Physics - Kinetic Theory PDF

## LEGENDS' (MGDN) STUDY MATERIAL - I PUC PHYSICS (NCERT)-Kinetic Theory ### CHAPTER - 13 ### KINETIC THEORY **Introduction**, Molecular nature of matter, behavior of gases, Kinetic theory of an ideal gas, Law of equipartition of energy, specific heat capacity, Mean free path. 5 hours **Kinetic t...

## LEGENDS' (MGDN) STUDY MATERIAL - I PUC PHYSICS (NCERT)-Kinetic Theory ### CHAPTER - 13 ### KINETIC THEORY **Introduction**, Molecular nature of matter, behavior of gases, Kinetic theory of an ideal gas, Law of equipartition of energy, specific heat capacity, Mean free path. 5 hours **Kinetic theory - ಅಧ್ಯಾಯದಲ್ಲಿ ಬರುವ ಕೆಲವು ಮುಖ್ಯವಾದ ಅಂಗ್ಲ ಪದಗಳಿಗೆ ಕನ್ನಡ ಅರ್ಥ** | English | Kannada | |---|---| | instantaneous | ತಕ್ಷಣದ | | obeys | ಪರಿಪಾಲಿಸು | | assumptions | ಊಹೆಗಳನ್ನು | | behave | ವರ್ತಿಸು | | collision | ಘರ್ಷಣೆ | | consistent | ಸ್ಥಿರ | | container | ಪಾತ್ರೆ | | degrees | ಅ೦ಶ | | hypothesis | ಕಲ್ಪನೆ | | diatomic | ದಿನ | | diffusion | ಹರಡುವಿಕೆ | | enclosed | ಸುತ್ತುವರಿದ | | features | ವೈಶಿಷ್ಟ್ಯಗಳು | | freedom | ಸ್ವಾತಂತ್ರ್ಯ | | freedom | ಸ್ವಾತಂತ್ರ್ಯ | | hit | ಹೋಡೆ | | interaction | ಪರಸ್ಪರ ಕ್ರಿಯೆ | | interpretation | ಅರ್ಥ ವಿವರಣೆ | | mono | *single* | | origin | ಮೂಲ | | random | ಗೊತ್ತುಗುರಿಯಿಲ್ಲದೆ | | successive | ಅನುಕ್ರಮದ | | vessel | ಪಾತ್ರೆ | ### ADDITIONAL INFORMATION (For CET and understanding the subject) The kinetic theory was developed by Maxwell, Boltzmann and others in the nineteenth century. It explains the behavior of gases based on the idea the gases consists of a large number of atoms or molecules in a state of continuous random motion. The interatomic forces binding the atoms are negligible. The theory is consistent with various gas laws and Avogadro's hypothesis. It gives molecular interpretation of temperature, pressure, specific heat capacities, viscosity, conduction and diffusion of gases. We shall study some of the features of kinetic theory of gases in this chapter. ### Perfect gas or Ideal gas It is the gas which satisfies the relation PV = n RT. It is the gas which obeys Boyle's law and Charles' law. ### Equation of state of perfect gas or Perfect gas equation **PV = nRT** for n moles of gas, **PV = RT** for n =1 mole of gas, where P is the pressure of one mole of ideal gas V is the volume of ideal gas at temperature *T* R is the universal gas constant. ### NOTE 1) This equation is applicable for real gases at low pressures or at high temperatures. This is because, at low pressures or at high temperatures, the molecules are far apart and molecular interactions are negligible. Without interactions the gas behaves like an ideal gas. 2) The value of universal gas constant, R = 8.31 J mol-1 K-1 3) If P₁, V₁ and T₁ are the initial values and P₂,V₂ and T₂ are the final values, then **P₁V₁/T₁ = P₂V₂/T₂** = constant 4) Number of moles in terms of molecular mass and mass of the gas. **n = M/M₀** where, n is the number of moles of gas, M is the molecular mass of the gas and M₀is the mass of the gas. 5) Number of moles in terms of number of molecules and Avogadro number. **n = N/NA** where, n is the number of moles of gas, N is the number of molecules and NA is the Avogadro number = 6.023×10²³. ### Give the molecular picture of matter related to kinetic theory of gases **OR** ### State the assumptions of kinetic theory of gases 1) A given amount of gas is a collection of large number of molecules 2) All the molecules of gas are in incessant random motion. 3) The average distance between molecules is 10 times or more than the size of the molecule (2A°). 4) The gas molecules do not exert any force of attraction or repulsion on each other, except during collision. 5) The gas molecules move in straight lines and obey Newton's first law of motion. 6) The gas molecules change their velocities when they collide with each other or with the walls of the container. 7) The molecular collisions are elastic collisions. 8) The molecular collisions are instantaneous. ### NOTE The molecular collisions are elastic collisions. That is, the total KE and momentum of gas molecules are conserved. The molecular collisions are instantaneous. That is, the time of collision between two molecules is negligible compared to time interval between two successive collisions. ### Mean free path It is the average distance covered by the molecule between two successive collisions. ### Mention an expression for mean free path of a gas molecule in a gas and explain the symbols used. **l = 1 / √2 ηπα²** where n is the number of molecules per unit volume, d is the diameter of a molecule, ### Derive an expression for mean free path of a gas molecule in a gas. Consider a gas molecule (say A in the figure) in the form of a sphere of diameter *d* moving with average speed *vav*. This molecule will collide any molecule (say B or C) comes within a distance *d* between the centres of the molecules as indicated in the figure. The volume (V) swept by the molecule A in moving from the point *Q* to the point *Q* in a time *Δt* is given by, V = surface area of the molecule (A)× Distance traveled by the molecule (x) V = πd² × vavΔt ----->(1) If n is [the] number of molecules per unit volume of the gas, the molecule A collides with nV number of molecules in time *Δt*. Therefore, the number of collisions executed by the molecule A, N = nV ----->(2) (1) in (2) ⇒ N = n πd² × vavΔt N Rate of collision of the molecule A, = η πd² × vav Δt Reciprocal of rate collision is the time (τ) between the two successive collisions. .. τ = Δt / N = 1 / ηπα² × vav ----->(3) But, mean free path (l) = average distance covered by the molecule between two successive collisions in a time τ. .. l = vavτ ------>(4) In deriving the above equation, we assumed that the other molecules to be at rest. But actually all molecules are moving and collision rate is determined by the average relative velocity of the molecules. Therefore, *vav* should be replaced by (Vav)relative. (4) ⇒ l = (Vav)relativet ------>(5) But, (Vav) relative = Vav √2 -->(6) (6) in (5)⇒ l = Vavτ / √2 ----->(7) (3) in (7) ⇒ l = 1 / √2 ηπα² From the above equation it is observed that the mean free path depends inversely on number of molecules per unit volume (n) and size of the molecule (d). ### Derive the relation, P = 1/3 mnv² using kinetic theory of gases. Consider a gas enclosed in a cubical vessel of side *l*. The area of each face of the vessel is *l²* and the volume of the gas inside the vessel is *l³*. Let *m* be the mass of each molecule of gas and *n* be the number of molecules per unit volume of the gas. Let the corner *O* be the origin of the three dimensional rectangular coordinate system. Consider a molecule *A* of the gas moving with a velocity *v*. If *vx*, *vy* and *vz* are the components of *v* along X, Y and Z axes respectively, Before collision with the face-1, then v² = vx²+ Vy²+ V₂² Initial momentum of the molecule along X-axis = mvx After collision with the face-1, Final momentum of the molecule along X-axis = -mvx ..Change in momentum of the molecule along X-axis = (-mvx)-(mvx)=-2mvx Momentum transfered to the face-1 = +2mvx. In a small time interval *dt*, a molecule with x-component of velocity *vx* will hit the face-1 if it is within a distance *vx dt* from the face-1. That means, all the molecules within the volume *A vx dt* only can hit the wall in a time *dt*. But, on an average, half of these molecules are moving towards the wall and the other half move away from the wall. Thus, the number of molecules hitting the face-1 in a time *dt* = 1/2 n Avx dt Total momentum transferred to the wall in a time *dt* is *dpx* = 1/2 (2mvx)( 1/2 n A vx dt) = nm A v² dt The force *F* on the face-1 is the rate of change of momentum transferred. That is, F = *dpx*/ dt = nmAv₂²/ dt = nmAv₂² The pressure *P* exerted by the molecules is given by, P = F/A = nmAv₂²/ A = nm v₂² ----->(1) In a gas, all the molecules do not have the same velocity. There is a distribution of velocities. Equation (1) stands for the pressure due to the group of molecules with velocity *v* along the *x*-direction. Therefore, total pressure exerted by all the molecules on the face-1 is given by, P = nmv₂² --->(2) where *v₂²* is the average of *v₂²* Inside the vessel, the molecules are in random motion. Let *vy²* is the average of *vy²* along Y-axis and *vz²* is the average of *vz²* along Z-axis. Then we have,* v² = 1/3 v₂² + 1/3 vy² + 1/3 vz²* where *v²* is the mean of the squared speed of the gas molecules. (2) ⇒ P = 1/3 nm v² ### Kinetic interpretation of temperature How does an average kinetic energy of gas molecule depend on the absolute temperature? Average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas. ### Mention of expression for average kinetic energy of a molecule in terms of absolute temperature and explain the symbols. KE = 1/2 KT m v² = 3/2 KT where *kB* is the Boltzmann constant and *T* is the absolute temperature. ### Derive the relation between kinetic energy of a gas molecule and its absolute temperature. We have, Average kinetic energy of a gas molecule, P = 1/3 nm v² ⇒ PV = 1/3 nV mv² ----->(1) Divide and multiply by 2 for RHS ⇒ PV = 2/3 N mv² where N = nV=number of molecules in a sample of gas where KE = 1/2 mv² = average kinetic energy. Also, for 1 mole, PV = RT ---->(3) On comparing (1) and (3), we get, N(KE) = RT Where *T* is the absolute temperature ⇒ KE = 3/2 RT / N ⇒ KE = 3/2 kB T where *kB* = R/N is the Boltzmann constant. Thus the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas. Average KE of a gas molecule is independent of pressure, volume and the nature of an ideal gas. kB=1.38 X10⁻²³ J/K ### Internal energy The sum of average kinetic energy of all the molecules is the internal energy of a gas. ### Definition of rms speed of gas molecule rms speed of gas molecule is the square root of the mean of the squares of the velocities of individual molecules of gas. ### Mention the formula for rms speed of gas molecule and explain the symbols. Vrms = √3kBT / m where *kB* is the Boltzmann's constant, *T* is the absolute temperature and *m* is the mass of the gas. ### Degrees of freedom These are the number of co-ordinates required to specify the configurations of a gas molecule. no. of independent motions of a partide. ### Formula used to calculate degrees of freedom for non-linear molecule. n= 3x - I where n = number of degrees of freedom x = number of atoms in a molecule I = number of independent relations. ### Calculation of degrees of freedom We have n = 3x - l ------>(1) a) For mono atomic molecule, x = 1 (1) ⇒ n = 3x 1-0 = 3 All the three degrees of freedom are due to three translatory motions. b) For diatomic molecule, n = 3x 2-1 = 5 x = 2 and I=1 Three degrees of freedom are due to three translatory motions and two are due to two rotatory motions. c) For Triatomic molecule, x = 3 and I=3 (1) ⇒ n = 3x3-3 =6 Three degrees of freedom are due to three translatory motions and three are due to three rotatory motions.. ### Law of equipartition of energy In thermal equilibrium of gas, the total energy of a gas is equally distributed among all the degrees of freedom. The energy per molecule per degree of freedom is equal to 1/2 kT. where k is the Boltzmann's constant and T is the absolute temperature. ### Expressions for the average energy of a monatomic /diatomic/triatomic molecule In general, the average energy of a molecule with n degrees of freedom is given by, For monatomic molecule, n = 3, (1) ⇒ E = 1/2 kTn ----->(1) E = 1/2 kT (3) ⇒ E = 3/2 kT For 1 mol of monatomic gas of N molecules, E = 3NkT⇒ E = 3RT : Nk = R For diatomic molecule, n = 5 (1) ⇒ E = 1/2 kT (5) ⇒ E = 5/2 kT For 1 mol of monatomic gas of N molecules, E = 5 / 2 NkT ⇒ E = 5/2 RT For triatomic molecule, n = 6, (1) ⇒ E = 1/2 kT (6) ⇒ E = 3kT For 1 mol of triatomic gas of N molecules, E = 3 N ķ T → E = 3R T ### Ratio of Specific heats (γ) It is the ratio of specific heat of gas at constant pressure to the specific heat of gas at constant volume. That is, γ = Cp / Cv ### Importance of γ The value of γ helps in the determination of number of atoms in the gas molecule. ### Calculation of γ Consider the equation, γ = 1 + 2/n --->(1) where n = number of degrees of freedom. 1) For monatomic molecule. n = 3(1) ⇒ γ = 1 + 2/3 = 5/3 = 1.67 2) For diatomic molecule n = 5 (1) ⇒ γ =1 + 2/5 = 7/5 = 1.4 3) For triatomic molecule n = 6 (1) ⇒ γ = 1 + 2/6 = 8/6 = 1.3 There is no unit and dimensions for γ. It is a number ### Derive an expression for specific heat capacity of a solid in terms of universal gas constant. We know that, The average energy of a molecule with *n* degrees of freedom is given by, E = 1/2 kTn---->(1) where k is the Boltzmann's constant and *T* is the absolute temperature. Consider a solid of *N* monatomic molecules (ie molecule=atom) each vibrating about its mean position. The atom of solid oscillating in one dimension has two degrees of freedom(n=2). One degree of freedom is due to potential energy *mode* and the other due to kinetic energy mode. .. (1) E = 2(KT) .. E = KT---->(2) If the atom vibrates in three dimensions, E = 3 kT ---->(3) For 1mol of solid, consisting of N atoms, E = 3NkT But Nk = R, the universal gas constant, .. E = 3RT But E = U, the internal energy of a solid, U = 3RT On differentiating above equation w.r.t T, we get dU/dT = 3 R ---->(4) From first law of thermodynamics we have dQ = dU + P dV---->(5) The pressure inside the solid do not change with heat and hence change in volume is negligible (dV ≈ 0). ..(5) dU = dQ Specific heat of a solid, C = dQ / dT = dU / dT --->(6) On comparing (4) in (6) we get, C = 3 R ### Derive an expression for specific heat of water in terms of universal gas constant. For 1mol of solid, consisting of N monatomic molecules, average internal energy of a solid is given by E = 3RT where R is the universal gas constant and *T* is the absolute temperature. A water molecule(H2O) contains 3 atoms (2 hydrogen atoms and 1 oxygen atom). Assume the water as a solid with triatomic molecules. Now the total internal energy of 1mol of water is E E = 3(3 R T) = 9RT But E= U, the internal energy of a water ⇒ U = 9RT On differentiating above equation w.r.t *T*, we get dU/dT = 9 R ---->(1) From first law of thermodynamics we have dQ = dU + P dV---->(2) The pressure inside the solid do not change with heat and hence change in volume is negligible (0), .:. (2)⇒ dU = dQ Specific heat of a solid, C = dQ / dT = dU / dT --->(3) On comparing (1) in (3) we get, C = 9 R