Gases

Unit 6:Gases & The Kinetic

Molecular Theory

Dr. Jorge L. AlonsoMiami-Dade College –

Kendall CampusMiami, FL

CHM 1045: General Chemistry and Qualitative Analysis

Textbook Reference:

•Module # 8

Gases

Characteristics of Gases

Condensed phases

• Unlike liquids and solids, gases . . . . Are highly compressible. Expand to fill their containers. Have extremely low densities.

Gases

Characteristics of Gases

• Variables affecting the behavior of gases Amount = number of moles () Pressure (P) Volume (V) Temperature (T in Kelvin)

{PropGases*}

Gases

• Pressure is the amount of force applied to an area.

Pressure

• Atmospheric pressure is the weight of air per unit of area.

P =FA

Force = mass x acceleration

Newton = 1kg . m/sec2

Approx. 12 miles105 Newtons

meter2= = 101.325 kPa

= (104kg)(10 m/sec2)105 Newtons

Gases

Units of PressureTorricelli’s

(Normal atmospheric pressure at sea level).

• Atmosphere1.00 atm = 760 mm Hg (torr) = 101.325 kPa

760 mm Hg = weight of equal surface area of the atmosphere

Gases

Barometer

33 ft H2O = weight of equal surface area of the atmosphere

Gases

Manometerinstrument used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

{Manometer}

Gases

Manometer

Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

Pgas = 760 torr Pgas = 760 + 6 torrs Pgas = 760 - 6 torrs

Gases

Gas Laws

1. Boyle’s Law Compared: P versus V ( & T are held constant).

2. Charles’s Law Compared: V versus T ( & P are held constant).

3. Avogadro’s Law Compared: V versus η (P & T are held constant).

4. Combined Gas Law Compared: P vs V vs. T ( is held constant).

5. Ideal Gas Law Compared: P vs V vs. η vs T (no variable held constant).

6. Dalton’s Law of Partial Pressure Compared: individual pressures of gases in a mixture

Variables affecting gases: moles (η), pressure (P), volume (V) and Temperature (T)

Gases

Boyle’s Law: Pressure-Volume Relationship

The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

{Boyle’s Law}

( & T are held constant).

V 1P

2 x

V x ½

V ?

Gases

P & V: inversely proportional

Also, P ↑ V ↓ = k

{PV.Graphs}

This means a plot of V versus 1/P will be a straight line.

V 1P

V =kP

OR

Gases

Boyle’s Law

Gases

• The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

( & P are held constant).

{*Charles’s Law Liq N2}

Charles’s Law: Temp. – Volume Relationship

V T

T 2 x

V 2 x

V ?

Gases

Charles’s Law• The volume of a gas is directly proportional to its

absolute temperature.

A plot of V versus T will be a straight line.

orVT

= k

V = k T

V T

Gases

Charles’s Law

Gases

Avogadro’s Law: Moles-Volume Relationship

• The volume is directly proportional to the number of moles of the gas.

{AvogLaw}

(P & T are held constant).

{Avogadro’s Law}

V n

2 x

V 2 x

V ?

Gases

Avogadro’s Law

or, V = k n

• Mathematically, this means V n

{*Avogadro’s Law in Reactions}

Gases

Standard Temperature & Pressure (STP) and Molar Volume

• Standard Temperature: 00C or 273 K• Standard Pressure: 760 torr (1 atm)

At STP the Molar Volume of any gas is 22.4 L

1 mole = 6.022 x 1023 part. = gMM = 22.4 L

(11.1 in)3 or (28.2cm)3

Gases

Standard Temperature & Pressure (STP) and Molar Volume

CO g 75 mL ? 2

g 44.0

CO 1 2

At STP the Molar Volume of any gas is 22.4 L

1 mole = 6.023 x 1023 part. = gMM = 22.4 L

Problem: At STP, what volume in mL would 75g of CO2 occupy?

H2 = 2.0g

O2 = 32.0g

CO2 = 44.0 g

2CO 1

L4.22

L1

mL 1000 mL 10x 3.8 4

Gases

Ideal-Gas Equation

V 1/P (Boyle’s law)V T (Charles’s law)V n (Avogadro’s law)

The Gas Laws:

V nTP

k =nTP Vor

Combining these, we get

V =knTP

or

Gases

Ideal-Gas Equation

The relationship

then becomes

PV = nRT

R =P VnT

k =nTP V

=1R

Gases

PV = nRT

Ideal-Gas Equation:

(torr) (L) = (mol) R (K)Units:

{PV= nRT RapVideo}

Kmol

torrL

RapVideoLinkYouTube

Useful for pure gas under one set of conditions.

Gases

Ideal Gas Law Problems

What volume (in mL) would a 2.20 g sample of hydrogen gas (H2) at 50.00C occupying at 443 torr?

PV = nRT V = nRTP

V =

g 2.0

1 g 2.20

K36.62

torr L K0.50 273

torr 443

R= 50.0 L

L1

mL1000

Gases

Ideal-Gas Equation: Densities of Gases

RT MW

g# PV

MW-g

mole 1 g)(# (n) moles

MW V

TR g# P

PV = nRTSince

Then

For Ideal Gas Equation:

Where d = Density of Gas

Dividing both sides of the equation on the left by V we get

( ) MW

TR d P

PV

RT g# MW and

If we solve the equation for density, we get……..

Gases

T

(MW) P d

R

P

T d MW

R

Problem: What is the density of the oxygen in a tank in an AC room (25°C) and whose pressure gauge reads 25.0 atm

Problem: A gas whose density is 0.0131 g/mL and is in a container at room temperature and whose pressure gauge reads 1.9 x 104 mmHg. What is its MW?

Ideal-Gas Equation: Densities and Molecular Weigh of Gases

K) (298

)(32.0g/ atm) (25.0

KatmL 0.0821 L

g7.32

torrs10 x 1.9

( L

mL 1000 mL

g0.0131

4

K) 298

K

torrL36.62

g8.12

Gases

Ideal-Gas Equation: Densities & Molecular Weigh of Gases Problems

What is the density (in g/mL) of SO2 at STP?

RT MW

g# PV

MW V

TR g# P ( )

PV = nRT

MW

TR d P

T

(MW) P d

R =

K0273

K0

torr L 36.62

torr 760 )g 0.64( = 2.62 g/L

mL1000

L1

mLg00262.0

Gases

Ideal-Gas Equation: Densities & Molecular Weigh of Gases Problems

What is the molecular weight of a gas whose density @ STP is 7.78 g/L?

RT MW

g# PV

MW V

TR g# P ( )

PV = nRT

MW

TR d P

=

K0273

K0

torr L 36.62

torr 760P

T d MW

R

)Lg .787(g 741

Gases

2006 A

Gases

Gases

Combined Gas Law Equation

V 1/P (Boyle’s law)

V T (Charles’s law)

The Gas Laws:

Combining, we can get

P1V1

T1

P2V2

T2

The Combined Gas Law

Useful for a constant amount of a pure gas under two different conditions.

k=

Gases

Combined Gas Law EquationConstant

P1V1

T1

=P2V2

T2

Gases

Combined Gas Law Problem

P1V1

T1

=P2V2

T2

A scuba diver takes a gas filled 1.0 L balloon from the surface where the temperature is 34 0C down to a depth of 66 ft (33 ft H2O = 1 atm). What volume will the gas balloon have at that depth if the temperature is 15 0C?

V2

K273 15

) ( atm 3

K273 34

L 1 atm 100

L .313

atm 3 K307

K288 L 1 atm 1

0

0

V2

Gases

Dalton’s Law ofPartial Pressures

• The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone.

• In other words,

Ptotal = P1 + P2 + P3 + …

Pair = P N2 + PO2 + PH2O + …

Gases

Partial Pressures• When one collects a gas over water, there is water vapor mixed

in with the gas.

• To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

P of atmP of gas

Gases

Vapor Pressure of Water

Vapor Press  (torr)

    Vapor

Press (torr)

-10 2.15 40 55.3

0 4.58 60 149.4

5 6.54 80 355.1

10 9.21 95 634

11 9.84 96 658

12 10.52 97 682

13 11.23 98 707

14 11.99 99 733

15 12.79 100 760

20 17.54 101 788

25 23.76 110 1074.6

30 31.8 120 1489

37 47.07 200 11659

Ptotal = Pgas + PH2O

Pgas = Ptotal - PH2O

• To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

• Daltons Law:

{Press on can}

Gases

Evaporation vs Boiling in terms of Vapor Pressure

Vapor Pressure (v.p. or Pvap)

++

=

Patm

Pvap

Patm

Pvap

Patm

Pvap

• Caused by the tendency of solids & liquids to evaporate to gaseous form. It is temperature (K.E.) dependent.

Gases

Stoichiometry with Gases

L 2.0

torr) 24 - 775(

298

36.62H η082.0 0

2

K

Kmol

torrL

Vo

Mg g 2.0 H η ? 2

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

Problem: If 2.0 g of Mg are reacted with excess HCl, what volume of H2 will be produced at 250C and 775 torr? At STP?

PV = nRT

Mg 3.24

Mg 1

g

Mg 1

H 1 2

2H 082.0

Gases

Kinetic-Molecular TheoryA model that aids in our understanding of what happens to gas particles as environmental variables change.

1. Gases consist of large numbers of molecules that are in continuous, random motion.

Main Tenets:

2. Collisions between gas molecules and between gas molecules and the walls of the container must be completely elastic (energy may be transferred between molecules, but none is lost).

Gases

Kinetic-Molecular Theory

Main Tenets:

3. Attractive and repulsive forces between gas molecules are negligible.

4. The combined volume of all the molecules of the gas is negligible (excluded volume) relative to the total volume in which the gas is contained.

Gases

Kinetic-Molecular TheoryMain Tenets:

{KE T(K)}

6. The average kinetic energy (KE=½mv2) of the molecules is proportional to the absolute temperature.

@ 100 0C5. Energy can be transferred

between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.

Gases

EffusionThe escape (diffusion) of gas molecules through a tiny hole into an evacuated space.

DiffusionMovement of

molecules from an area of high concentration to an area of low concentration until equilibrium is reached (homogeneity).

Gases

Effect of Molecular Mass on Rate of Effusion and Diffusion

AMW

1 ARate

Thomas Graham (1846): rate of diffusion is inversely proportional to the square root of its molar mass

2

2

1mvKE

Kinetic Energy per individual molecule:

Gases

Rate of Diffusion & Effusion

{BrDiffusion}

Comparing the rates of two gases:

Graham’s Law of Diffusion and Effusion of Gases

A

B

MW

MW

B

A

Rate

Rate

HCl Rate

NH Rate 3 {GasDiff}

Dropper with Br (l)

AMW

1 ARate

Thomas Graham (1846): rate of diffusion is inversely proportional to the square root of its molar mass

g/mol 17

g/mol 361.4 2.1

Gases

Effusion and Diffusion• This is the most widespread uranium

enrichment method. Uranium is reacted with fluorine to make uranium hexafluoride gas: 235UF6 & 238UF6

• The physical principle is that the diffusion speed of a gas molecule depends on the mass of the molecule: the lighter ones diffuse faster and get through a porous material easier.

• In gas diffusion units, uranium-

hexafluoride gas diffuses through an etched foil made of either an aluminum alloy or teflon, due to artificially maintained difference in pressure. The lighter molecules (i.e. those containing 235U) get through easier to the other side, therefore the gas accumulating there will be richer in 235U.

Gases

Gas Centrifugation

The gas centrifuge is essentially a bowl, in which there is a rotor spinning round at a very high speed. The gas (UF6) directed to the centrifuge is forced to spin by the rotor. Due to the centrifugal force the heavier molecules (those which contain 238U) will accumulate near the wall of the bowl, while the lighter molecules containing 235U will

stay closer to the center of the centrifuge.  

Gases

Boltzmann DistributionsThe Maxwell–Boltzmann distribution is the statistical distribution of molecular speeds in a gas. It corresponds to the most probable speed distribution in a collisionally-dominated system consisting of a large number of non-interacting particles. {Boltzman Plot}

Gases

Kinetic Energy of Gas Molecules

2 2

1vmKE

Kinetic Energy per individual molecule:

Kinetic Energy per mole:

m

rms M

RTu

3

Combining above equations and solving for velocity we get:

• The root-mean square velocity of gases is a very close approximation to the average gas velocity.

• To calculate this correctly: The value of R = 8.314 kg m2/s2 K mol Mm = molar mass, and it must be in kg/mol.

RTKE2

3

☺

☺

☺

Gases

The Kinetic-Molecular Theory

• Example: What is the root mean square velocity of N2 molecules at room T, 25.0oC?

kg/mol 028.0

K 298molK sec

m kg8.3143

u2

2

rms

mi/hr 1159=m/s 515

m

rms M

RTu

3

• To calculate this correctly: The value of R = 8.314 kg m2/s2 K mol And M must be in kg/mol.

Gases

The Kinetic-Molecular TheoryProblem: What is the root mean square velocity of He atoms

at room T, 25.0oC?

You do it!You do it!

m

rms M

RTu

3

• To calculate this correctly: The value of R = 8.314 kg m2/s2 K mol And M must be in kg/mol.

kg/mol

K mol K sec

m kg8.3143

u2

2

rms 004.0

298

mi/hr 3067= m/s 1363

• Can you think of a physical situation that proves He molecules have a velocity that is so much greater than N2 molecules?

• What happens to your voice when you breathe He(g) or SF6 (g)?

mi/hr 1159=m/s was 515N 2 Gas MWHe 4N2 28SF6 146

Gases

Ideal vs.

Real Gases

In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

Gases

Deviations from Ideal Behavior

Two particular assumptions made in the kinetic-molecular model break down at high pressure and/or low temperature:

(1) attractive forces and (2) excluded volume.

Gases

2003 A

Gases

Corrections for Non-ideal Gas Behavior

• The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account.

• The corrected ideal-gas equation is known as the van der Waals equation.

Gases

The van der Waals Equation

) (V − nb) = nRTn2aV2(P +

Gases

Real Gases:Deviations from Ideality

• van der Waals’ equation accounts for the behavior of real gases at low temperatures and high pressures.

nRT nVV

n + P

2

2

• The van der Waals constants a and b take into account two things:1. a accounts for intermolecular attraction2. b accounts for volume of gas molecules

• At large volumes a and b are relatively small and van der Waal’s equation reduces to ideal gas law at high temperatures and low pressures.

a b

Gases

Real Gases:Deviations from Ideality

• Example: Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the ideal gas law.

You do it!You do it!

mol g 17.0

mol NH g 84.0 = n 3 94.4

1

V

nRT = P

L 5.00

K K mol

atm Lmol 4730821.094.4

atm 4.38

Gases

Real Gases:Deviations from Ideality

molL0.0371=b

mol

atm L 4.17 =a mol 4.94 = n

2

2

• Example: Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the van der Waal’s equation You do it!You do it!

nRTnb-VV

an + P

2

2

2

2

V

an

nb-V

nRT=P

2

2

00.5

17.494.4

)0371.0)(94.4(00.5

4730821.094.4

L

mol

mol L

K mol P

2

2

molatm L

molL

K molatm L

atm atm atm atm L

atm L P 7.35)1.48.39(07.4

817.4

8.191

behavior ideal from difference 7.6% a is thisatm, 38.4 toCompared