2013 General Chemistry I 1 Chapter 4. THE PROPERTIES OF GASES 2013 General Chemistry I THE NATURE OF GASES THE GAS LAWS 4.1 Observing Gases 4.2 Pressure

2013 General Chemistry I 1

Chapter 4.THE PROPERTIES OF GASES

2013 General Chemistry I

THE NATURE OF GASES

THE GAS LAWS

4.1 Observing Gases4.2 Pressure4.3 Alternative Units of Pressure

4.4 The Experimental Observations4.5 Applications of the Ideal Gas Law4.6 Gas Density4.7 The Stoichiometry of Reacting Gases4.8 Mixtures of Gases


Prelude to Chapters 4 and 5. The Three States of Matter


An Overview of the Physical States of Matter

The Distinction of Gases from Liquids and Solids

1. Gas volume changes greatly with pressure.

2. Gas volume changes greatly with temperature.

3. Gases have relatively low viscosity.

4. Most gases have relatively low densities under normal conditions.

5. Gases are miscible.


THE NATURE OF GASES (Sections 4.1-4.3)

4.1 Observing Gases4.1 Observing Gases

Many of physical properties of gases are very similar, regardless of the identity of the gas. Therefore, they can all be described simultaneously.Samples of gases large enough to study are examples of bulk matter – forms of matter that consist of large numbers of molecules

Compressibility – the act of reducing the volume of a sample of a gasExpansivity - the ability of a gas to fill the space available to it rapidly

Two major properties of gases:


4.2 Pressure4.2 Pressure

-

– SI unit of pressure is the pascal (Pa)

– Pressure arises from the collisions of gas molecules on the walls of the container.


Barometer – A glass tube, sealed at one end, filled with liquid mercury, and inverted into a beaker also containing liquid mercury (Torricelli)

where h = the height of a column, d = density of liquid, and g = acceleration of gravity (9.80665 ms-2)

Measurement of Pressure


Manometer

-Two types of Hg manometer:(a) open-tube and (b) closedtube system

This is a U-shaped tube filled with liquid and connected to an experimental system, whose pressure is being monitored.


Self-Test 4.1B

The density of water at 20 oC is 0.998 g.cm-3. What heightwould the column of liquid be in a water barometer at 20 oCwhen the atmospheric pressure corresponds to 760. mm ofmercury?

Solution

P = dwhwg = dHghHgg

Hence hw =dHghHg

dw

=(13.595 g cm-3)(760. mm)

(0.998 g cm-3)

= 1.04 x 104 mm (10.4 m)


Self-Test 4.2A

What is the pressure in kilopascals in a system when themercury level in the system-side column in an open-tubemercury manometer is 25 mm lower than the mercury levelin the atmosphere-side column and the atmospheric pressurecorresponds to 760. mm of mercury at 15 oC?

P = dhg = (13595 kg m-3)(0.785 m)(9.80665 m s-1)

= 104657 Pa = 105 kPa

Solution


Example 4.3

135s

A student attaches a glass bulb containing neon gas to an open-tube

manometer and calculates the pressure of the gas to be 0.890 atm.

(a) If the atmospheric pressure is 762 Torr, what height difference

between the two sides of the mercury in the manometer did the

student find?

(b) Which side is higher, the side of the manometer

attached to the bulb or the side open to the

atmosphere?

(c) If the student mistakenly switches the numbers

for the sides of the manometer when recording

the data in the laboratory notebook, what

would be the reported pressure in the gas bulb?

135


Solution to Exercise 4.3


4.3 Alternative Units of Pressure4.3 Alternative Units of Pressure

- 1 bar = 105 Pa = 100 kPa- 1 atm = 760 Torr = 1.01325×105 Pa (101.325 kPa)- 1 Torr ~ 1 mmHg

mbar

Weather map


Self-Test 4.3A

The US National Hurricane Center reported that the eyeof Hurricane Katrina (2005) fell as low as 902 mbar. Whatis the pressure in atmospheres?

1 atm = 1.01325 bar, hence 0.920 bar =

Solution

0.920 bar x 1 atm

1.01325 bar= 0.890 atm


THE GAS LAWS (Sections 4.4-4.6)

4.4 The Experimental Observations4.4 The Experimental Observations

Boyle’s law: For a fixed amount of gas at constant temperature, volume is inversely proportional to pressure.

This applies to an isothermal system (constant T) with a fixed amount of gas (constant n).


- For isothermal changes between two states (1 and 2),


Self-Test 4.4B

In a petroleum refinery a 750.-L container containing ethylenegas at 1.00 bar was compressed isothermally to 5.00 bar. Whatwas the final volume of the container?

Solution

Isothermally means at constant temperature, hence Boyle'slaw can be used.

P1V1 = P2V2, (1.00 bar)(750. L) = (5.00 bar)V2

V2 = 150. L


Charles’s law: For a fixed amount of gas under constant pressure, the volume varies linearly with the temperature.

This applies to an isobaric system (constant P) with a fixed amount of gas (constant n).


- Kelvin temperature scale

T = 0 K = -273.15 oC,

when V → 0.

- Celsius temperature scale

t (oC) = T (K) - 273.15

0 oC = 273.15 K

The Kelvin Scale of Temperature

If a Charles’ plot of V versus T (at constant P and n) is extrapolated to V = 0, the intercept on the T axis is ~-273 oC.


Another aspect of gas behavior (Gay-Lussac’s Law)

This applies to an isochoric system (constant V) with a fixed amount of gas (constant n).


Self-Test 4.5A

A rigid oxygen tank stored outside a building has a pressureof 20.00 atm at 6:00 am when the temperature is 10. oC. Whatwill be the pressure in the tank at 6:00 pm, when the temperatureis 30. oC?

Solution

Volume is constant, hence Gay-Lussac's Law can be used.

P1

T1

= P2

T2

(20.00 atm)

(283.15 K)

= P2

(303.15 K)

P2 = 21.41 atm


Avogadro’s Principle

Under the same conditions of temperature and pressure, a given number of gas molecules occupy the same volume regardless of their chemical identity.

- This defines molar volume


Self-Test 4.6A

A helium weather balloon was filled at -20. oC and acertain pressure to a volume of 2.5 x 104 L with 1.2 x 103

mol He. What is the molar volume of helium under theseconditions?

Solution

Molar volume =Volume

No. moles= (2.5 x 104 L)

(1.2 x 103 mol)

= 21 L mol-1


This is formed by combining the laws of Boyle, Charles, Gay-Lussac and Avogadro.

The ideal gas law:

Gas constant, R = PV/nT.It is sometimes called a “universal constant” andhas the value 8.314 J K-1 mol-1 in SI units, althoughother units are often used (Table 4.2).

The Ideal Gas Law


-The ideal gas law, PV = nRT, is an equation of state that summarizes the relations describing the response of an ideal gas to changes in pressure, volume, temperature, and amount of molecules; it is an example of a limiting law.(it is strictly valid only in some limit: here, as P 0.)

Table 4.2. The Gas Constant, R


- Standard ambient temperature and pressure (SATP)

298.15 K and 1 bar, molar volume at SATP = 24.79 L·mol-1

- Standard temperature and pressure (STP)

0 oC and 1 atm (273.15 K and 1.01325 bar)

- Molar volume at STP

- For conditions 1 and 2,

- Molar volume

4.5 Applications of the Ideal Gas Law4.5 Applications of the Ideal Gas Law


EXAMPLE 4.4

In an investigation of the properties of the coolant gas used in an air-conditioning system, a sample of volume 500 mL at 28.0 oC wasfound to exert a pressure of 92.0 kPa. What pressure will the sampleexert when it is compressed to 30 mL and cooled to -5.0 oC?


Self-Test 4.7B

A idling, badly tuned automobile engine can release asmuch as 1.00 mol of CO per minute into the atmosphere.At 27 oC, what volume of CO, adjusted to 1.00 atm, isemitted per minute?Solution

This question requires the use of PV = nRT, whereP = 1.00 atm, n = 1.00 mol (min-1), and T = 300.15 K.(R = 8.206 x 10-2 L atm K-1 mol-1)

V = (1.00 mol (min-1)(0.08206 L atm K-1 mol-1)(300.15 K)

(1.00 atm)= 24.6 L (min-1)

Calculating the pressure of a given sample


Self-Test 4.8A

A sample of argon gas of volume 10.0 mL at 200. Torr isallowed to expand isothermally into an evacuated tubewith a volume of 0.200 L. What is the final pressure of theargon in the tube?

P1V1

T1

P2V2

T2

Solution

The volume is increased by a factor of 20, so we expect adecrease in pressure by the same factor, under isothermalconditions.

n1 n2

= , where T1 = T2 and n1 = n2(reduces to Boyle's Law)

(200. Torr)(10.0 mL) = P2(Torr)(200 mL)

P2 = 10.0 Torr

Using the combined gas law when one variable is changed


Self-Test 4.9A

A parcel of air (the technical term in metereology for asmall region of the atmosphere) of volume 1.00 x 103 Lat 20. oC and 1.00 atm rises up the side of a mountainrange. At the summit, where the pressure is 0.750 atm,the parcel of air has cooled to -10. oC. What is the volumeof the parcel of air at that point?

Solution

P1V1

n1T1=

P2V2

n2T2

where n1 = n2

(1.00 atm)(1.00 x 103 L)

(293.15 K)= (0.750 atm)V2(L)

(263.15 K)

V2 = 1.20 x 103 L

Using the combined gas law when two variables are changed


Self-Test 4.10A

Calculate the volume occupied by 1.0 kg of hydrogen at 25 oCand 1.0 atm.

Solution

We can use the ideal gas equation PV = nRT, after first findingthe number of moles of H2 in 1.0 kg.

n = Mass

Molar mass

1.0 x 103 g

2.016 g/mol= = 496 mol

(1.0 atm)V(L) = (496 mol)(8.206 x 10-2 L atm K-1mol-1)(298.15 K)

V = 1.2 x 104 L


Example 4.25 143s

A sample of methane gas, CH4, was slowly heated at a constant pressure

of 0.90 bar. The volume of the gas was measured at a series of different

Temperatures and a plot of volume vs. temperature was constructed.

The slope of the line was 2.88×10-4 L K-1. What was the mass of the

sample of methane?

Solution


4.6 Gas Density4.6 Gas Density

Molar concentration of a gas at STP (where molar volume is 22.4141 L):

Molar concentration of a gas is the number moles divided by the volumeoccupied by the gas.

Density, however, does depend on the identity of the gas.

This value is the same for all gases, assuming ideal behavior.


Density at STP

• For a given P and T, the greater the molar mass, the greater its density.

• At constant T, the density increases with P. In this case, P is increased either by adding more material or by compression (reduction of V).

• Raising T allows a gas to expand at constant P, increases V and therefore reduces its density.

Gas Density Relationships


Self-Test 4.11A

The oil produced from eucalyptus leaves contains the volatileorganic compound eucalyptol. At 190. oC and 60 Torr, a sampleof eucalyptol vapor had a density of 0.320 g L-1. Calculate themolar mass of eucalyptol.

Solution

Density =MP

RTwhere M is molar mass

0.320 g L-1 = M(g mol-1)(60 Torr)

(62.324 L Torr K-1 mol-1)(463.15 K)

M = 154 g mol-1


-Molar volumes of gases are generally > 1000 times those of liquids and solids. e.g. Vm (gases) = ~ 25 L mol-1; Vm (liquid water) = 18 mL mol-1

-Reactions that produce gases from condensed phases can be explosive.

e.g. sodium azide (NaN3) for air bags

4.7 The Stoichiometry of Reacting Gases4.7 The Stoichiometry of Reacting Gases


EXAMPLE 4.6

The carbon dioxide generated by the personnel in the artificial atmosphere of submarines and spacecraft must be removed form the air and theoxygen recovered. Submarine design teams haveinvestigated the use of potassium superoxide, KO2,as an air purifier because this compound reacts withcarbon dioxide and releases oxygen:

4 KO2 (s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)

Calculate the mass of KO2 needed to react with 50 Lof CO2 at 25 oC and 1.0 atm.

Vm = 24.47 Lmol-1; 1 mol CO2 -> 2 mol KO2; MKO2 = 71.10 gmol-1


Solution

6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)

Self-Test 4.12A

Calculate the volume of carbon dioxide, adjusted to 25 oC and 1.0 atm,that plants need to make 1.00 g of glucose, C6H12O6, by photosynthesisin the reaction

From the equation, the stoichiometry of CO2:glucose is 6:1.The molar mass of glucose is 180 g/mol.The molar volume of CO2 at 25 oC and 1 atm is 24.47 L mol-1

Volume of CO2 =24.47 L

1 mol CO2

x6 mol CO2

1 mol glucosex1.00 g glucose

180 g glucose

1 mol glucosex

= 0.82 L


Example 4.59

148s

A 15.0-mL sample of ammonia gas at 1.00×102 Torr and 30 oC is mixed

with 25.0 mL of hydrogen chloride gas at 1.50×102 Torr and 25 oC, and

the following reaction takes place:

NH3(g) + HCl(g) NH4Cl(s)

(a) Calculate the mass of NH4Cl that forms.


148s


(b) Identify the gas in excess and determine the pressure of the excess

gas at 27 oC after the reaction is complete (in the combined volume

of the original two flasks).

148s


4.8 Mixtures of Gases4.8 Mixtures of Gases

– A mixture of gases that do not react with one another behaves like a single pure gas.

Partial pressure: The total pressure of a mixture of gases is the sum of the partial pressures of its components (John Dalton).

P = PA + PB + … for the mixture containing A, B, …

- Humid gas: P = Pdry air + Pwater vapor (Pwater vapor = 47 Torr at 37 oC)

mole fraction: the number of moles of molecules of the gas expressed as a fraction of the total number of moles of molecules in the sample.


EXAMPLE 4.7

Air is a source of reactants for many chemical processes. To determinehow much air is needed for these reactions, it is useful to know thepartial pressures of the components. A certain sample of dry air oftotal mass 1.00 g consists almost entirely of 0.76 g of nitrogen and0.24 g of oxygen. Calculate the partial pressures of these gases whenthe total pressure is 0.87 atm.


Self-Test 4.14A

A baby with a severe bronchial infection is in respiratory distress.The anesthetist administers heliox, a mixture of helium and oxygenwith 92.3% by mass O2. What is the partial pressure of oxygen beingadministered to the baby if the atmospheric pressure is 730 Torr?

n(He)0.077 g

4.00 g mol-1n(O2) =

0.923 g

32.0 g mol-1=

Solution

= 0.0193 mol = 0.0288 mol

x(He) =0.0193 mol

0.0481 mol= 0.401

x(O2) =0.0288 mol

0.0481 mol= 0.599

[= 1.00 - 0.401 (only two components)]

P(O2) = x(O2)P = 0.599 x 730 Torr = 437 Torr


Chapter 4.THE PROPERTIES OF GASES

2012 General Chemistry I

MOLECULAR MOTION

REAL GASES

4.9 Diffusion and Effusion4.10 The Kinetic Model of Gases4.11 The Maxwell Distribution of Speeds

4.12 Deviations from Ideality4.13 The Liquefaction of Gases4.14 Equations of State of Real Gases


MOLECULAR MOTION (Sections 4.9-4.11)

4.9 Diffusion and Effusion4.9 Diffusion and Effusion

Diffusion: gradual dispersal of one substance through another substance

Effusion: escape of a gas through a small hole into a vacuum


Graham’s law: At constant T, the rate of effusion of a gas is inversely proportional to the square root of its molar mass:

Strictly,Graham’s law relates to effusion, but it can also be used for diffusion.

For two gases A and B with molar masses MA and MB,


Rate of effusion and average speed increase as the square root of the temperature:

Combined relationship: The average speed of molecules in a gas is directly proportional to the square root of the temperature and inversely proportional to the square root of the molar mass.


Self-Test 4.15A

It takes 30. mL of argon 40. s to effuse through a porousbarrier. The same volume of vapor of a volatile compoundextracted from Caribbean sponges takes 120. s to effusethrough the same barrier under the same conditions. Whatis the molar mass of the compound?

Solution

Time for Ar to effuse

Time for unknown to effuse

M(Ar)

M(unknown)=

40 (s)

120 (s)

39.95 (g mol-1)

M(unknown)

M(unknown)

=

= 360 g mol-1


Example 4.73

152s

A sample of argon gas effuses through a porous plug in 147 s.

Calculate the time required for the same amount of (a) CO2, (b) C2H4,

(c) H2, and (d) SO2 to effuse under the same conditions of pressure

and temperature.


4.10 The Kinetic Model of Gases4.10 The Kinetic Model of Gases

Kinetic molecular theory (KMT) of gases makes four assumptions:

1. A gas consists of a collection of molecules in continuous random motion.2. Gas molecules are infinitesimally small points.

3. The molecules move in straight lines until they collide.4. The molecules do not influence one another except during collisions.

- Collision with walls: consider molecules traveling only in one dimensional x with a velocity of vx.


The change in momentum (final – initial)of one molecule: -2mvx = 2mvx momentumchange for the wall

All the molecules within a distance vxt of the walland traveling toward it will strike the wall during theInterval t.

If the wall has area A, all the particles in a volumeAvxt will reach the wall if they are traveling toward it.


The number of molecules in the volume Avxt is thatfraction of the total volume V, multiplied by the totalnumber of molecules:

The average number of collisions with the wall duringthe interval t is half the number in the volume Avxt:

The total momentum change = number of collisions × individual molecule momentum change


Force = rate of change of momentum = (total momentum change)/t

for the average value of <vx2>

Mean square speed:

Pressure on wall:


where vrms is the root mean square speed,

or

- The temperature is proportional to the mean square speed of the molecules in a gas.- This was the first acceptable physical interpretation of temperature: a measure of molecular motion.


EXAMPLE 4.7

What is the root mean square speed of nitrogenMolecules in air at 20 oC?


Self-Test 4.16A

Solution

Estimate the root mean square speed of water moleculesin the vapor above boiling water at 100. oC.

Molar mass of water is 18.01 g mol-1 or 0.01801 kg mol-1.

From vrms = (3RT/M)1/2,

vrms =3 x (8.3145 J K-1 mol-1) x (373 K)

0.01801 kg mol-1

= 719 m s-1


A molecular description of Boyle’s Law

153s


A molecular description of Charles’s Law

153s


A molecular description of Avogadro’s Law

153s


A molecular description of Dalton’s law of partial pressures153s


4.11 The Maxwell Distribution of Speeds4.11 The Maxwell Distribution of Speeds

v = a particle’s speed

N = the number of molecules with speeds in the range between v +v

N = total number of molecules; M = molar mass

f(v) = Maxwell distribution of speeds

For an infinitesimal range,

average speed

Maxwell derived equation 22, for calculating the fraction of gas moleculeshaving the speed v at any instant, from the kinetic model.


- Molar mass (M) dependence:as M increases, the fraction of molecules withspeeds greater than a specific speed decreases.

- Temperature dependence:as T increases, the fraction of molecules with speeds greater than a specific speed increases.


REAL GASES (Sections 4.12-4.14)

4.12 Deviations from Ideality4.12 Deviations from Ideality

- Gases condense to liquids when cooled or compressed (attraction).- Liquids are difficult to compress (repulsion).

Deviation from ideal gases

- Deviations from the ideal gas law are significant at high pressures and low temperatures (where significant intermolecular interactions exist).


Compression factor (Z): the ratio of the actual molar volume of the gas to the molar volume of an ideal gas under the same conditions.

For an ideal gas, Z = 1

Long range attractions; smaller Z, condensation of gases

Short range repulsions; larger Z, low compressibility of liquids and solids,finite molecular volume


- For many gases, attractions dominate at low pressure (Z < 1), while repulsive interactions dominate at high pressure (Z > 1).


4.13 The Liquefaction of Gases4.13 The Liquefaction of Gases

Joule-Thomson effect: when attractive forces dominate, a real gas cools as it expands.

– In this case expansion requires energy, which comes from the kinetic energy of the gas, lowering the temperature. – The effect is used in some refrigerators and to effect the condensation of gases such as oxygen, nitrogen, and argon.


The Linde refrigerator for the liquefactionof gases

i.e. Adiabatic cooling; temperature decrease under isentropic expansionof any gas (w 0)


4.14 Equations of State of Real Gases4.14 Equations of State of Real Gases

Virial equation:

van der Waals equation:

or

pressure reduced due to attractions between pairs of molecules

–nb volume excluded since molecules cannot overlap (repulsions)

b volume excluded by 1 mol ~ molar volume in the liquid state


The effect of intermolecular attractions on

measured gas pressure

Pideal = P + a(n/V)2(P: actual pressure)

actual pressure is smaller than the ideal pressure

161s


The effect of molecular volume on measured gas volume

Videal = V - nb

actual volume is greater than the ideal volume

161s


Virial expansion of the van der Waals equation

At low particle densities


Table 4.5 Van der Waals Parameters for some Common Gases


Model of gas

1. A large number of gas molecules in ceaseless, random, and straight motion.

2. The average speed and the spread of speeds increase with T and decrease with m.

3. Molecules travel in straight lines until they collide with other molecules or the container wall.

4. Widely separated. Intermolecular forces have only a weak effect on the properties.

5. Repulsions increase the molar volume, whereas attractions decrease the molar volume.


EXAMPLE 4.9 Refrigerant gas (a = 16.2 L2 atm mol–2, b = 0.084 L/mol), 1.50 mol in 5.00 L at 0 oC; Estimate the pressure.

Use expanded form of the van der Waals equation:P = nRT/(V – nb) - an2/V2


Self-Test 4.17A

Solution

A 10.0-L tank containing 25 mol of O2 is stored in a diving supplyshop at 25 oC. Use the data in table 4.5 and the van der Waalsequation to estimate the pressure in the tank.

From P = nRT/(V - nb) -an2/V2,

P =(25 mol) x (0.08206 L atm K-1 mol-1) x (298 K)

10.0 L (25 mol) x (3.19 x 10-2 L mol-1)_

_ (1.364 L2 atm mol-2) x(25 mol)2

(10.0 L)2

= 58 atm

Documents

2013 General Chemistry I 1 Chapter 4. THE PROPERTIES OF GASES 2013 General Chemistry I THE NATURE OF GASES THE GAS LAWS 4.1 Observing Gases 4.2 Pressure