Skip to main content

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

 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.  

 Δ 

 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 

 m  

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  
 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 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)  
 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 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.”  
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 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

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=900

α=β=γ=900

α=β=γ=900

α=β=γ≠900

α=β=γ=900

α-β=900

β ≠ 900

α≠β≠γ≠900

NaCl, ZnS 

SnO2, BaSO4. 4H2

KNO3, ZnSO4.7H2O  AgNO3, KNO3

SiO2

CuSO4

CuSO4,5H2O

 


Comments

Popular posts from this blog

Free Download Notes XI & XII Commerce Notes & Book In Pdf

Free Download Notes XI & 1st Year Notes Download In Pdf XI Accounting Book  Economics Notes Pdf Free Download Urdu Notes for 1st Year (Class 11)  Another Urdu Notes For XI  Islamiat For XI  Principle Of Commerce 1st year XI POC Notes (Best) Server 01 XI POC ( Principle of Commerce ) Guess Paper Solved 2021 Free Download Notes  XII Commerce Notes & Book In Pdf XII Accounts Book With Solutions 2nd-year Urdu Notes Free Download Pdf XII English Notes Pdf 2nd Year English Notes Adam jee  2nd Year Commercial Geography  2nd Year Pak Studies Notes XII Pak Studies Notes In English Adam Jee XII Pak Studies Solved Paper XII Banking Notes (Shah Commerce) Another 2nd Year Banking Notes XII  CG Solved Papers Commerce Online Notes By Sir sultan hamid hussain (All credit goes to Sir Sultan) Server 01 Commercial Geography Notes In English By Sir Sultan Server 01 Commercial Geography Notes Notes In Urdu By Sir Sultan Server 01 XII Banking Notes Short Q

Questions And Answers The Count’s Revenge By J.H.Walsh

THE COUNT’S REVENGE (J.H.WALSH) Q: 1 What Arab custom is referred in the short play “The Count’s Revenge”? An Arab custom mentioned by the Countess of Morcerf to Albert and the Count of Morcerf when the count of Monte Cristo leaves their house without eating anything. The custom runs as “Never to eat food at the house of a deadly enemy”. She strongly believes that since the Count of Monte Cristo has a faith in that custom and thinks them to be his enemy, and hence not eat anything at their place.   Q: 2 What do you know about the reaction, plans or intention of Albert? Albert, the brave young son of the count of Morcerf, was deeply shocked by the disgrace of his father and family. As a man of honour, he showed severe emotional reaction to the unhappy incident. Albert made his mind to trace about the unknown enemy of his family and avenge the family honour. On his request, Beauchamp, a close friend of Albert, discovered the name of the enemy. It was Albert;s

English Essay Problems of Karachi

Problems of Karachi Karachi is the biggest city in Pakistan and one of the most thickly populated cities in the world. Its population has increased rapidly and accordingly has given rise to many social problems. People of this metropolis are becoming more and more concerned about solving these serious problems, some of which are discussed below. The ever-increasing rush of heavy traffic on the roads is resulting in heavy loss of human life. One day or the other, people suffer from accidents due to reckless driving. Some lose their vehicles and some go to the police. This is due to lack of civic sense in the citizens and violation of traffic rules. Traffic jams, road quarrels, untidiness and damage of public property are also the results of this problem. The government has not done any planning to control this situation in the past two decades. In the same manner, the government has never emphasized upon population distribution. As a result, slum areas are rapidly being built, wher

Short Questions Answers of The Prisoner of Zenda ~Drama Novel Prisoner of Zenda

 N OVEL Question 1)Tell in your own words how the first meeting came off between the two distant cousins? Answer) Rudolf leaves the inn one day as he is given an opportunity to stay at Jahan’s sister at Strelsau. Instead of going, he decides to walk through the forest and have a look at the castle of Zenda. He sits down in the forest to have some rest as well as smoke a cigar. After smoking his cigar, he unintentionally falls asleep. Shouts and sound of laughter wake him up. On opening his eyes, he sees two men standing near him. They are Fritz Von Tarlenheim and Colonel Sapt. They tell him that he looks exactly like their king except that he has a beard. At that moment King Rudolf appears. Rassendyll greatly surprises to see king Rudolf in the forest of Zenda. He gives a cry when he finds that Rudolf is just like him. Rudolf’s face and appearance are quite like his own. Rudolf’s height appears to be slightly less than his. Rassendyll bows respectfully before the king. In a happy m