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CHAPTER 9

GASES: THEIR PROPERTIES AND BEHAVIOUR

The Air Around Us

Measuring Gases

Any sample of gas has , and . Unlike solids and liquids, gases expand to fill all of the space available. Gas mixture are always homogeneous at equilibrium.They are also very much more compressible than solids and liquids.

Under the same conditions carbon dioxide, CO2(g) o, is heavier than air (higher density) and , H2 (g)mis lighter (lower density) than air. So, in pouring carbon dioxide it must be poured down and hydrogen would have to be poured up! Most of a sample of a gas at atmospheric is empty space.

9.1 Amount of Gas

There are various properties that we can measure for a sample of gas:

Mass m Volume V Pressure P T

To be useful in our calculations we need to be able to relate these to the number of moles present in the sample.

9.2 of Pressure

Torricelli (1608 - 1647) - Mercury Barometer The atmospheric pressure varies from day to day but can normally support a column of mercury about 760 mmHg high. Consider a column 760 mm tall x 1 cm2 cross-section. What pressure does it exert in Pascal the SI pressure unit? The densityof mercury, Hg = 13.534 g cm-3.. Mass of the column = 76.0 cm x 1 cm2 x 13.534 g cm-3

= 1028.84 g = 1.028584 kg So, Force = mass x acceleration

-2 = 1.029 kg x 2.81 m s = 10.09 kg m s-2 The pressure is force per unit area so now we need to convert to the SI base unit Since this is the force per cm2 we must convert to the force per m2. 1 cm2 = ? m2

1 2 1 2 -4 2 = ( 100 m) = 1000 m = 10 m 10.09 kg m s-2 5 -1 -2 5 -2 Pressure = 10-4 m2 = 1.009 x 10 kg m s or 1.009 x 10 N m = 101 kPa (since 1 N m-2 = 1 Pa pascal)

The Standard Atmosphere

The standard atmosphere is defined as = 760 mmHg = 101.3 kPa = 1 atm. 1 standard atm = 101.3 kPa = 760 mmHg. Since Hg expands with temperature, correction must be applied (density changes!) Since mercury is recognized as a poisonous substance (remember the Mad Hatter!) mercury barometer are not often seen today. One might ask why water is not used instead of mercury. However, because the density of water is only 1.0 g cm-3 a water barometer would have to be about ~ 33.7 feet or ~10 tall not ideal for the dining room!

9.3 Note: Standard pressure for gas measurements has been redefined to be 1 bar, or 100.000 kPa rater than 101.3 kPa (1 atm) .

Most household barometers are aneroid barometers like the one in the picture opposite.

Measuring Gas Pressure

The Gas Laws

9.4 Early Study 1660

Robert Boyle - Boyle’s Law How does the volume of a gas change as the pressure changes (keeping the temperature and the number of moles constant)? This can be studied quite easily using something a simple as a bicycle pump and give results similar to the following graphs:

Volume is inversely proportional to pressure 1 1 V } P or V = constant x P so PV = constant at constant n and T

Amount of substance Amount of Substance/Volume

V varies directly with n so V = k1n where V k1 is a constant. P & T constant Double moles and the volume doubles (if P and T constant)! n, number of moles

Charles’ Law V directly proportional to T absolute V Charles, (1746 - 1823) and (Gay-Lussac 1778 - 1850)

If we use the Scale - add 273.1 to -273 the Celsius temp. - get new expression o O t C V is directly proportional to T V = Vo + k t 9.5 V = kT

So we have V = k n and V = k T The laws can be combined PV = k (amount of substance constant) PV = k n T m The new k is a constant for any sample of gas and this we call R the gas const. Law PV = nRT The law fits behaviour of well-behaved gases quite well but there are significant deviations for most real gases especially near the boiling point.

Avogadro Law (1776 - 1856)

Equal volumes of gases at the same temperature and pressure contain equal numbers of

Consider the reaction

N2(g) + 3 H2(g) ® 2 NH3(g)

Say we have 1 volume of N2(g) and 5 volumes of H2(g) and the reaction goes to completion. What will the final volume be at the same temperature and pressure?

N2(g) + 3 H2(g) ® 2 NH3(g) Total volume Initial 1 volume 5 volumes 0 volumes 6 volumes Final 0 volumes 2 volumes 2 volumes 4 volumes (limiting) (excess)

9.6 Standard Temperature & Pressure STP

Standard Pressure 1 atm 760 mmHg (760 torr) 101.3 kPa

Standard Temperature 273.15 K (0 ºC)

At STP, the for a well behaved (ideal) gas is about 22.4 L mol–1 Using these values, we get R by rearranging PV = nRT PV 760 mm Hg x 22.4 L R = nT = 1 mol x 273.15 K R = 62.3 L mmHg K–1 mol–1

or if the pressure is measured in atmospheres

1 atm x 22.4 L R = 1 mol x 273.15 K

R = 0.0820 L atm K-1 mol-1

or if the pressure is measured in kilopascals, kPa 101.3 kPa x 22.4 L R = 1 mol x 273.15 K

R = 8.31 L kPa K-1 mol-1 but 1 L kPa = 1 N m = 1 J so R = 8.314 J K-1 mol-1

9.7 Simple Calculations

Example If 3.66 L of a gas is collected at 0.886 atm and 100 ºC, what would the volume be at 1.000 atm pressure? Keeping n & T constant. PV PV = nRT but at constant n and T we can say T = constant

P1V1 P2V2 P1V1 = constant = P2V2 or = T 1 T 2

P1V1 V2 = will expect the volume to decrease! P2 0.886 atm x 3.66 L = 1.000 atm

Ans: 3.24 L

Example A rigid car tyre is pumped to 28.0 lbs in-2 at –10.0 ºC. What is the pressure in the tyre at 20.0 ºC? Here we have constant V and n P V P V P P Again PV = nRT and 1 1 = 2 2 but V is constant 1 = 2 T 1 T 2 T1 T2

This the pressure should increase.... We must convert the to Kelvin

T1 = –10.0 + 273.15 K = 263.15 K and T2 = 20.0 + 273.15 K = 293.15 K

P1T2 293.15 K P = = 28.0 lbs in-2 x = 31.2 lbs in-2. 2 T1 263.15 K

This is enough to be a problem, so be careful if you plan to drive fast!

Example: Making O2 in the Laboratory

What volume of “dry” O2 collected at 25.0ºC and 1.0 atm when 50.0 mL of 5.0% H2O2 is decomposed (density 1.0 g mL-1)

9.8 2 H2O2 t 2 H2O + O2(g) 1 1 1/2

-1 5.0% Mass H2O2 = 50.0 mL x 1.0 g mL x 100%

= 2.50 g 2.50 g available moles of H2O2 = = 0.07350 mol 34.01 g. mol-1

mol O2 = ½ x 0.07350 mol = 0.0368 mol nRT 0.0368 mol x 0.0821 L atm K-1 mol-1 x 298.1K V = P = 1.0 atm = 0.900 L

Ans: The volume of dry O2 = 0.90 L

Stoichiometry in Gas Reactions

Once we can calculate moles of gases, we can use gas volumes in calculations. Consider the reaction:

N2(g) + 3 H2(g) ® 2 NH3(g)

What volume of hydrogen, H2, measured at 3.00 atm and 28.0 ºC is required to make 1.00 kg of ammonia, NH3? 1.00 kg x 1000 g kg-1 Moles NH3 required = 17.03 g mol-1 = 58.719 mol. Divide by 2 to focus on the ammonia: 1 3 2 N2(g) + 2 H2(g) ® 1 NH3(g) 3 Moles of H2 required = 2 x 58.719 mol = 88.079 mol. Using the PV = nRT nRT and rearranging gives V = P and substituting 88.079 mol x 0.0821 L atm K-1 mol-1 x 301.15 K V = 3.00 atm = 726 L Ans: 726 L

9.9 250 mL of dry HCl(g) at 25.0ºC and 790 mmHg is bubbled into 50.0 mL of ice cold water so that it all dissolves. The solution is then made up to 100.0 mL. What is the of HCl (molarity) in the final solution? mol PV CHCl = V b PV = nRT n = RT

790 mmHg x 0.250 L 760 mmHg atm-1 n = 0.0821 L atm -1 K-1 mol-1 x 298.15 K =0.0106145 mol, so 0.0106145 mol -1 cHCl = 100.0 x 10-3 L = 0.106 mol L Ans: 0.106 mol L-1 Notice the concentration of the HCl could also be found by titration

What volume of the CO2(g) + H2O(g) gases produced when 1.00 g of octane, C8H18 , burns - measured at, say, 130 ºC at a car exhaust at 100.0 kPa? So first write the balanced equation (complete combustion of octane gives only carbon dioxide and water)

C8H18(g) + 12½ O2(g) ® 8 CO2(g) + 9 H2O(g)

1.00 g -3 Moles of octane burned = 114.22 g mol-1= 8.755 x 10 mol

-3 so total moles of gas formed = 17 x 8.755 x 10 mol = 0.1488 mol nRT 0.1488 mol x 8.314 L kPa K-1 mol-1 x 403.15 K now, V = P = 100.0 kPa = 4.99 L Ans: 4.99 L

Mixtures of Gases - ’s Law of Partial

Recall - Equal volumes of gases at the same T & P contain equal moles. This means that it is the number of moles that - not the mass or what the gas is - as long as they don’t react. Dalton extended this to deal with mixtures of non-reacting gases. The total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gas.

Ptotal = pA + pB + pC ......

ntotalRT nART nBRT nBRT Ptotal = V = V + V + V ......

9.10 So, if we have a mixture of gases, we can treat it as if it was one pure gas - as long as they don’t react!

Example What is the total pressure inside a 2.00 L container that contains 1.00 g of Ne, 1.00 g

of N2 and 1.00 g of CO2? What is the partial pressures of neon in the mixture. All pressure measured at 25.0º C. Calculate the number of moles of each gas: 1.00 g mol Ne 20.18 g mol-1 = 0.04955 mol 1.00 g mol N2 28.01 g mol-1 = 0.03570 mol 1.00 g mol CO2 44.01 g mol-1 = 0.02272 mol

Total moles 0.1080 mol

PV = nRT

nRT 0.1080 mol x 0.0821 L atm K-1mol-1x 298.15 K P = V = 2.00 L Answer: total pressure = 1.32 atm

The is the pressure that the neon gas would exert on 0.04955 mol x 0.0821 L atm K-1mol-1x 298.15 K = 2.00 L = 0.606 atm. or in proportion to the FRACTION of the moles that are Ne 0.04955 mol = 0.1080 mol

xNe = 0.459

pNe = 0.459 x 1.322 atm = 0.606 atm This number is called the and usually given the symbol x. Notice that it has no units.

9.11 To review:

Mole fraction We have seen that the mole fraction is simply the fraction of the total moles that is the mol Ne substance of interest. Thus xNe = total mol and it is not difficult to see that the partial pressure, pNe = XNe x Ptotal Since the above is true

pNe xNe % Ptotal XNe = = Ptotal Ptotal In the above question .02272 mol x = = 0.2103 CO2 0.1080 mol so PCO2 = 1.32 atm x 0.2103 = 0.278 atm.

A common circumstance where a gas mixture is obtained - though it might not be immediately obvious - is the collection of a gas over water. Zinc is an active metal that reacts with dilute acid to give hydrogen gas:

2+ - Zn(s) + 2 HCl(aq) ® Zn (aq) + 2 Cl (aq) + H2(g)

+ - 2+ - Zn(s) + 2 H (aq) + Cl (aq) ® Zn (aq) + 2 Cl (aq) + H2(g)

+ 2+ Zn(s) + 2 H (aq) ® Zn (aq) + H2(g) The zinc is oxidized and the hydrogen are reduced.

The H2(g) obtained is a mixture of H2O(g) and H2(g) so the total pressure is

Ptotal = pH2 + pH2O pH2O, the vapour pressure of water depends on temperature:

9.12 The table shows how the vapour pressure (mmHg) changes with temperature. Something to think about: Why/when does a liquid boil? If we collect hydrogen over water at 25.0ºC and 770.0 mmHg, the gas collected is saturated with water vapour (to the saturated vapour pressure).

Ptotal = 770.0 mmHg = Ptotal = pH2 + pH2O

= pH2 + 23.76 mmHg so, pH2 = 770.0 mmHg - 23.76 mmHg

If we were to dry the gas, keeping other things constant, the pressure should drop to this value.

So, if we collect 600 mL of H2 over water, how many moles of H2 do we really have?

PV = nRT

770.0 mmHg - 23.76 mmHg Moles of PV 760.0 mmHg x 0.600 L = RT = hydrogen, n 0.0821 L atm-1 K-1 mol-1 x (25.0 + 273.15 K)

n = 0.02407 mol

What volume would the gas occupy if collected under the same conditions DRY? -1 -1 nRT 0.02407 mol x 0.0821 L atm K mol x 298.15K V = P = 770.0 mmHg 760 mmHg atm-1

= 0.5815 L or 582 mL

9.13 So, as expected, the volume collected is a little greater than it should have been because the gas was wet. p H2 746.24 mmHg Using the mole fraction of H2 = = = 0.9691 Ptotal 770.0 mmHg

so fraction of volume = 600 mL x 0.9691 = 581 mL, just as it should be ☺ (notice the effect of the borderline significant digit)

A 0.4480 g sample of an alkali metal was allowed to react with water and the H2(g) produced was collected over water at 1.00 atm at 25.0ºC. the volume collected was 246 mL. What was the of the alkali metal. The vapour pressure of water, p at 25.0ºC = 0.0313 atm. H2 First we must write the equation ....

2 M + 2 H2O ® 2 MOH + H2(g)

- 2 M + 2 H2O ® 2 M+(aq) + 2 OH (aq) + H2(g)

p = 1.00 atm – 0.0313 atm = 0.9687 atm H2 PV PV=nRT n = RT 0.987 atm x 0.246 L -3 Mole of H2 produced = 0.0821 L atm K-1 mol-1 x 298.15 K = 9.735 x 10 mol -3 Moles of the alkali metal, M = 2 x 9.735 x 10 mol = 0.01947 mol 0.4480g -1 â molar mass = 0.01947mol = 23.0 g mol So what was the alkali metal?

Most gases don’t behave quite as predicted by the gas laws.

However, we can define exactly what a perfect gas should be by considering what we call Kinetic Theory

Kinetic Theory

The basic postulates of the Kinetic Theory are

9.14 1. Gases consist of particles in continuous random motion. They collide with one another and with the walls of the container. Gas Pressure is the result of the collisions. (for those doing physics: it is the change in momentum of the gas particles per unit surface area per ) 2. Collisions must be perfectly elastic. All Kinetic energy is retained (not converted to heat) 3. The particles must be so small that the volume they occupy can be ignored (To be ideal they should occupy no volume!). A gas is mostly empty space. 4. The gas particles do not react or interact. (attractive forces)

5. Average translational (motion) kinetic energy, Et, of gas particles is directly proportional to the absolute temperature (ie. T µ K) or the Et = constant x T where T is the absolute temperature.

6. At a given temperature, all gases have the same average kinetic energy , Et and the constant is the same (universal) for any gas. Since, the kinetic energy of the particles is given by E = ½ mv2, heavy particles move more slowly on average.

Graham’s Law: and Effusion of Gases

Since the Kinetic Energy, K.E., of the gas particles depends only on the absolute temperature (Kelvin) of the gas we can easily guess that heavy particles will travel more slowly than light particles. The higher the temperature the higher the average speeds. Graham observed that the rate of effusion of a gas is inversely proportionally to the square root of its molar mass, M.

Rate 1 ∞ Molar Mass So that for two gases 1 and 2

Rate1 M2 Rate2 ∞ M1 We can see that this comes from the the observation that the K.E. depends on temperature. At a constant temperature

9.15 1 2 1 2 2 m1u1 gas 1 = 2 m2u2 gas 2

u1 m2 u = 2 m1 which is similar to Graham’s Law. The faster the gas particles move (on average) the faster they would be expected to effuse or diffuse. This property has been use separate mixtures of gases. It can also be used to determine the molar mass of a gas.

The Behaviour of Real Gases

Real gases have that size and occupy part of the volume of a the gas. At low pressure the percentage of the volume used in this way would be very small (at STP about 0.05%) but a high pressure it can become significant (500 atm about 20%). This leads to deviations from the gas laws at high pressures. Similarly, if there are attractive forces between the particle the molecular collisions are not quite perfectly elastic. To get around these problems attempts have been made to correct for these deviations.

The van der Waals equation. an2 P + V2 (V - nb) = nRT The van der Walls equation attempts to do this with corrections for the pressure and volume.

Gas Density

Since a gas is mostly empty space, gas are much lower than those of solids and liquids. The density of water at room temperature is 1.0 g mL-1 or 1000 g L-1 so the volume of a 18.0 g mol-1 -1 -1 mole is 1.0 g mL-1 =18.0 mL mol or 0.0180 L mol .

-1 The volume of a mole of carbon dioxide, CO2(g), at STP = 22.4 L mol (about) so the 44.0 g mol-1 mass -1 density is = volume = 22.4 L mol-1 =1.964 g L . Because gas density is so small we normally quote it in g L-1 rather than g mL-1.

9.16 We normally quote gas density as g L-1 rather than g mL-1

The gas law given by PV = nRT can be expressed in terms of density if we recall that mass density = volume and that mass = molar mass (M) x number of moles (n) or mass = nM MP n x M, so density, d = nRT = RT P

Gas density measurements were once used to determine properties of gases. Eg. What is the molar mass of the gas particles, are the monatomic or diatomic etc. dRT molar mass of gas, M = P

Example The density of sulfur vapour at 445 ºC and 755 mmHg is found to be 4.33 g L-1. Use this information to find the molecular formula of the sulfur in the vapour. This is not difficult if we think the problem through carefully.

First we will assume that the vapour is molecules of sulfur with formula SY where Y is a whole number. The density tells us that the mass of 1 = 4.33 g so the number of moles of sulfur , n, in 1 litre of the gas is 4.33 g 0.1350 mol PV = Y x 32.066 g mol-1 = Y . So, because PV = nRT, and n = RT PV 0.1350 mol RT Combining the relationships, RT = Y so Y = 0.1350 mol % PV 0.0821 L atm K-1 mol-1 x 718.15 K = 0.1350 mol % 755 mm Hg = 8.006(no units, just a number) So the % 1 L 760 mm Hg atm-1

formula is S8 and we know from other studies the looks like

S S S S S S S S

9.17