4 Session Objectives Graham’s law of diffusion/effusion Postulates of kinetic theory of gasesKinetic gas equation and kinetic energy of gasesVelocity of gas moleculesMaxwell-Boltzmann velocity distributionExplanation of gas laws on the basis of kinetic theory of gases
6 Illustrative example 7Which of the following compounds is steam volatile?
7 SolutionThe compound which forms intramolecular H-bonding has lower boiling point and hence, steam volatile.Hence, the answer is (b).
8 Illustrative example 1A mixture CO and CO2 is found to have a density of 2gL–1 at 250C and 740 torr. Find the composition of the mixtureSolution:
9 Illustrative example 2A 1500 ml flask contains 400 mg of O2 and 60 mg H2 at 1000C and they are allowed to react to form water vapour. What will be the partial pressure of the substances present at that temperature?Solution:
17 Illustrative example 4A balloon filled with ethylene (C2H4) is pricked with a sharp point & quickly dropped in a tank full of hydrogen at the same pressure. After a while the balloon will have(a) Shrunk (b) Enlarged(c) Completely collapsed (d) Remain unchanged in sizeSolutionSince, molar mass of H2 is much less than C2H4.H2 will diffuse into the balloon.Hence, answer is (b).
18 Kinetic molecular theory All gases are made up of very large number of extremely small particles called molecules.The actual volume of the molecules is negligible as compared to the total volume of the gas.The distances of separation between the molecules are so large that the forces of attraction or repulsion between them are negligible.
19 Kinetic molecular theory The molecules are in a constant state of motion in all directions. During their motion, they collide with one another and also the walls of the container.The molecular collisions are perfectly elastic.The pressure exerted by the gas is due to bombardment of the gas molecules on the walls of the container.
20 Kinetic Theory of Gases The kinetic gas equation,Where,m = mass of each gas moleculen’ = number of gas moleculesc = velocity of gas moleculeKE of one molecule =
21 Kinetic Theory of Gases (where n = number of moles, n’ = number of molecules)(for one mole of a gas, n =1 and n’ = NA )= Average kinetic energy per mole
22 What will be the average kinetic energy of one molecule? Ask yourselfWhat will be the average kinetic energy of one molecule?Average kinetic energy of one molecule =k = Boltzmann constant= 1.38 × 10–16 ergs k–1 molecule–1
29 Illustrative example 6The gas molecules have root mean square velocity of 1000 m/s. What is its average velocity?(a) 1000 m/s (b) m/s (c) 546 m/s (d) 960 m/sSolution:From (1) and (2)
30 Illustrative example 7Calculate the temperature at which root mean square velocity of SO2 molecules is same as that of O2 molecules at 27° C.Solution:For O2 at 27° C,For SO2 at t° C,
31 Solution Since both these velocities are equal, or 600 = 273 + t or t = 600 – 273 = 327° C
32 Maxwell-Boltzmann velocity distribution A very small fraction of molecules has very low or very high speeds.The fraction of molecules possessing higher and higher speed goes on increasing till it reaches a peak. The fraction with still higher speed then goes on decreasing.The peak represents maximum fraction of molecules at that speed. This speed, corresponding to the peak in the curve,is known as the most probable speed.On increasing temperature, the value of the most probable speed also increases.
33 Characteristic features of Maxwell’s distribution curve A very small fraction of molecules has very low or very high speeds.The fraction of molecules possessing higher and higher speed goes on increasing till it reaches a peak. The fraction with still higher speed then goes on decreasing.The peak represents maximum fraction of molecules at that speed. This speed, corresponding to the peak in the curve,is known as the most probable speed.On increasing temperature, the value of the most probable speed also increases.
34 Explanation of Boyle’s law on the basis of kinetic theory According to kinetic gas equation
35 Explanation of Boyle’s law on the basis of kinetic theory
36 Explanation of charl’s law on the basis of kinetic theory As deduced from the kinetic gas equation, we haveHence, if P is kept constant,= constant which is Charles’ law.
37 Explanation of Dalton’s law on the basis of kinetic theory Let us consider only two gases. According to kinetic gas equation,Now, if only the first gas is enclosed in the vessel of volume V, the pressure exerted would beAgain, if only the second gas is enclosed in the same vessel (so that V is constant), then the pressure exerted would be
38 Explanation of Dalton’s law on the basis of kinetic theory Lastly, if both the gases are enclosed together in the same vessel then since the gases do not react with each other, their molecules behave independent of each other. Hence, the total pressure exerted would be= P1 + P2Similarly, if more than two gases are present, then it can be proved that P = P1 + P2 + P
39 Explanation of Avogadro’s hypothesis on the basis of kinetic theory Let us assume equal volume of two gases at the same temperature and pressure, then from kinetic gas equation.Since average kinetic energy per molecule depends on temperature,
40 Explanation of Avogadro’s hypothesis on the basis of kinetic theory Dividing equation (i) by (ii), we haveThis is Avogadro’s hypothesis.
41 Explanation of Graham’s law on the basis of kinetic theory From kinetic gas equation,at constant pressureThis is in accordance with Graham’s law.
42 Illustrative example 8One mole of N2 at 0.8 atm takes 38 seconds to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 1.6 atm takes 57 seconds to diffuse through the same hole. Calculate the molecular mass of the compound. (Xe = 131, F = 19)
45 Class exercise 120 dm3 of SO2 diffuses through a porous partition in 60 s. What volume of O2 will diffuse under similar conditions in 30 s?Solution:For diffusion,1 for O2, 2 for SO2Ans dm3
46 Class exercise 2180 cm3 of an organic compound diffuses through a pinhole in vacuum in 15 minutes, while 120 cm3 of SO2 under identical condition diffuses in 20 minutes. What is the molecular weight of the organic compound?Solution:
47 Class exercise 3The ratio of rates of diffusion of gases A and B is 1 : 4, if the ratio of their masses present in the mixture is 2 : 3, calculate the ratio of their mole fractions.Solution:Let WA and WB are the weights of two gases in the mixtureWA : WB = 2 : 3
50 Class exercise 4 Calculate the root mean square velocity of (i) O2 if its density is g ml–1 at 1 atm.(ii) ethane at 27° C and 720 mm of HgSolution:P = 1 × 76 × 13.6 × 981 dyne cm–2
51 Solution= 4.99 × 104 cm sec–1Ans. (i) 1.94 × 104 cm sec–1, (ii) 4.99 × 104 cm sec–1
52 Class exercise 5At what temperature will H2 molecules have the same root mean square velocity as N2 gas molecules at 27° C?Solution:Ans K
53 Class exercise 6If a gas is expanded at constant temperature, the kinetic energy of the molecules(a) remains same (b) will increase (c) will decrease (d) None of theseSolution:Hence, the answer is (a).
54 Class exercise 7The relation between PV and kinetic energy of an ideal gas is PV = ____ KESolution:From kinetic gas equation,(for one molecule)
55 Class exercise 8Can we use vapour densities in place of densities in the formula?Solution:Now, if we replace density with vapour density of two gases, thenWhich is also valid.So, we can replace density with vapour density.Ans. Yes
56 Class exercise 9The ratio of root mean square velocities of SO2 to He at 25° C is (a) 1 : 2 (b) 1 : 4 (c) 4 : 1 (d) 2 : 1Solution:
58 Class exercise 10There is no effect of ____ on the motion of gas molecules. (a) temperature (b) pressure (c) gravity (d) densitySolution:According to kinetic theory of gases, velocity of gas molecules will not be affected by the gravity as these are considered to be point masses.Hence, the answer is (c).