XI Chemistry Chapter 02 Solid Liquid Gas Solved Numerical, Question Answer & Exercise

Friday, September 17, 2021

CHAPTER #2

SOLID, GAS, LIQUID

Any thing which occupies space or has mass is called MATTER. Each and every moment, we come in contact with matter, e.g. air, water, food.

CLASSIFICATION

Matter can be classified on the basis of pure and impure substances.

MATTER

PURE SUBSTANCES IMPURE SUBSTANCES (MIXTURE)

COMPOUND ELEMENTS HOMOGENOUS HETROGENOUS MIXTURE MIXTURE

METALS NON-METALS

PURE SUBSTANCES

Pure substances are those substances which have same properties throughout their bulk, e.g. salt, sugar, and gold. They are further divided into compound and element.

Compounds

A compound is a pure substance. Its composition is fixed. Elements forming compound loose their original properties.

CaCO3 CaO + CO2 (g)

Heat

2H2 + O 2H2O

Elements

An element is a substance in which the atoms are chemically identical having same atomic numbers. Iron = 26

Copper = 29

Elements can be divided into metals and non metals.

MIXTURE

It is an impure substance. Its composition is not fixed. It can be separated into two or more components. or

A mixture is a substance which consists of two or more substances which are chemically not combined with each other, e.g. air and soil.

Homogenous Mixture

A homogenous mixture is a mixture in which we have uniform composition. A homogenous mixture is also known as a solution, e.g. salt, sugar and water.

Hetrogenous Mixture

A heterogenous mixture is a mixture which does not have uniform composition throughout its mass. They have visible boundaries of separation between the substances, e.g. food, air and rock. Matter exists in three physical states.

a) SOLID

b) LIQUID

c) GAS

Q. Define properties of solid, liquid and gas. Differentiate between solid, liquid and gaseous state.

SOLID

Solids are composed of atoms, ions or molecules. They are held together by strong electrostatic forces. The force of attraction between the molecules is very strong which does not allow the molecules to move freely. The space between the molecules is very less, which is the reason behind the definite shape and volume of solids. Solids cannot be compressed or squeezed appreciably by high pressure.

LIQUIDS

Liquid molecules are held together by weak forces. A liquid has no definite shape. It takes the shape of its container. However, it occupies a definite volume. It can be compressed to a negligible extent by high pressure.

GASES

The molecules of gases are apart from each other. Intermolecular forces are very weak. A gas has no shape of its own; rather, it takes the shape of its container. It has no definite volume but can be compressed or squeezed into small volume.

Q) Differentiate between solid, liquid and gases.

	Gas	Liquid	Solid
Shape and Volume	Gases have no shape of their own and have no definite volume.	Liquids have no shape of its own but takes the shape of container. They occupy definite volume.	Solids have definite shape and volume.
Nature	Gases consist of molecules.	Liquids consist of ions or molecules.	Solids consist of atom molecules or ions.
Arrangement	According to kinetic theory, gas molecules are randomly arranged.	Liquids are fairly randomly arranged and consist of clusters that are very close together.	The particles are arranged in a fixed pattern.

Intermolecular Space

Gases have very large spaces between particles.

Liquids have less space between molecules or ions.

Solids have no space between particles. They are tightly packed.

Compressibility

It can be easily

compressed due to the large spaces between the molecules.

It can be compressed to a small extent.

It cannot be compressed. They deform..

BASIC POSTULATES OF KINETIC THEORY OF GASES

The basic postulates of kinetic theory of gases are as follows:

1. MOLECULES – CONSTITUENT PARTICLES OF A GAS

The gases consist of small particles called molecules except inert gases (He, Ne, Ar, Rn). They exist in atomic form, therefore, gases are classified as:

a. Monoatomic gases, e.g. He, Ar

b. Diatomic gases for e.g., H2 O2, N2

c. Polyatomic gases for e.g., CO2, SO2

2. MOLECULAR VOLUME AND DISTANCE

The gaseous state is the one in which the molecules are widely separated from one another but having negligible volume. The gases are easy to compress due to large empty spaces.

3. MOLECULAR MOTION

The gas molecules are in a continuous motion. They travel in straight line until they collide with another or with the walls of container as a result of which the direction of their motion is changed.

4. ELASTIC COLLISION OF MOLECULES

The molecules collide with one another and with the walls of container, but these collisions are perfectly elastic (result in no loss or gain of energy).

5. GAS PRESSURE

Gas pressure is the result of the collision of gas molecules and the walls of the container.

6. NO ATTRACTIVE OR REPULSIVE FORCES

In an ideal gas, there are no attractive or repulsive forces between the molecules, and each molecule acts quite independently of the others.

7. KINETIC ENERGY OF MOLECULES

The molecules of the gases posses average kinetic energy due to their motion.

K.E = ½ mv2

The average kinetic energy of the gas molecules is directly proportional to the absolute temperature. K.E α T

BASIC POSTULATES OF KINETIC THEORY OF LIQUIDS

This theory is based on the following assumptions:

1. The particles of a liquid are quite close to each other due to which a liquid has fixed volume.

2. The particles in a liquid are free to move so they have no definite shape.
3. During the motion, these molecules collide with each other and with the walls of the container.

4. These molecules possess kinetic energy which is directly proportional to its temperature. K.E α T

BASIC POSTULATES OF KINETIC THEORY OF SOLIDS

The assumptions of kinetic theory for solids are as follows:

1. The particles in a solid are closely packed due to strong attractive forces between molecules.

2. These molecules are present at a fixed position and are unable to move.
3. They have definite shape because the particles are arranged in a fixed pattern.

4. They possess only vibration energy.

Behaviour of Gases

Gases exhibit the following behavior and kinetic theory provides the explanation as follows:

SHAPE AND VOLUME

Gases do not have definite shape and volume. According to kinetic theory, this is because there is no force of attraction among gas molecules. So they are free to move. Hence, they do not have any shape and occupies space for available volume.

DIFFUSION

It is a dispersion process. The distribution and spreading of the gas molecules throughout the vessel is called diffusion and this property is called diffusibility.

According to kinetic theory, there is large distance between the gas molecules. Due to absence of any force of attraction, the molecules move freely and diffuses easily, e.g. when NH3 gas is liberated in a room, its molecules easily spread throughout the room.

EFFUSION This is the opposite of diffusion. In this process, the gas molecules are allowed to escape or pass through the tiny holes or pores, e.g. the air effuses from the tire as a result of which the tire looses pressure gradually.

Effusion of gas through a small hole

COMPRESSIBILITY

Gases can be compressed easily or volume of gases is highly affected by change in pressure. According to kinetic theory, there is large distance between the molecules of gases. They come closer by application of pressure, and hence, when the volume is decreased when pressure is increased, e.g. air is squeezed into automobile tires.

EXPANSION (Effect of temperature)

Reverse process of compressibility is expansibility. Expansion and contraction of gases results largely due to heating or cooling. According to kinetic theory, when temperature of a gas is increased, the average K.E and velocity of molecules are increased so gases expand fast at constant pressure. When the temperature is decreased, or when cooling is done, the K.E of molecules decrease as the molecules come closer that decreases the volume also, e.g. when the tire is punctured, the air rushes out.

LIQUIFICATION

Gases are liquefied at low temperature. According to kinetic theory, when temperature of gases is sufficiently lowered, average K.E and velocity of molecules are decreased. This is why molecules come closer and gases change into liquids state.

PRESSURE

All gases exert pressure. According to kinetic theory, the pressure of gases is due to the collision of gas molecules with the walls of container.

Due to the collision, molecules exert a force (F) on some area of the wall,

Pressure = Force / Area

P = F/A

= Newton’s / m2

But Newton is kg.m/s2. Therefore, the unit of pressure is also expressed as.

P = kg m \ s2 = kg m-1 sec2

The other units of pressure are:

a) Atmospheric which is 760mm of Hg level or 1 atmosphere = 760mm = 760 torr b) Pound / sq inch or P.S.I

PSI, 4.7 Psi = 760 torr

Q) What is the pressure in atmosphere correspond to 950 torr?

DATA

Given

Pressure in torr = 950 torr

Requirement.

Pressure in atmosphere = ?

Solution.

As we know that

760 torr = 1 atm

1 = 1/760

950 = 1 x 950 / 760

= 1.25 atm.

DATA

Given.

Atmospheric pressure.

P = 3.5 atm

Requirement

Pressure in torr = ?

Solution

As we know that

1 atm = 760 torr

1 = 760/1

3.5 = 760 x 3.5

= 2660 torr.

What is the pressure in torr of 6.5 atmosphere?

DATA

Given

Atmospheric pressure = 6.5 atm

Requirement

Pressure in torr = ?

Solution

As we know that

1 atm = 760 torr

1 = 760/1

6.5 = 760 x 6.5/1

= 4940 torr.

GAS LAWS

The mole, volume, temperature and pressure of gases explain a simple mathematical relationship to each other. The precise statement of these relationship are known as the gas laws. Gas laws consist of: 1. Boyle’s Law

2. Charle’s Law

3. Avogadro’s Law

4. Graham’s Law of Diffusion

5. Dalton’s Law of Partial Pressure.

BOYLE’S Law

INTRODUCTION

The relationship between pressure and volume of a gas at constant temperature was discovered by Robert Boyle in 1662. This relationship is known as BOYLE’s LAW.

STATEMENT

“At constant temperature, the volume of a given mass of gas is inversely proportional to the pressure exerted on it.”

MATHEMATICAL EXPRESSION

Mathematically, Boyle’s law can be represented as:

V α 1/P (T constant)

V = K. 1/P

K=PV

The product of pressure and volume of given mass of a gas at constant temperature is always constant.

Now for initial state of a gas,

P1V1 = K

final State of Gas,

P2V2 = k

P2V2 = P1V1

GRAPHICAL REPRESENTATION OF BOYLE’S LAW

When the volume of a gas is plotted against the total pressure, a parabolic curve is obtained. A graph of volume with 1/p gives a straight line.

Explanation of the Boyle’s Law

Robert designed a J. shaped glass tube with its shorter arm closed. He poured mercury in the longer arm so as to enclose some amount of air in the shorter arm. The equal level of mercury in the two arms indicated that the confined air was subjected to the atmospheric pressure.

Boyle added more mercury as a result of which the mercury level raised above the original level (h) and the increased pressure reduced the volume of the air. The total pressure was calculated as: P total = P atm + Pn

Where Pn = Excess pressure

Boyle compared the volume of air with total pressure and observed the inverse relationship b/w them.

EXPLANATION OF BOYLE’S LAW IN TERMS OF KINETIC THEORY If the temperature remains constant, the volume of a gas is decreased because the average velocity of molecules remain constant. Molecules come close to each other; hence, they collide more frequently with the wall of the vessel producing higher pressure.

LIMITATIONS OF THE LAW

This law is not obeyed by those gases which are under conditions of high pressure and low temperature.

NUMERICALS

1) What volume does 200 cm3 of a gas at 0.92 atm pressure is changed into 2 atm at constant temperature?

Data

Given

Initial volume v1 = 200 cm3

Initial pressure P1 = 0.92 atm.

Final pressure P2 = 2 atm

Requirement: Final volume V2 = ?

Solution:

According to equation

P1V1 = P2V2

V2 = P1V1/ P2

V2 = 0.92 x 200/ 2

V2 = 184/ 2

V2 = 92 cm3

2) What volume does 400 cm3 of a gas at 700 torr occupy when the pressure is changed to 2 atm?

Data:

Given:

Initial volume V1 = 400 cm3

Initial pressure P1 = 700 torr.

Pressure is change into atm pressure P1 = 700/760 = 0.92 atm

Final pressure P2 = .2 atm.

Requirement:

Final volume V2 = ?

Solution:

According to equation,

P1V1 = P2V2

V2 = P1V1/ P2

V2 = 0.92 X 400/ 2

V2 = 184.2 cm3

3) Find out the change in pressure if volume of a gas in 2.80 dm3 at 670 torr and the change in volume is 4.60 dm3.

Data

Given

Initial volume V1 = 2.80 dm3

Final volume V2 = 4.60 dm3

Initial pressure P1 = 670 torr.

Solution

According to equation

P1V1 = P2V2

P2 = P1V1/ V2

P2 = 670 X 2.80/ 4.60

P2 = 407.82 torr.

CHARLE’S LAW

The solids, liquids and gases show expansion due to heat and contraction due to cooling. This process is very small in case of solids and liquids due to their compact nature and lack of intermolecular spaces. However, expansion and contraction are widely exhibited by gases due to large intermolecular spaces. The statement of temperature and volume relationship is known as Charle’s law which was presented in 1746 – 1823. This law was governed by J.A.C Charles.

STATEMENT

“At constant pressure, the volume of a given mass of a gas is directly proportional to the absolute temperature.”

MATHEMATICAL REPRESENTATION

Mathematically, Charle’s law can be represented as:

V α T (at constant Pressure)

V = KT

V/T = K 1

Mass of a gas to its absolute temperature is always constant provided the pressure is constant. For initial stage of gas

V1/T1 = K

And

For final stage of gas

V2/T2 = K

Hence V1/T1 = V2/T2

EXPLANATION OF CHARLES’S LAW ON THE BASIS OF K.M. THEORY The average kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas. Thus, the effect of raising gas temperature is to raise the average kinetic energy of molecules. Due to rise in the kinetic energy, molecules move more energetically and collide frequently with the container walls. This increases the pressure of the gas. However, if pressure is kept constant under the above mentioned condition, volume is raised. Molecules with a greater force will move through a greater distance and the gas expands to a greater volume. Hence, moving away from each other.

ABSOLUTE ZERO

The temperature – 273 0C is called absolute zero and is defined as that theoretical temperature at which volume of all gases should become zero, and all motion should cease to exist so the gas changees into liquid or solid. It is denoted by T.

T = 0C + 273 = K or A

ABSOLUTE OR KELVIN TEMPERATURE

Steam Point It is the temperature obtained by adding 273 to 0C, i.e. 1000C 373A

0C + 273 = A or K

Ice-point 0C0 273A Example

27 0C is equal to

27 + 273 = 300 K or A

Celcius scale -2730C OA

Absolute scale

ABSOLUTE SCALE

Absolute scale is the scale of temperature in which:

1. Lowest temperature is 00A or K

2. Ice point is 273 K

3. Steam point is 373 K

1) 3.0 dm3 of H2 gas at –200C is warmed at a new temperature at 270C at constant pressure. Find out the new volume of gas.

Data

Given

Initial volume of gas V1 = 3.0 dm3

Initial temperature of gas T1 = -200C + 273 = 253k

Final temperature of gas T2 = 27 + 273 = 300k

Requirement: Find volume of gas V2 = ?

Solution

We know the formula

V1/T1 = V2/T2

V2 = V1 x T2/T1

V2 = 3.0 x 300/253

V2 = 3.55 dm3

2) The volume of a gas is 4.00 dm3 at 150C at constant. Pressure. If the volume is changed into 8.00 dm3, find the change in temperature.

Data

Given

Initial volume of gas V1 = 4.00 dm3

Initial temperature of gas T1 = 15C + 273 = 288k

Final volume of gas V2 = 8.00 dm3

Requirement Find temperature of gas T2 = ?

Solution

We know that the formula

V1/T1 = V2/T2

T2 = V2 x T1/ V1

T2 = 8.00 x 288/4.00

T2 = 576k

3) A balloon has volume 3.80 dm3, when the temperature is 350C. If the balloon is put into a refrigerator and cooled to 50C assuming that pressure inside the balloon is equal to atmospheric pressure at all times. Find the volume?

Data

Given

Initial volume of balloon V1 = 3.8 dm3

Initial temperature T1 = 35 + 273 = 308k

Final temperature T2 = 5 + 273 = 278k

Pressure = constant

Requirement Find volume of balloon = ?

Solution:

We know the formula

V1/T1 = V2/T2

V2 = V1 T2/T1 = 3.8 x 278/308

= 3.42 dm3

AVAGADRO’S LAW

This law was governed by Amado Avogadro’s in 1811. He gave the relation between the volume and the number of molecules of a gas.

STATEMENT

“Equal volume of all gases at same pressure and temperature contains the same number of molecules.” Thus, the volume of a gas is directly proportional to the number of molecules of the gas at constant temperature and pressure.”

MATHEMATICAL REPRESENTATION

v α n

v = kn

v = volume in dm3

n = no of molecules of a gas

EXPLANATION

When hydrogen and chlorine gases react chemically to give hydrochloric acid, according to Avagadro’s law.

X molecules of H2 + X molecules of Cl2

= 2 molecules of HCl.

X is common so

1 molecule of H2 + 1 molecule of Cl2

= 2 molecule of HCl

i.e., H2 + Cl2 ⮴ 2HCl

“A definite volume of a gas at certain temperature and pressure contains definite number of molecules.”

After performing a series of experiments, Avogadro calculated that one mole of a substance contains 6.02 x 1023 atoms / molecules. If gas exist in molecular form as O2, Cl2, CO2, then one mole contains 6.02 x 1023 molecules, but if gas exists in atomic form as He, Nc, and other inert gases, then its one mole contains 6.02 x 1023 atoms.

AVOGADRO’S NUMBER

The number of particles (atoms or molecules) in one mole of a substance is 6.023 x 1023. This is known as Avogadro’s number.

Avogadro’s also proved that one mole of a gas at S.T.P. occupy 22.4 dm3. This volume is also known as molar volume.

Results of Avagadro’s Law

1. It helps in determining a relationship between the volume and weight of the gases.

2. The weight in gm of 22.4 dm3 (Molar Volume) of a gas at S.T.P is the weight of one mole of that gas.

22.4 dm3 of O2 at S.T.P = 32 gm = 1 mole

22.4 dm3 of SO2 at S.T.P = 80 gm = 1 mole

22.4 dm3 of CO2 at S.T.P = 44 gm = 1 moles

3. It also helps in determining the relative molecular mass of gases. Under same temperature and pressure the relation between weight and molecular weight of two gases are:

Wt of gas A = Wt of gas B

Mol. weight of gas Mol. weight of B

GENERAL GAS EQUATION

Given:

The equation gives relation b/w pressure, temperature ,volume and no. of moles of a gas. It is derived as follows:

According to Boyle’s law: (statement)

V α 1/P (at const T)

According to Charle’s law: (statement)

V α T (at const P)

According to Avogadro’s law : (statement)

V α n (at const T,P)

Combining all the three laws we get

V α nT/P

Or PVα nT

Or PV = nRT 1 R= Gases Constant The numerical values of ‘R’ depend on the units of P and V.

a) When Pressure is expressed in atm and vol in dm3.

Data:

P = 1atm PV = nRT

V = 22.4 dm3 R = PV/nT

n = 1 mole R = 1 atm x 22.4 dm3

T = 00C + 273 1 mole x 273k

= 273 0A

R = ? R = 0.0821 dm3.atms.K-1.mol-1

b) When Pressure is expressed in N/m2 and vol in m3.

P = 101300 N/m2

V = 0.0224 m3 (1dm3 = 10-3m3)

T = 00C + 273

= 273 K

n= 1 mole

R = ?

R = PV/ T

R= (101300) x (0.0224)

(1mol) x (273K)

= 8.3143 N.m.K-1. mol-1

= 8.3142 J.K-1.mol-1

The constant R is called general gas constant.

Now

For 1 mole of a gas n = 1

So we can write

PV = RT

Or R = PV/T

For initial state of gas

P1V1/T1 = R

For final state of gas

P2V2/T2 = R

Hence

P2V2/T2 = P1 V1/T1 2

This is another farm of general Gas equation.

DIMENSIONS OF R

Unit of R is J mole-1 k-1.

J (joule) is the unit of energy, hence the dimensions of R as same as those of energy i.e. M.L2.T-2 Now

Joules = N x M M = Mass = kg m/sec2 x m L=Volume = kg m2/sec2 T=Temperature Joule = kg x m2 x sec-2 = M.L2.T-2

Thus dimensions of R are

M.L2.T-2

A mathematical expression which describes the state of a gas, giving the relation b/w pressure, volume, temperature and no. of moles of gas, is called EQUATION OF STATE OF GAS. General gas equation PV = nRT is an equation of state because when we specify the four variables i.e. Pr, Temp, vol and no. of moles.

GENERAL GAS EQUATION IS ALSO CALLED IDEAL GAS

General gas equation PV = nRT is called ideal gas equation b/c it specifies the state of and ideal gas i.e. a gas whose behavior can be predicted precisely on the basis of kinetic molecular theory and gas law:

IDEAL GAS

An ideal gas is that whose behavior predicted precisely on the basis of k.T one gas is ideal. It is a hypothetical gas.

Molecules do not attract each other. Vol of the molecules are negligible.

NON – IDEAL OR REAL GAS

A non-ideal or real gas behavior can not be predict precisely on the basis of k.T. All gases are real.

Molecules attract each other.

Vol of molecules are not negligible.

The main cause of deviation of the real gases from the ideal behavior is the presence of intermolecular forces like vander waals forces of attraction b/w the molecules.

Gases like H2 & O2 do not greatly from the ideal behavior at moderate temperature all gases show marked deviation from ideality.

1) What will be the volume occupied by 14 gm of nitrogen at 200C and 740 torr pressure? Data:

Given: P = 740 torr

= 740/760 = 0.974 atm

wt of nitrogen = 14 gm

so it can be converted into mole now

n= 14/28 = 0.5 mole

. / 3 1 dm a m

R = 0.0821 mol kelvin

. Temperature T = 20 + 273

= 293 k0

REQUIREMENT: Volume occupied by nitrogen gas V = ?

Solution:

According to eq we have

PV = nRt

V = nRT/P

= 0.5 x 0.0821 x 293/0.974

V = 12.345 dm3

2. A certain mass of nitrogen gas at 200C and at 740 torr pressure occupies 12.345 dm3. Calculate the volume that it will occupy at S.T.P.

Data:

Given:

Initial volume V1= 12.345 dm3

Initial pressure p1 = 740 torr

Final pressure p2 = 760 torr

Initial temp T1 = 20+273=293k

Final temp T2 = 273k

Requirement: Find Volume v2 = ?

Solution:

According to general gas equation.

V2 = P1 V1 x T2 / T1 x P2

V2 = 740 x 12.345 x 273/293 x 760

V2 = 11.12 dm3

3. Calculate the vol that will be occupied by 0.8 moles of oxygen gas taken at 300C and at 800 torr of Hg pressure ?

Data:

Given:

R = 0.0821 dm3 atm/deg.mol.

n = 0.8 mol

T = 30+273=303k

P = 800/760 = 1.053 atm

Requirement: Volume occupied by oxygen gas v = ?

Solution:

According to equation

PV = nRT

V = nRT/P

V = 0.8 x 0.0821 x 303/1.053

= 18.899 dm3

GRAHAM’S LAW OF DIFFUSION

Statement

The rate of diffusion of the gases are inversely proportional to the square root of their densities at constant temperature and pressure.

MATHEMATICAL REPRESENTATION OF THE LAW

Suppose two gases with densities Cl1 and Cl2 diffuse into each other at the same condition of temperature and pressure. If the rate of diffusion of the two gases are r1 and r2 respectively, then according to the law,

r1 α 1 / d1 (at constant temperature)

r1 = k / d1 1

similarly for the second gas

r2 1 / d2 (at constant temp)

r2 = k / d2 2

dividing equation 1 by 2

r1/r2 = k / d1

k / d2

r1/r2 = d2/d1 3

As the density of the gas is proportional to its molecular mass, therefore equation 3 can be written as: r1/r2 = M2 / M1

NUMERICALS

1) The ratio of the rates of diffusion of two gases A and B 1.5 : 1. If the relative molecular mass of gas A is 16, find the relative molecular mass of gas B.

Data:

Given:

Ratio rates of two gases

rA/rB are = 1.5 : 1

Molecular mass of gas A = 16

Requirement:

Relative molecular mass of gas B =?

Solution:

According to Graham’s law

RA/rB = MB/MA

1.5/1 = MB/16

(1.5/1)2 = ( MB/16 )2

= 2.25 x 16

MB = 36

2. Compare the rates of Helium and Sulphar dioxide.

Solution:

The molecular masses of Helium and Sulphar dioxide are 4 and 64 respectively. Hence, RHe/RSO2 = MSO2/MHe

= 64/4 = 8/2

RHe = 4 RSO2

DALTON’S LAW OF PARTIAL PRESSURE

STATEMENT

At constant temperature and volume, the total pressure of the mixture of gases are always equal to the sum of partial pressure of all the gases present in the mixture.

MATHEMATICAL REPRESENTATION OF THIS LAW

Let us consider a mixture of three different gases say A, B and C. The partial pressure of these gases be PA, PB, and PC respectively. Now according to the law, the total pressure Pt is given by, Pt = PA + PB + PC

PARTIAL PRESSURE OF A GAS FROM THE MOLECULAR FRACTION

Suppose a mixture of three gases is enclosed in a cylinder of volume ‘v’. Let the number of moles of there gases n1, n2 and n3 respectively. If nt is the total number of moles, then,

nt = n1 + n2 + n3

From general gas equation

PV = nRT

P = nRT/V

So the partial pressure of each gas would be

For gas # 1

P1 = n1 RT / V 1

For gas # 2

P2 = n2 RT / V 2

For gas # 3

P3 = n3 RT / V 3

From Dalton’s law

Pt = P1 + P2 + P3

Therefore

Pt = n1RT + n2RT + n3RT

V V V

or Pt = (n1 + n2 + n3) RT/V

Pt = nt RT/V 4

Dividing equation (1) by (4)

P1/Pt = n1RT/V

ntRT/V

or P1/Pt = n1/nt

Hence Partial pressure of gas = No of moles of gas

Total pressure of mix Total no of moles of gases

i.e. the ratio of the partial pressure of a gas to the total pressure of the mixture of gases is equal to the mole fraction.

APPLICATION OF THE LAW

When a gas is collected over water, then it contains trace of water vapours. Thus, the pressure of this mixture would be the partial pressure of the gas and the partial pressure of water vapours, i.e. Pt = Pgas + PH2O Therefore, Pgas = Pt – PH2O

Hence the volume of dry gas can be calculated by applying general gas equation at S.T.P. P1V1 / T1 = P2V2 / T2

NUMERICALS

1. For the preparation of oxygen gas in the laboratory, 500cm3 of the moist gas was collected over water at 250C and 724 torr. What volume of dry oxygen gas at S.T.P was produced? (Pressure of water vapors at 250C = 24 torr)

Data:

Given:

Pressure of moist gas = 724torr

Pressure of water = 24torr

P of dry gas = Pmoist – PH2O

= 724 – 24

P1 = 700 torr.

Pressure P2 = 760 torr.

Temperature T1 = 25+273=298k

Temperature T2 = 273 k

Initial volume V1 = 500cm3

Requirement:

Vol of dry oxygen

V2 = ?

Solution:

According to gas equation

P1V1 / T2 = P2V2 / T2

V dry gas = P1V1T2 / P2T1

= 700 x 500 x 273

760 x 298

= 421.9 cm3

2. A mixture of gases at 760 torr contains 2 moles nitrogen gas and 4 moles carbon dioxide gas. What is the partial pressure of each gas in torr?

Given:

Total moles n = 6 moles moles of CO2 = 2mol

Total pressure pt = 760 torr moles of N2 = 4mol

Requirement:

Partial pressure of each gas in torr?

Solution:

Pgas = Ptotal x ngas

n(total)

PNr = 760 x 2 / 6 = 253.33 torr.

PCO2 = 760 x 4 / 6 = 506.67 torr.

Total pressure = sum of partial pressure

= 760 torr.

BEHAVIOUR OF LIQUIDS

KINETIC THEORY OF LIQUIDS

1) According to kinetic theory the molecules of liquid are arranged randomly, but consist of cluster in which they are close together.

2) A strong attractive force present to them.

3) Their particles are fairly free to move.

4) They make the volume definite.

DIFFUSIBILITY

The molecules of liquids can diffuse into each other but the rate of diffusion in liquid is slower than the gases, e.g. a drop of a dye or any oily solution diffuses through water.

COMPRESSIBILITY

According to kinetic molecular theory, the intermolecular spaces in liquids are lesser than gas molecules. Hence, they cannot be compressed easily. However, a high pressure is required to compress a liquid to a very little extent.

EXPANSION AND CONTRACTION

A liquid normally expands on heating and contract on cooling. When a liquid is heated, the kinetic energy of its molecules increases. As a result, the attractive forces between the molecules become weak and these molecules move far apart from each other. Hence, the liquid expands. Similarly on cooling, the kinetic energy is lowered and the molecules come closer to each other and the liquid contracts.

VISCOSITY

DEFINATION

“The internal resistance to flow of a liquid is called viscosity.”

EXPLANATION

The internal resistance to flow of a liquid is due to internal friction between the two layers of the liquid. This could easily be understood by taking the liquid in a tube consists of concentric layers. When this liquid is allowed to flow, the layers which are in contact with the wall of the tube are stationary; while, the layer at the center has high viscosity. The intermediate layers move with medium viscosity. Hence each layer exerts a drag to the next layer which causes the liquid to flow.

UNIT OF VISCOSITY

Viscosity is expressed in poise 1 poise = 1 gm/cm. The S.I unit of viscosity is Newton second per sq metre. (N.S m-2)

FACTORS AFFECTING VISCOSITY

1. TEMPERATURE: Viscosity decreases with the increases in temperature because at high temperature the attractive forces between the molecules become weak.

2. INTERMOLECULAR ATTRACTIVE FORCES: Those liquids having strong molecular attraction between the molecules are more viscous than the liquids with weaker molecular attraction.

3. SHAPE OF MOLECULES: Molecules with regular shapes have greater tendency to flow over one another, hence, the liquids having such molecules are less viscous than the liquids containing irregular shape of molecules.

VAPOUR PRESSURE

DEFINATION

The pressure applied by the vapors of a liquid at a particular temperature and at equilibrium state is called vapor pressure. OR

The pressure exerted by the vapors when they are in equilibrium with the liquid phase is called vapor pressure.

EXPLANATION

Consider a liquid in a closed container. Initially, there are no vapors in the empty space. When the liquid is heated, it starts evaporating and its vapors gather above the surface of the liquid. The vapors convert into the liquid by loosing a part of their kinetic energy. Since the rate of evaporation is higher than the rate of condensation; therefore, with the passage of time, more and more vapors are accumulated in the space. This will finally equalize the two rates. Hence, the equilibrium is achieved. At this state, the vapor molecules due to their continuous state of random motion, collide with the wall of the container and exert pressure on it. This is the vapor pressure. evaporation Liquid Vapors Condensation

FACTORS AFFECTING THE VAPOUR PRESSURE

1. TEMPERATURE: The vapor pressure increases with the increases in temperature because at high temperature, the attractive forces between the molecules become weak.

2. INTERMOLECULAR ATTRACTIVE FORCES: The liquids having strong molecular attraction have lower vapor pressure then the liquid with weaker molecular attraction.

3. PRESENCE OF IMPURITIES: By adding impurities, the vapor pressure decreases. It means that the vapor pressure of pure liquid is always higher then impure liquid.

SURFACE TENSION

DEFINATION

“The force acting per unit length on the surface of a liquid at right angle direction is called SURFACE TENSION.”

EXPLANATION

Consider a liquid present in a beaker. The molecules inside the liquid are surrounded by the other molecules of the liquid so the force of attraction present in all directions. But the force of attraction acting on the molecules of the surface from the lower layer molecules is not balanced. The molecules lie on the surface are attracted by the molecules present below the surface. Due to this downward pull, a membrane is formed which tends to contract to a smaller area and causes a tension on the surface of the liquid which is known as S.T.

UNIT OF SURFACE TENSION

The S.T unit of surface tension is Newton per metre.

FACTORS AFFECTING THE S.T

1. TEMPERATURE: The surface tension of a liquid decreases with the increases in temperature because at high temperature, the attractive forces between the molecules become weak.

2. INTERMOLECULAR ATTRACTIVE FORCES: The liquid having strong molecular attraction have high values of surface tension than the liquids with weaker molecular attraction.

3. PRESENCE OF IMPURITIES: By adding the impurities, the S.T decreases. It means the S.T of pure liquid is always higher than impure liquid.

BOILING POINT

Definition

Temperature at which vapour pressure of a liquid becomes equal to the atmospheric pressure is called BOILING POINT.

EXPLANATION

When a liquid is heated, the rate of evaporation of the molecules also increases with the increase in temperature. When the pressure of the vapors become equal to the atmosphere pressure, the liquid starts boiling and this temperature is known as boiling point.

If the external pressure on a liquid is changed, the boiling point of the liquid is also changed. The increase in external pressure on liquid increases the boiling point while decrease of external pressure decreases the boiling point.

CAPILLARY ACTION

DEFINITION

“Rising or falling of a liquid in a capillary tube is called capillary action.”

When a glass tube with small bore is dipped in water or any other liquid, the level rises in capillary tube depending upon nature of the substance used (wetting or non-wetting liquid). This process is called CAPILLARY ACTION.

EXPLANATION

When a capillary tube is dipped into a wetting liquid like water and alcohol, the liquid rises in the tube. This is due to low surface tension of water. Forces of alcohol are greater than the forces of water. Hence, liquid rises upward and balances the gravitational pull.

BEHAVIOUR OF SOLIDS

KINETIC THEORY OF SOLIDS

According to the kinetic theory, the molecules of solid are closely packed together. The attractive forces between them are very strong due to which the molecules of solid are unable to move freely. Hence, the shape and the volume of solid are definite.

DIFFUSIBILITY

In solids, the molecules are very close to each other which do not allow the molecules to move. Hence, the rate of diffusion in solid is very low.

COMPRESSIBILITY

Solids are almost incompressible because the molecules of solids are very close to each other and there are no empty spaces among them.

DEFORMITY

Solids are deformed by high pressure because the

molecules have strong binding forces strong and the

molecules are held equally well to their neighbours.

MELTING

When a solid is heated, the kinetic energy of the molecules increases and it overcomes the intermolecular attraction. As a result, solid starts converting into liquid by melting.

SUBLIMATION

There are some solids which do not melt on heating but directly converted to vapors. Such solids are known as sublimes and the process is called sublimation.

CLASSIFICATION OF SOLIDS

Solid are mainly classified into the following.

1. Crystalline solids 2. Amorphous solids

CRYSTALLINE SOLIDS

The solids which have particular geometrical shape due to highly ordered three dimensional arrangement of the particles are called crystalline solids.

AMORPHOUS SOLIDS

The solids which do not have a particular geometrical shape are called Amorphous solids. Q) Differentiate between Crystalline and Amorphous Solids.

Crystalline Solid

1. Geometry: Particles of crystalline solids are arranged in an order of three dimensional network called crystal lattice. Hence they have uniform shape.

2. Melting Point: They have sharp melting point.

3. Cleavage and Cleavage Plane: They can be cleaved at fixed cleavage plane.

4. Anisotropy and Isotropy: They are anisotropy b/w their physical properties are different in different directions.

5. Symmetry: They are symmetrical, i.e. their appearance do not change by rotating them around an axis.

Amorphous Solids

Geometry: Particles of amorphous solids are not arranged in a definite pattern. They do not have a uniform shape.

Melting Point: They do not have sharp melting point.

Cleavage and Cleavage Plane: They cannot be cleaved at fixed cleavage plane.

Anisotropy and Isotropy: They are isotropic, i.e. their physical properties are same in all direction.

Symmetry: They are unsymmetrical

CLASSIFICATION OF CRYSTALS

Crystals are classified into the following four types which are based upon the bonding between the particles (atoms, ions or molecules) constituting the crystal:

ATOMIC CRYSTALS

Crystals consisting of atoms packed together by metallic bonding are called atomic crystals.

PROPERTIES OF ATOMIC CRYSTALS

1. They are good conductors of heat and electricity.

2. They are lustrous.

3. They possess high melting points.

4. They are ductile and malleable.

IONIC CRYSTAL

Crystals consisting of oppositely charged ions are called ionic crystal. These oppositely changed ions are held together by electrostatic force of attraction.

PROPERTIES OF IONIC CRYSTALS

1. They are hard and brittle.

2. They have high melting points.

3. They conduct electricity in molten state or in solution state.

COVALENT CRYSTALS

Crystal consisting of atoms which held together by covalent bonding are called covalent crystal. These covalent bonds are strong and require high energy to break.

PROPERTIES OF COVALENT CRYSTALS

1. They are usually non-conductor of heat and electricity.

DEFINITION

The phenomenon in which two or more compounds occur in the same crystalline structure is called isomorphism.

ISOMORPHOUS SUBSTANCES

These are the substances that have same crystalline structure.

PROPERTIES

1. They have different physical and chemical properties.

2. They have same empirical formula.

3. When their solutions are mixed, they form mixed crystal.

4. They show the property of overgrowth.

EXAMPLES

1. CaCO3 & NaNO3 (They have trigonal structure)

2. ZnSO4 and NiSO4 (They have orthorhombic structure)

POLYMORPHISM

DEFINITION

It is the phenomenon in which a solid substance exists in more than one crystalline form under a given set of temperature and pressure.

POLYMORPHOUS SUBSTANCES

It is a substance which can exist in more than one crystalline forms.

PROPERTIES

1. They have same formula having same types of composition.

2. They have same properties.

3. There crystalline systems are different.

EXAMPLES

2. They have high melting points.

3. They are high value of refraction index.

MOLECULAR CRYSTALS

Crystals consisting of molecules that are held together by weak attractive forces are called molecular crystals. In these crystals, following attractive forces are present:

a) Hydrogen bonding: It is the electrostatic force of attraction between partially positive hydrogen of one molecule and the electronegative atom of other molecule.

b) Vanderwaal’s forces: It is the attractive force developed between the atomic nuclei of one atom and electrons of other molecules.

PROPERTIES OF MOLECULAR CRYSTAL

1. They are soft.

2. They have low melting point.

3. They are usually non-conductors of heat and electricity.

ISOMORPHISM

Calcium carbonate exists in two different crystalline structure i.e.

a. Calcite: It has trigonal structure.

b. Aragonite: It has orthorhombic structure.

UNIT CELL

The smallest volume of crystal which shows all the characters of its parent is called unit cell. A unit cell is the basic structure unit which when repeats in three dimensions, generates the crystals. A unit cell is described by:

1. Length of its edges, denoted by a, b,c and

2. Angles between the edges devoted by α β γ

CRYSTAL SYSTEMS

There are seven groups of crystal system. These are:

S.No

CrystalSystem

Axes

Angles

Example

Cubic system

Tetragonal system

Orthorhombic system

Trigonal / rhombohedral

Hexagonal system

Mono clinic system

Triclinic system

a=b=c

a=b≠c

a≠b≠c

a=b=c

a=b≠c

a≠b≠c

α=β=γ=90⁰

α=β=γ≠90⁰

α=β=γ=90⁰

α-β=90⁰

β ≠ 90⁰

α≠β≠γ≠90⁰

NaCl, ZnS

SnO_2,BaSO_4. 4H₂O

KNO₃, ZnSO₄.7H₂O AgNO₃, KNO₃

SiO₂

CuSO₄

CuSO₄,5H₂O