Chapter 12 Gases and Their Properties

Three phases of matter: Gas Liquid Solids

The phase of a material is determined by balance between the kinetic energy in the particles and their interactions

Condensed phases: intermolecular interaction overcome the kinetic energy holding the molecules together

Gas phase: intermolecular interactions are too weak to overcome the interaction and the molecules are separated and move around freely

Properties of Gasses

To completely describe the state of a gaseous system, we need to know:

How many particles?

How much space they occupy?

On average how fast are they moving?

Moles

Volume

Temperature Pressure

PK.E α T

Conservation of momentum

V

K.E

PressurePressure is defined as a force experienced over a given surface area

P = F/A

N/m2Units = Pa (Pascal)

J/m3 = Pa (Pascal) Energy/volume !!!

Recall 1 kPa = 1000 Pa

Other Units

1 atm = 101.3 kPa

760 mm Hg = 1 atm

1 mm Hg = 1 Torr

= 760 mm Hg = 760 Torr

PV = energy

1 atm = 101.3 kPa

Gas LawsIn the 17th and 18th century many observations were made on gasses relating T, V, P and n (number of particles).

Robert Boyle (1662) demonstrated that the pressure of a gas varies inversely proportional to the volume of a gas (using air)

For example, the gas-tight syringe is placed in a sealed flask, then compressed air is applied to the flask.

As the applied pressure increases, the volume of the syringe decreases

1V

PBoyle’s Law

(absolute)V T“Extrapolation”

Gas LawsJacques Charles (~1787) demonstrated that at constant pressure the volume of a gas is a direct linear function of its temperature

By graphing the volume of the gas against temperature, 0C had no significance, but all measured gases seem to converge to a common zero point: absolute zero

We must always use absolute temperatures (Kelvin scale) with the gas laws!

Charles’ law

Charles

V n

Gas LawsAmedeo Avogadro (1811) demonstrated that the volume of a gas depends directly on the quantity of gas present.

His work with gases lead directly to the development of the molecular hypothesis and to the determination of Avogadro’s Number

Consider a gas forming reaction:

If you use twice as much reactants you get twice the volume of gas, under constant pressure and temperature

Ex) HOOH(l) → H2O(l) + ½ O2(g)

Ideal Gas Law

PV nRT

The modern gas law is a combination of all three (Boyle, Charles, Avogadro) into one combined equation, the ideal gas law

P = pressure V = volume

n = moles (quantity) T = absolute temperature (K)

Why is it called the ideal gas law?

Serious deviations occur at (a) very low temperature or (b) very high pressure

Air obeys the ideal gas law to within 1% under “normal” conditions

R is the ideal gas constant which is unit dependent.

– P(Pascal), V(m3), n(mol): 8.3145 Pa m3 mol-1 K-1 = 8.3145 J mol-1 K-1

– P(atm), V(Litre), n(mol): 0.0820574 L atm mol-1 K-1

Ideal Gas Law

PV = nRT

Ex) Compute the number of moles of 1.00 L an ideal gas under standard temperature and pressure (STP).

Standard temperature is room temperature at 0.00 oC

Standard pressure is ambient pressure of 1.000 atm

P = 1.000 atm = 101.3 kPa = 101300 Pa

T = 0.00 oC = 273.15 K

n =PV

RTn =(101300 Pa)( 0.00100 m3)/(8.314 J/mol K)(273.15 K)

V = 1.00 L = (1.00 dm3)(0.1 m/dm)3 = 0.00100 m3

= 0.0446 mol

Ex) Compute the volume of 1.000 mol of gas at STP

=nRT

VP

V = (1.000 mol)(8.314 J/molK)(273.15 K)/(101300 Pa)

= 0.02242 m3

= (0.002242 m3)(10 dm/m)3 = 22.42 dm3= L

m (H2) = 1.0 g; m (Ar) = 8.0 g

Ideal Gas Law1) Divide by V

nRT

PV

1PV

Boyle’s Law

2) Divide by P

V TCharles’ law

nRT

VP V n Avogadro’s law

If n and T is constant

If n and P is constant

If T and P is constant

Ideal Gas Law

3) Devide by nT

PV

RnT

Constant

Therefore we can directly compare any two (different) systems

1 1 2 2

1 1 2 2

PV PV

n T n T

We can compare a system to itself which has undergone a change.

Ideal Gas Law

If the number of particles remains the same:

1 1 2 2

1 2

PV PV

T T

We can change any two of P, V and T and predict the remaining variable

Closed System

We can change any one of P, V and T, keep one constant and predict the remaining variable

Change T, with fixed V, compute P Change P, with fixed V, compute T

Change V, with fixed T, compute P Change P, with fixed T, compute V

Change V, with fixed P, compute T Change T, with fixed P, compute V

1 2 2

2

1

1

P V P V

T T1 2 1

2

1

1

P V P V

T T

Volume remains the same

Solve for P21

2 21

PP T

T

1 2

2

1

P P

T T

P2 = (507000 Pa)(333 K)/(373 K) = 463000 Pa

Changing Temperature With Constant Volume.

Ex) A gas is in a closed 3.00 L container at 100. oC with a pressure of 5.00 atm. Predict the pressure in Pa when the system has cooled to 50. oC

P1 = 5.00 atm

V1 = 0.00300 m3

= (5.00 atm)*(101.3 kPa/atm) = 507000 Pa

T1 = 100. oC = 100. + 273.15 K= 373 K

T2 = 50. oC = 100. + 273.15 K = 333 K

Decreasing T increases K.E. If the V does not increase, since the amount of K.E. per unit volume is decreased, P must decrease.

Changing Volume Under Constant Temperature. Ex) A gas is in a closed 3.00 L container at 100. oC with a pressure of 5.00 atm. Predict the pressure in Pa when the system has expanded to 50.0 L.

P1 = 5.00 atm

V1 = 0.00300 m3

= (5.00 atm)*(101300 Pa/atm) = 507000 Pa

T1 = 100. oC = 100. + 273.15 K = 373 K = T2

V2 = 0.0500 m3

1 2 2

2

1

1

P V P V

T TTemperature Is constant

1 2 2

1

1

1

P V P V

T T

1 2 21P V P V Solve for P21

22

1P VP

VP2 = (507000 Pa)(0.00300 m3)/(0.0500 m3) = 30400 Pa

Increasing the volume with out changing the temperature, means that the gas has to less energy per unit volume to generate pressure. Therefore P decreases.

Changing Volume Under Constant Pressure.

Ex) A gas in a closed 3.00 L container at 100. oC with a pressure of 5.00 atm. Predict the temperature in K when the system has increased in volume to 50.0 L.

P1 = 5.00 atm

V1 = 0.00300 m3

= (5.00 atm)*(101.3 kPa/atm) = 507000 Pa

T1 = 100. oC = 100. + 273.15 K = 373 K

V2 = 0.0500 m3

1 2 2

2

1

1

P V P V

T TPressure constant

1 1 2

2

1

1

P V P V

T T

= P2

2

2

1

1

V V

T TSolve for T2

22 1

1

T VT

VT2 = (373 K)(0.0500 m3)/(0.00300 m3) = 6220 K

Increasing the volume, while the pressure is maintained, requires the gas to have much more energy to maintain the pressure. Therefore T increases.

Changing Volume and Pressure Simultaneously

Ex) A gas in a closed 3.00 L container at 100. oC with a pressure of 5.00 atm. Predict the temperature in K when the system has decreased in volume to 0.300 L and increased in pressure to 10.00 atm.

1 2 2

2

1

1

P V P V

T T

P1 = 5.00 atm

V1 = 0.00300 m3

= (5.00 atm)*(101.3 kPa/atm) = 507000 Pa

T1 = 100. oC = 100. + 273.15 K = 373 K

V2 = 0.000300 m3

P2 = 10.00 atm = (10.00 atm)*(101.3 kPa/atm) = 1013000 Pa

Solve for T22 2

21

11

P VT T

P V

T2 = (1013000 Pa)(0.000300 m3)(373 K)/(507000 Pa)(0.00300 m3) = 66.3 K

P has been doubled, but V has been decreased to 10 %. The increase in P does not compensate for the decrease in V, therefore, T should decrease to 2*10% = 20 % its original value

Density of ideal gases

m/M m = mass & M = molar massn = Where

Densities of solids and liquids are ~1 g mL-1 (up to a high of about 15 g ml-1 Hg)

Densities of gases at “normal” conditions are about 1000 times lower

Hence are reported in units of g L-1

Solids and liquids are very difficult to compress: density almost does not vary, and hence are regarded as a fundamental property

Gases are compressible and thus gas density is variable and not regarded as a fundamental property as with solids and liquids

Ex) Compute the density of an N2(g) under STP.

PV = nRT

PV = mRT/M P = mRT/MV

P = (m/V)RT/M = RT/M =PM/RT = (101300 Pa)(28.02 g/mol)/(8.314 J/mol K)(273.15 K)

= 1249 g/m3 = 1.249 g/L

=PM/RT M = RT/P

Chloroform is a common liquid used in the laboratory. It readily vaporizes. If the pressure of CHCl3 is 195 mmHg in a flask at 25C and = 1.25 g L–1, what is its molar mass?

From the density of a gas (or vapor) one can determine the molecular mass.

Density of ideal gases

P = [(195 mm Hg)/(760 mm Hg/atm)](101300 Pa/atm) = 26000 Pa

T = 25 + 273 = 298 K

M = (1250 g/m3)(8.314 J/molK)(298 K)/(26000 Pa)

= (1.25 g/L)(1000 L/m3) = 1250 g/m3

= 119 g/mol

ii

i

where sum of moles over each componentn i

i

i

moles of i nX = =

total moles of gas n

A BA B A B

A B A B

n nX = X = X + X = 1

n +n n +n

A B CA B C

A B C A B C A B C

n n nX = X = X =

n +n +n n +n +n n +n +n

A B CX + X + X = 1

Dealing with Gas MixturesAvogadro’s Law states that the number of particles is the key to the behaviour of gases

The mole fraction expresses the relative number of particles (atoms, molecules) of specific kinds of gases in a mixture of gases

For a two component mixture: A + B:

For a three component mixture: A + B + C

Mole fraction is a new unit of concentration

A A B Bp V = n RT and p V = n RT

A Btot A B A B tot

n RT n RT RT RTP = p +p = + = (n +n ) = n

V V V V

totAA tot

n RTn RTp = P =

V V

A

A AA

tottot tot

n RTp nV= = = X

n RTP nV

A A totp = X P i i totp XP

Dalton’s Law of Partial PressuresThe total pressure of a gas mixture equals the sum of the partial pressures of the individual gases (when the Ideal Gas Law is valid)

Consequences of Dalton’s Law: consider two gases A and B in the same container

Now combine the equation for the total mixture and that of one component A

Rearranging gives: In general:

total ii

P p

Illustrating Dalton’s Law

Consider the two flasks, one containing nitrogen, the other containing half as many molecules of oxygen

Since the flasks are the same size and the temperature is the same, the pressure in the oxygen flask is half that of the nitrogen

If we combine the oxygen with the nitrogen (making a sort of synthetic air), we find that the pressure obtained is the sum of the individual pressures

3

ArAr

(0.200mol)(8.314J / molK)(300K)

(0.0030m )

n RTP

V= =

2H ArP P P 412 166 kPa 578kPa

Dalton’s LawWhat is the total pressure in atmospheres of a gas mixture that contains 1.00 g of H2 and 8.00 g of Ar in a 3.0 L container at 27C? What are the partial pressures of the two gases?

V = 3.0 L = 0.0030 m3 T = 27+273.15 K = 300 K m (H2) = 1.00 g; m (Ar) = 8.00 g

&2

2

H ArH Ar

n RT n RTp = p =

V VWe need nH2 and nAr

nH2 = (1.00 g/2.016 g/mol) = 0.496 mol nAr = (8.00 g/39.95 g/mol) = 0.200 mol

2H

3

n RT (0.496mol)(8.314J/molK)(300K)P

V 0.0030m

2H = = 412000 Pa

= 166000 Pa

silane oxygen silane

oxygen

1 2 120mmHgX = ; X = ; p = = 40mmHg

3 3 32 ×120mmHg

p = = 80mmHg3

tot oxygenP p 80mmHg

Dalton’s LawSilane, SiH4, reacts with O2 to give silicon dioxide and water

SiH4(g) + 2 O2(g) SiO2(s) + 2 H2O(g)

If you mix SiH4 with O2 in the correct stoichiometric ratio, and if the total pressure of the mixture is 120 mm Hg, what are the partial pressures of SiH4 and O2? When the reactants have been completely consumed, what is the total pressure in the flask

Ratio is 1:2

After reaction (e.g. by sparking mixture)

Because SiO2 is solid it does not contribute to gas pressure

Notice that the number of H20 molecules produced is equal to the number of O2 consumed.

Kinetic Molecular Theory of GasesGases are the best understood among the four phases of matter

Their behaviour has been predicted by a mathematical theory developed by Clausius, Maxwell and Boltzmann.

The theory is based on some simplifying assumptions:

3. The molecules of a gas exert no forces on one another except during collisions, so that between collisions they move in straight lines with constant velocities

2. The molecules of a gas are constantly moving in random directions with a distribution of speeds

1. A pure gas consists of a large number of identical molecules separated by distances that are large compared with their size

4. The collisions of the molecules with each other and with the walls of the container are elastic; no energy is lost during a collision

From these postulates, it is possible to calculate a model that leads mathematically to the ideal gas law

P (impulse per collision)

× (rate of collision with the walls)

P mass of particles

1

PV

2P (velocity of particles)

Pressure and kinetic energyConsider the consequences of this theory on collisions with the walls of the container

(A) Pressure is created by the impact of the gas molecules on a solid surface, e.g. the container walls

(B) If each particle doubles in mass, the impulses are twice as large

(C) If the gas density is doubled, then twice as many collisions occur with the walls

(D) If the average molecule speed is doubled, there are twice as many collisions per time interval and each impulse has twice the momentum

2kinetic

1E mv

2

The distribution of velocitiesIncreases with increasing

temperature

Kinetic energyWhen these postulates are expressed mathematically, the kinetic energy of each gas molecule is calculated, using the usual definition

random motion it gas moleculeshave different velocities and kinetic energy

The Maxwell-Boltzmanndistribution indicates a statisticaldistribution of molecule velocities

Even describing the average speedcauses difficulties, and there arethree common ways to do so:v(mp) – most probable speedv(av) – average speed

v(rms) – root mean square speed

2N2 2 i1

3i

Average of speed squaredwherev

K.E. Nmv vN

PV nRT

2 21 13 3 A

A

NNmv nRT n N mv RTN

2rms

3RTv v

M R = 8.3145 J mol-1 K-1 because J =

kg2m2/s2, M must be in kg/mol!!!!!

Maxwell-Boltzmann Statistical MechanicsThe mathematical model that takes all of the above factors into account results in the following prediction:

Recall that P V = energy (in units: N / m2 m3 = N m = J)

Compare the M-B equation to the ideal gas law (R in Joules):

We now combine and rearrange the terms, and convert from number of particles to moles of particles. (NA = Avogadro’s number)

In practice we will always use the root-mean-square velocity, for which by rearrangement: (m*NA = M = molar mass)

K.E. PV

2rms

3RTv v

M

2 21 21 2

3RT 3RTv ; v

M M

2 22 1 21

2 2 21 1

T v 240.15 (1.1v )T 240.15 1.1 290.58 291K

v v

12

1 12

22 2

3RTv TM

3RT TvM

Maxwell-Boltzmann Statistical MechanicsYou have a sample of helium gas at –33C, and you want to increase the average speed of helium atoms by 10.0%. To what temperature should the gas be heated to accomplish this?

T1 = -33 + 273.15 = 240 K; M (He) = 0.0040026 kg/mol; v2 = 1.1v1