chemistry form 6 sem 1 04

7/27/2019 Chemistry Form 6 Sem 1 04

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CHEMISTRY FORM 6CHEMISTRY FORM 6CHEMISTRY FORM 6CHEMISTRY FORM 6

PHYSICAL CHEMISTRYPHYSICAL CHEMISTRYPHYSICAL CHEMISTRYPHYSICAL CHEMISTRY

:STATE OF MATTER

SOLID, LIQUID AND GAS


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2.1 THE KINETIC THEORY OF MATTER

The Gaseous State A gas is composed of atoms or molecules that are separated from each

other by distances far greater than their own size.

The gas particles can be considered as point particles they possess

mass but have negligible volume There are no forces between the gas particles

Gas particles can vibrate, rotate and move anywhere within the container

where the gas is placed.

A gas has no fixed shape or volume and can be easily compressed Particles of gas are in constant random motion, moving in straight line

unless they collide with the wall of container or with other gas particles

As the particles collide with the wall, they exert a pressure on container

The collision are perfectly elastic. Theres no loss of kinetic energy duringcollision

The average kinetic energy of the particles is directly proportional to the

absolute temperature (Kelvin scale) of the gas


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2.2 The Gas Laws

2.2.1 Boyles Law Stated by Robert Boyle (1662) during his investigation of

relationship between volume occupied by gas to the pressure of

gas using the apparatus below.

The total pressure on the gas is the sum of the atmospheric

pressure and the pressure exerted due to the difference in the

height (h) of the mercury in the column


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A plot of the result yields the following graph

Based on the results, he formulated Boyles Law which state thatThe volume occupied by a fixed mass of gas at constant

temperature is inversely proportional to its pressure

P 1/V Px V= constant

P1 x V1 = P2 x V2


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Ideal gas obeys Boyles Law under all conditions of temperature andpressure.

Ideal laws does not exist as real gases obeys Boyles Law closely only at lowpressures and high temperatures. Real gases do not obey Boyles Law closelyat high pressures and low temperatures.

Pressure = force per unit area, SI unit of pressure = Newton (N) persquare metre

1 Nm-2 is called a pascal (Pa) 1 kPa = 1000 Pa or 1000 N m-2

Atmospheric pressure is expressed in units of milimetres of mercury (mmHg). Standard atmospheric pressure is 760 mm Hg or 101325 Pa or 101325N m-2.

The SI Unit of volume is cubic metre, m3.

1 m3 = 103 dm3 = 106 cm3 ; 1dm3 = 1 litre

The deviation of gases from Boyles Law is called non-ideal behaviour


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(a) 1/P against V

1/P

V

(b) V against 1/P

V

1/P

(c) PV against PPV

P

(d) V against PV

P


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2.2.2 CHARLES LAW

JAC Charles investigated the relationship between the volumeoccupied by a gas, at constant pressure with temperature which

is describe by the apparatus setting below.

As Tincreases Vincreases


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He found that the relationship is a linear, as shown in the graph

VT

V= constant x T

V1/T1 = V2/T2 T(K) = t(0C) + 273

Temperature must

Kelvin (absolute scale)


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When the temperature of a sample of gas is increased, the

kinetic energy of the gas particles will increase, thus increasing,

the rate of collision between the gas particles and the wall of

container. At the same time, the collision are more energetic

and this lead to an increase in the pressure exerted by gas.

Hence, in order to maintain the original pressure, the volumeoccupied by the gas must now increase.


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Avogadros Law

V number of moles (n)

V= constant x

n

V1/n1 = V2/n2

Constant temperature

Constant pressure

5.3


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2.2.3 Avogadros Law

~ state that equal volumes of all gases at the same temperature

and pressure contain equal number of atoms/molecules.

The volume occupied by 1 mole of any gas (molar volume)

depends on the temperature and pressure of the gas. The molar

volume of gas at standard temperature & pressure is 22.4 dm3. The condition of s.t.p are :-

Temperature : 0 oC (273 K)

Pressure : 101 kPa (1 atm) Under room temperature and pressure, molar volume of gas is

24.4 dm3/ mol.

The condition of r.t.p are :-Temperature : 25 oC (298 K)

Pressure : 101 kPa (1 atm)


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Ideal Gas Equation

Charles law: VT (at constant n and P)

Avogadros law: V n (at constant Pand T)

Boyles law: V (at constant n and T)1P

VP

V= constant x = RnT

P

nT

P

Ris the gas constant

PV= nRT


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2.2.4 Ideal Gas Equation

For a fixed mass of gas P V / T = constant

Pressure, P Volume, V Gas constant, R Temperature, TPa or N m-2 m3 8.31 J K-1 mol-1 K

atm dm3 0.0821 dm3 atm K-1 mol-1 K

2211 VPVP=

21Density (d) Calculations

d =m

V=

PM

RT

m is the mass of the gas in g

Mis the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRT

PM= dis the density of the gas in g/L


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1. Calculate the volume of the following gases at s.t.p.

(at 0oC and 1 atm = 101 kPa = 760 mm Hg)

a) 200 cm3 hydrogen at 30C and 2.00 x 104 Pa

b) 4.25 dm3 hydrogen chloride at 50C and 650 mm Hg

P1V1 / T1 = P2V2 / T2(2.00 x 104)(200) / (273+30) = (101 x 103)(V2) / (0 + 273)

V2 = 35.7 cm3

P1V1 / T1 = P2V2 / T2(650 mmHg)(4.25) / (273+50) = (760 mmHg)(V2) / (0 + 273)

c) 600 cm3 oxygen at 308 K and 2.20 atm

d) 1.70 dm3 neon at 95C and 1.45 atm

V2 = 3.07 dm

P1V1 / T1 = P2V2 / T2(2.20)(600) / (308) = (1.00)(V2) / (0 + 273)

V2 = 1170 cm3

P1V1 / T1 = P2V2 / T2(1.45)(1.70) / (273+95) = (1.00)(V2) / (0 + 273)

V2 = 1.83 dm3


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2. An aerosol can containing helium gas at 30C and 1.8 x 103 Pa isheated to 60C. What is the pressure of helium in the can now?

3. Determine the density of SO2

gas at 25C and 101 kPa.

P1 / T1 = P2 / T2(1.8 x 103) / (273+30) = P / (60 + 273)

P2 = 2.0 x 103 Pa

PV = nRT @ P = mRT / MRV => P = dRT / MRd = (101 x 103) (32 + 2(16) / 8.31 (25 +273)

d = 2610 g / m3 @ 2.61 x 10-3 g / cm3

4.


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4. A 20.0 m3 steel tank was constructed to store liquefied natural gas (LNG) whichcontained mainly methane at 160C and 101 kPa

(a) How many grams of methane can be stored in the container if the density of theliquid is 416 kg m-3?

(b) Calculate the volume of a storage tank capable of holding the same mass LNG as agas at 28C and 101 kPa.

Since density = 416 kg m-3 ;

Mass = 20.0 x 416 = 8320 kg @ 8.32 x 106 g

nce e empera ure o gas ecrease

V1 / T1 = V2 / T220.0 / 160 + 273 = V2 / 28 + 273

V2 = 13.9 m3


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5. An organic compound consists of 24.24% carbon, 4.04% hydrogenand 71.72% chlorine by mass. If a 1.803 g sample occupies 546 cm3 at

84.2C and 745 mm Hg (Given 760 mm Hg = 101 kPa)

(a) calculate its molar mass

Pressure in kPa = 745 mm Hg x 101 kPa / 760 mm Hg

Pressure in kPa = 99.0 kPaPV = nRT

(99.0 x 103)(546 x 10-6) = (1.803 / MR) 8.31 (84.2 + 273)

MR = 99.0

e erm ne s mo ecu ar ormu a

Element C H Cl

Mass 24.24 4.04 71.72

Mol

24.24

12= 2.02

4.04

1= 4.04

71.72

35.5= 2.02

Ratio2.02 / 2.02= 1

4.04 / 2.02= 2

2.02 / 2.02= 1

Empirical formula = CH2Cl

(CH2Cl)n = 99.0

[(12(1) + 2(1) + 35.5(1)]n = 99

n = 2Molecular formula = (CH2Cl)2

= C2H4Cl2


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2.3 Daltons Law Partial Pressure

Partial Pressure In a mixture of gases which do not interact

with one another, each gas in the mixture will exert its own

pressure independent of the other gases.

Daltons Law of Partial PressureDaltons Law of Partial PressureDaltons Law of Partial PressureDaltons Law of Partial Pressure

In a mixture of gases which do not interact with one another, the totalpressure of the mixture is the sum of the partial pressure of the

constituent gases

T a b c ..

The partial pressure of a gas is also given by the following expression:

Pa = mole fraction of A in mixture (Xa) X Total pressure

Mole fraction of A (Xa) = moles of A

Total moles


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Daltons Law of Partial Pressures

Vand T

areconstant

P1 P2 Ptotal = P1 + P2


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Consider a case in which two gases,A and B, are in a

container of volume V.

PA =nART

V

PB =n

BRT

V

nA is the number of moles ofA

nB is the number of moles of B

PT = PA + PB XA =nA

nA + nBXB =

nB

nA + nB

PA

=XA

PT

PB

=XB

PT

Pi=X

iPT


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A sample of natural gas contains 8.24 moles of CH4, 0.421

moles of C2H6, and 0.116 moles of C3H8. If the total pressure of

the gases is 1.37 atm, what is the partial pressure of propane(C3H8)?

Pi=X

iPT PT = 1.37 atm

Xpropane = .8.24 + 0.421 + 0.116= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm


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Example 2

A mixture of gases at 200 kPa contains the following composition of gases by

volume. (20% CO2 ; 50% CO ; 30% O2)

a) What is the partial pressure of each gas in the mixture?b) If CO2 is removed from the vessel, what would be the partial pressure of O2and CO

For CO2

PCO2 = xCO2 . PtotPCO2 = (20) / (20 + 50 + 30) x 200 kPa

PCO2 = 40 kPa

For CO

PCO = xCO . PtotPCO = (50) / (20 + 50 + 30) x 200 kPa

PCO = 100 kPa

For O2

PO2 = xCO . PtotPO2 = (30) / (20 + 50 + 30) x 200 kPa

PO2 = 60 kPa

Partial pressure of O2 and CO will remain as 60 kPa and 100 kPa

respectively, since removing CO2

will not affect the partial pressure of

each gas in the vessel


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Example 3 :

5.00 dm3 of H2 at 200 kPa, 12.0 dm3 of N2 at 300 kPa and 1.50 dm

3 of Cl2 at

120 kPa were forced into a vessel of capacity 10.0 dm3 at constant temperature.

Calculate the pressure in the vessel as a result of the mixture of these gases.

For hydrogen

P1V1 = P2V2(200 kPa)(5.00) = P2(10.00)

P2 = 100 kPa

For nitrogen

P1V1 = P2V2(300 kPa)(12.00) = P2(10.00)

P2 = 360 kPa

Total pressure = PH2 + PN2 + PCl2= 100 kPa + 360 kPa + 18 kPa

= 478 kPa

P1V1 = P2V2(120 kPa)(1.50) = P2(10.00)

P2 = 18 kPa


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Nitrogen monoxide and oxygen gas are mixed in the vessel as in the diagram below

Calculate its partial pressure of each gas and total pressure if the stopper is liftedup from the vessel.

P1V1 = P2V2(0.5 atm)(4) = P2(4 + 2))

P2 = 0.333 atm

P1V1 = P2V2(1.0 atm)(2) = P2(4 + 2)

P2 = 0.333 atm

Total pressure = PNO + PO2= 0.333 + 0.333

= 0.666 atm


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2.5 THE DISTRIBUTION OF MOLECULAR SPEEDS IN A GAS

Speed of various gas molecules differ widely and are constantlychanging. This is due to their frequent collision and the resultantchanges in energy.

At any time, the speed instantly increase and spread widely. This spread

of molecular speeds is called Maxwell-Boltzmann distribution. There are 2 factors which can only affect the distribution of the graph

. .

At constant pressure, the Maxwell Boltzmann distribution graph can bevaries with temperature

The graph below shows the distribution of the graph at 3 different

temperature


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Distribution of molecular speeds at three temperatures.


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THE DISTRIBUTION OF MOLECULAR SPEEDS IN A GAS

Speed of various gas molecules differ widely and are constantly

changing. This is due to their frequent collision and the resultantchanges in energy.

At any time, the speed instantly increase and spread widely. This

spread of molecular speeds is called Maxwell-Boltzmann distribution.

Important points to note about the distribution curves for molecular

speed in a gas.

Area directly proportional to the total number of molecules in the gas

At any temp., a small proportion have high/low speed mostlyaverage

T , distribution curve shift to right peak become lower number

of gas molecules with high speed increase ; average speed decrease.

When heat is supplied, the energy is used to increase the averagekinetic energy and to increase the motion of the gas molecules.

M-B apply not only to molecular speeds but also to molecular energy.

2 M l l


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2. Molecular mass

At constant temperature and pressure, the molecular gas distribution

can also be influenced by the mass of gas itself. Distribution graph below shows the distribution of different gases

molecules with different molar mass


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Important points to note about the distribution curves formolecular speed in a gas. Area directly proportional to the total number of molecules in the gas

At constant temperature, a small proportion of molecules have high/lowspeed but mostly average

Smaller the molecular mass of the gaseous molecules, distribution curveshift to right & at the same time, peak become lower. This indicated lessermolecules travel at average speed while more molecules travel at high speed.

This is due to lighter molecules can move faster than a heavier molecule.More light molecules are able to travel at high speed, causing lessermolecules to travel at average speed (represent by lowered peak)


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4.5 Real gases Deviations from ideality

Real gases area gases that do not obey the ideal gas equation, PV = nRT

Real gases behave ideally at _____________ & ________________

At high pressure and low temperature, gases would ______________

The deviation of real gases from ideal gas can be explained usingPV/RT against P. The graph below shows how 3 real gases behave

At very low pressure, the gas molecules are far apart and attraction

low pressure high temperature

liquefied

orces e ween e mo ecu es _____________________________

However, when pressure increased, the attraction forces become__________ since the molecules are now __________. Thus, whenmolecules are at _________ deviation, these attractions make the gas

more compressible. Volume of the gas _________ more thanexpected. Thus, PV nRT and the gas exhibit _______________deviation from ideality.

significant closer

negative

lower

< negative deviation

G h f PV / RT i t P


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Graph of PV / RT against PPV

RT

1

H2

CH4

H2

NH3

P

A hi h h l l h l h d


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At higher pressure, the molecules are pushes very close together, andnow repulsive forces operate between them. These repulsive forces

make the gas harder to compress. As a result, a higher pressure has tobe used to compress the gas. Thus, PV . nRT. Moreover, as highpressure, volume of the gas become smaller and hence the actual

volume of the gas molecules in comparison with the volume of thecontaining vessel cannot be ignored. Thus, the gas exhibit _________deviation from ideality.

>

positive

The same curve could also be given at different temperature.Example, for carbon dioxide gas at 100 K, 200 K and 300 K

G h f PV / RT g i t P


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Graph of PV / RT against PPV

RT

1

700 K

400 K

100K

P

A l h l l i b llid i h h ll f


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At low temperatures, when a molecule is about to collide with the wall ofcontainer, the intermolecular forces of attraction between molecules will the

force exerted by the impact. As a result, the pressure exerted by the gas isreduced. This will cause the value of PV RT (negative deviation)

At very high temperatures, the kinetic energy of the gas molecules is veryhigh and there are no chances for molecules to interact with each othercausing no intermolecular forces occur.

Negative deviation caused by polar bond

chemistry form 6 sem 1 04

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