chapter 3gas

8/6/2019 CHAPTER 3GAS

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4.1 INTRODUCTION

4.2 KINETIC MOLECULAR THEORY 4.3 GAS PRESSURE

4.4 THE IDEAL GAS LAW

4.5 GASES IN REACTION STOICHIOMETRY

4.6 DALTON’S LAW OF PARTIAL PRESSURE4.7 GRAHAM’S LAW OF EFFUSION ANDDIFFUSION

4.8 REAL GASES



4.1 INTRODUCTION

Pressure is defined as the force exerted per unitarea of surface

Gas pressure are commonly measured inatmospheres (atm) or in mmHg

The S.I unit in which pressure is the Pascal(N/m2)

P = F A

1 atm = 760 mmHg = 760 torr = 101 325 Pa (Nm-2) = 101.325 kPa



4.2 KINETIC MOLECULAR THEORY,KMT

The molecules are in constant random motion.They can easily move when force is applied tothe gas

These molecules completely fill the container The collisions between molecules are elastic,

there is no gain or loss of energy duringcollisions

The molecules move faster as the temperature is

raised, temperature increases the averagekinetic energy Gas pressure results from collisions of particles

with the walls of the container



4.3 GAS PRESSURE

Pressure is defined as force per unit area

The S.I unit of pressure is the newton (N) per squaremetre. A pressure of 1 Nm-2 is called a pascal (Pa)

1 kPa = 1000 Pa or 1000 Nm

-2

The atmospheric pressure is usually expressed in theunits of milimetres of mercury (mmHg)

Standard atmospheric pressure is 760 mmHg or 101 325Pa or 101 325 Nm-2

1 atm = 760 mmHg = 760 torr = 101 325 Pa (Nm-2) = 101.325 kPa

P = F (force) = N A (area) = m2



4.4 THE IDEAL GAS LAW

An ideal gas obeys the ideal gas law

P = pressure (usually atm)

V = volume (usually L)

n = moles

T = temperature (K)R = ideal gas constant = 0.082058 L atm mol-1 K-1

Standard Temperature = 0oC or 273.15K

Standard Pressure = 1 atm

PV = nRT



Uses of the Ideal Gas Law:

(a) Combined Gas Law Problems:

(i) Boyle’s law

(ii) Charles’s law

(iii) Avogadro’s law

(b) Solving for one variables in PV = nRT

(c) Determination of molar mass, Mr(d) Gas density problem



(a) Combined Gas Law

(i) Boyle’s law- State that at constant temperature(T), the volume (V) of

a fixed mass of an ideal gas is inversely proportional

to its pressure(P)

- the mathematical expression of Boyle’s Law is shown below

v 1/P (at constant temperature)

or PV = c (a constant)

or P1 V 1 = P2 V 2

P = pressure V = volume

Boyle’s lawCharles’s law

Avogadro’s law



Example A sample of chlorine gas occupied a volume of 0.50 dm3 at

a pressure of 1 atm. What would be its volume if thepressure is reduced to 0.4 atm at constant temperature?

Solution

P1 V 1 = P2 V 2

The volume at a pressure of 0.4 atm = 1.0 x 0.50

0.4

= 1.25 dm3



(ii) Charles’s law

- State that at constant pressure(P), the volume(V) of a fixed mass of an ideal gas is directly proportional to its absolute temperature(T)

- the mathematical expression of Charles’s Law is shown below

v T

or V = c (a constant)

T

or V 1 / T 1 = V 2 / T2

T = temperature in K

V = volume

NOTE: if T in oCneed to change to K



Example(a) A sample of gas occupies a volume of 125.0 cm3 at 27oC.

Calculate its volume at 35oC, assuming the pressure

remains constant(b) 100 cm3 of a gas at a pressure of 100 kPa is compressed to

250 kPa at a constant temperature. What is the final volume of the gas

Solution(a) Charles’ Law

V 1/T1 = V 2/T2 (at constant pressure)

125 = V 2

V 2 = 128.3 cm3

(27 + 273) (35 + 273)



(b) Boyle’s Law

p1 V 1 = p2 V 2 (at constant temperature)

100 x 100 = 250 x V 2 V 2 = 40 cm3



(iii) Avogadro’s Law

- State that at a fixed temperature and pressure, the volume of gas is directly proportional to the amount of gas

- The molar volume ( 1 mol gas) of any gas at standardtemperature and pressure (s.t.p) is 22.4 dm3

- The conditions of s.t.p are

Temperature : 0oC (273 K), Pressure : 101.3 kNm-2 (1atm)

V n an d V=c n



Molar volume of a gas is the volume of 1 mol gas undercertain condition. (1 mol of particles = 6.02 x 1023)

At the same temperature and athmospheric pressure,all gases occupy the same amount of volume.

1 mol of any gas occupies 242.4 dm3 at standardtemperature and pressure (s.t.p). This volume isknown as= molar volume of gas.

Molar volume of gas at s.t.p= 22.4 dm3/ mol

Molar volume of gas at room condition = 24 dm3/ mol

No of moles = Volume (dm3)Molar volume (dm3/mol)



ExampleCalculate the volume at s.t.p of

(a) 0.20 moles of carbon dioxide gas

(b) 6.0 g of chlorine gas

Solution

(a) Volume of 0.20 mol of CO2

at s.t.p

= 0.2 x 22.4 = 4.48 dm3

(b) 6.0 g of chlorine gas (CI2) = 6.0 = 0.0845 mol

Volume = 0.0845 mol x 22.4 dm3 mol-1

= 1.89 dm3

35.5 x 2



Example

The density of a gas at s.t.p is found to be 1.960 g dm-3.

What is the relative molecular mass of the gas?

Solution

Density = Mass

Volume

Mass of 1 dm-3 of gas = 1.960 g

Mass of 1 mol (22.4 dm-3

) of gas = 1.960 x 22.4 = 43.9 gRelative molecular mass of the gas = 43.9



(iii) Gay Lussac’s Law

- At constant volume and moles, The pressure exertedby a gas at constant volume is directly proportional to itsabsolute temperature

P T (at constant volume )

or P = c T(c is constant)

or P1= P2

T1 T2P = pressure

T= temperature

P T



(a) Combined Gas Law

(i) Boyle’s law - describes the effect of pressure V 1/P

(ii) Charles’s law - describes the effect of temperature V T

(iii) Avogadro’s law - describes the effect of the amount of gas V n

Combination of (i)Boyle’s law and (ii)Charles’s law form General gas Law.

Boyle’s lawCharles’s law

Avogadro’s law

V α T

P V = c T

PPV = cT

General Gas LawP1V1 = P2V2

T1 T2



We can combine these 3 laws (i), (ii) and (iii) into a single

equation – the ideal gas equation –that includes

all four gas variables volume, pressure, temperature and

amount of gas

From the 3 simple gas laws, it seems reasonable that

the V of gas should be: directly proportional to the amount of gas, n

directly proportional to the Kelvin temperature, T

inversely proportional to pressure, P

- That is

- Which R = gas constant

V nT and V = RnT , PV = nRTP P



The numerical value of R can be derived using

Avogadro’s law, which states that one mole of any gas at

STP will occupy 22.4 liters. PV = nRT R = PV

nT

R = ( 1 atm) (22.4 L) = 0.08205 atm L mol-1 K-1

(1 mol) (273 K)

Or

R = (101325 Nm-2) (22.4 L)(1 mol) (273K) = 8.31 Nm mol-1 K-1

= 8.31 Pa L mol-1 K-1

= 8.31 J mol-1 K-1



THE IDEAL GAS LAW

An ideal gas obeys the ideal gas law

P = pressure (usually atm)

V = volume (usually L)

n = moles

T = temperature (K)R = ideal gas constant = 0.082058 L atm mol-1 K-1

Standard Temperature = 0oC or 273.15K

Standard Pressure = 1 atm

PV = nRT



(b) Determination of Molar Mass, M

- one application of the ideal gas equation is the determinationof the molar mass of a gas or a volatile liquid

- the ideal gas equation can be rearranged as follows

PV = nRT = m x R x T m = mass of gasM=molar mass of gas

M = m x R x TP V

From above equation, the molar mass) of any gas can be determined if the values of m, p, V and T for the gas are known. Notice that m/V represents the

density () of a gas. Thus equation can be written as

M = RT = density of a gas

P

M



Example 5

Calculate the relative molecular mass of a gas from the following data

Mass of empty flask = 25.385 gMass of flask fully filled with gas = 26.017 g

Mass of fully filled with water = 231.985 g

Temperature of the experiment = 32oC

Atmospheric pressure = 101 kNm-2

SolutionMass of gas = (26.017 – 25.385) = 0.632 g

Volume of the flask = (231.985 – 25.385) = 206.6 x 10-6 m3

Temperature = (273 + 32) = 302 KPressure = 101 kNm-2 = 101 x 103 Nm-2

Mr = m x RT = 0.632 x 8.31 x 305

pV 101 x 103 x 206.6 x 10-6

= 76.8



Exercise 1 and 2:

1. At 750 torr and 27 C, 0.60 g of certain gas occupies

o.50 L. Calculate it molecular weight?

2. What volume will 1.216 g of SO2 gas occupy at 18 Cand 755 torr.



Dalton’s Law of Partial Pressures: the sum of all thepressures of all the different gases in a mixture equals thetotal pressure of the mixture.

4.6 DALTON’S LAW OF PARTIAL PRESSURE

....32

1 PPPPtot



- In a mixture of gases which do not interact with one another, eachgas in the mixture will exert its own pressure independent of the

other gases.- This pressure is known as the partial pressure of that particular gas

in the mixture.

- Dalton`s law of partial pressure state in a mixtureof gases which do not interact with one another, thetotal pressure of the mixture is the sum of thepartial pressure of the constituent gases.

- Dalton’s Law of partial pressures can be

Ptotal = P1 + P2 + P3 +.... = n1RT/V + n2RT/V + n3RT/V = (ntotal)RT/V

P1, P2, P3….are the partial pressures of different gasesin the mixture

ntotal = n1 + n2 + n3 +…….



MOLE FRACTION ()

-definition: amount of one substance / total amount

- The composition of a gaseous mixture can be expressed interms of a mole fraction or percentage (%)

- The mole fraction of gas A in a mixture of gas A and gas B

is

Mole fraction of a gas A( A ) = number of moles of gas A______________

number of moles of gas A + number of moles of gas B

= n A = n A

n A + nB nT

Partial pressure, P = mole fraction x total pressure

P = x PT



If the mixture is made up of two components A and B, then A + B = 1

The partial pressure of a component of a gas mixture is equal to the mole

fraction of the gas times the total pressure of the mixture

P A = A x PT

where PT = total pressure, P A = partial pressure of gas A

and A = mole fraction of gas A



Example

A gases mixture contains 0.274 mol methane, 0.072 mol of ethane and

0.011 mol of propane. The mixture exerts a pressure of 1.60 atm. Calculate

the partial pressure of each component in the mixture

Partial pressure, P = mole fraction x total pressure

nt = 0.274 + 0.072 + 0.011 = 0.357 mol

Pmethane = 0.274 x 1.60 = 1.22 atm

0.357

Pethane = 0.072 x 1.60 = 0.32 atm0.357

Ppropanee = 0.011 x 1.60 = 0.049 atm

0.357



Example

A mixture of 20% nitrogen, 35% oxygen and 45% carbon dioxide has a

pressure of 9.5 x 104 Nm-2. What is the partial pressure of nitrogen in

this mixture

Solution

Partial pressure of nitrogen = 9.5 x 104 x 20/100

= 1.90 x 104 Nm-2



Example

0.60 g of oxygen and 1.30 g of carbon dioxide are introduced into a

Container. If the total pressure of the gaseous mixture is 52.0 Nm -2,

what are the partial pressures of oxygen and carbon dioxide?

Solution

Number of moles of oxygen = 0.60/32 = 0.0188 mol

Number of moles of carbon dioxide = 1.3/44 = 0.0295 mol

Mole fraction of oxygen = 0.0188 = 0.39

0.0188 + 0.0295

Partial pressure of oxygen = 52.0 x 0.39 = 20.28 Nm-2

Partial pressure of carbon dioxide = 52.0 – 20.28 (Dalton’s Law)

= 31.72 Nm-2



4.7 GRAHAM’S LAW OF EFFUSION AND

DIFFUSION

- Effusion refers to the passage of a substance

through a small hole a container or through

a small orifice

-The faster the speed of a molecule, the faster it will effuse. Let'scompare two gases at the same temperature and pressure. Which gas will effuse faster?

- Graham’s Law of Effusion states that the effusion of a gas through asmall hole a container is inversely proportional to its density

or since the density of a gas is proportional to its molar mass ormolecular weight, M w

Effusion rate 1/ M w where M w = molecular weight /molar mass

Effusion rate 1/ d where d = density



- Equivalently, the relative rates of effusion of two gases at thesame pressure and temperature are given by the inverse squareroots of their densities or molecular weight

Rate A = MB

Rate B M A

`

t A = M A

t B MB



- Diffusion refers to the passage of one substance through another. An example for gases would be the passage of an aroma, such as aperfume through still air.

- Comparing two gases at the same pressure and temperature, lowermolecular mass molecules diffuse faster than higher molecularmass molecules.

- Graham’s Law of Diffusion states that the rate of diffusion of agas is inversely proportional to the square root of its density

or since the density of a gas is proportional to its molecular weight,M w

Diffusion rate 1/ d where d = density

Diffusion rate 1/ M w where M w = molecular weight



Eg; An unknown gas effuses through a small hole in94.0 s. Under the same quantity of neon gas effuses in

54.0 s. Calculate the molar mass of the unknown gas.(Molar mass Ne = 20.0 g/mol)

t unknown = Munknown

t Ne MNe

Munknown = Mneon x t unknown

tneon

= 20.0 x 94.0

54.0

Munknown = 7.78

Munknown = 7.782

= 60.61 g/mol



4.8 REAL GASES

- We know that all of the assumptions of the kinetic-moleculartheory (KMT) cannot be exactly true:

(i) if there were no forces of attraction between molecules,then molecules would not stay together in liquids or

solids(ii) If molecules truly have no volume, then liquids and

solids would have no volume either

- The Van Der Waals equation takes account of both of these

objections



The Van Der Waals equation takes account of both of theseobjections:

(i)

(ii)

(iii) This gives the Van der Waals equation:

V correction = V - nb b = the volume occupied by themolecules

P correction = P + an2

V 2a = the attractives forces between

gas molecules

P + an2 (V – nb) = nRT V 2

P V= nRT

the ideal gas lawthe real gas law

P V



- Scientists have attempted to discover an equation which is moreclosely related to the behaviour of real gases than the ideal gasequation. The most well-known is the Van der Waals equation

P + a n2 ( V – nb) = nRT

V 2

where a and b are constants for each gas

The terms an2/V 2 is added to the pressure, p, to take intoaccount the intermolecular attractions

So, corrected pressure = P + a n2

V 2



- The volume available for the molecules to move about in thecontainer is always less than the observed volume because of thespace occupied by the molecules themselves. This tends to make pV greater than RT, and causes over-perfect behaviour. The Van der Waals equation takes into account to co-volume effect of gas by substracting a correcting term, b, from the observed volume

- Corrected volume = V – nb

where b (co-volume of the gas) is a constant, depending onthe nature of the gas

- If the ideal gas equation is modified by introducing the correctpressure and the corrected volume in the place of p and V, theresults is Van der Waals equation

P + a n2 ( V – nb) = nRT V 2

This equation is obeyed by all gases, with reasonable

accuracy, up to quite high pressures



SUMMARY

1 Boyle’s Law states that, at constant temperature, the volumeoccupied by a given mass of ideal gas is inversely proportional tothe applied pressure

2 Charles’s Law states that, at constant pressure, the volume of agiven mass of ideal gas is proportional to its temperature expressedin kelvin

3 Avogadro’s Law states that, under the same conditions of temperature and pressure, equal volumes of all ideal gases containthe same number of molecules

4 Dalton’s Law of partial pressures states that, in a mixture of idealgases, the total pressure is equal to the sum of the partial pressures

of all the gases making up the mixture5 The partial pressure of a gas is the pressure that the gas would

exert if it alone occupied the total volume of the vessel at the sametemperature.



6 An ideal gas or perfect gas is a gas that obeys the gas laws or theideal gas equation exactly

7 A real gas is a gas that does not obey the gas laws or the ideal gasequation exactly

8 The molar volume (that is, the volume occupied by one mole) of any gas at s.t.p is 22.4 dm3

9 The kinetic theory of gases states that

(a) a gas is composed of tiny molecules, the molecules are inrapid and random motion, moving in straight lines

(b) the molecules have negligible volume

(c) there are no intermolecular forces, that is the particles exertno attraction on one another

(d) the molecules will frequently collide with each other and with

the walls of the container(e) the average kinetic energy of gas particles is directly

proportional to the absolute temperature, and remainsconstant at any one temperature

chapter 3gas

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