaVery small unit, thus it is not commonly used

ExampleExample

A. What is 475 mm Hg expressed in atm?A. What is 475 mm Hg expressed in atm?

485 mm Hg x 485 mm Hg x 1 atm 1 atm = 0.625 atm = 0.625 atm

760 mm Hg760 mm Hg

B. The pressure of a tire is measured as 29.4 psi.B. The pressure of a tire is measured as 29.4 psi.

What is this pressure in mm Hg?What is this pressure in mm Hg?

29.4 psi x 29.4 psi x 1.00 atm1.00 atm x x 760 mmHg760 mmHg = 1.52 x 10 = 1.52 x 1033 mmHg mmHg

14.7 psi 1.00 atm14.7 psi 1.00 atm

ManometerManometer Device for Measuring the Pressure of a Gas in a ContainerDevice for Measuring the Pressure of a Gas in a Container

5.25.2 The Gas Laws of Boyle, CharlesThe Gas Laws of Boyle, Charles and Avogadro and Avogadro

Boyle’s Law:Boyle’s Law: PV = constPV = const at constant n, Tat constant n, T

Charles’ Law:Charles’ Law: V/T = constV/T = const at constant n, Pat constant n, P

Avogadro’s Law:Avogadro’s Law: V/n = constV/n = const at constant P, Tat constant P, T

Boyle’s LawBoyle’s Law

Boyle's Law

0

100

200

300

400

500

600

700

800

0 50 100 150 200Volume (L)

Pre

ss

ure

(T

orr

)

Boyle's Law

0

0.002

0.004

0.006

0.008

0.01

0.012

0 50 100 150 200Volume (L)

1/P

res

su

re (

1/T

orr

)

P1V1 = P2V2

PV =k(at constantT and n)

Slope= 1/kV p

1

A Plot of PV Versus P for Several Gases at A Plot of PV Versus P for Several Gases at Pressure Below 1 atmPressure Below 1 atm

Boyle’s holds Only at very Low pressures

A gas strictly obeys Boyle’s law is called Ideal gas

Charles’s LawCharles’s Law

Charles' Law

0

5

10

15

20

25

30

35

0 100 200 300 400 500Temperature (K)

Vo

lum

e (

L)

•V/T = b•V = bT

(constant P & n)

•V1/T1 = V2/T2

Plots of V Versus T(Celsius) for Several GasesPlots of V Versus T(Celsius) for Several Gases

Volume of a gasChanges by

When the temp. Changes by 1oC.I.e., at -273oC , V=0 ???

273

1

All gases will solidify or liquefy before reaching zero volume.

Avogadro’s LawAvogadro’s Law

Avogadro's Law

0

20

40

60

80

100

120

0 1 2 3 4 5moles

Vo

lum

e (

L)

• Vn• V = an (constant P& T)

2

2

1

1

n

V

n

V

5.35.3 The Ideal Gas LawThe Ideal Gas Law

p

kV

bTV anV

Boyle’s law

Charles's law

Avogadro’s law

)(P

TnRV nRTPV

Universal gasconstant

The Ideal Gas LawThe Ideal Gas Law

PV = nRT

R = 0.0821 atm L mol-1 K-1

P

nR

T

V V

nR

T

P P

RT

n

V

The Ideal Gas Law can be used to derivethe gas laws as needed!

Molar Volume

At STP

4.0 g He 16.0 g CH4 44.0 g CO2

1 mole 1 mole 1mole (STP) (STP) (STP)

V = 22.4 L V = 22.4 L V = 22.4 L

The value of RThe value of R What is What is RR for 1.00 mol of an ideal gas at for 1.00 mol of an ideal gas at

STP (25 STP (25 ooC and 1.00 atm)?C and 1.00 atm)?Given thatGiven that

V of 1 mol of gas at STOP= 22.4LV of 1 mol of gas at STOP= 22.4L

nRT PV nT

PV R

)273(00.1

4.2200.1R

Kmol

LatmKmol

Latm

.

.0821.0R

ExampleExample

A reaction produces enough COA reaction produces enough CO22(g) (g)

to fill a 500 mL flask to a pressure of to fill a 500 mL flask to a pressure of 1.45 atm at a temperature of 22 1.45 atm at a temperature of 22 ooC. C. How many moles of COHow many moles of CO22(g) are (g) are

produced?produced?

K

L

Latm

Kmolatm

295

500.0

0821.045.1

PV = nRTRT

PVn

The Ideal Gas Law: Final and initial state problemsThe Ideal Gas Law: Final and initial state problems

equation) gas (Ideal T

VP

2

22

1

11 T

VP

ConstantnRT

PV

Ideal Gas LawIdeal Gas Law

The ideal gas law is an The ideal gas law is an equation of stateequation of state.. Independent of how you end up where Independent of how you end up where

you are at. Does not depend on the path.you are at. Does not depend on the path. The state of the gas is described by: The state of the gas is described by: P,

V, T and n Given 3 you can determine the fourth.Given 3 you can determine the fourth. Ideal gas equation is an empirical Ideal gas equation is an empirical

equation - based on experimental equation - based on experimental evidenceevidence..

Example

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Data Table

Set up Data Table

P1 = 0.800 atm V1 = 0.180 L T1 = 302 K

P2 = 3.20 atm V2= 90.0 mL T2 = ????

Solution

Solve for T2

Enter data

T2 = 302 K x atm x mL = K

atm mL

T2 = K - 273 = °C

Calculation

Solve for T2

T2 = 302 K x 3.20 atm x 90.0 mL = 604 K

0.800 atm 180.0 mL

T2 = 604 K - 273 = 331 °C

ExampleExample

A gas has a volume of 675 mL at 35°C and A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the 0.850 atm pressure. What is the temperature in °C when the gas has a temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 volume of 0.315 L and a pressure of 802 mm Hg?mm Hg?

SolutionSolution

TT11 = 308 K = 308 K TT22 = ? = ?

VV11 = 675 mL = 675 mL VV22 = 0.315 L = 315 mL = 0.315 L = 315 mL

PP11 = 0.850 atm = 0.850 atm PP22 = 802 mm Hg = 802 mm Hg

= 646 mm Hg = 646 mm Hg

TT22 = 308 K x = 308 K x 802 mm Hg 802 mm Hg x x 315 mL315 mL

646 mm Hg 675 mL646 mm Hg 675 mL

= 178 K - 273 = = 178 K - 273 = - 95°C - 95°C

5.4 Gas Stoichiometry5.4 Gas Stoichiometry

Reactions happen in molesReactions happen in moles At Standard Temperature and Pressure At Standard Temperature and Pressure

(STP, 0ºC and 1 atm) 1 mole of gas (STP, 0ºC and 1 atm) 1 mole of gas occuppies 22.4 L.occuppies 22.4 L.

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 volume calculate moles of reactant or volume of product.of product.

ExampleExample

A.What is the volume at STP of 4.00 g of CHA.What is the volume at STP of 4.00 g of CH44??

4.00 g CH4.00 g CH44 x x 1 mole CH1 mole CH44 x x 22.4 L (STP) 22.4 L (STP) = 5.60 L = 5.60 L

16.0 g CH16.0 g CH44 1 mole CH 1 mole CH44

B. How many grams of He are present in 8.0 L of gas at STP? B. How many grams of He are present in 8.0 L of gas at STP?

8.00 L x 8.00 L x 1 mole He 1 mole He x x 4.00 g He 4.00 g He = 1.43 g He = 1.43 g He

22.4 He 1 mole He 22.4 He 1 mole He

A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure in the cylinder.

PV = nRT

P =V =

n =

R =

T =

?

12.25 L

75.5 g = mol20.18 g

mol 3.74

0.082 L•atm mol•K24.5 + 273 = 297.5 K

P = nRT V

= (3.74 mol)(0.082L•atm)(297.5K) (12.25 L) mol•K

= 1.009 atm

= 5670 torr

Example

ExampleExample

30.2 mL of 1.00 M HCl are reacted 30.2 mL of 1.00 M HCl are reacted with excess FeS. What volume of with excess FeS. What volume of gas is generated at STP?gas is generated at STP?

STP means standard temperature and pressure . . . 0 oC and 1 atm.

HCl + FeS FeCl2 + H2S2 HCl + FeS FeCl2 + H2S

Now . . . Go for moles,

2 HCl + FeS FeCl2 + H2S

atm

KKmol

Latmmol

molL

molL

00.1

15.2730821.0

HCl2SHHCl00.1

0302.0 2

K

Kmol

Latm

mol

mol

L

molL 15.273

0821.0

HCl2

SHHCl00.10302.0 2

Kmol

Latm

mol

mol

L

molL

0821.0

HCl2

SHHCl00.10302.0 2

L

molL

HCl00.10302.0 L0302.0

HCl2

SHHCl00.10302.0 2

mol

mol

L

molL

L339.0PV = nRTV = nRT/P

ExampleExample

The decomposition of sodium azide, NaN3, at high temperatures produces N2(g). What volume of N2(g), measured at 735 mm Hg and 26°C, is produced when 70.0 g NaN3 is decomposed.

2 NaN3(s) → 2 Na(l) + 3 N2(g)

Determine moles of N2:

Determine volume of N2:

nN2 = 70 g NaN3

1 mol NaN3

65.01 g N3/mol N3

3 mol N2

2 mol NaN3

= 1.62 mol N2

= 41.1 L

P

nRTV = =

(735 mm Hg)

(1.62 mol)(0.08206 L atm mol-1 K-1)(299 K)

760 mm Hg1.00 atm

X X

2 NaN3(s) → 2 Na(l) + 3 N2(g)

Molar mass of a gasMolar mass of a gas

P P xx V = V = m m xx R R xx T T M M

m = mass, in gramsm = mass, in grams M = molar mass, in g/molM = molar mass, in g/mol Molar mass = Molar mass = m R T m R T

P V P V

P x V = n x R x TP x V = n x R x T

M

mn

DensityDensity Density Density (d)(d) is mass divided by volume is mass divided by volume P P xx V = V = m m xx R R xx T T

MM

P = P = m m xx R R xx T T V V xx M M

P = P = d d xx R R xx T T

M M

d = d = m m

VV

RT

PMd

ExampleExampleA glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.2410 when filled with water at 25°C (d=0.9970 g cm-3) and 40.2959 g when filled with propylene gas at 740.3 mm Hg and 24.0°C. What is the molar mass of polypropylene?

nRTPV RTM

mPV

PV

mRTM

Volume of the vessel

Vflask =

= 0.1654 g

mgas = mfilled - mempty = (40.2959 g – 40.1305 g)

= 98.41 cm3 = 0.09841 L

OH

OH

2

2

d

m)cm g (0.9970

g 40.1305– g 138.24103-

PV = nRT PV = mM

RT M = m

PVRT

M = (0.9741 atm)(0.09841 L)

(0.6145 g)(0.08206 L atm mol-1 K-1)(297.2 K)

M = 42.08 g/mol

ExampleExample

Calculate the density in g/L of OCalculate the density in g/L of O22 gas at STP. gas at STP.

From STP, we know the P and T.From STP, we know the P and T.

P = 1.00 atm T = 273 KP = 1.00 atm T = 273 KRearrange the ideal gas equation for moles/LRearrange the ideal gas equation for moles/L

d = d = PXM PXM R R xx T T

The density of OThe density of O22 gas at STP is gas at STP is

1.43 grams per liter1.43 grams per liter

RXT

PXMd

(273K) )X

mol.K

L.atm (0.0821

)mol

g(32.0 X ) atm (1.00

d Lg /43.1

ExampleExample

2.00 g sample of SX2.00 g sample of SX66(g) has a volume of 329.5 (g) has a volume of 329.5 CmCm33 at 1.00 atm and 20 at 1.00 atm and 20ooC. Identify the C. Identify the element X. Name the compoundelement X. Name the compound

P= 1.00 atmP= 1.00 atm

T = 273+20 = 293KT = 273+20 = 293K

M =

mPVRT

L3295.01000Cm3

1LXCm3 392.5 V

)3295.0)(00.1(

)293)(.

.0821.0)(00.2(

Latm

KKmol

atmLg

MM

f

= 146 g SX6 /mol

Molar mass of (X6 )= 146- 32 = 114 g/mol

Molar mass of X = (114 g/mol X6) /6 = 19

X = with a molar mass of 19 = F

The compound is SF6

5.55.5 Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

For a mixture of gases in a container, For a mixture of gases in a container, the total pressure is the sum of the the total pressure is the sum of the pressure each gas would exert if it were pressure each gas would exert if it were alone in the container.alone in the 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 gas:For each gas: V

nRTP

PPTotalTotal = n = n11RT + nRT + n22RT + nRT + n33RT +...RT +...

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

the same.the same.

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

V V

Thus, )(nPTotalTotal V

RT

•Partial pressure–Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone

A 250.0 mL flask contains 1.00 mg of He and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres.The total pressure is due to the partial pressures of each of these gases.

so:

V

RT)nn(PPP HHeHHetotal 22

For He:

_____________________ = mol He1.00 x 10-3 g He4.00 g

mol 2.50 x 10-4

For H2:

______________________ = mol H22.00 x 10-3 g H2

2.016 gmol 9.92 x 10-4

A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres.

so:

V

RT)nn(PPP HHeHHetotal 22

For He: _____________________ = mol He1.00 x 10-3 g He4.00 g

mol 2.50 x 10-4

For H2: ______________________ = mol H22.00 x 10-3 g H2

2.016 gmol 9.92 x 10-4

And: Ptotal = (2.50 x 10-4 + 9.92 x 10-4)(RT/V)

= (0.001242 mol)(0.0821 L•atm)(25 + 273)K mol•K (0.2500 L)

Ptotal= 0.1216 atm

A 250.0 mL flask contains 1.00 mg of He and and 2.00 mg of H2 at 25.0oC. Calculate the total gas pressure in the flask in atmospheres.

so:

V

RT)nn(PPP HHeHHetotal 22

For He: _____________________ = mol He1.00 x 10-3 g He4.00 g

mol 2.50 x 10-4

For H2: ______________________ = mol H22.00 x 10-3 g H2

2.016 gmol 9.92 x 10-4

Calculate the pressure due just to He (???):

V

RTn P He

He = 0.0245 atm

and Phydrogen= ? 0.1216 - 0.0245 = 0.0971 atm

Magnesium is an active metal that replaces hydrogen from an acid by the following reaction:

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

How many g of Mg are needed to produce 5.0 L of H2 at a temperature of 25 oC and a pressure of 745 mmHg?

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) 5.0 L? g

Hint: find moles of H2 using PV = nRT then work as a stoichiometry problem.

n = PV RT

n = 0.20 mol

=____________________________________745 mmHg 5.0 L62.4 L•mmHg

mol•K298 K

The mole fractionThe mole fraction

Mole fractionMole fraction: number of moles of one : number of moles of one component in a mixture relative to the component in a mixture relative to the total number of moles in the mixturetotal number of moles in the mixture

symbol is Greek letter chi symbol is Greek letter chi

Totaln

n1

1

....321

1

nnn

n=

Mole fraction expressed in pressuresMole fraction expressed in pressures

....321

11

nnn

n

n

n

Total

.......... );( );(2211 RT

VPn

RT

VPn

.....)()()(

)(

321

1

1

1

RTV

PRTV

PRTV

P

RT

VP

n

n

Total

....))((

)(

321

1

1

PPPRT

VRT

VP

TotalP

P

PPP

P1

321

1

1 ....)(

TotalTotalP

P

n

n11

1

TotalPP

11

ExampleExample A 1.00 L sample of dry air at 786 Torr and A 1.00 L sample of dry air at 786 Torr and

25 25 ooC contains 0.925 g NC contains 0.925 g N22 plus other plus other

gasses (such as Ogasses (such as O22, Ar and CO, Ar and CO22.) a) What .) a) What

is the partial pressure of Nis the partial pressure of N22? b) What is ? b) What is

the mole fraction of Nthe mole fraction of N22??

molg

molg 0330.0

0.28925.0

atm

L

KKmolLatmmol807.0

00.1

2980821.00330.0 11

Torratm

Torratm 613

760807.0 780.0

786

6132

Torr

TorrX N

Collecting gas over waterCollecting gas over waterAn insoluble gas is passed into a container of water, the gas rises because its density is much less than that of water and the water must be displaced

KClO3

O2 gas

Collection of Gases over Water

Assuming the gas is saturated with water vapor, the partial pressure of the water vapor is the vapor pressure of the water.

Pgas = Ptotal – PH2O(g)

Ptotal = Pgas + PH2O(g)

Oxygen gas generated was collected over water. If the Oxygen gas generated was collected over water. If the volume of the gas is 245 mL and the barometric pressure volume of the gas is 245 mL and the barometric pressure is 758 torr at 25is 758 torr at 25ooC, what is the volume of the “dry” oxygen C, what is the volume of the “dry” oxygen gas at STP? gas at STP? (P(Pwaterwater = 23.8 torr at 25 = 23.8 torr at 25ooC)C)

PPO2O2 = P = PTotalTotal - P - Pwaterwater = (758 - 23.8) torr = 734 torr = (758 - 23.8) torr = 734 torr

Example

P1= PO2 = 734 torr; P2= SP = 760. torrV1= 245mLT1= 298K; T2= 273K; V2= ?

2

22

1

11

T

VP

T

VP

12

211

2 TP

TVPV

(245mL)(734torr)(273K) (298K)(760.torr)

V2 =

=217 mL

5.65.6 The Kinetic Molecular Theory of GasesThe Kinetic Molecular Theory of Gases

It explains why ideal gases behave the way they do.It explains why ideal gases behave the way they do. Assumptions are made to simplify the theory, but Assumptions are made to simplify the theory, but

don’t work in real gases.don’t work in real gases.

Postulates of the kinetic Theory:Postulates of the kinetic Theory: Gas particles (Gas particles (atoms or moleculesatoms or molecules) are ) are

so small compared with distances so small compared with distances between them, thus we can ignore their between them, thus we can ignore their volume.volume.

The particles are in constant motion The particles are in constant motion and their collisions with walls cause and their collisions with walls cause pressure exerted by the gas pressure exerted by the gas

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

neither attracting or repelling neither attracting or repelling The average kinetic energy The average kinetic energy (KE )(KE ) is is

proportional to the Kelvin temperature.proportional to the Kelvin temperature. KE = 1/2 muKE = 1/2 mu22

m=mass of gas particlem=mass of gas particle v=average velocity of particlesv=average velocity of particles

KMT explains ideal gas lawsKMT explains ideal gas laws

• P&V: P = (nRT) . (1/V) P 1/V– # collisions increases when V decreases

• P & T: P = (nR/V).T P T– When T increases hits with walls become stronger and more frequent

• V & T: V=(nR/P).T V T– When T increases hits with walls become stronger and more frequent.

To keep P constant, V must increase to compensate for particles speeds

• V & n: V= (RT/P). N V n – When n increase P would increase if the volume to be kept constant. V

must increase to return P to its original value

• Dalton’s law: Individual particles are independent of each other and their volumes are negligible. Thus identities of gas particles do not matter

Driving the ideal gas law from KMTDriving the ideal gas law from KMT

• The following expression was derived for pressure of an ideal gas:

])2/1(

[3

2 2

V

umnNP A

particles of s velocitiesquare of average u

particleeach of mass m

number '

2

sAvogadroN

A

)(1/2N(KE) 2

Aavgum

](KE)

[3

2 avg

V

nP

T )(3

2 avg

KEn

PV

T n

PV

T Rn

PV

The meaning of temperatureThe meaning of temperature

T Rn

PVavg

KE)(3

2=

avgKE)( = RT

2

3

Root mean square velocityRoot mean square velocity

(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

Combine these two equationsCombine these two equations

Molecular speed for same gassame gas at two different temperatures

2/1

12

21 )(2

1

MT

MT

u

u

T

T

2/11 )3

(1 M

RTu

T

2/12 )3

(2 M

RTu

T

2/1

2

1 )(2

1

T

T

u

u

T

T

Molecular speed for two different gases at two different temperatures

2/1

1

1 )3

(1 M

RTu

T

2/1

2

2 )3

(2 M

RTu

T

2/1

12

21 )(2

1

MT

MT

u

u

T

T

Effects of Molar Mass on Effects of Molar Mass on uurmsrms

At constant T, 273 K, the most probable speed for O2 > CH4 > H2

EOS

urms M–1/2

so smaller molar masses result in higher molecular speeds

Range of velocitiesRange of velocities

The average distance a molecule The average distance a molecule travels before colliding with another is travels before colliding with another is called the mean free path and is small called the mean free path and is small (near 10(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.

Effects of Temperature on Effects of Temperature on uurmsrms

EOS

urms T1/2 so higher temperatures result in higher molecular speeds

At constant mass the most probable speeds for O2 increase with temperature

Summary of Behaviors

num

ber

of p

arti

cles

Molecular Velocity

273 K

num

ber

of p

arti

cles

Molecular Velocity

273 K

1273 K

num

ber

of p

arti

cles

Molecular Velocity

273 K

1273 K

1273 K

Molecular VelocityMolecular Velocity

Average increases as temperature Average increases as temperature increases.increases.

Spread increases as temperature Spread increases as temperature increases.increases.

Smaller molar masses result in higher molecular speeds

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.

5.7 Effusion and Diffusion5.7 Effusion and Diffusion

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 LawGraham’s Law : the rate of effusion is : the rate of effusion is inversely proportional to the square inversely proportional to the square root of the root of the mass of its particlesmass of its particles..

Rate of effusion for gas 1

Rate of effusion for gas 2

M

M

2

1

DerivingDeriving

The rate of effusion should be The rate of effusion should be proportional to uproportional to urmsrms

Effusion Rate 1 Effusion Rate 1 uurms rms for gas 1for gas 1

Effusion Rate 2 Effusion Rate 2 u urms rms for gas 2for gas 2=

effusion rate 1

effusion rate 2

u 1

u 2

3RT

M

3RT

M2

M

Mrms

rms

1 2

1

DiffusionDiffusion

The spreading of a gas through a room.The spreading of a gas through a room. Slow considering molecules move at Slow considering molecules move at

100’s of meters per second.100’s of meters per second. Collisions with other molecules slow Collisions with other molecules slow

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

DiffusionDiffusion

The spreading of a gas through a room.The spreading of a gas through a room. Slow considering molecules move at Slow considering molecules move at

100’s of meters per second.100’s of meters per second. Collisions with other molecules slow Collisions with other molecules slow

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

5.85.8 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.

Volume CorrectionVolume Correction

The actual volume free to move in is The actual volume free to move in is less because of particle size.less 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 gasb is a constant that differs for each gas..

nbV

nRTP

'

Pressure correctionPressure correction

Because the molecules are attracted to Because the molecules are attracted to each other, the pressure on the 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 effect since two molecules interact, the effect must be squared.must be squared.

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 squaredeffect must be squared

2)('V

naPPobserved

Corrected Corrected CorrectedCorrected Pressure Pressure Volume Volume

P + an

V x V - nb nRTobs

2

Van der Wall’s equationVan der Wall’s equation

a and b are determined by experiment.a and b are determined by 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 polarity.a depends on both size and polarity.

ExampleExample

Calculate the pressure exerted by Calculate the pressure exerted by 0.5000 mol Cl0.5000 mol Cl22 in a 1.000 L container at in a 1.000 L container at

25.0ºC25.0ºC 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