KINETIC THEORY

MADE BY-MEGHA PATELAND PALLAVI JOSHI

INTRODUCTIONINTRODUCTION The actual atomic theory got established more The actual atomic theory got established more

than 150 years later.than 150 years later. Kinetic theory explains the behaviour of gases Kinetic theory explains the behaviour of gases

based on the idea that the gas consists of rapidly based on the idea that the gas consists of rapidly moving atoms or molecules.moving atoms or molecules.

The kinetic theory was developed in the nineteenth The kinetic theory was developed in the nineteenth century by Maxwell,Boltzmann and others.century by Maxwell,Boltzmann and others.

It gives a molecular interpretation of pressure and It gives a molecular interpretation of pressure and temperature of a gas laws and Avagadro’s temperature of a gas laws and Avagadro’s hypothesis.hypothesis.

It also relates measurable properties of gases such It also relates measurable properties of gases such as viscosity,conduction and diffusion .as viscosity,conduction and diffusion .

It correctly explains specific heat capacities of It correctly explains specific heat capacities of many gases.many gases.

BEHAVIOUR OF GASES

From above fig.we come to know that properties of gases are easier to understand than those of solids and liquids. This is mainly because in a gas ,molecules are far from each other and their mutual interactions are negligible except when two molecules collides

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EXPANSION OF GASES &GAS LAWS

In the case of gases,thermal expasion is very large as compared to that of solids and liquids.It is found that the volume of a solid or liquid practically remains the same,even when a large pressure is applied.however,the volume of a gas changes appreciably,when it’s pressure is changed even by a small amount. Therefore to describe the condition of a gas its volume,pressure and temperature must be specified. A change of one affects the others

The gas laws are the study of the relation The gas laws are the study of the relation between any two of these quantities,when between any two of these quantities,when the third kept constant.The three Gas laws the third kept constant.The three Gas laws are the studies of are the studies of

(1)effect of pressure on the volume of a (1)effect of pressure on the volume of a given mass of a gas at constant given mass of a gas at constant temperature.temperature.

(2)effect of temperature on the volume of a (2)effect of temperature on the volume of a given mass of a gas at constant pressure given mass of a gas at constant pressure andand

(3)effect of temperature on the pressure of (3)effect of temperature on the pressure of a given mass of gas at constant volume. a given mass of gas at constant volume.

BOYLE’S LAWBOYLE’S LAW This law was discovered by Robert Boyle in This law was discovered by Robert Boyle in

1662 and it is most fundamental laws1662 and it is most fundamental laws It states-”It states-”the temperature remaining the temperature remaining

constant,the volume of a given mass of a constant,the volume of a given mass of a gas is inversely proportional to thegas is inversely proportional to the pressure.pressure.It is called Boyle’s lawIt is called Boyle’s law

mathematically, if temperature of the gas mathematically, if temperature of the gas is constant,thenis constant,then

PV =CONSTANT ORPV =CONSTANT OR V=1 V=1 P P

In general if P1 and V1 are the initial

pressure and volume of a gas and P2 and V2,their final values,then

P1V1=P2V2

Fig1.01 shows graphs between P and V at constant temperatures T1 and T2,

SUCH THAT T2>T1

CHARLES LAWCHARLES LAW

Around 1787,A.C.Charles studied the variation Around 1787,A.C.Charles studied the variation in volume of a gas with temperature and in volume of a gas with temperature and formulated the following law:formulated the following law:

““The pressure remaining constant, the The pressure remaining constant, the volume of a given mass of a gas increases or volume of a given mass of a gas increases or decreases by a constant fraction of it’s decreases by a constant fraction of it’s volume at 0 C for each rise or fall of 1 C in volume at 0 C for each rise or fall of 1 C in temperature.”temperature.”

V T=CONSTANTV T=CONSTANT

The above equation tells that the graph betweenV and T for a given mass of gas at constant pressure must be straight line as shown dotted in fig.

GAY LUSSAC’S LAW

Joseph Gay Lussac studied the variation of pressure of a gas with temperature and arrived at the following result:

“The volume remaining constant ,the pressure of a given mass of a gas increases or decreases by a constant fraction of its pressure at 0 C for each rise or fall of 1 C in temperature”.

p/t = constant

PERFECT GAS EQUATION In practice, the gases do not obey the gas laws at all values of

pressure and temperature. It is because of the intermolecular forces between the gas molecules.

Perfect gas equation

consider one mole of a perfect gas. Let P1,V1,T1,be its initial pressure,volume and temperature respectively and P2,V2,andT2 be their final values. The change of the state of gas from the state (P1,V1,T1,)to(P2,V2,T2)may be supposed to be taking place in the following two steps:

STEP 1

Suppose that the temperature of the gas remains constant at T1 and the pressure is changed from P1 to P2 so that volume of the gas changes from V1 to, say V’.Since the temperature of the gas remains constant , from Boyle’s law, we have

P1V1=P2V2 ………….. (1.09)

STEP 2 Now,the pressure,volume and the

temperature of the gas are P2,V’,and T1 respectively.Suppose that the pressure of the gas is kept constant at P2 and temperature of the gas is changed from T1 to T2,so that volume changes from V’ to V2.Since the pressure of the gas remains constant, from charles’ law , we have

V’/T1=V2/T2 OR V’=V2*T1/T2

Substituting for V’in the equation(1.09),Substituting for V’in the equation(1.09),

we havewe have

PP11VV11=P=P22*(V*(V22*T*T11/T/T22) OR) OR

PP11VV11/T/T11=P=P22VV22/T/T22 OR OR

PV/T=constant …….(1.10)PV/T=constant …….(1.10)

in above equation,for 1 mole of gas, the in above equation,for 1 mole of gas, the constant is denoted by R and is called constant is denoted by R and is called universal gas constant.universal gas constant.

PV=RT ………..(1.11)PV=RT ………..(1.11)

this equation called this equation called perfect gas equation.perfect gas equation.

KINETIC THEORY OF GASES

A gas is a collection of a large number of a molecules,which are in continuous rapid motion. The gas molecules have a size of

two into ten rest to minus ten meters.But it’s distance between gas molecules is much bigger than it’s size.Therefore the gas molecules can be considered as to be moving freely with respect to each other.Due to change in velocities the gas molecules collides.

Rudolph Claussius and JamesClark Maxwell developed Rudolph Claussius and JamesClark Maxwell developed the Kinetic theory of gases in order to explains gas laws the Kinetic theory of gases in order to explains gas laws in terms of the motion of the gas molecules.in terms of the motion of the gas molecules.

The theory is based on following assumptions as regards The theory is based on following assumptions as regards to the motion of molecules and the nature of gases:to the motion of molecules and the nature of gases:All gases consist of molecules.The molecules of a gas All gases consist of molecules.The molecules of a gas are all alike and differ from those of other gases.are all alike and differ from those of other gases.The molecules of a gas are very small in size as The molecules of a gas are very small in size as compared to the distance between them.compared to the distance between them.The molecules of a gas behaves as perfect spheresThe molecules of a gas behaves as perfect spheresThe molecules are always in random motion.The molecules are always in random motion.The molecules do not exert any force on each other The molecules do not exert any force on each other except during collisions. except during collisions.

PRESSURE OF A IDEAL GAS

A gas exerts pressure on the walls of the containing vessels due to continuous collisions of the molecules against the wall.

consider a gas enclosed in a cube .suppose that the edges OA,OC,OE of the vessels are along X,Y,andZ axis respectively shown in fig

FIG.13.4 Elastic collision of a gas molecule with the wall of the container.

A molecule with velocity (VX.VY,VZ) hits the planer wall parallel to yz-plane of area A.Since the collision is elastic,the molecule rebound with the same velocity;its y and z components of velocity does not change in the collision but the x-component reverses sign. That is, after collision is (-VX,VY,VZ).The change in momentum of the molecule is: -mvx-(mvx)=-2mvx.By the principle of conservation of momentum, the momentum imparted to the wall in the collision=2mvx.

To calculate the force on the wall, we need to calculate momentum imparted to the wall per unit time.In small interval t

A molecule with x-component of velocity Vx will the wall if it is within the distance Vx t from the wall.That is ,all molecules within the volume AVx t only can hit the wall in time.But ,on the average,half of these are moving towards the wall and the other half away from the wall.Thus the number of molecules with velocity(Vx,Vy,Vz)hitting the wall in time t is

1/2AVx tn where n is the no. of molecules per unit volume.The totol momentum transferred to the wall by these molecules in t is:

Q=(2mVx)(1/2nAVx t)……….(13.10)

The force on the wall is the rate of momentum transfer Q/ t and pressure is force per unit area:

P=Q/(A t)=nmVx2……………………(13.11)

P=nmV2x……………….(13.12)

where V2x is the average of V2x. Now gas is isotropic.

V2x=V2y=v2z=(1/3)v2x+V2y+v2x=(1/3)v2

Where V is the speed and V2x denotes the mean of squared speed.Thus

P=1/3nmv2

KINETIC INTERPRITATION OF KINETIC INTERPRITATION OF TEMPERATURETEMPERATURE

Equation (13.14) can be written as Equation (13.14) can be written as PV=(1/3)nVmV2………..(13.15A)PV=(1/3)nVmV2………..(13.15A) PV=(2/3)NXPV=(2/3)NX1/21/2mV2………(13.15B)mV2………(13.15B) Where N(=nV)is the number of Where N(=nV)is the number of

molecules in the sample.molecules in the sample. The quantity in the bracket is the average The quantity in the bracket is the average

translational kinetic energy of the molecules in translational kinetic energy of the molecules in the gas . Since the internal energy E of an ideal the gas . Since the internal energy E of an ideal gas is purely kinetic,gas is purely kinetic,

E=N*(1/2)mv2…………………..(13.16)E=N*(1/2)mv2…………………..(13.16)

Equation (13.15) then gives:

PV=(2/3)E…………(13.17)

we now ready for the kinetic interpritation of temperature. Combining Eq(13.17) with the ideal gas Eq(13.3), we get

E=(3/2)KbNT …..(13.18)

i.e., the average kinetic energy of a molecule is proportional to the the absolute temperature of the gas.The two domains are connected by the Boltzmann constant.

we note that eq(13.18) tell us that internal energy of an ideal gas depend upon only on temperature, not on volume or pressure.

For a mixture of non reactive ideal gases,the total pressure gets contribution from each gas in the mixture.Eq(13.14) becomes

P=(1/3)[n1m1V21+n2m2v22+…] (13.20)

1/2m1v21=1/2m1v22=(3/2KbT)

SO THAT P=(n1+n2+…)KbT………(13.21)

Which is Dalton’s Law of Partial pressures.

LAW OF EQUIPARTITON OF LAW OF EQUIPARTITON OF ENERGYENERGY

The kinetic energy of a single molecule isThe kinetic energy of a single molecule is EETT=1/2mv=1/2mv22x+1/2mvx+1/2mv22y+1/2mvy+1/2mv22zz (13.22)(13.22) For a gas in thermal equilibrium at For a gas in thermal equilibrium at

temperature T the average value of temperature T the average value of energy denoted by<energy denoted by<EETT>IS>IS

<<EETT>=<1/2mV>=<1/2mV22X>+<1/2mVX>+<1/2mV22Y>+<1/2mvY>+<1/2mv22

z>=3/2KbT (13.23)z>=3/2KbT (13.23) <1/2mV<1/2mV22Z>=1/2KbTZ>=1/2KbT

Molecules of monoatomic gas like argon have only translational degrees of freedom.But what about a diatomic gas such as o2 or N2?A molecule of o2 has three translational degrees of freedom. But in addition it can also rotate about it’s centre of mass.Fig 13.6 shows two independent axes of rotation 1 and 2,normal to the axis joining the two oxygen atoms about which the molecule can rotate.

Eachquadratic term occuring in the expression for the energy is a mode of absorption of energy by the molecule .We have seen that in thermal equilibrium at absolute temperature T,for each translation mode of motion,the average energy is 1/2kbT. A most elegant principle of classical statistical mechanics states that this is so for each mode of energy:translational,rotational and vibrational.that is equally distributed with all energy modes this is known as law of equipartition of energy.

SPECIFIC HEAT CAPACITY

MONOATOMIC GASES The molecule of monoatomic

gas has only three translational degrees of freedom. Thus ,the average energy of a molecule at temperature T is (3/2)KbT.The total internal energy of a mole of such a gas is

U=3/2KbT*Na=3/2RT

The Molar specific heat at constant volume,Cv is

Cv(monoatomic gas= dU/dT=3/2RT

for an ideal gas

Cp-Cv=R

where Cp is the molar specific heat at constant pressure,thus,

Cp=5/2R

DIATOMIC GASESDIATOMIC GASESAs explained earlier ,a diatomic molecule As explained earlier ,a diatomic molecule treated as a rigid rotator like a dumbbell treated as a rigid rotator like a dumbbell has 5 degrees of freedom: 3 translational has 5 degrees of freedom: 3 translational and 2 rotational.and 2 rotational.

using the law of equipartition of using the law of equipartition of energy, the total internal energy of a mole energy, the total internal energy of a mole of such a gas isof such a gas is

U=5/2KbT*Na=5/2RTU=5/2KbT*Na=5/2RT THE molar specific heat is given by THE molar specific heat is given by

Cv=5/2R,Cp=7/2R Cv=5/2R,Cp=7/2R

POLYATOMIC GASESPOLYATOMIC GASES

In general a polyatomic molecule has In general a polyatomic molecule has 3 translational,3 rotational degrees of 3 translational,3 rotational degrees of freedom and a certain number(f) of freedom and a certain number(f) of vibritional modes. According to law of vibritional modes. According to law of equipartition of energy,it is easily equipartition of energy,it is easily seen that one mole of such a gas hasseen that one mole of such a gas has

U=(3/2kbT+3/2KbT+FKbT)NA U=(3/2kbT+3/2KbT+FKbT)NA

Predicted values of specific Predicted values of specific heat capacities of gasesheat capacities of gases

Nature Nature of of gasesgases

CvCv

(j/mol/k(j/mol/k))

CpCp

(j/mol/k(j/mol/kCp-CvCp-Cv

(j/mol/(j/mol/k)k)

??

MonoatMonoatomicomic

12.512.5 20.820.8 8.318.31 1.671.67

diatomidiatomicc

20.820.8 29.129.1 8.318.31 1.401.40

triatomitriatomicc

24.9324.93 33.2433.24 8.318.31 1.331.33

SPECIFIC HEAT CAPACITY OF SPECIFIC HEAT CAPACITY OF SOLIDSSOLIDS

We can determine the law of equilipartition of We can determine the law of equilipartition of energy to determine specific heat of solids. energy to determine specific heat of solids. Consider a solid of Natoms, each vibrating Consider a solid of Natoms, each vibrating about in its mean position. For a mole of about in its mean position. For a mole of solid,N=Na,the total internal energy issolid,N=Na,the total internal energy is

U=3KbT*Na=3RTU=3KbT*Na=3RT Now at constant pressure Now at constant pressure

c=deltaQ/deltaT=deltaU/deltaT=3R c=deltaQ/deltaT=deltaU/deltaT=3R

SUBSTANCE

SPECIFIC HEAT

<J/KG/K>

MOLAR SPECIFIC HEAT(J/KG/K)

ALUMINIUM

900.0 24.4

CARBON 506.5 6.1

LEAD 127.7 26.5

COPPER 386.4 24.5

SPECIFIC HEAT CAPACITY OF SOLIDS

MEAN FREE PATHMEAN FREE PATH

Molecules in a gas have rather large Molecules in a gas have rather large speeds of the order of the speed of speeds of the order of the speed of sound.The top of a cloud of smoke sound.The top of a cloud of smoke holds together for hours.this happens holds together for hours.this happens because molecule in a gas have a finite because molecule in a gas have a finite though small size so they are bound to though small size so they are bound to undergone collisions.as a result move undergone collisions.as a result move straight unhindered their path keep straight unhindered their path keep getting incessantly deflected.getting incessantly deflected.