Physics Unit 1 - Kinetic Theory of Gases PDF

Physics : Unit 1 - Kinetic Theory of Gases Kinetic theory of gases The kinetic theory of gases is the study that relates the microscopic properties of gas molecules (like speed, momentum, kinetic energies etc..)with the macroscopic properties of gas molecules (like pressure,...

Physics : Unit 1 - Kinetic Theory of Gases Kinetic theory of gases The kinetic theory of gases is the study that relates the microscopic properties of gas molecules (like speed, momentum, kinetic energies etc..)with the macroscopic properties of gas molecules (like pressure, temperature and volume). Fundamental postulates of kinetic theory 1. The molecules of a gas are considered to be rigid, perfectly elastic, identical in all respects. They are solid spheres. Their size is negligible compared to intermolecular distances. 2. The molecules are in random motion in all directions with all possible velocities. 3. The molecules collide with each other and with the walls of the container. At each collision, velocity changes but the molecular density is constant in steady state. 4. As the collisions are perfectly elastic, there is no force of attraction or repulsion between the molecules. Thus the energy is only kinetic. 5. Between any two successive collisions, molecules travel with uniform velocity along a straight line. Expression for Pressure of the gas : Consider a gas contained in a cubical vessel of side l with perfectly elastic walls F B containing a large number of molecules. A E Let c1 be the velocity of a molecule in a Y Z 𝑣1 direction as shown. This can be resolved 𝑤1 X into three components u1, v1 and w1 along 𝐶1 𝑢1 X, Y and Z directions as shown. Then H C 𝑐12 = 𝑢12 + 𝑣12 + 𝑤12 ……..(1) The momentum of this molecule that G D strikes the wall ABCD of the vessel is equal to mu1where m is its mass. As the collision is elastic, the molecule will rebound with same momentum i.e. mu1. The change in momentum due to impact is equal to mu1 – ( - mu1) = 2mu1. It strikes the wall EFHG and returns back to ABCD after travelling a distance 2l. 2𝑙 The time between the successive collisions (time for one collision) on ABCD is. 𝑢1 B.Sc. - II Semester Dr. K S Suresh Page 1  Physics : Unit 1 - Kinetic Theory of Gases 𝑢1 Thus the number of collisions per second, this molecule makes with ABCD is. 2𝑙 Hence the rate of change of momentum = change in momentum × number of collisions per 𝑢1 𝑚𝑢12 second is equal to 2𝑚𝑢1 × =. 2𝑙 𝑙 From Newton’s second law, rate of change of momentum = impressed force. 𝑚𝑢12 If f1 is the force, then 𝑓1 = along X – direction. 𝑙 Similarly, the force on another molecule of velocity c2 whose components are u2, v2 and w2 due 𝑚𝑢22 to impact is 𝑓3 = along X – direction. 𝑙 Hence the total force Fx on the face ABCD due to impacts of all the n molecules in the X – 𝑚 direction is given by 𝐹𝑥 = (𝑢12 + 𝑢22 + 𝑢32 … … ….. + 𝑢𝑛2 ) ………..(2) 𝑙 Since pressure is force per unit area, the pressure Px on ABCD is given by 𝐹𝑥 𝑚 𝑃𝑥 = = (𝑢12 + 𝑢22 + 𝑢32 … … ….. + 𝑢𝑛2 ) ………..(3) ( since P = F/A and A = l2) 𝑙2 𝑙3 Similarly, if Py and Pz are the pressures on faces EFBA and FBCH, then 𝐹𝑦 𝑚 𝑃𝑦 = = ( 𝑣12 + 𝑣22 + 𝑣32 … … ….. + 𝑣𝑛2 ) ………(4) 𝑙2 𝑙3 𝐹𝑧 𝑚 𝑃𝑧 = = (𝑤12 + 𝑤22 + 𝑤32 … … ….. + 𝑤𝑛2 ) ………(5) 𝑙2 𝑙3 As the pressure exerted by the gas is same in all directions, the average pressure P of the gas is 𝑃𝑋 +𝑃𝑦 + 𝑃𝑧 𝑃=.......(6) 3 𝑚 𝑃= 3 (𝑢12 + 𝑢22 + 𝑢32 … … ….. + 𝑢𝑛2 ) + ( 𝑣12 + 𝑣22 + 𝑣32 … … ….. +𝑣𝑛2 ) 3𝑙 + (𝑤12 + 𝑤32 … … ….. + 𝑤𝑛2 ) 𝑚 𝑃 = 3 (𝑢12 + 𝑣12 + 𝑤12 ) + ( 𝑢22 + 𝑣22 + 𝑤22 ) … 3𝑙 + (𝑢𝑛2 + 𝑣𝑛2 + 𝑤𝑛2 ) …. (7) Since l3 = V, the volume of the cube and 𝑐12 = 𝑢12 + 𝑣12 + 𝑤12 𝑐22 = 𝑢22 + 𝑣22 + 𝑤22 and so on. 𝑚 𝑃= (𝑐12 + 𝑐22 + 𝑐32 + ⋯ … ….. + 𝑐𝑛2 )….(8) 3𝑉 1 𝑚𝑛 2 2 𝑐12 + 𝑐22 + 𝑐32 +⋯……..+ 𝑐𝑛2 or 𝑃 = 𝐶 where 𝐶 = known as the mean square velocity of the 3 𝑉 𝑛 molecules. 𝟏𝑴 If M is the total mass of the gas, ie. M = n m, then 𝑷 = 𝑪𝟐 ……(9) 𝟑 𝑽 B.Sc. - II Semester Dr. K S Suresh Page 2  Physics : Unit 1 - Kinetic Theory of Gases 𝟏 If  is the density of the gas, then pressure of the gas is 𝑷 = 𝟑  𝑪𝟐 C is called the root mean square velocity of the molecules and it is equal to the square root of the mean of the squares of the velocities of individual molecules. 𝟑𝑷 It is given by = √. 𝝆 𝟑 To derive the relation 𝑼 = 𝑹𝑻 𝟐 The pressure exerted by a gas of n molecules occupying volume V is given by 1 𝑚𝑛 1 𝑃= 𝐶 2 or 𝑃𝑉= 𝑚 𝑛 𝐶2 3 𝑉 3 If V is the volume occupied by a gram molecule of the gas and M is the molecular weight of the gas, then M = m NA where NA is the Avogadro number. 1 𝑃𝑉= 𝑀 𝐶 2 …..(1) 3 From the perfect gas equation PV = RT ………(2) 1 From (1) and (2) we get 𝑀 𝐶2 = 𝑅 𝑇 3 or 𝑀 𝐶 2 = 3 𝑅 𝑇 1 3 Dividing the above equation on both the sides by 2 we get 𝑀 𝐶2 = 𝑅 𝑇 ……..(3) 2 2 𝟑 1 or 𝑼 = 𝑹 𝑻 ……(4) where 𝑈 = 𝑀 𝐶 2 is called the internal energy of the gas. 𝟐 2 Dividing both sides of equation (3) by NA, which is the number of molecules in one gram 1 𝑀 3 𝑅 molecule of the gas or one mole, called Avogadro number, we get 𝐶2 = 𝑇 2 𝑁𝐴 2 𝑁𝐴 As M/𝑁𝐴 = m and R/𝑁𝐴 = k, where k is Boltzmann constant. 1 3 𝑚 𝐶2 = 𝑘 𝑇 …….(5) 2 2 Thus the mean kinetic energy per molecule in a given mass of gas is proportional to the absolute temperature of the gas. Deduction of perfect gas equation From kinetic theory of gases, the expression for pressure of a gram molecule of the gas is 1 𝑚𝑛 1 𝑃 = 𝐶 2 or 𝑃𝑉= 𝑚 𝑁𝐴 𝐶 2 ……(1) 3 𝑉 3 2 1 2 1 Or 𝑃 𝑉 = × 𝑚 𝑁𝐴 𝐶 2 = 𝑁𝐴 × 𝑚 𝐶 2 ……..(2) 3 2 3 2 B.Sc. - II Semester Dr. K S Suresh Page 3  Physics : Unit 1 - Kinetic Theory of Gases The average kinetic energy of 1 gram molecule of a gas at absolute temperature T is given by 1 1 𝐾𝐸 = 𝑀 𝐶2 = 𝑚 𝑁𝐴 𝐶 2 2 2 1 3 Average kinetic energy of a molecule is 𝑚 𝐶2 = 𝑘 𝑇 …..(3) 2 2 2 3 Comparing equations (2) and (3) 𝑃 𝑉 = 𝑁𝐴 × 𝑘 𝑇 3 2 or 𝑃 𝑉 = 𝑁𝐴 𝑘 𝑇 or 𝑷 𝑽 = 𝑹 𝑻 where R = 𝑁𝐴 k R is called universal gas constant given by R = 8.31 J mol-1 K-1 Derivation of Gas laws (1) Boyle’s law – From the kinetic theory of gases, the pressure exerted by a gas is given by 1𝑀 1 𝑃= 𝐶2 or 𝑃𝑉= 𝑀 𝐶2 3 𝑉 3 At a constant temperature C2 is a constant. Thus for a given mass of a gas, from the above equation, P V = constant. Hence Boyle’s law. 1𝑀 1 𝑚𝑛 (2) Charle’s law – The pressure of a gas is 𝑃 = 𝐶 2 or 𝑉= 𝐶2 3 𝑉 3 𝑃 Thus for a given mass of gas, at constant pressure 𝑉 ∝ 𝐶 2 As 𝐶 2 ∝ 𝑇, we get 𝑽 ∝ 𝑻. Hence the Charle’s law. Similarly it can be shown that 𝑃 ∝ 𝑇 at constant volume called the Regnault’s law. (3) Avogadro’s law – This law states that at the same temperature and pressure equal volumes of all gases contain the same number of molecules. Let n1 and n2 be the number of molecules of two different gases, m1 and m2 their masses and C1 and C2 the respective root mean square velocities. Since the two gases have the same pressure and for unit volume the gases, 1 1 𝑃 = 𝑚1 𝑛1 𝐶12 = 𝑚2 𝑛2 𝐶22 …….(1) 3 3 As the temperature of the two gases are same, there is no change in temperature when they are mixed. This is possible only if the mean kinetic energy per molecule in the two gases is the 1 1 same. ie. 𝑚1 𝐶12 = 𝑚2 𝐶22 …(2) 2 2 Based on the equation (2) equation (1) reduces to the condition n1 = n2. Hence the Avogadro law. B.Sc. - II Semester Dr. K S Suresh Page 4  Physics : Unit 1 - Kinetic Theory of Gases Mean free path The average distance travelled by a molecule in a gas between any two successive collisions is called mean free path of the molecule. It is denoted by . 𝑺 If the total path travelled in N collisions is S, then the mean free path is given by 𝝀 = 𝑵 Expression for mean free path  Consider n as the number of molecules per unit v volume of a gas and let  be the diameter of each of  these molecules. The assumption made here is that only the molecule under consideration is in motion, while all other molecules are at rest. The moving molecule will collide with all those molecules whose centres lie within a distance  from its centre as shown in the figure. If v is the velocity of the molecule, in one second it will collide with all the molecules the centres of which lie in a cylinder of radius  and length v, and hence in a volume 2 v. The number of molecules in this cylindrical volume is 2 v n. Thus the number of collisions N made by the moving molecule is also   2 v n. or N = 2 v n. As the distance S traversed by the molecule in one second is its velocity v, the mean free path  𝑺 𝒗 𝟏 is given by 𝝀 = = = 𝑵 𝝅 𝝈𝟐 𝒗 𝒏 𝝅 𝝈𝟐 𝒏 𝟏 Thus 𝝀= …….(1) This equation connecting the mean free path with the molecular 𝝅 𝝈𝟐 𝒏 diameter and the number of molecules per unit volume was deduced by Clausius. Boltzmann, assuming that all the molecules have the same average speed deduced the equation 𝟑 𝝀= ………(2) 𝟒 𝝅 𝝈𝟐 𝒏 Maxwell, based on the exact law of distribution of velocities, obtained a more correct equation 𝟏 𝝀= ……..(3) √𝟐 𝝅 𝝈𝟐 𝒏 From the above equation it is clear that mean free path is inversely proportional to the square of the molecular diameter. From the perfect gas equation P V = R T or P V = NAk T where R = NA k 𝑁𝐴 𝑘 𝑇 𝑃 We get 𝑃 = =𝑛𝑘𝑇 or 𝑛 = 𝑉 𝑘𝑇 B.Sc. - II Semester Dr. K S Suresh Page 5  Physics : Unit 1 - Kinetic Theory of Gases 𝒌𝑻 Substituting for n in equation (3) 𝝀 =. Thus mean free path is directly proportional √𝟐 𝝅 𝝈𝟐 𝑷 to the absolute temperature and inversely proportional to the pressure. Maxwell’s law of distribution of velocity among the molecules The molecules of a gas are in random motion. There is a continuous change in the magnitude and direction of their velocities (speeds) due to random motion and collisions between the molecules. Maxwell analysed the distribution of velocities by the statistical method. Maxwell’s law – According to this law, the number of molecules (dn) possessing velocities between c and c + dc is given by 2 𝑑𝑛 = 4 𝜋 𝑛 𝑎3 𝑒 − 𝑏 𝑐 𝑐 2 𝑑𝑐 ……….(1) 𝑏 𝑚 Where n is the number of molecules per unit volume and 𝑎 = √ = √ 𝜋 2𝜋𝑘𝑇 Relation (1) is called Maxwell’s law of distribution of velocities. let b c2 = x2, differentiating 2bc dc = 2x dx or dc = x dx/bc 𝑏 𝑏 3/2 Also 𝑎 = √ Thus 𝑎3 = ( ) Also 𝑐 = √𝑏 𝑥 𝜋 𝜋 Substituting the above terms in equation (1) we get 𝑑𝑛 𝑏 3/2 − 𝑥 2 𝑥 = 4𝜋( ) 𝑒 𝑑𝑥 𝑛 𝜋 𝑏𝑐 𝑑𝑛 Simplifying the above equation = 4 𝜋 − 1/2 𝑒 − 𝑛 Plotting Maxwell speed distribution function y versus x (y = f(v) = dn/n and x the molecular speed) we get the graph as shown. 1. The shaded region between x and x + dx gives the total number of molecules (dn/n) whose velocities lie between c and c + dc. 2. The area under the graph gives the total number of molecules n with velocities between zero and infinity. 3. The co ordinate y corresponding to any value of x gives the number of molecules having velocity represented by x. 4. dn/n is maximum at x = 1 which represents the most probable velocity. B.Sc. - II Semester Dr. K S Suresh Page 6  Physics : Unit 1 - Kinetic Theory of Gases To find root mean square (rms) velocity Let dn be the number of molecules having velocities between c and c + dc. If n is the total number of molecules, the mean square velocity is 1 ∞ 𝑑𝑛 2 2 𝑐𝑚𝑠 = ∫ 𝑐 2 𝑑𝑛 As = 4 𝜋 𝑎3 𝑒 − 𝑏 𝑐 𝑐 2 𝑑𝑐 𝑛 0 𝑛 2 ∞ 2 3 𝜋 𝑐𝑚𝑠 = 4 𝜋 𝑎3 ∫0 𝑒 − 𝑏 𝑐 𝑐 4 𝑑𝑐 = 4 𝜋 𝑎3 √𝑏 8 𝑏2 ∞ 2 3 𝜋 where the standard integral used is ∫0 𝑒 − 𝑏 𝑐 𝑐 4 𝑑𝑐 = √𝑏 8 𝑏2 2 𝑏 3/2 3 𝜋 3 By simplifying the above equation, we get 𝑐𝑚𝑠 = 4𝜋( ) √𝑏 = 𝜋 8 𝑏2 2𝑏 2 = √ 3 The root mean square velocity 𝑐𝑟𝑚𝑠 = √𝑐𝑚𝑠 2𝑏 𝑚 𝟑𝒌𝑻 𝒌𝑻 As 𝑏 = we get 𝒄𝒓𝒎𝒔 = √ = 𝟏. 𝟕𝟑𝟐√ ………(1) 2𝑘𝑇 𝒎 𝒎 To find mean or average velocity 1 ∞ The average velocity of the molecule is 𝑐̅ = ∫ 𝑐 𝑑𝑛 𝑛 0 ∞ 2 1 𝑐̅ = 4 𝜋 𝑎3 ∫0 𝑒 − 𝑏 𝑐 𝑐 3 𝑑𝑐 = 4 𝜋 𝑎3 2 𝑏2 ∞ 2 1 where the standard integral used is ∫0 𝑒 − 𝑏 𝑐 𝑐 3 𝑑𝑐 = 2 𝑏2 𝑏 3/2 1 8𝑘𝑇 𝟖𝒌𝑻 𝒌𝑻 𝑐̅ = 4 𝜋 ( ) = √ ̅=√ Thus 𝒄 = 𝟏. 𝟓𝟗𝟔 √ …..(2) 𝜋 2 𝑏2 𝜋𝑚 𝝅𝒎 𝒎 To find most probable velocity It is the velocity possessed by maximum number of molecules in the gas. Therefore the probability of molecules having velocities in the range c and c + dc must be 𝑑 𝑑𝑛 maximum or ( )=0 𝑑𝑐 𝑛 𝑑 2 (4 𝜋 𝑎3 𝑒 − 𝑏 𝑐 𝑐 2 𝑑𝑐) = 0 𝑑𝑐 2 2 4 𝜋 𝑎3 (𝑒 − 𝑏 𝑐 2𝑐 − 𝑐 2 𝑒 − 𝑏 𝑐 2𝑏𝑐) = 0 2 4 𝜋 𝑎3 2𝑐 𝑒 − 𝑏 𝑐 (1 − 𝑏 𝑐 2 ) = 0 B.Sc. - II Semester Dr. K S Suresh Page 7  Physics : Unit 1 - Kinetic Theory of Gases 1 𝟏 𝟐𝒌𝑻 or 1 − 𝑏 𝑐 2 = 0 or 𝑐2 =. Thus 𝒄𝒎𝒑 = √ = √ 𝑏 𝒃 𝒎 𝟐𝒌𝑻 𝒌𝑻 or 𝒄𝒎𝒑 = √ = 𝟏. 𝟒𝟏𝟒 √ …..(3) 𝒎 𝒎 Comparison of the three equations shows 𝑐𝑚𝑝 ∶ 𝑐̅ ∶ 𝑐𝑟𝑚𝑠 = 1.414 ∶ 1.596 ∶ 1.732 Thus 𝑐𝑚𝑝 < 𝑐̅ < 𝑐𝑟𝑚𝑠 Degrees of freedom It is defined as the number of independent variables required to specify the position and configuration of a molecule completely. It is also the number of independent components of velocities needed to describe the motion of molecules completely. The degrees of freedom of the system is given by 𝒇 = 𝟑 𝑵 − 𝒓 where N is the number of particles in the system and r is the number of independent relations among the particles. A monoatomic gas (one atom per molecule like neon, argon, helium etc..) has only three translational kinetic energy along the three coordinate axes x, y and z. Thus N = 1 and r = 0. f = 3. (three degrees of freedom) A diatomic gas (two atoms per molecule like N2, O2. H2, CO etc….) has both translational and rotational kinetic energies. If there are rotational kinetic energy along two perpendicular axes along with three translational kinetic energy, then N = 2 and r = 1. Thus f = 5.(five degrees of freedom) A triatomic (three atoms per molecule like, H2S, H2O, SO2) or polyatomic gas there are three translational and three rotational kinetic energies or degrees of freedom for a non linear molecule. Thus N = 3 and r = 3. This give f = 6. For triatomic linear molecule like CO2, CS2, HCN, two atoms lie along the side of the central atom. N = 3, r = 2 and f = 7. Principle of Equipartition of energy Statement According to this law, the total energy of a dynamic system in thermal equilibrium is shared equally by all its degrees of freedom, the energy associated per molecule per degree of 1 freedom being a constant equal to 𝑘 𝑇 where k is the Boltzmann constant and T is the 2 absolute temperature of the system. Consider one mole of a monoatomic gas in thermal equilibrium at temperature T. It has three degrees of freedom. According to the Maxwell’s theory of equipartition of energy in the steady B.Sc. - II Semester Dr. K S Suresh Page 8  Physics : Unit 1 - Kinetic Theory of Gases state, ̅̅̅ 𝑢2 = ̅̅̅ 𝑣 2 = ̅̅̅̅ 𝑤 2 where 𝑢̅, 𝑣̅ and 𝑤 ̅ are the average values of the component velocities of a molecule. 1 As 𝑐 2 = ̅̅̅ ̅̅̅2 + ̅̅̅̅ 𝑢2 + 𝑣 𝑤 2 Thus ̅̅̅ 𝑢2 = ̅̅̅ 𝑣 2 = ̅̅̅̅ 𝑤2 = 𝑐2 3 1 ̅̅̅2 = 𝑚 ̅̅̅ 1 1 11 Thus 𝑚𝑢 𝑣 2 = 𝑚 ̅̅̅̅ 𝑤2 = 𝑚 𝑐2 2 2 2 32 1 According to kinetic theory of gases, the average kinetic energy of a gas molecule is 𝑚 𝑐2 = 2 3 𝑘𝑇 2 1 ̅̅̅2 = 1 𝑚 ̅̅̅ 1 13 1 Thus 𝑚𝑢 𝑣 2 = 𝑚 ̅̅̅̅ 𝑤2 = 𝑘𝑇= 𝑘𝑇 2 2 2 32 2 1 Therefore, the average kinetic energy per degree of freedom i equal to 𝑘 𝑇. 2 𝟑 Derivation of 𝑼 = 𝑹𝑻 𝟐 According to the equipartition of energy, the average kinetic energy associated with each 1 degree of freedom is equal to 𝑘 𝑇 where k is the Boltzmann constant and T is the absolute 2 temperature. 3 The energy associated with the three degrees of freedom is equal to 𝑘 𝑇. 2 The energy associated with one gram molecule of a gas is given by 3 3 𝑈 = 𝑁𝐴 × 𝑘𝑇= (𝑁𝐴 𝑘) 𝑇 2 2 𝟑 Thus 𝑼 = 𝑹 𝑻 where R = NA k and U is the internal energy. 𝟐 Specific heat of gases For a monoatomic gas, the total energy U of 1 gram molecule of the gas at the absolute 1 3 temperature T is given by 𝑈 = 3 × 𝑘 𝑇 × 𝑁𝐴 = 𝑅𝑇 2 2 If CV is the molecular specific heat at constant volume, (the quantity of heat required to raise one gram molecule through one degree), then 𝑑𝑈 3 𝐶𝑉 = = 𝑅 …….(1) 𝑑𝑇 2 If CP is the molecular specific heat at constant pressure, then CP – CV = R 5 CP = CV + R = 𝑅 ………(2) 2 From (1) and (2) it is observed that molecular specific heats of monoatomic gas are constants independent of the temperature and nature of the gas. B.Sc. - II Semester Dr. K S Suresh Page 9  Physics : Unit 1 - Kinetic Theory of Gases The theoretical predictions agree with the experimental results. 5 𝑑𝑈 5 7 For a diatomic gas 𝑈 = 𝑅 𝑇. Hence 𝐶𝑉 = = 𝑅 and CP = 𝑅. 2 𝑑𝑇 2 2 6 𝑑𝑈 For triatomic or polyatomic gas, 𝑈 = 𝑅 𝑇. Hence 𝐶𝑉 = = 3 𝑅 and CP = 4 𝑅. 2 𝑑𝑇 Ratio of specific heat of gases The ratio of specific heat of a gas at constant pressure to the specific heat at constant volume is 𝐶𝑃 called the atomicity of the gas denoted by 𝛾 = 𝐶𝑉 5 𝐶𝑃 ( )𝑅 5 2 For a monoatomic gas 𝛾 = = 3 = = 1.67 𝐶𝑉 ( 2) 𝑅 3 7 𝐶𝑃 ( )𝑅 7 2 For a diatomic gas 𝛾 = = 5 = = 1.41 𝐶𝑉 ( 2) 𝑅 5 𝐶𝑃 4𝑅 4 For a polyatomic gas 𝛾 = = = = 1.33 𝐶𝑉 3𝑅 3 Importance of - (1) Its value lies between 1 and 1.67, (2) It gives the atomicity of a molecule, (3) Its value decreases with increasing atomicity and (4) It helps in finding the molecular structure of the gas. Relation between  and degrees of freedom 1 The average kinetic energy per molecule is 𝑓𝑘𝑇 2 The total energy associated with 1 mole of the gas for f degrees of freedom is 𝑈 = 1 1 𝑓 𝑘 𝑇 𝑁𝐴 = 𝑓𝑅𝑇 2 2 𝑑𝑈 𝑓 𝐶𝑉 = = 𝑅 …….(1) 𝑑𝑇 2 If CP is the molecular specific heat at constant pressure, then CP – CV = R 𝑓 CP = CV + R = ( + 1) 𝑅 ………(2) 2 𝑓 𝐶𝑃 ( +1) 𝑅 2 𝑓 𝟐 2 Thus 𝛾 = = 𝑓 = ( + 1) or 𝜸 = 𝟏 + 𝐶𝑉 𝑅 𝑓 2 𝒇 2 B.Sc. - II Semester Dr. K S Suresh Page 10  Physics : Unit 1 - Kinetic Theory of Gases Real gases Gases obeying the gas laws (Boyle’s law, charle’s law, perfect gas equation) at all temperatures and pressures are called the perfect or the ideal gases. Gases which do not obey the gas laws are called real gases. Study on the properties of vapours indicated that deviations from gas laws applicable to ideal gases was intimately connected with the process of liquefaction of a gas, as the deviation become more marked when the gas was nearing liquefaction. Andrew’s experiment on carbon dioxide Principle – The principle involved in this experiment is to measure the volume of a fixed mass of the gas at various pressures at a given temperature and draw curves of variation of pressure with volume called isothermals. From this it is possible to investigate the deviation of a gas from Boyle’s law. Apparatus – Andrews' apparatus consist of two identical tubesA and B, initially open at both the ends, with thickwalled capillary at the upper side. The capillary tubes are graduated to measure their volume and length correctly. Carbon dioxide is passed through tube A fora long time and its two ends are properly sealed. Then the lower end of the tube is immersed in mercury and opened. When the tube is warmed and allowed to cool, a small column of mercury (M) enters the tube and encloses the gas in the capillary portion of the tube. To measure pressure, tube B is enclosed with dry air above the mercury column. The two tubes are mounted side by side inside two big copper cylinders as shown in figure. Procedure - The copper cylinders are filled with water. The water in the cylinders is in communication through C and the pressure is equalized. The lower end of the copper cylinders are provided with screw plungers (S). When the plungers are turned, the gases are forced into the capillaries and are compressed and both carbon B.Sc. - II Semester Dr. K S Suresh Page 11  Physics : Unit 1 - Kinetic Theory of Gases dioxide and dry air are subjected to high pressures.The projecting portions of the capillaries are enclosed in two separate water baths, one around B is maintained at constant temperature and that around A can be maintained between 0°C to 100°C. The plungers (S) are turned into the cylinders to increase pressure. Then the volume of the gas and that of dry air are measured from the graduations on A and B.Maintaining the temperature of CO2 at a particular value, the volume ofthe gas is measured at different pressures. This is repeated for different constant temperaturesof the gas. The graphs between pressures and volumes of carbon dioxide are drawn called isothermals as shown. Interpretation of the isothermals 1. At 13.10C, the isothermal starts from A. Till B it obeys Boyle’s la representing gaseous state. B to C indicates liquefaction in which liquid and vapour at in equilibrium. CD indicates purely liquid state. 2. At 21.50C, the horizontal portion is shorter, ie. Vapour liquid region is smaller. 3. At 31.10C the horizontal portion is just a point, called the point of inflexion or critical point and the isothermal is called critical isothermal. The corresponding temperature is called critical temperature. The gas cannot be liquefied above this temperature even at high pressure. 4. At higher temperatures, the behaviour of the graph is that of a permanent gas and obeys boyle’s law. Results of Andrew’s experiment 1. For every substance there is a critical temperature TC above which liquefaction is not possible. The corresponding pressure and volume is called critical pressure PC and VC respectively. These values are called critical constants. 2. At TC the densities and refractive indices are the same for the liquid and vapour.But the surface tension and latent heat of vapourisation of the liquid are zero. 3. Gases like oxygen nitrogen and hydrogen have TC as low as – 118.80C, - 147.1 0C and – 239.90C respectively. 4. Andrew concluded that liquid and gaseous state of a substance are only the distinct stages of a long series of continuous physical changes. Van der waals’ equation of state The two fundamental assumptions of kinetic theory of gases are (1) the molecules are point masses having no dimensions and (2) no intermolecular forces exists between the B.Sc. - II Semester Dr. K S Suresh Page 12  Physics : Unit 1 - Kinetic Theory of Gases molecules, had justified the perfect gas equation in case of ideal gases. D Experiments like Andrew’s experiment and Joule Thomson effect have shown that real gases or actual gases deviated from the perfect gas equation. The A C B deviation being small at low pressure and high temperature, large at high pressure and low temperature. Van der waals’ said that the modification in the perfect gas equation to cover real gases requires the changes in assumptions in kinetic theory of gases. Accordingly, the finite size of the molecules and intermolecular forces between the molecules cannot be neglected and must be considered. 1 Correction for finite size of the molecule Consider V to be the volume occupied by a gas having molecules each of diameter . As the molecules are assumed to be solid spheres, every molecule has a sphere of influence into which other molecules cannot enter. D is the sphere of influence of molecule C of radius  which is the diameter of a molecule. 4 The volume of the sphere D is 𝑉𝑆 = 𝜋 𝜎3 3 4 𝜎 3 1 4 The volume of a molecule 𝑉𝑚 = 3 𝜋( ) = 2 ( 𝜋 𝜎 3) 8 3 𝑉𝑆 Thus 𝑉𝑚 = or 𝑉𝑆 = 8 𝑉𝑚 ……(1) 8 If V is the volume available for the first molecule, then the volume available for the second molecule is V – VS. Thus the volume available for the nth molecule is V – (n – 1) VS. 𝑉𝑆 (1+2+⋯….+(𝑛−1)) The average volume available for each molecule is 𝑉 − 𝑛 𝑉𝑆 𝑛 (𝑛−1) 𝑛 (𝑛−1) =𝑉− × [ ∵ 1 + 2 + ⋯ …. + (𝑛 − 1) = ] 𝑛 2 2 𝑛 𝑉𝑆 𝑉𝑆 =𝑉− + 2 2 𝑉𝑆 𝑉𝑆 𝑛 𝑉𝑆 𝑉𝑆 =𝑉− 𝑛 [∵ < ] 2 2 2 2 𝑛 =𝑉− 8 𝑉𝑚 or average volume = 𝑉 − 4 𝑛 𝑉𝑚 = 𝑉 − 𝑏 2 B.Sc. - II Semester Dr. K S Suresh Page 13  Physics : Unit 1 - Kinetic Theory of Gases where b = 4 n Vm. Substituting for average volume in ideal gas equation P V = R T, we get P (V – b) = R T ……..(2) 2 Correction for intermolecular attraction A molecule of the gas present deep inside the gas experience zero force as it is pulled by other molecules around it with same force. But a molecule near the wall of vessel containing the gas, experiences net inward force pulling the molecule back, away from the wall.Thus the velocity and momentum of the molecule impinging on the wall decreases. The resultant pressure exerted by the gas on the wall of the container is less as compared to that in the absence of intermolecular forces. Therefore, the decrease in pressure depends on (1) the number of molecules striking unit area of the walls of the vessel and (2) the number of molecules attracting them backwards. Each of these factors depend on the density  of the gas. Thus the decrease in pressure is directly proportional to 2 For a given mass density is inversely proportional to the volume of the gas. Thus decrease in 1 pressure  𝑉2 𝑎 or decrease in pressure is equal to where a is a constant. 𝑉2 Therefore the correct pressure exerted by the gas in the absence of intermolecular attraction is 𝑎 obtained by adding this term to the reduced pressure P i.e. (𝑃 + ). Substituting this in 𝑉2 𝒂 equation (2) we get (𝑷 + 𝑽𝟐 ) (𝑽 − 𝒃) = 𝑹 𝑻. This equation is called the Van der Waals’ equation of state. The graph showing the variation of P with V plotted using van der waals equation is as shown. Comparison between the graphs of Andrew’s experiment and the van der waals equation – Similarities 1 The graphs are similar at higher temperatures. 2 At lower temperatures the curved portions are sharper in both the graphs. 3 At 31. 10C, both the graphs show horizontal tangent indicating the critical temperature. 4 At lower temperatures the gas does not behave as a perfect gas. B.Sc. - II Semester Dr. K S Suresh Page 14  Physics : Unit 1 - Kinetic Theory of Gases Dissimilarities 1 Below the critical temperature, van der waal curve show turning point whereas Andrew curve show horizontal position. 2 In the van der waal curve, the regions AC and DE represent unstable, supercooled and super heated liquid states respectively and region CBD show decrease in pressure which has no meaning. Expression for critical constants The limiting temperature at which a gas can be liquified by increasing pressure only and above which it cannot be liquefied is called critical temperature TC. The pressure applied to the gas at critical temperature so that gas can be liquefied is called critical pressure PC. The volume of the gas at critical temperature so that gas can be liquefied is called critical volume VC. The critical temperature and the corresponding values of pressure and volume at the critical point are called critical constants. 𝑎 Consider the Van der Waals’ equation (𝑃 + 𝑉2 ) (𝑉 − 𝑏) = 𝑅 𝑇…(1) 𝑅𝑇 𝑎 𝑃= − …..(2) (𝑉−𝑏) 𝑉2 𝑑𝑃 𝑅𝑇 2𝑎 Differentiating equation (2) =− + …….(3) 𝑑𝑉 (𝑉−𝑏)2 𝑉3 At critical point, the rate of change of pressure with volume is zero 𝑑𝑃 𝑑2𝑃 ie. = 0 and also = 0. Also T = TC and V = VC 𝑑𝑉 𝑑𝑉 2 𝑅 𝑇𝐶 2𝑎 Equation (3) reduces to 0 = − + (𝑉𝐶 −𝑏) 2 𝑉𝐶3 2𝑎 𝑅 𝑇𝐶 or = …..(4) 𝑉𝐶3 (𝑉𝐶 −𝑏)2 Differentiating (3) with respect to V again 𝑑2𝑃 2𝑅 𝑇 6𝑎 = − 𝑑𝑉 2 (𝑉 − 𝑏)3 𝑉 4 𝑑2𝑃 2𝑅𝑇𝐶 6𝑎 At critical point = 0, thus, 0 = − 𝑑𝑉 2 (𝑉𝐶 −𝑏) 3 𝑉𝐶4 3𝑎 𝑅𝑇𝐶 or = ……(5) 𝑉𝐶4 (𝑉𝐶 −𝑏)3 2 𝑉𝐶 Dividing (4) by (5) we get = 𝑉𝐶 − 𝑏 or 𝑽𝑪 = 𝟑𝒃 ….(6) 3 B.Sc. - II Semester Dr. K S Suresh Page 15  Physics : Unit 1 - Kinetic Theory of Gases 2𝑎 𝑅 𝑇𝐶 Substituting for VC from (6) in (4) = (3𝑏)3 (3𝑏 −𝑏)2 𝟖𝒂 Or 𝑻𝑪 = ……(7) 𝟐𝟕 𝑹 𝒃 Substituting for VC and TC from (6) and (7) in (2) 𝑅 8𝑎 𝑎 𝒂 𝑃𝐶 = × − or 𝑷𝑪 = …..(8) (3 𝑏−𝑏) 27 𝑅 𝑏 9𝑏2 𝟐𝟕 𝒃𝟐 The expressions for the critical constants are 𝒂 𝟖𝒂 𝑷𝑪 = , 𝑻𝑪 = and 𝑽𝑪 = 𝟑𝒃 𝟐𝟕 𝒃𝟐 𝟐𝟕 𝑹 𝒃 Coefficients of Van der waals’ constants or critical constants 𝑇𝐶 𝟖𝒂 𝟐𝟕 𝒃𝟐 8𝑏 Dividing the expression for TC by PC = × = 𝑃𝐶 𝟐𝟕 𝑹 𝒃 𝒂 𝑅 𝑹 𝑻𝑪 Thus 𝒃 = 𝟖 𝑷𝑪 𝟖𝒂 𝟖 𝑷𝑪 Substituting for b in the equation for TC 𝑻𝑪 = × 𝟐𝟕 𝑹 𝑹 𝑻𝑪 𝟐𝟕 𝑹𝟐 𝑻𝟐𝑪 Thus 𝒂= 𝟔𝟒 𝑷𝑪 𝑹 𝑻𝑪 𝟖 The critical coefficient of the gas is given by. Its value is equal to. This value is same 𝑷𝑪 𝑽𝑪 𝟑 for all gases. Limitations of van der waals equation 1. According to van der waals equation the critical coefficient is 8/3. But experimentally it varies from 3.27 to 4.99. 2. The value of b must be theoretically is Vc/3. But experimentally its value is Vc/2. 3. The isothermals obtained theoretically from van der waals equation do not agree with the isothermals found from Andrews experiment. 4. The value of a and b do not remain constant at all temperatures even for the same gas. Differences between Ideal gases and Real gases S. Ideal gas Real gas No. Obeys Boyles’ law, Charles’ law and Does not obey Boyles’ law, Charles’ law and 1 perfect gas equation PV = RT perfect gas equation Equation of state PV = RT 𝑎 2 (𝑃 + 2 ) (𝑉 − 𝑏) = 𝑅 𝑇 𝑉 B.Sc. - II Semester Dr. K S Suresh Page 16  Physics : Unit 1 - Kinetic Theory of Gases Their specific heat is independent of Their specific heat is dependent on 3 temperature temperature𝐶𝑉 ∝ 𝑇 Their internal energy depends only on Their internal energy depends on 4 temperature temperature as well as volume At high temperatures and low pressures At low temperature and high pressure 5 behave as ideal gas behave as real gas Transport Phenomenon The phenomenon that can be explained on the basis of movement of molecules of a gas is called transport phenomenon. The phenomenon of viscosity and thermal conductivity take place due to the transfer of momentum and transfer of heat energy respectively. Viscosity of gases When different layers of a fluid have different velocities, in order to attain steady state, momentum is transferred from faster moving molecules to slower moving molecules. This phenomenon is called viscosity. Consider a gas moving over a horizontal solid surface. The gas is imagined to be made up of different layers, having velocity 𝑑𝑣 gradient 𝑑𝑦 with increasing velocities with distance from the surface. Thus, a relative motion exists between layers of the fluid resulting in opposing force between 𝑑𝑣 them. This force is 𝐹 = 𝜂 …….(1) 𝑑𝑦 where is called the coefficient of viscosity. 𝑁 Consider different layers of fluid as shown in the diagram. Let 𝑛 = be the number of 𝑉 molecules per unit volume and 𝑐̅ be their average velocity. Due to thermal agitation, the molecules move in all possible directions. If the three coordinate axes are considered with both positive and negative directions, the number of molecules crossing unit area of layer B in one second in any one direction is (ie. upwards or downwards) 1 1 = 6 𝑛 𝑐̅. The mass of these molecules 𝑀 =6 𝑛 𝑐̅ 𝑚 where m is mass of each molecule. Let P and Q be the two layers of gas at distance  called mean free path on either side of layer B between which the molecules do not collide with each other. Let v be the velocity of gas in layer B. B.Sc. - II Semester Dr. K S Suresh Page 17  Physics : Unit 1 - Kinetic Theory of Gases 𝑑𝑣 The velocity of gas in layer P = 𝑣 + 𝜆 𝑑𝑦 𝑑𝑣 The velocity of gas in layer Q = 𝑣 − 𝜆 𝑑𝑦 The momentum transferred from P to Q per unit area of the plane B per second is given by 𝑑𝑣 1 𝑑𝑣 𝑃1 = 𝑀 × (𝑣 + 𝜆 𝑑𝑦 ) = 6 𝑛𝑚 𝑐̅ (𝑣 + 𝜆 𝑑𝑦 ) Similarly the momentum transferred from Q to P per unit area of plane B per second is given by 𝑑𝑣 1 𝑑𝑣 𝑃2 = 𝑀 × (𝑣 − 𝜆 𝑑𝑦 ) = 6 𝑛𝑚 𝑐̅ (𝑣 − 𝜆 𝑑𝑦 ) Thus change in momentum per unit area per second is ∆𝑃 = 𝑃1 − 𝑃2 1 𝑑𝑣 𝑑𝑣 ∆𝑃 = 𝑛𝑚 𝑐̅ {(𝑣 + 𝜆 ) − (𝑣 − 𝜆 )} 6 𝑑𝑦 𝑑𝑦 1 𝑑𝑣 or ∆𝑃 = 𝑛𝑚 𝑐̅ {2𝜆 } 6 𝑑𝑦 1 𝑑𝑣 This term is equal to force F from Newton’s second law. Thus 𝐹 = 𝑛𝑚 𝑐̅ 𝜆 ……(2) 3 𝑑𝑦 𝑑𝑣 1 𝑑𝑣 Comparing equations (1) and (2) we get 𝜂 = 𝑛𝑚 𝑐̅ 𝜆 𝑑𝑦 3 𝑑𝑦 𝟏 𝑚𝑎𝑠𝑠 𝑁 Or 𝜼 = 𝒏𝒎 𝒄̅ 𝝀 Also 𝜌 = = 𝑛𝑚 where 𝑛 =. 𝟑 𝑣𝑜𝑙𝑢𝑚𝑒 𝑉 𝟏 1 1 1 𝑚 𝑐̅ Thus 𝜼 = 𝝆 𝒄̅ 𝝀 As 𝜆 = √2 𝑛 𝜋 𝜎2 , 𝜂= 𝑛 𝑚 𝑐̅ = 𝟑 3 √2 𝑛 𝜋 𝜎2 3√2 𝜋 𝜎 2 Thus the coefficient of viscosity depends on mass, velocity and diameter of the molecules. As 𝑐̅ ∝ √𝑇 , 𝜼 ∝ √𝑇 Thermal conductivity of gases When different layers of a fluid have different temperatures, in order to attain steady state, heat energy is transferred from higher temperature layer to lower temperature layer. This phenomenon is called thermal conduction. Consider a gas moving over a horizontal solid surface. The gas is imagined to be made up of different layers, having 𝑑𝑇 temperature gradient with increasing temperatures with 𝑑𝑦 distance from the surface. Thus, there is a heat transfer from a layer of higher temperature to the layer of lower 𝑑𝑇 temperature. If Q is the transfer of heat per unit area per second then 𝑄 = 𝐾 …….(1) 𝑑𝑦 where K is called the coefficient of thermal conductivity. B.Sc. - II Semester Dr. K S Suresh Page 18  Physics : Unit 1 - Kinetic Theory of Gases 𝑁 Consider different layers of fluid as shown in the diagram. Let 𝑛 = be the number of 𝑉 molecules per unit volume and 𝑐̅ be their average velocity. Due to thermal agitation, the molecules move in all possible directions. If the three coordinate axes are considered with both positive and negative directions, the number of molecules crossing unit area of layer B in one second in any one direction is (ie. upwards or downwards) 1 1 = 6 𝑛 𝑐̅. The mass of these molecules is 𝑀 = 𝑛 𝑐̅ 𝑚 where m is mass of each molecule. 6 Let P and Q be the two layers of gas at distance  called mean free path on either side of layer B between which the molecules do not collide with each other. Let T be the temperature of layer 𝑑𝑇 B. The temperature of gas in layer P = 𝑇 + 𝜆 𝑑𝑦 𝑑𝑇 The temperature of gas in layer Q = 𝑇 − 𝜆 𝑑𝑦 The heat transferred from P to Q per unit area of plane B per second is given by 𝑑𝑇 1 𝑑𝑇 𝑄1 = 𝑀 𝐶𝑉 × (𝑇 + 𝜆 𝑑𝑦 ) = 6 𝑛𝑚𝐶𝑉 𝑐̅ (𝑇 + 𝜆 𝑑𝑦 ) Similarly the heat transferred from Q to P per unit area of plane B per second is given by 𝑄2 = 𝑑𝑇 1 𝑑𝑇 𝑀 𝐶𝑉 × (𝑇 − 𝜆 𝑑𝑦 ) = 6 𝑛𝑚𝐶𝑉 𝑐̅ (𝑇 − 𝜆 𝑑𝑦 ) where CV is the specific heat of gas at constant volume. Thus change in heat per unit area per second is ∆𝑄 = 𝑄1 − 𝑄2 1 𝑑𝑇 𝑑𝑇 ∆𝑄 = 𝑛𝑚𝐶𝑉 𝑐̅ {(𝑇 + 𝜆 ) − (𝑇 − 𝜆 )} 6 𝑑𝑦 𝑑𝑦 1 𝑑𝑇 or ∆𝑄 = 𝑛𝑚𝐶𝑉 𝑐̅ {2𝜆 } 6 𝑑𝑦 1 𝑑𝑇 This term is equal to net heat energy Q transferred. Thus 𝑄 = 𝑛𝑚𝐶𝑉 𝑐̅ 𝜆 ……(2) 3 𝑑𝑦 𝑑𝑇 1 𝑑𝑇 Comparing equations (1) and (2) we get 𝐾 = 𝑛𝑚𝐶𝑉 𝑐̅ 𝜆 𝑑𝑦 3 𝑑𝑦 𝟏 𝑚𝑎𝑠𝑠 𝑁 Or 𝑲 = 𝒏𝒎𝑪𝑽 𝒄̅ 𝝀 Also 𝜌 = = 𝑛𝑚 where 𝑛 =. 𝟑 𝑣𝑜𝑙𝑢𝑚𝑒 𝑉 𝟏 Thus 𝑲 = 𝝆𝑪𝑽 𝒄̅ 𝝀 𝟑 Thus the coefficient of thermal conductivity is independent of pressure and directly proportional to the temperature. (As 𝑐̅ ∝ √𝑇 , 𝑲 ∝ √𝑇) Note : Relation between 𝜼 and K 𝟏 𝟏 As 𝜼 = 𝝆 𝒄̅ 𝝀 and = 𝝆𝑪𝑽 𝒄̅ 𝝀. Thus 𝑲 = 𝜼 𝑪𝑽 𝟑 𝟑 B.Sc. - II Semester Dr. K S Suresh Page 19  Physics : Unit 1 - Kinetic Theory of Gases B.Sc. - II Semester Dr. K S Suresh Page 20