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 B of side l with perfectly elastic walls F containing a large number of molecules. A E Let c1 be the velocity of a molecule in a Y Z 푣1 direction as shown. This can be resolved 푤1 X 퐶 푢 into three components u1, v1 and w1 along 1 1 X, Y and Z directions as shown. Then C 2 2 2 2 H 푐1 = 푢1 + 푣1 + 푤1 ……..(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. 2푙 The time between the successive collisions (time for one collision) on ABCD is . 푢1 B.Sc. - II Semester Dr. K S Suresh Page 1 Physics : Unit 1 - Kinetic Theory of Gases 푢 Thus the number of collisions per second, this molecule makes with ABCD is 1. 2푙 Hence the rate of change of momentum = change in momentum × number of collisions per 2 푢1 푚푢1 second is equal to 2푚푢 × = . 1 2푙 푙 From Newton’s second law, rate of change of momentum = impressed force. 2 푚푢1 If f is the force, then 푓 = along X – direction. 1 1 푙 Similarly, the force on another molecule of velocity c2 whose components are u2, v2 and w2 due 2 푚푢2 to impact is 푓 = along X – direction. 3 푙 Hence the total force Fx on the face ABCD due to impacts of all the n molecules in the X – 푚 2 2 2 2 direction is given by 퐹 = (푢 + 푢 + 푢 … … … . + 푢 ) ………..(2) 푥 푙 1 2 3 푛 Since pressure is force per unit area, the pressure Px on ABCD is given by 퐹푥 푚 2 2 2 2 푃 = = (푢 + 푢 + 푢 … … … . + 푢 ) ………..(3) ( since P = F/A and A = l2) 푥 푙2 푙3 1 2 3 푛 Similarly, if Py and Pz are the pressures on faces EFBA and FBCH, then 퐹푦 푚 2 2 2 2 푃 = = ( 푣 + 푣 + 푣 … … … . + 푣 ) ………(4) 푦 푙2 푙3 1 2 3 푛 퐹푧 푚 2 2 2 2 푃 = = (푤 + 푤 + 푤 … … … . + 푤 ) ………(5) 푧 푙2 푙3 1 2 3 푛 As the pressure exerted by the gas is same in all directions, the average pressure P of the gas is 푃푋+푃푦+ 푃푧 푃 = .......(6) 3 푚 푃 = (푢2 + 푢2 + 푢2 … … … . + 푢2) + ( 푣2 + 푣2 + 푣2 … … … . +푣2) 3 푙3 1 2 3 푛 1 2 3 푛 2 2 2 + (푤1 + 푤3 … … … . + 푤푛 ) 푚 푃 = (푢2 + 푣2 + 푤2) + ( 푢2 + 푣2 + 푤2) … 3 푙3 1 1 1 2 2 2 2 2 2 + (푢푛+ 푣푛 + 푤푛 ) … . (7) 3 2 2 2 2 Since l = V, the volume of the cube and 푐1 = 푢1 + 푣1 + 푤1 2 2 2 2 푐2 = 푢2 + 푣2 + 푤2 and so on. 푚 푃 = (푐2 + 푐2 + 푐2 + ⋯ … … . + 푐2)….(8) 3 푉 1 2 3 푛 1 푚 푛 푐2+ 푐2 + 푐2 +⋯……..+ 푐2 or 푃 = 퐶2 where 퐶2 = 1 2 3 푛 known as the mean square velocity of the 3 푉 푛 molecules. ퟏ 푴 If M is the total mass of the gas, ie. M = n m, then 푷 = 푪ퟐ ……(9) ퟑ 푽 B.Sc. - II Semester Dr. K S Suresh Page 2 Physics : Unit 1 - Kinetic Theory of Gases ퟏ If is the density of the gas, then pressure of the gas is 푷 = 푪ퟐ ퟑ 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. ퟑ 푷 It is given by = √ . 흆 ퟑ To derive the relation 푼 = 푹 푻 ퟐ The pressure exerted by a gas of n molecules occupying volume V is given by 1 푚 푛 2 1 2 푃 = 퐶 or 푃 푉 = 푚 푛 퐶 3 푉 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 푃 푉 = 푀 퐶2…..(1) 3 From the perfect gas equation PV = RT ………(2) 1 From (1) and (2) we get 푀 퐶2 = 푅 푇 3 2 or 푀 퐶 = 3 푅 푇 1 3 Dividing the above equation on both the sides by 2 we get 푀 퐶2 = 푅 푇 ……..(3) 2 2 ퟑ 1 or 푼 = 푹 푻 ……(4) where 푈 = 푀 퐶2 is called the internal energy of the gas. ퟐ 2 Dividing both sides of equation (3) by NA, which is the number of molecules in one gram 1 푀 3 푅 molecule of the gas or one mole, called Avogadro number, we get 퐶2 = 푇 2 푁퐴 2 푁퐴 As M/푁퐴 = m and R/푁퐴 = k, where k is Boltzmann constant. 1 2 3 푚 퐶 = 푘 푇 …….(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 푚 푛 2 1 2 푃 = 퐶 or 푃 푉 = 푚 푁 퐶 ……(1) 3 푉 3 퐴 2 1 2 1 Or 푃 푉 = × 푚 푁 퐶2 = 푁 × 푚 퐶2 ……..(2) 3 2 퐴 3 퐴 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 2 1 2 퐾퐸 = 푀 퐶 = 푚 푁 퐶 2 2 퐴 1 3 Average kinetic energy of a molecule is 푚 퐶2 = 푘 푇 …..(3) 2 2 2 3 Comparing equations (2) and (3) 푃 푉 = 푁 × 푘 푇 3 퐴 2 or 푃 푉 = 푁퐴 푘 푇 or 푷 푽 = 푹 푻 where R = 푁퐴 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 1 푀 2 1 2 푃 = 퐶 or 푃 푉 = 푀 퐶 3 푉 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. 1 푀 2 1 푚 푛 2 (2) Charle’s law – The pressure of a gas is 푃 = 퐶 or 푉 = 퐶 3 푉 3 푃 Thus for a given mass of gas, at constant pressure 푉 ∝ 퐶2 As 퐶2 ∝ 푇, we get 푽 ∝ 푻 . Hence the Charle’s law. Similarly it can be shown that 푃 ∝ 푇 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 2 1 2 푃 = 푚 푛 퐶 = 푚 푛 퐶 …….(1) 3 1 1 1 3 2 2 2 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 2 1 2 same. ie. 푚 퐶 = 푚 퐶 …(2) 2 1 1 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 . 푺 If the total path travelled in N collisions is S, then the mean free path is given by 흀 = 푵 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.