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Chemistry

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**States of matter – Session 2**

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Session Objectives

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**Session Objectives Graham’s law of diffusion/effusion**

Postulates of kinetic theory of gases Kinetic gas equation and kinetic energy of gases Velocity of gas molecules Maxwell-Boltzmann velocity distribution Explanation of gas laws on the basis of kinetic theory of gases

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Questions

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Illustrative example 7 Which of the following compounds is steam volatile?

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Solution The compound which forms intramolecular H-bonding has lower boiling point and hence, steam volatile. Hence, the answer is (b).

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Illustrative example 1 A mixture CO and CO2 is found to have a density of 2gL–1 at 250C and 740 torr. Find the composition of the mixture Solution:

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Illustrative example 2 A 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:

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**Animation for diffusion**

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**Graham’s Law of Diffusion**

(i) For same volume of two gases (ii) For two gases with different volumes and same diffusion time

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**Effusion It happens under pressure through a Small aperture.**

Rate of effusion A=area of aperture Normally considered against vacuum

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Question

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Illustrative example 3 4:1 molar mixture of He and CH4 is effusing through a pinhole at a constant temperature.What is the composition of the mixture effusing out initially? Solution:

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**Solution Let x mole of CH4 is effusing in t sec.**

Then, 8x mole of He is effusing in same time Composition of He in the mixture = 89% Composition of CH4 in the mixture = 11%

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Question

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Illustrative example 4 A 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 size Solution Since, molar mass of H2 is much less than C2H4. H2 will diffuse into the balloon. Hence, answer is (b).

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**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.

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**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.

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**Kinetic Theory of Gases**

The kinetic gas equation, Where, m = mass of each gas molecule n’ = number of gas molecules c = velocity of gas molecule KE of one molecule =

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**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

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**What will be the average kinetic energy of one molecule?**

Ask yourself What 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

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Questions

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Illustrative example 5 Calculate the average kinetic energy per molecule and total kinetic energy of 2 moles of an ideal gas at 25oC. Solution: Average KE per molecule of the gas = = 6.17 × 10–21 J

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**Solution Average KE per mole of the gas = RT = x 8.314 x 298**

= 3.72 kJ/mole Total KE of 2 moles of the gas = RT x 2 = 7.44 kJ

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Molecular velocity Average velocity Most probable velocity (CMP)

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**Molecular velocity Root mean square velocity (CRMS)**

Interrelation of molecular velocities CAvg : CRMS = CMP : CRMS = CAvg: : CMP =

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Questions

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Illustrative example 6 The 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/s Solution: From (1) and (2)

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Illustrative example 7 Calculate 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,

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**Solution Since both these velocities are equal, or 600 = 273 + t**

or t = 600 – 273 = 327° C

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**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.

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**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.

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**Explanation of Boyle’s law on the basis of kinetic theory**

According to kinetic gas equation

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**Explanation of Boyle’s law on the basis of kinetic theory**

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**Explanation of charl’s law on the basis of kinetic theory**

As deduced from the kinetic gas equation, we have Hence, if P is kept constant, = constant which is Charles’ law.

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**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 be Again, if only the second gas is enclosed in the same vessel (so that V is constant), then the pressure exerted would be

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**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 + P2 Similarly, if more than two gases are present, then it can be proved that P = P1 + P2 + P

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**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,

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**Explanation of Avogadro’s hypothesis on the basis of kinetic theory**

Dividing equation (i) by (ii), we have This is Avogadro’s hypothesis.

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**Explanation of Graham’s law on the basis of kinetic theory**

From kinetic gas equation, at constant pressure This is in accordance with Graham’s law.

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Illustrative example 8 One 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)

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Solution

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Class Exercise

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Class exercise 1 20 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 SO2 Ans dm3

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Class exercise 2 180 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:

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Class exercise 3 The 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 mixture WA : WB = 2 : 3

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Solution Similarly, mole fraction of B,

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Solution \ xA : xB = 1 : 24 Ans. 1 : 24

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**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 Hg Solution: P = 1 × 76 × 13.6 × 981 dyne cm–2

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Solution = 4.99 × 104 cm sec–1 Ans. (i) 1.94 × 104 cm sec–1, (ii) 4.99 × 104 cm sec–1

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Class exercise 5 At what temperature will H2 molecules have the same root mean square velocity as N2 gas molecules at 27° C? Solution: Ans K

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Class exercise 6 If 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 these Solution: Hence, the answer is (a).

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Class exercise 7 The relation between PV and kinetic energy of an ideal gas is PV = ____ KE Solution: From kinetic gas equation, (for one molecule)

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Class exercise 8 Can we use vapour densities in place of densities in the formula? Solution: Now, if we replace density with vapour density of two gases, then Which is also valid. So, we can replace density with vapour density. Ans. Yes

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Class exercise 9 The 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 : 1 Solution:

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Solution Hence, the answer is (b).

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Class exercise 10 There is no effect of ____ on the motion of gas molecules. (a) temperature (b) pressure (c) gravity (d) density Solution: 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).

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Thank you

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CHAPTER 12 GASES AND KINETIC-MOLECULAR THEORY

CHAPTER 12 GASES AND KINETIC-MOLECULAR THEORY

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