Gases 2009

Gases

• Properties of Gases

• Gas Laws (pressure, volume, temperature, moles)

• Gases in Chemical Reactions

• The Kinetic Model of Gases

Gaseous elements

Pressure = Force/Area

V ~ 1/P

If a fixed amount of gas is released into a larger container, how much force does it exert?

Boyle’s Law

P ~ 1/V

Boyle’s Law:P1V1 = P2V2

If the volume of a gas is constrained to a smaller container, how much force does it exert?

Decrease volume more collisions

V1

V2

P1

P1V1 = P2V2

Relationshipbetween volume andpressure.

• Ex. 1 A sample of gas occupies 21 liters at a pressure of 2.2 atm. What would be the volume if the pressure was increased to 6.2 atm?

• Ex. 2 A sample of O2 occupies 10.0 L at 785 torr. At what pressure would it occupy 14.5 L?

Charle’s Law:

V1/T1 = V2 / T2

If the temperature of gas is increased, how will its volume respond (if the pressure is kept constant)?

Gas 1

Gas 2

Gas 3

Gas 4

The point where a gaswould have zero volume!

P ~ T

If a gas is heated, how much force does it exert (if the volume is kept constant)?

Combined Gas Law (Boyle and Charle’s)

P1 V1 = P2 V2

T1 T2

Re-cap of Gases:

Boyle’s Law: Charles’s Law:

P1 V1 = P2 V2 V1/T1 = V2 / T2

P1 V1 = P2 V2

T1 T2

1) Gas equations should make sense: volume, pressure, temperature2) Watch your units! (K, L, atm)

One standardized set of conditions =

“STP”

• 0o C (273 K)

• 1 atm (760 Torr)

• Ex. 3 A balloon filled with He occupies 413 mL at 100.oC. At what temperature would it occupy 577 mL if the pressure was constant?

• Ex. 4 A sample of hydrogen sulfide (H2S) occupies 210 L at 27oC at 1200 T. What volume would it occupy at STP?

How much volume does 1 mole of gas occupy at 0o C, 1 atm?

If one mole of Argon occupies 22.1 L, how much volume will be occupied by:

one mole CO2

one mole N2

one mole O2

one mole H2

Liters

How much volume does 1 mole of gas occupy at 0o C, 1 atm?

Ideal gas: 22.41 L per mole

Ex. 5 What is the density of a gas that has a molar mass of 44.01 g/mol at STP?

What is “R”?

Ideal Gas Law: PV = nRT(where n = moles)

“STP” = 0 o C and 1 atm

• Ex. 6 What is the volume of a balloon filled with 32.02 grams of Helium when the atmospheric pressure is 722 torr and the temperature is 40o C?

• Ex. 7 The Goodyear blimp must be inflated with Helium prior to a football game. Its volume is 7601 ft3. How many grams of He are needed for a pressure of 740 torr at 22o C?

(1 ft3 = 28.3 L)

• Ex. 9 A 0.723 g sample of a gas occupies 176 mL at 100.o C and 750. torr. What is its molar mass?

Stoichiometric Calculations Involving Gases

C3H8 + 5O2 3CO2 + 4H2O

25.0 g of propane produces how many moles of CO2? 25.0 g of propane produces how many liters of CO2 at STP? How many grams of propane are need to react with 50.0 L of O2 at 25o C and 1.0 atm?

Key Concepts:

• Ideal Gas Law PV = nRT

• STP (0o C and 1 atm)

• Standard molar volume: 22.4 L/mole at STP

• Gases in stoichiometric calculations

An airbag inflates in less than 50 msec by the reaction of NaN3 to produce Na and nitrogen gas:

NaN3 Na + N2

The volume of the airbag is about 30 L when inflated, and it is filled to a pressure of 1.4 atm. How many grams of NaN3 must be used for each air bag? The molar mass of NaN3 is 65.1 g/mole. Assume the process occurs at room temperature.

Gas mixtures• Dalton’s Law of partial pressures

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

PT=P1+P2+P3+….Pn

Exercise: A gaseous mixture is made from 6.00g oxygen and 9.00g methane placed in a 15L vessel at 0oC. What is the partial pressure of each gas and the total pressure in the vessel?

Dalton’s law ofpartial pressures

PT = Pa + Pb + Pc + Pd = …

Remember that each pressureis also equal to: nXRT/V

Mole Fractions

• The ratio n1/nT is called the mole fraction (denoted x1), a dimensionless number between 0 and 1.

TT

TTT

Pn

nP

n

n

VRTn

VRTn

P

P

11

111

/

/

Mole fraction of N2 in air is 0.78, therefore if the total barometric pressure is 760 torr, the partial pressure of N2 is (0.78)(760) = 590 torr.

Kinetic –Molecular TheoryTheory describing why gas laws are obeyed (explains both pressure and

temperature of gases on a molecular level).• Complete form of theory, developed over 100 years or so, published by

Clausius in 1857. Gases consist of large numbers of molecules that are in continuous,

random motion Volume of all molecules of the gas is negligible, as are

attractive/repulsive interactions Interactions are brief, through elastic collisions (average kinetic energy

does not change) Average kinetic energy of molecules is proportional to T, and all gases

have the same average kinetic energy at any given T.

Because each molecule of gas will have an individual kinetic energy, and thus individual speed, the speed of molecules in the gas phase is usually characterised by the root-mean-squared (rms) speed, u,(not the same though similar to the average speed). Average kinetic energy є = ½mu2

Application to Gas Laws

• Increasing V at constant T:Constant T means that u is unchanged.

But if V is increased the likelihood of collision with the walls decreases, thus the pressure decreases (Boyle’s Law)

• Increasing T at constant V:Increasing T increases u, increasing

collisional frequency with the walls, thus the pressure increases (Ideal Gas Equation).

• Ex. 10 A 20.5 L bulb contains 0.200 moles of methane, 0.300 moles of hydrogen, and 0.400 moles of nitrogen at 20.0 o C. It is stinky and explosive. What is the pressure inside the bulb? How much pressure is contributed by each of the three gases?

• Ex. 12 One tank of gas contains 5.00 L of N2 at 32.0 atm. A second tank contains 3.00 L of O2 at 24.0 atm. What pressure is attained when the valve between the tanks is opened?

Mole fraction: the portion of a specific substance within a mixture

What is the portion (mole fraction) of red spheres?

Kinetic model of gases

each dot isone gasmolecule

The Kinetic-Molecular Theory• The basic assumptions of kinetic-molecular theory

are:• Postulate 1

– Gases consist of discrete molecules that are relatively far apart.

– Gases have few intermolecular attractions.– The volume of individual molecules is very small compared

to the gas’s volume.

• Proof - Gases are easily compressible.

The Kinetic-Molecular Theory

• Postulate 2– Gas molecules are in constant, random, straight

line motion with varying velocities.

• Proof - Brownian motion displays molecular motion.


• Postulate 3– Gas molecules have elastic collisions with

themselves and the container.– Total energy is conserved during a collision.

• Proof - A sealed, confined gas exhibits no pressure drop over time.

The Kinetic-Molecular Theory• Postulate 4

– The kinetic energy of the molecules is proportional to the absolute temperature.

– The average kinetic energies of molecules of different gases are equal at a given temperature.

• Proof - Brownian motion increases as temperature increases.

The Kinetic-Molecular Theory• The kinetic energy of the molecules is proportional

to the absolute temperature. The kinetic energy of the molecules is proportional to the absolute temperature.• Displayed in a Maxwellian distribution.

The Kinetic-Molecular Theory• The gas laws that we have looked at earlier in this chapter are

proofs that kinetic-molecular theory is the basis of gaseous behavior.

• Boyle’s Law– P 1/V – As the V increases the molecular collisions with container

walls decrease and the P decreases.• Dalton’s Law

– Ptotal = PA + PB + PC + .....– Because gases have few intermolecular attractions, their

pressures are independent of other gases in the container.

• Charles’ Law– V T – An increase in temperature raises the molecular velocities,

thus the V increases to keep the P constant.


uRT

Mrmsm

3

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

• Calculating the root-mean square velocity is simple:

• To calculate this correctly:– The value of R = 8.314 kg m2/s2 K mol

– And M must be in kg/mol.


• Example 12-17: What is the root mean square velocity of N2 molecules at room T, 25.0oC?

u

3 8.314kg m

sec K mol K

kg / mol

m / s = 1159 mi / hr

rms

2

2

298

0 028

515

.


• What is the root mean square velocity of He atoms at room T, 25.0oC?


u

3 8.314kg m

sec K mol K

kg / mol

m / s = 3067 mi / hr

rms

2

2

298

0 004

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?

Diffusion and Effusion of Gases• Diffusion is the intermingling of gases.• Effusion is the escape of gases through

tiny holes.

Diffusion and Effusion of Gases

• This is a demonstration of diffusion.


• The rate of effusion is inversely proportional to the square roots of the molecular weights or densities.

1

2

2

1

1

2

2

1

D

D

R

R

or

M

M

R

R


• Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SO2, at the same temperature and pressure.

R

R

M

M

g / mol4.0 g / mol

R R

He

SO

SO

He

He SO

2

2

2

641

16 4 4

.


• Example 12-16: A sample of hydrogen, H2, was found to effuse through a pinhole 5.2 times as rapidly as the same volume of unknown gas (at the same temperature and pressure). What is the molecular weight of the unknown gas?

g/mol 54 = g/mol) 0.2(27M

g/mol 0.2

M27

g/mol 0.2

M2.5

M

M

R

R

unk

unk

unk

H

unk

unk

H

2

2

Real Gases: Deviations from Ideality

• Real gases behave ideally at ordinary temperatures and pressures.

• At low temperatures and high pressures real gases do not behave ideally.

• The reasons for the deviations from ideality are:1. The molecules are very close to one another, thus

their volume is important.

2. The molecular interactions also become important.

Real Gases:Deviations from Ideality

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

P + n a

VV nb nRT

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.


• What are the intermolecular forces in gases that cause them to deviate from ideality?

1. For nonpolar gases the attractive forces are London Forces

2. For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds.


• Example 12-19: 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.

atm 4.38PL 5.00

K 473K mol

atm L0821.0mol 94.4

V

nRT = P

mol 94.4g 17.0

mol 1NH g 84.0 =n 3


• Solve Example 12-19 using the van der Waal’s equation.

2

2

2

2

2

2

V

an

nb-V

nRT=P

nRTnb-VV

an + P

molL0.0371=b

mol

atm L 4.17 = a mol 4.94 =n


ideal from difference 7.6% a is which atm 7.35P

)atm 1.4atm 8.39(atm 07.4L 817.4

atm L 8.191P

L 00.5

17.4mol 94.4

)0371.0)(mol 94.4(L 00.5

K4730821.0 mol 94.4P 2

molatm L2

molL

K molatm L

2

2

“Real” gases• Deviations from ideal behavior as gases are

exposed to:

– high pressures

– low temperatures

(What happens at high pressures and low temperatures???)

• The molar volume is not constant as is expected for ideal gases. • These deviations due to an attraction between some molecules. • Finite molar molecular volume.• For compounds that deviate from ideality the van der Waals

equation is used:

where a and b are constants that are characteristic of the gas. • Applicable at high pressures and low temperatures.

nRT=nb)-(VV

an+P2

2

Ideal: PV = nRT

Real: (P + a n2/V2) (V – nb) = nRT

• a is a molecular attractivity factor

• b is a molecular volume factor

Ideal: P = nRT/V

Real:

P = nRT/(V – nb) - a n2/V2

“Real” gases

Ex. 13 Calculate the pressure exerted by 5.00 moles

of NH3 in a 1.00 L vessel at 25.0oC assuming ideal and non-ideal behavior.

Compression Factor

RT

VpZ

gasperfectvolumemolar

volumemolarZ

Variation of Z with Pressure

At 0 C

Low pressures, Z=1, all gases ideal

At high pressures, Vreal >Videal ; Z>1 repulsive forces dominate

Low pressure, Vreal <Videal, Z< 1

attractive forces dominate

Low temperatures -> molecules moving less rapidly more influenced by attractive forces

Van der Waals Equation (1873)2

V

na

nbV

nRTp2

mm V

a

bV

RTp

a reflects how strongly molecules attract each other

b corrects for the molecule’s size

“Derivation” of vdw Eq. State

Repulsive interactions cause molecules to behave as impenetrable spheres

Molecules restricted to smaller volume V-nb, where nb is volume molecules take up

Pressure depends on frequency of collisions with walls and force of each collision – both reduced by attractive forces proportional to molar concentration (n/V)

Pressure is then reduced according to a(n/V)2

Phase Transition: CondensationImagine you are compressing gas with piston

At C, piston slides without further change in pressure

Liquid begins to appear – Two phases coexist

Going to E – amount of liquid increases

Pressure along CDE is vapour pressure

At E, sample is liquid – now the compressibility changes ….

Similar behave happens in lipidic or polymeric systems, which exhibit a rich variety of phase transitions …

Experimental isotherms of CO2 at several temperatures.

Critical ConstantsThe `critical isotherm', the isotherm at the critical temperature Tc, is at 31.04 C. The critical point is marked with a star.

Critical pressure Pc

Critical molar volume Vc

Tc, Pc,Vc critical constants

If you compress along Tc, liquid does not appear – these highly dense materials are called “supercritical fluids” and their properties are a subject of intense current research

Virial Equation of State

...)()(

12

32 V

TB

V

TB

RT

VPZ VV

...)()(1 232 PPBPPB

RT

VPZ PP

)(2 TB V= 0 at Boyle temperature

Most fundamental and theoretically sound

Polynomial expansion – Viral Expansion

Used to summarize P, V, T data

Also allow derivation of exact correspondence between virial coefficients and intermolecular interactions

Gases 2009

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