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
Icepoint 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.K1.mol1
b) When Pressure is expressed in N/m2 and vol in m3.
P = 101300 N/m2
V = 0.0224 m3 (1dm3 = 103m3)
T = 00C + 273
= 273 K
n= 1 mole
R = ?
R = PV/ T
R= (101300) x (0.0224)
(1mol) x (273K)
= 8.3143 N.m.K1. mol1
= 8.3142 J.K1.mol1
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
So
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 mole1 k1.
J (joule) is the unit of energy, hence the dimensions of R as same as those of energy i.e. M.L2.T2 Now
Joules = N x M M = Mass = kg m/sec2 x m L=Volume = kg m2/sec2 T=Temperature Joule = kg x m2 x sec2 = M.L2.T2
Thus dimensions of R are
M.L2.T2
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 nonideal 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)
or
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 m2)
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.”
or
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 nonwetting 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 nonconductor 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 nonconductors 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

01
02
03
04
05
06
07

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
a≠b≠c

α=β=γ=90^{0}
α=β=γ=90^{0}
α=β=γ=90^{0}
α=β=γ≠90^{0}
α=β=γ=90^{0}
αβ=90^{0}
β ≠ 90^{0}
α≠β≠γ≠90^{0}

NaCl, ZnS
SnO_{2, }BaSO_{4.} 4H_{2}O
KNO_{3}, ZnSO_{4}.7H_{2}O AgNO_{3}, KNO_{3}
SiO_{2}
CuSO_{4}
CuSO_{4},5H_{2}O

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