KINETIC THEORY OF GASES

Transcription

1 Kineti Theory of Gases 10 KINETIC THEORY OF GASES As you have studied in the previous lessons, at standard temperature and pressure, matter exists in three states solid, liquid and gas. These are omposed of atoms/ moleules whih are held together by intermoleular fores. At room temperature, these atoms/moleules have finite thermal energy. If thermal energy inreases, moleules begin to move more freely. This state of matter is said to be the gaseous state. In this state, intermoleular fores are very weak and very small ompared to their kineti energy. Under different onditions of temperature, pressure and volume, gases exhibit different properties. For example, when the temperature of a gas is inreased at onstant volume, its pressure inreases. In this lesson you will learn the kineti theory of gases whih is based on ertain simplifying assumptions.you will also learn the kineti interpretation of temperature and its relationship with the kineti energy of the moleules. Why the gases have two types of heat apaities and onept of thermal expansion will also be explained in this lesson. OBJECTIVES After studying this lesson, you should be able to : define heat apaity and speifi heat; state priniple of aorimetry; explain thermal expansion; derive relation between α, β and γ; state the assumptions of kineti theory of gases; derive the expression for pressure P = 1 3 ρ ; explain how rms veloity and average veloity are related to temperature; derive gas laws on the basis of kineti theory of gases; 77

2 Kineti Theory of Gases give kineti interpretation of temperature and ompute the mean kineti energy of a gas; explain degrees of freedom of a system of partiles; explain the law of equipartition of energy; explain why a gas has two heat apaities; and derive the relation p V = R/J THERMAL ENERGY During a year, the spring (Basant) season, when the temperature is not as high as in summer and not as low as in winter, is very pleasant. How does this hange in temperature affet our day to day ativities? How do things hange their properties with hange in temperature? Is there any differene between temperature and heat? All suh questions will be disussed in the following setions. The term temperature and heat are often used interhangeably in everyday language. In Physis, however, these two terms have very different meaning. Supply of heat does often inrease the temperature but does it happen so when water boils or freezes? Why the wind in the oastal areas often reverses diretion after the Sun goes down? Why does ie melt when kept on the palm of hand and why does the palm feel ool? All these fats will be explained in this hapter Heat Capaity and Speifi Heat When heat is supplied to a solid (or liquid), its temperature inreases. The rise in temperature is found to be different in different solids in spite of having the same mass and being supplied the same quantity of heat. This simply implies that the rise in the temperature of a solid, when a ertain amount of heat is supplied to it, depends upon the nature of the material of the solid. The nature of the solid is haraterized by the term speifi heat apaity or speifi heat of the solid. The speifi heat of the material of a solid (or a liquid) may be defined as the amount of heat required to raise the temperature of its unit mass through 1 C or 1K. If an amount of heat ΔQ is required to raise the temperature of a mass m of the solid (or liquid) through Δθ, then the speifi heat may be expressed as C= Δ Q mδθ Thus, the amount of heat required to raise the temperature of a substane is given by: Δ = Δθ Q mc SI unit of speifi heat is J kg 1 K 1 78

3 Kineti Theory of Gases Calorimetry When two bodies at different temperatures are kept in ontat, transfer of heat takes plae from the body at higher temperature to the body at lower temperature till both the bodies aquire the same temperature. The speifi heat of a material and other physial quanties related to this heat transfer are measured with the help of a devie alled alorimeter and the proess of the measurement is alled alorimetry Priniple of Calorimetry Let two substanes of mass m 1 and m, of speifi heat apaities C 1 and C and at temperatures θ 1 and θ (θ 1 > θ ), respetively be kept in ontat. Then, the heat will be transferred from the higher to the lower temperature and the substanes will aquire the same temperature θ. (say) assuming that no energy loss takes plae to the surroundings and applying the law of onservation of energy, we an say Heat lost = Heat gained. mc ( θ θ) = mc ( θ θ) This is the priniple of alorimetry. By using this relation the resultant temperature θ an be determined. Also, by knowing θ 1, θ and θ the speifi heat apaity of a substane an be determined if the speifi heat apaity of the other substane is known Thermal Expansion When heat is given to a substane it expands in length, area or volume. This is alled thermal expansion. The expansion in length, area and volume are alled linear, superfiial and ubial expansion, respetively. In linear expansion, the hange in length is diretly proportional to the original length and hange in temperature. Δl l 0 Δθ or Δ l =α l0 Δθ where α is the oeffiient of linear expansion or temperature oeffiient of linear expansion. It is given by If, 1 Δθ = C and l 0 = 1m Then α=δl α= Δl l Δθ 0 79

4 Kineti Theory of Gases Thus, α is defined as the hange in length of unit length of the substane whose temperature is inreased by 1 C. In superfiial expansion, the hange in area is diretly proportional to the original area and hange in temperature: ΔA A 0 Δθ or Δ A=βA0 Δθ where β is the temperature oeffiient of superfiial expansion. In ubial expansion, the hange in volume is diretly proportional to the hange in temperature and original volume: ΔV V 0 Δθ or Δ V = γv0 Δθ where γ in the temperature oeffiient of ubial expansion. If V 0 = 1m 3 and Δθ = 1 C, then γ =ΔV Thus, oeffiient of ubial expansion is defined as the hange in volume of a unit volume of a substane whose temperature is inreased by 1 C. Relation between α, β and γ Let there be a ube of side l whose temperature is inreased by 1 C. The hange in length: Δl = αδθ l = αl ( Δθ = lc ) or, new length l = l+δ l = l+α l = l(1 +α) = 1 C l Fig

5 Kineti Theory of Gases Thus, Δ α= l l And Δ A l (1 +α) l β= = A l = 1+α + α 1 Sine α is very small therefore α may be negleted. We therefore have β = α similarly, Δ V l (1 +α) l γ= = V l3 or, γ = l +α + 3α + 3α l As α is very small, the term α and α 3 may be negleted. We, therefore, have γ = 3α Anomalous expansion in water and its effet Generally, the volume of a liquid inreases with inrease in temperature. The oeffiient of expansion of liquids is about 10 times that of solids. However the volume of water does not inrease with temperature between 0 to 4 C. As the temperature inreases from 0 C to 4 C, the water ontrats and hene the density of water reahes a maximum value of 1 g ml 1 or 1000 kg m 3 at 4 C. After that the volume starts inreasing (while the density dereases) as shown in Fig density (g ml ) ( C) Fig. 10. temperature 81

6 Kineti Theory of Gases Now, it an be understood why a pond or lake freezes at its surfae whereas water may remain below it in liquid state. As the pond ools, the older, denser water at the surfae initially sinks to the bottom. When the temperature of the entire water body reahes 4 C, this flow stops. The temperating of surfae water keeps on dereasing and freezes ultimately at 0 C. As water freezes at the surfae, it remains there sine ie is a bad ondutor of heat; and sine ie is less denser than water the ie ontinues to build up at the surfae whereas water near the bottom remains at 4 C. If this had not happened fish and all the marine life would not have survived Thermal Expansion in Gases When heat is supplied to the gases they also expand. This expansion is very large as ompared to solids and liquids. But in ase of a gas pressure and volume both may hange simultaneously with rise in temperature. Hene we have to onsider either expansion of the gas with temperature at onstant pressure or the inrease in its pressure at onstant volume. Thus the oeffiient of volume expansion of a gas at onstant pressure is given by V V γ v = V 1 1Δθ Δ p= 0 and similarly p p1 γ p = p Δθ 1 Δ= v KINETIC THEORY OF GASES You now know that matter is omposed of very large number of atoms and moleules. Eah of these moleules shows the harateristi properties of the substane of whih it is a part. Kineti theory of gases attempts to relate the marosopi or bulk properties suh as pressure, volume and temperature of an ideal gas with its mirosopi properties suh as speed and mass of its individual moleules. The kineti theory is based on ertain assumptions. (A gas whose moleules an be treated as point masses and there is no intermoleular fore between them is said to be ideal.) A gas at room temperature and atmospheri pressure (low pressure) behaves like an ideal gas Assumptions of Kineti Theory of Gases Clark Maxwell in 1860 showed that the observed properties of a gas an be explained on the basis of ertain assumptions about the nature of moleules, their motion and interation between them. These resulted in onsiderable simplifiation. We now state these. 8

7 Kineti Theory of Gases (i) (ii) A gas onsists of a very large number of idential rigid moleules, whih move with all possible veloities randomly. The intermoleular fores between them are negligible. Gas moleules ollide with eah other and with the walls of the ontainer. These ollisions are perfetly elasti. (iii) Size of the moleules is negligible ompared to the separation between them. (iv) Between ollisions, moleules move in straight lines with uniform veloities. (v) Time taken in a ollision is negligible as ompared to the time taken by a moleule between two suessive ollisions. (vi) Distribution of moleules is uniform throughout the ontainer. To derive an expression for the pressure exerted by a gas on the walls of the ontainer, we onsider the motion of only one moleule beause all moleules are idential (Assumption i). Moreover, sine a moleule moving in spae will have veloity omponents along x, y and zdiretions, in view of assumption (vi)it is enough for us to onsider the motion only along one dimension, say x-axis.(fig. 10.1). Note that if there were N (= moleules m 3 ), instead of onsidering 3N paths, the assumptions have redued the roblem to only one moleule in one dimension. Let us onsider a moleule having veloity C in the fae LMNO. Its x, y and z omponents are u, v and w, respetively. If the mass of the moleule is m and it is moving with a speed u along xaxis, its momentum will be mu towards the wall and normal to it. On striking the wall, this moleule will rebound in the opposite diretion with the same speed u, sine the ollision has been assumed to be perfetly elasti (Assumption ii). The momentum of the moleule after it rebounds is (mu). Hene, the hange in momentum of a moleule is mu (mu) = mu If the moleule travels from fae LMNO to the fae ABCD with speed u along x axis and rebounds bak without striking any other moleule on the way, it overs a distane l in time l/u. That is, the time interval between two suessive ollisions of the moleules with the wall is l/u. Aording to Newton s seond law of motion, the rate of hange of momentum is equal to the impressed fore. Therefore y L B Rate of hange of momentum at ABCD = Change in momentum Time M w u O A C x = mu / l u = mu l z N D Fig : Motion of a moleule in a ontainer 83

8 Kineti Theory of Gases This is the rate of hange of momentum of one moleule. Sine there are N moleules of the gas, the total rate of hange of momentum or the total fore exerted on the wall ABCD due to the impat of all the N moleules moving along x-axis with speeds, u 1, u,..., u N is given by Fore on ABCD = m ( u 1 + u + u un) l We know that pressure is fore per unit area. Therefore, the pressure P exerted on the wall ABCD of areas l by the moleules moving along x-axis is given by P = m u u u l l ( N ) = m ( u 1 + u u 3 N ) (10.1) l If u represents the mean value of the squares of all the speed omponents along x-axis, we an write u = or Nu = Substituting this result in Eqn. (10.1), we get u + u + u u N 1 3 N u + u + u u 1 3 N Nmu P = 3 l It an be shown by geometry that = u + v + w (10.) sine u, v and w are omponents of along the three orthogonal axes. This relation also holds for the mean square values, i.e. = u + v + w Sine the moleular distribution has been assumed to be isotropi, there is no preferential motion along any one edge of the ube. This means that the mean value of u, v, w are equal : u = w v = so that u = 3 84

9 Kineti Theory of Gases Substituting this result in Eqn. (10.), we get P = 1 3 Nm 3 l But 3 l defines the volume V of the ontainer or the volume of the gas. Hene, we get PV = 1 3 Nm = 1 3 M (10.3) Note that the left hand side has marosopi properties i.e. pressure and volume and the right hand side has only mirosopi properties i.e. mass and mean square speed of the moleules. Eqn (10.3) an be re-written as P = 1 3 Nm V If ρ = mn is the density of the gas, we an write V P = 1 3 ρ or = 3P ρ (10.4) If we denote the ratio N/V by number density n, Eqn. (10.3) an also be expressed as P = 1 3 mn (10.5) The following points about the above derivation should be noted: (i) From Eqn. (10.4) it is lear that the shape of the ontainer does not play any role in kineti theory; only volume is of signifiane. Instead of a ube we ould have taken any other ontainer. A ube only simplifies our alulations. (ii) We ignored the intermoleular ollisions but these would not have affeted the result, beause, the average momentum of the moleules on striking the walls is unhanged by their ollision; same is the ose when they ollide with eah other. (iii) The mean square speed is not the same as the square of the mean speed. This is illustrated by the following example. 85

10 Kineti Theory of Gases Suppose we have five moleules and their speeds are 1,, 3, 4, 5 units, respetively. Then their mean speed is = 3 units 5 Its square is 9 (nine). On the other hand, the mean square speed is = = 11 Thus we see that mean square speed is not the same as square of mean speed. Example 10.1 : Calulate the pressure exerted by 10 moleules of oxygen, eah of mass kg, in a hollow ube of side 10 m where the average translational speed of moleule is 500 m s 1. Solution : Change in momentum m u = ( kg) (500 m s 1 ) = kg m s 1. Time taken to make suessive impats on the same fae is equal to the time spent in travelling a distane of 10 m or 10 1 m. Hene Time = 10 m ms = s Rate of hange of momentum = 5 10 kgms s = N The fore on the side due to one third moleules and f = = N pressure = Fore Area = 417 N m = N m INTEXT QUESTIONS (i) A gas fills a ontainer of any size but a liquid does not. Why? (ii) Solids have more ordered struture than gases. Why?. What is an ideal gas? 3. How is pressure related to density of moleules? 86

11 Kineti Theory of Gases 4. What is meant by speifi heat of a substane? 5. Define oeffiient of ubial expansion. 6. A steel wire has a length of m at 0 C. Its length beomes.01 m at 10 C. Calulate oeffiient of linear expansion α of the material of wire KINETIC INTERPRETATION OF TEMPERATURE From Eqn. (10.3) we reall that P V= 1 3 m N Also, for n moles of a gas, the equation of state is PV = n RT, where gas onstant R is equal to 8.3 J mol 1 K 1. On ombining this result with the expression for pressure, we get n R T = 1 3 m N Multiplying both sides by 3 n we have 3 R T = 1 Nm n = 1 m N A where N n = N is Avogadro s number. It denotes the number of atoms or moleules A in one mole of a substane. Its value is per gram mole. In terms of N A, we an write 3 R N A T = 1 m But 1 m is the mean kineti energy of a moleule. Therefore, we an write 1 m = 3 R N A T = 3 k T (10.6) R where k = N (10.7) A is Boltzmamn onstant. The value of k is J K 1. In terms of k, the mean kineti energy of a moleule of the gas is given as ε = 1 m = 3 k T (10.8) 87

12 Hene, kineti energy of a gram mole of a gas is 3 R T Kineti Theory of Gases This relationship tells us that the kineti energy of a moleule depends only on the absolute temperature T of the gas and it is quite independent of its mass. This fat is known as the kineti interpretation of temperature. Clearly, at T = 0, the gas has no kineti energy. In other words, all moleular motion eases to exist at absolute zero and the moleules behave as if they are frozen in spae. Aording to modern onepts, the energy of the system of eletrons is not zero even at the absolute zero. The energy at absolute zero is known as zero point energy. From Eqn.(10.5), we an write the expression for the square root of, alled root mean square speed : rms = = 3kT m = 3RT M This expression shows that at any temperature T, the rms is inversely proportional to the square root of molar mass. It means that lighter moleule, on an average, move faster than heavier moleules. For example, the molar mass of oxygen is 16 times the molar mass of hydrogen. So aording to kineti theory, the hydrogen moleules should move 4 times faster then oxygen moleules. It is for this reason that lighter gases are in the above part of our atmosphere. This observed fat provided an early important evidene for the validity of kineti theory DEDUCTION OF GAS LAWS FROM KINETIC THEORY (i) Boyle s Law We know that the pressure P exerted by a gas is given by Eqn. (11.3) : P V = 1 3 M When the temperature of a given mass of the gas is onstant, the mean square speed is onstant. Thus, both M and on the right hand side of Eqn. (10.3) are onstant. Thus, we an write P V = Constant (10.9) This is Boyle s law, whih states that at onstant temperature, the pressure of a given mass of a gas is inversely proportional to the volume of the gas. 88

13 Kineti Theory of Gases (ii) Charle s Law From Eqn. (10.3) we know that P V = 1 3 M or V = 1 3 P M i.e, V, if M and P do not vary or M and P are onstant. But T V T (10.10) This is Charle s law : The volume of a given mass of a gas at onstant pressure is diretly proportional to temperature. Robert Boyle ( ) British experimentalist Robert Boyle is famous for his law relating the pressure and volume of a gas (PV = onstant). Using a vauum pump designed by Robert Hook, he demonstrated that sound does not travel in vauum. He proved that air was required for burning and studied the elasti properties of air. A founding fellow of Royal Soiety of London, Robert Boyle remained a bahalor throughout his life to pursue his sientifi interests. Crater Boyle on the moon is named in his honour. (iii) Gay Lussa s Law Aording to kineti theory of gases, for an ideal gas M P = 1 3 V For a given mass (M onstant) and at onstant volume (V onstant), P But T P T (10.11) whih is Gay Lussa s law. It states that the pressure of a given mass of a gas is diretly proportional to its absolute temperature T, if its volume remains onstant. (iv) Avogadro s Law Let us onsider two different gases 1 and. Then from Eqn. (10.3), we reall that P 1 V 1 = 1 3 m N

14 and P V = 1 3 m N If their pressure and volume are the same, we an write Kineti Theory of Gases P 1 V = P V Hene 1 3 m 1 N 1 = m N Sine the temperature is onstant, their kineti energies will be the same, i.e. 1 m 1 = 1 1 m Using this result in the above expression, we get N 1 = N. (10.1) That is, equal volume of ideal gases under the same onditions of temperature and pressure ontain equal number of moleules. This statement is Avogadro s Law. (v) Dalton s Law of Partial Pressure Suppose we have a number of gases or vapours, whih do not reat hemially. Let their densities be ρ 1, ρ, ρ 3... and mean square speeds, 1, 3... respetively. We put these gases in the same enlosure. They all will have the same volume. Then the resultant pressure P will be given by P = 1 3 ρ ρ ρ Here 1 3 ρ, ρ, 1 3 ρ signify individual (or partial) pressures of different gases or vapours. If we denote these by P 1, P, P 3, respetively we get P = P 1 + P + P (10.13) In other words, the total pressure exerted by a gaseous mixture is the sum of the partial pressures that would be exerted, if individual gases oupied the spae in turn. This is Dalton s law of partial pressures. (vi) Graham s law of diffusion of gases Graham investigated the diffusion of gases through porous substanes and found that the rate of diffusion of a gas through a porous partition is inversely proportional to the square root of its density. This is known as Graham s law of diffusion. 90

15 Kineti Theory of Gases On the basis of kineti theory of gases, the rate of diffusion through a fine hole will be proportional to the average or root mean square veloity rms. From Eqn. (10.4) we reall that = 3P ρ or = rms = 3P ρ That is, the root mean square veloities of the moleules of two gases of densities ρ 1 and ρ respetively at a pressure P are given by Thus, ( rms ) 1 = 3P ρ and ( rms ) = 1 3P ρ Rate of diffusion of one gas Rate of diffusion of other gas = ( rms) ( ) rms 1 = ρ ρ 1 (10.14) Thus, rate of diffusion of gases is inversely proportional to the square root of their densities at the same pressure, whih is Graham s law of diffusion. Example 10. : Calulate the root mean square speed of hydrogen moleules at 300 K. Take m(h ) as kg and k = J mol 1 K 1 Solution : We know that rms = 3 1 3kT m = 3 ( J K ) (300 K) kg = 197 m s 1 Example 10.3 : At what temperature will the root mean square veloity of hydrogen be double of its value at S.T.P., pressure being onstant (STP = Standard temperature and pressure). Solution : From Eqn. (10.8), we reall that rms α T Let the rms veloity at S.T.P. be 0. If T K is the required temperature, the veloity = 0 as given in the problem 0 = 0 0 = T T 0 91

16 Squaring both sides, we get Kineti Theory of Gases 4= T T 0 or T = 4T 0 Sine T 0 = 73K, we get T = 4 73K = 109K = C Example 10.4 : Calulate the average kineti energy of a gas at 300 K. Given k = JK 1. Solution : We know that 1 M = 3 k T Sine k = J K 1 and T= 300 K, we get E = 3 ( J K 1 ) (300 K) = J INTEXT QUESTIONS Five gas moleules hosen at random are found to have speeds 500 m s 1, 600 m s 1, 700 m s 1, 800 m s 1, and 900 m s 1. Calulate their RMS speed.. If equal volumes of two nonreative gases are mixed, what would be the resultant pressure of the mixture? 3. When we blow air in a balloon, its volume inreases and the pressure inside is also more than when air was not blown in. Does this situation ontradit Boyle s law? Degrees of Freedom Degrees of freedom of a system of partiles are the number of independent ways in whih the partiles of the system an move. Suppose you are driving along a road and several other roads are emanating from it towards left and right. You have the freedom to be on that road or to turn to the left or to the right you have two degrees of freedom. Now, say the 9

17 Kineti Theory of Gases road has a flyover at some point and you take the flyover route. Now, you do not have any freedom to turn left or right, whih means that your freedom has get restrited. You an move only along the flyover and we say that your degree of freedom is 1. Refer to Fig A string is tied in a taut manner from one end A to other end B between two opposite walls of a room. An ant is moving on it. Then its degree of freedom is 1. A B Fig Now suppose it falls on the floor of the room. Now, it an move along x or y diretion independently. Hene its degrees of freedom is two. And if the ant has wings so that it an fly. Then it an move along x, y or z diretion independently and its degree of freedom is 3. A monatomi moleule is a single point in spae and like the winged ant in the above example has 3 degrees of freedom whih are all translational. A diatomi moleule whih is made up of two atoms, in addition to translator y motion an also rotate about two mutually perpendiular axes. Hene a diatomi moleule has (3 + = 5) degrees of freedom: three translational and two rotational THE LAW OF EQUIPARTITION OF ENERGY We now know that kineti energy of a moleule of a gas is given by 1 3 = kt m. Sine the motion of a moleule an be along x, y, and z diretions equally probably, the average value of the omponents of veloity (i.e., u, v and w) along the three diretions should be equal. That is to say, for a moleule all the three diretions are equivalent : u = v = w and u = v = w = 1 3 Sine = u + v + w = u + v + w 93

18 Kineti Theory of Gases Multiplying throughout by 1 m, where m is the mass of a moleule, we have 1 m u = 1 m v = 1 m w But 1 m u = E = total mean kineti energy of a moleule along xaxis. Therefore, 3 E x = E y = E z. But the total mean kineti energy of a moleule is k T. Hene, we get an important result : E x = E y = E z = 1 k T Sine three veloity omponents u, v and w orrespond to the three degree of freedom of the moleule, we an onlude that total kineti energy of a dynamial system is equally divided among all its degrees of freedom and it is equal to 1 k T for eah degree of freedom. This is the law of equipartition of energy and was dedued by Ludwing Boltzmann. Let us apply this law for different types of gases. So far we have been onsidering only translational motion. For a monoatomi moleule, we have only translational motion beause they are not apable of rotation (although they an spin about any one of the three mutually perpendiular axes if it is like a finite sphere). Hene, for one moleule of a monoatomi gas, total energy E = 3 k T (10.15) A diatomi moleule an be visualised as if two spheres are joined by a rigid rod. Suh a moleule an rotate about any one of the three mutually perpendiular axes. However, the rotational inertia about an axis along the rigid rod is negligible ompared to that about an axis perpendiular to the rod. It means that rotational energy onsists of two terms suh as 1 I ω y and 1 I ω. z Now the speial desription of the entre of mass of a diatomi gas moleules will require three oordinates. Thus, for a diatomi gas moleule, both rotational and translational motion are present but it has 5 degrees of freedom. Hene 1 E = 3 kt + 1 kt = 5 k T (10.16) 94

19 Kineti Theory of Gases Ludwing Boltzmann ( ) Born and brought up in Vienna (Austria), Boltzmann ompleted his dotorate under the supervision of Josef Stefan in He also worked with Bunsen, Kirhhoff and Helmholtz. A very emotional person, he tried to ommit suiide twie in his life and sueeded in his seond attempt. The ause behind these attempts, people say, were his differenes with Mah and Ostwald. He is famous for his ontributions to kineti theory of gases, statistial mehanis and thermodynamis. Crater Bolzmann on moon is named in his memory and honour HEAT CAPACITIES OF GASES We know that the temperature of a gas an be raised under different onditions of volume and pressure. For example, the volume or the pressure may be kept onstant or both may be allowed to vary in some arbitrary manner. In eah of these ases, the amount of thermal energy required to inrease unit rise of temperature in unit mass is different. Hene, we say that a gas has two different heat apaities. If we supply an amount of heat ΔQ to a gas to raise its temperature through ΔT, the heat apaity is defined as ΔQ Heat apaity = ΔT The heat apaity of a body per unit mass of the body is termed as speifi heat apaity of the substane and is usually denoted by. Thus heat apaity Speifi heat apaity, = m Eqns. (10.16) and (10.17) may be ombined to get (10.17) = ΔQ m ΔT (10.18) Thus, speifi heat apaity of a material is the heat required to raise the temperature of its unit mass by 1 ºC (or 1 K). The SI unit of speifi heat apaity is kilo alories per kilogram per kelvin (kal kg 1 K 1 ). It may also the expressed in joules per kg per K. For example the speifi heat apaity of water is 1 kilo al kg 1 K 1 = J kg 1 K 1. 95

20 Kineti Theory of Gases The above definition of speifi heat apaity holds good for solids and liquids but not for gases, beause it an vary with external onditions. In order to study the heat apaity of a gas, we keep the pressure or the volume of a gas onstant. Consequently, we define two speifi heat apaities : (i) Speifi heat at onstant volume, denoted as V. (ii) Speifi heat at onstant pressure, denoted as P. (a) (b) The speifi heat apaity of a gas at onstant volume ( v ) is defined as the amount of heat required to raise the temperature of unit mass of a gas through 1K, when its volume is kept onstant : ΔQ v = (10.19) ΔT The speifi heat apaity of a gas at onstant pressure ( P ) is defined as the amount of heat required to raise the temperature of unit mass of a gas through 1K when its pressure is kept onstant. V p = ΔQ (10.0) ΔT P When 1 mole of a gas is onsidered, we define molar heat apaity. We know that when pressure is kept onstant, the volume of the gas inreases. Hene in the seond ase note that the heat required to raise the temperature of unit mass through 1 degree at onstant pressure is used up in two parts : (i) heat required to do external work to produe a hange in volume of the gas, and (ii) heat required to raise the temperature of the gas through one degree ( v ). This means the speifi heat apaity of a gas at onstant pressure is greater than its speifi heat apaity at onstant volume by an amount whih is thermal equivalent of the work done in expending the gas against external pressure. That is p = W + v (10.1) 10.7 RELATION BETWEEN C P AND C V Let us onsider one mole of an ideal gas enlosed in a ylinder fitted with a fritionless movable piston (Fig. 10.5). Sine the gas has been assumed to be ideal (perfet), there is no intermoleular fore between its moleules. When suh a gas expands, some work is done in overoming internal pressure. 96

21 Kineti Theory of Gases A P V 1 V Fig : Gas heated at onstant pressure Let P be the external pressure and A be the ross setional area of the piston. The fore ating on the piston = P A. Now suppose that the gas is heated at onstant pressure by 1K and as a result, the piston moves outward through a distane x, as shown in Fig Let V 1 be the initial volume of the gas and V be the volume after heating. Therefore, the work W done by the gas in pushing the piston through a distane x, against external pressure P is given by W = P A x = P (Inrease in volume) = P (V V 1 ) We know from Eqn. (10.) that p v = Work done (W) against the external pressure in raising the temperature of 1 mol of a gas through 1 K, i.e. p v = P (V V 1 ) (10.) Now applying perfet gas equation to these two stages of the gas i.e. before and after heating, we have Substrating Eqn. (10.3) from Eqn.(10.4), we get PV 1 = RT (10.3) PV = R (T + 1) (10.4) P (V V 1 ) = R (10.5) Hene from Eqns. (10.19) and (10.) we get where R is in J mol 1 K 1 Converting joules into alories, we an write p v = R (10.6) p v = R J (10.7) where J = 4.18 al is the mehanial equivalent of heat. 97

22 Kineti Theory of Gases Example 10.5 : Calulate the value of p and v for a monoatomi, diatomi and triatomi gas moleules. Solution : We know that the average KE for 1 mol of a gas is given as E = 3 R T Now v is defined as the heat required to raise the temperature of 1 mole of a gas at onstant volume by one degree i.e. if E T denotes total energy of gas at T K and E T + 1 signifies total energy of gas at (T + 1) K, then v = E T+1 E T. (i) We know that for monoatomi gas, total energy = 3 R T monoatomi gas V = 3 R (T + 1) 3 R T = 3 R. Hene p = V + R = 3 R + R = 5 R. (ii) For diatomi gases, total energy = 5 R T V = 5 R (T + 1) R 5 R T = 5 R p = V + R = 5 R + R = 7 R. (iii) You should now find out V and p for triatomi gas. INTEXT QUESTIONS What is the total energy of a nitrogen moleule?. Calulate the value of p and V for nitrogen (given, R = 8.3J mol 1 K 1 ). 3. Why do gases have two types of speifi heat apaities? Brownian Motion and Mean Free Path Sottish botanist Robert Brown, while observing the pollen grains of a flower suspended in water, under his mirosope, found that the pollen grains were tumbling and tossing and moving about in a zigzag random fashion. The random motion of pollen grains, was initially attributed to live objets. But when motion of pollens of dead plants and partiles of mia and stone were seen to exhibit 98

23 Kineti Theory of Gases the same behaviour, it beame lear that the motion of the partiles, now alled Brownian motion, was aused by unbalaned fores due to impats of water moleules. Brownian motion provided a diret evidene in favour of kineti theory of matter. The Brownian displaement was found to depend on. (i) Size of the partiles of the suspension smaller the partiles, more the hanes of inbalaned impats and more pronouned the Brownian motion. (ii) The Brownian motion also inreases with the inrease in the temperature and dereases with the visosity of the medium. Due to mutual ollisions, the moleules of a fluid also move on zig-zag paths. The average distane between two suessive ollisions of the moleules is alled mean free path. The mean free path of a moleule is given by 1 σ = n π d where n is the number density and d the diameter of the moleules. WHAT YOU HAVE LEARNT The speifi heat of a substane is defined as the amount of heat required to raise the temperature of its unit mass through 1 C or 1 K. Aording to priniple of alorimetry: Heat lost = Heat gained Kineti theory assumes the existene of atoms and moleules of a gas and applies the law of mehanis to large number of them using averaging tehnique. Kineti theory relates marosopi properties to mirosopi properties of individual moleules. The pressure of a gas is the average impat of its moleules on the unit area of the walls of the ontainer. Kineti energy of a moleule depends on the absolute temperature T and is independent of its mass. At absolute zero of temperature, the kineti energy of a gas is zero and moleular motion eases to exist. Gas law an be derived on the basis of kineti theory. This provided an early evidene in favour of kineti theory. Depending on whether the volume or the pressure is kept onstant, the amount of heat required to raise the temperature of unit mass of a gas by 1ºC is different. Hene there are two speifi heats of gas : 99

24 Kineti Theory of Gases i) Speifi heat apaity at onstant volume ( V ) ii) Speifi heat apaity at onstant pressure ( p ) These are related as p = W + V p V = J R The degrees of freedom of a system of partiles are the number of independent ways in whih the partiles of the system an move. The law of equipartition of the energy states that the total kineti energy of a dynamial system is distributed equally among all its degrees of freedom and it is equal to 1 k T per degree of freedom. Total energy for a moleule of (i) a monatomi gas is 3 k T, (ii) a diatomi gas is 5, and (iii) a triatomi gas is 3 k T. TERMINAL EXERCISE 1. Can we use Boyle s law to ompare two different ideal gases?. What will be the veloity and kineti energy of the moleules of a substane at absolute zero temperature? 3. If the absolute temperature of a gas is raised four times, what will happen to its kineti energy, root-mean square veloity and pressure? 4. What should be the ratio of the average veloities of hydrogen moleules (moleular mass = ) and that of oxygen moleules (moleular mass = 3) in a mixture of two gases to have the same kineti energy per moleule? 5. If three moleules have veloities 0.5, 1 and km s 1 respetively, alulate the ratio between their root mean square and average speeds. 6. Explain what is meant by the root-mean square veloity of the moleules of a gas. Use the onepts of kineti theory of gases to derives an expression for the root-mean square veloity of the moleules in term of pressure and density of the gas. 7. i) Calulate the average translational kineti energy of a neon atom at 5 0 C. ii) At what temperature does the average energy have half this value? 300

25 Kineti Theory of Gases 8. A ontainer of volume of 50 m 3 ontains hydrogen at a pressure of 1.0 Pa and at a temperature of 7 0 C. Calulate (a) the number of moleules of the gas in the ontainer, and (b)their root-mean square speed. ( R= 8.3 J mol 1 K 1, N = mol 1. Mass of 1 mole of hydrogen moleule = kg mol 1 ). 9. A losed ontainer ontains hydrogen whih exerts pressure of 0.0 mm Hg at a temperature of 50 K. (a) At what temperature will it exert pressure of 180 mm Hg? (b) If the root-mean square veloity of the hydrogen moleules at 10.0 K is 800 m s 1, what will be their root-mean square veloity at this new temperature? 10. State the assumptions of kineti theory of gases. 11. Find an expression for the pressure of a gas. 1. Dedue Boyle s law and Charle s law from kineti the theory of gases. 13. What is the interpretation of temperature on the basis of kineti theory of gases?. 14. What is Avagardo s law? How an it be dedued from kineti theory of gases 15. Calulate the root-mean square of the moleules of hydrogen at 0 C and at C (Density of hydrogen at 0 C and 760 mm of merury pressure = 0.09 kg m 3 ). 16. Calulate the pressure in mm of merury exerted by hydrogen gas if the number of moleules per m 3 is and the root-mean square speed of the moleules is m s 1. Avogadro s number and moleular weight of hydrogen =.0). 17. Define speifi heat of a gas at onstant pressure. Derive the relationship between p and V. 18. Define speifi heat of gases at onstant volume. Prove that for a triatomi gas V = 3R 19. Calulate P and V for argon. Given R = 8.3 J mol 1 K 1. ANSWERS TO INTEXT QUESTIONS (i) Beause in a gas the ohesive fore between the moleules are extremely small as ompared to the moleules in a liquid. (ii) Beause the moleules in a solid are losely paked. The bonds between the moleules are stronger giving a ordered struture. 301

26 Kineti Theory of Gases. The gas whih follows the kineti theory of moleules is alled as an ideal gas. 3. P = 1 3 ρ 4. The speifi heat of a substane is the amount of heat required to raise the temperature of its unit mass through 1 C or 1K. 5. The oeffiient of ubial expansion is defined as the hange in volume per unit original volume per degree rise in temperature C Average speed = = 700 m s 1 Average value of = = 510,000 m s rms = = 510,000 = 714 m s 1 rms and are not same. The resultant pressure of the mixture will be the sum of the pressure of gases 1 and respetively i.e. P = P 1 + P. 3. Boyle s law is not appliable For eah degree of freedom, energy = 1 k T for 5 degrees of freedom for a moleule of nitrogen, total energy = 5 k T. 30

27 Kineti Theory of Gases. V for a diatomi moleule = 5 R V = J mol1 K 1 = 0.75 J mol 1 K 1. p = V + R = 9.05 J mol 1 C 1. Answers to Terminal Problem. zero 3. beomes 4 times, doubles, beomes 4 time : m s 1, 14 ºC , m s ºC, 560 m s m s 1, 088 m s N m J mol 1 K 1, 0.75 J mol 1 K