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Atomic Structure

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Atomic Structure

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  1. STATES OF MATTER INTRODUCTION The various kinds of substance that make up matter can be divided roughly into three categories, namely, gases, liquids and solids. These are called as three states of matter. The gaseous state is characterized by the following physical properties. Gases are highly compressible. Gases exert pressure equally in all directions. Gases have much lower density than the solids and liquids. The volume and the shape of gases are not fixed. These assume volume and shape of the container. Gases mix evenly and completely in all proportions without any mechanical aid. Basic parameters associated with qas The characteristics of gases are described fully in terms of four parameters (measurable properties) (i) Pressure: (ii) Volume : It is exerted by gas due to collision of randomly moving gas molecules with the walls of the container. Pressure can be measured in following units: Pascal: - It is Sl unit for pressure; Pascal is very small amount of pressure Atm: 2 (10mg weight on lcm area) o 2 0 1 Pa = 1 N/m atm is pressure exerted by mass of the atmosphere on the earth's surface Various units of pressure can be interrelated as follows 1 atm = 76 cm of Hg = 760 torr = 101325 Pa = 1.01325 bar It represents free volume available for motion of gas particles Various units of volume are interrelated as follows 1 L = 1000 ml = 103m3 = 1000 cc Units of temperature are interrelated as follws T K = TOC + 273 (iii) Temperature : (iv) Amount of gas : Generally measured in moles of gas. BOYLE'S LAW "At constant temperature the volume of the given mass of gas is inversely proportional to its pressure. Mathematically, V — when n & T are constant. log P + log V = log K k log P = —log V + log K Thus, Boyle's law can also be expressed as the product of pressure and volume of a given mass of gas at constant temperature is constant. i.e., PIM -P2V2 -P3V3 Graphical Representation: Boyle's law can be graphically represented in following ways
  2. Example 1: Solution: A 1.1 L flask containing nitrogen at a pressure of 710 mm is connected to an evacuated flask of unknown volume. The nitrogen, which acts ideally is allowed to expand into the combined system of both the flasks isothermally. If the final pressure of nitrogen is 583 mm, determine the volume of evacuated flask. Applying Boyle's law: PIVI = P2V2 Let V be the volume of evacuated flask X 1.1 = 583 (1.1 + V) V = 0.24 L CHARLES' LAW "At constant pressure, the volume of a definite mass of a gas is directly proportional to its temperature, that is V ocT; V=KT where K is a constant whose value depends upon the nature, mass and pressure of the gas. Hence Charle's law is stated as: The volume of a given amount of gas at constant pressure varies directly proportional to its absolute temperature. Mathematically Voc T if P and 'n' are constants or = constant if pressure and 'n' are constants. Thus, Graphical representation: Example 2 . Solution K (constant) (lines) O— T &lvin) A flask is of a capacity of 1 liter. What volume of air will escape from the flask if it is heated from 27 oc to 370C? Assume pressure is constant. . Since pressure is constant
  3. VI = 1.0333 litres Since capacity is 1 liter Thus Volume escaped = 1.0333-1 = 0.0333 L GAY - LUSSAC'S LAW Gay lussac's law state that pressure of a given mass of a gas at constant volume is directly proportional to its Kelvin temperature. P OCT or Example 3: A certain amount of ethane is confined in a bulb of 1 liter capacity. The bulb is so weak that it will burst if pressure exceeds 10 atm. Initially gas exerts 8 atm pressure at 270 C. Find temperature at which the bulb will burst? Solution: Considering limiting condition Since volume remain constant Thus 8/300=10/T2 375K AVOGADRO'S LAW At the same temperature and pressure equal volume of all gases contain equal number of moles or molecules. Mathematically Voc n at constant T and P IDEAL GAS EQUATION OF STATE Combination of Boyle's law, Charle's law and Avogadro's law gives the ideal gas equation. 1 Boyle's law V u — if T is constant. Charle's law Voc T if P is constant. And Avogadro's law Voc n if P and T are constant. Thus, voc i.e PVænT or PV=nRT PV = RT for 1 mole of gas Example 4 : What mass of ammonia will exert same pressure as 12 g of H2S(g) in the same container under the similar conditions of temperature? Solution : Under identical conditions of T and V, p n equal moles of ammonia as that of H2S(g) will exert same pressure, when confined in the same container Moles of H2S = 12/34 = moles of ammonia Mass of ammonia = (12/34) x 17=6g Dalton's law of Partial Pressure At a given temperature the total pressure exerted by two or more non reacting gases occupying a definite volume is equal to the sum of the partial pressure of the component gases, Mathematically, P +PB +Pc +PD +.... where P is the total pressure and m,pæpc,....... are the partial pressures of the gases A, B, C, . respectively. Partial Pressure The pressure that a component gas of the gaseous mixture would exert if it were the only gas present in the volume under consideration at a given temperature is the partial pressure of that component. Let ni and be the number of moles of two non-reacting gas A and B filled in a vessel of volume V at temp. T.
  4. Thus total pressure in the vessel P, may be calculated as PV - Partial pressure may be calculated as PAV = ril RT ... (i) and . .. .(ii) Thus, PA = XA . p Similarly i.e. Example 5: Solution: Partial Pr essure = Mole fraction x Total Pressure Calculate partial pressure of nitrogen and oxygen in air assuming it to be composed of mostly nitrogen and oxygen. Volume percentage of oxygen and nitrogen in air are 20 and 80 respectively, and atmospheric pressure to be 1.0. Mole fraction of N2(g) = 0.8 and Mole fraction of 02(g) = 0.2 Partial pressure of N2(g) = 0.8 x 1 = 0.8 atm Partial pressure of 02(g) = 0.2 x 1 = 0.2 atm DIFFUSION & EFFUSION Diffusion is the mixing of two or more gases due to partial pressure difference. A gas would move (if required against gravity also) from its region of high partial pressure to a region of low partial pressure. At the end of this process a gas would have the same partial pressure everywhere. Effusion is leaking of gas through a small orifice into vacuum. Graham's law of Diffusion 2TtRTM r = rate of effusion or diffusion through a small orifice (measured in mol/sec) P = Partial pressure difference of a gas (measured in Pascal) R = Universal gas constant (8.314 J mol-I K l) T = Temperature of the gas (measured in K) M = Molecular weight of the gas(measured in Kg mol-I) A = Area of orifice (measured in m2) At a constant temperature and pressure, the rates of diffusion of different gases are inversely proportional to the square root of their densities. Since the gas densities are directly proportional to their vapour densities and molecular weight Ml and (vapour density) 2 (vapour density)l r VI t2 where VI and are the volumes of gas diffused in time ti and t2. p When the pressure is not constant then r oc— or r2 Ml Example 6: Rate of effusion of ethane is 1.53 times faster than rate of a hydrocarbon containing 14.27% hydrogen by weight, under identical conditions. Deduce the molecular formula of hydrocarbon.
  5. r(ethane) Solution: r(hydrocarbon) 30 Empirical formula Wt.% Mol 0/0 SR Empirical formula = CH 2; Molecular weight = 70.23 .53 c 85.73 85.73 / 12 1 M - 70.23 14.27 14.27 2 = Empirical formula weight x n = 14 n n ; hence, molecular formula of hydrocarbon is C5H10. KINETIC THEORY OF GAS Postulates of Kinetic Theory of Gases 1. 2. 3. 4. 5. 6. Gases are made up of small structural units called atoms or molecules. Volume of individual atom or molecule is considered negligible. Gas molecules are always in rapid random motion colliding with each other and with the wall of the container. Collision among gas molecules is perfectly elastic, i.e. there is no loss in kinetic energy and momentum during such collision. Gas molecules neither attract nor repel each other. Pressure exerted by gas is due to collisions of gas molecules with the wall of the container. Kinetic energy of gas molecules depends only on absolute temperature. Kinetic energy absolute temperature Molecular Kinetic Gas Equation PV 2 —mnc 3 m = mass of one molecule n = number of gas molecules c = root mean square velocity Calculation of Kinetic Energy PV = —mnc2 3 For one gram mole of the gas, PV = RT and n = N or or 1 -mNc2 = RT 3 21 -mNc2 = RT 2 -.K.E. = RT or K E 3 1 —mNc2 = K.E.permol 2 3 2 Average Kinetic Energy The average kinetic energy (KE) is defied as (KE) = 2 2 —mu where k = R/NA and is known as the Boltzmann constant. The total kinetic energy for 1 mole of the gas is Example 7: Solution : E = NAV (KE) = -RT total Derive an expression relating to increase in I-Jrms of a gas for a relatively small temperature rise and calculate increase in ums for a sample of Ne(g) as the temperature is increased from 100 K to 101 K? 3RT durms u rms M 2M 1 3RT 2Murms
  6. 3RT durms 3 CIT 2Murms 8.314 x 100 20 Aurms AT u rms Aurms REAL GASES 3 x 8.134 _3 20 x 10 353 x 2Murms 353 ms x (101 -100) = 1.766 ms The ideal gas laws can be derived from the kinetic theory of gases which is based on the following two important assumptions: (i) The volume occupied by the molecules is negligible in comparison to the total volume of the gas. (ii) The molecules exert no forces of attraction upon one another. The molecules in gases also have weak forces of attraction (called van der Waals attraction) amongst themselves; as otherwise, the gases could never be liquefied and solidified. Correction for Forces of attraction: 2 an p = pre al + ideal Correction for molecular size: The container of volume V has n mole of gas molecules each of which possess a certain volume of their own. If the total volume blocked by 1 mole of molecules due to their sheer presence is taken as b, then the total volume blocked by them would be nb. We can now look at the container and see that the effective volume for the gas molecules to move is V — nb. This means that in the volume, V — nb the gas molecules have zero volume of their own. Since PI VI = nRT 2 'b' is called the excluded volume or co-volume. The numerical value of b is four times the actual volume occupied by 1 mole of gas molecules. This can be shown as follows. Deviation From Ideal Behavior If measurements of pressure, volume and temperature of a gas do not confirm to the ideal gas law (PV = RT), within the precision of measurements, the gas is said to deviate from ideal behaviour and exhibit non-ideal behaviour. Such gases are called real gases. At low pressures and moderately high temperatures, real gases obey ideal gas equation but as the pressure is increased or the temperature is decreased, a marked deviation from ideal behaviour is observed. To display deviations more clearly, the ratio of the observed molar volume (Vm) to the ideal molar volume (V = — ) is plotted as a function of pressure at constant temperature. This ratio is called m, ideal the compressibility factor (Z), which is expressed as z (for n mole gas) nRT or Z = EY (for 1 mole gas) For an ideal gas Z - 1 and is independent of pressure and temperature. Z = f(T,P), a function of both temperature and pressure. Note : For a real gas, 1. 2. 3. 4. Z = 1, the gas is ideal. Z > 1 means for positive deviation from ideal gas behaviour; usually at high P : Z > 1, which means PV > RT. Z < 1 means for negative deviation from ideal gas behaviour; usually at low P : Z < 1, which means PV < Z > 1 for 1-42 and He at all pressure
  7. van der Waal's Equation where, Note : (ii) a — [V (for 1 mole gas) n2a — [V (for n mole gas) a = van der Waal's constant for attraction b = van der Waal's constant for volume Normally for a gas a >> b Since 'b' is four times of the actual volume of gaseous molecules. b = 4NA x V = 4 x Avogadro's no. x —Ttr3 where, r is radius of one aseous molecule and V is volume of one molecule of gas in rest. -2 = atm dm6 mol-2. (iii) Units of a = atm litre mol Units of b = litre mol-I = dm3 mol-I Heat Capacity of Ideal Gases Specific heat c, of a substance is defined as the amount of heat required to raise the temperature of 1 g of substance through IOC, the unit of specific heat is calorie g-11
  8. Again Cav 8RT TM (iii) Most Probable Speed This is defined as the speed possessed by maximum number of molecules of a gas at a given temperature. 2RT c most: Relation among Cmost, Cav and Crms Most probable speed : Average speed : RMS speed 2RT M Graph : : 1.224 O Ump 300K rms U av 1800 K Molecular Speed— Virial equation : All real gas equations of state can be expressed approximately in one common form, called the Virial equation of state which has the following form for 1 mole of a gas. Example 8: Solution: where B, C, . are temperature dependent constants known as second, third, etc., virial coefficients. These coefficients must be evaluated experimentally at each different temperature. Calculate the pressure exerted by a mole of C02 gas confined in a 0.095 L container at 1000C. The 1 st and 2nd coefficients BV RT and cv RT are -2.26 (12 atm mol-I) and 0.161 (L 3 atm mol-I) respectively. Virial equation in volume is: pVm RT cvRT 0.0821 x 373 -2.26 0.161 = 259.32 atm 0.095 CRITICALCONSTANT : Critical Temperature (T c): (0.095)2 (0.095)3 The maximum temp. above which gases cannot be liquified whatever pressure may be applied, is called critical temperature of that gas.
  9. 27Rb Critical Pressure (Pc) : The minimum pressure which is applied at critical temperature in order to liquify the gas is called critical pressure. a pc 27b2 Critical Volume (V c) : The volume of one mole of any gas at critical temperature and critical pressure is called critical volume. Vc = 3b Relation among Pc,Vc & Tc : RTc 8 Boyle's temperature : The temperature at which real gases obey gas laws over a wide range of pressure is called Boyle's temperature (T b) a Inversion temperature : The temperature below which a gas on subjecting to Joule-Thomson effect shows cooling effect and above which it shows heating effect is called inversion temperature (Ti) 2a is Joule-Thomson coefficient expressed as — PJ.T. Gas Eudiometry : The relationship amongst gases, when they react with one another, is governed by two laws, namely Gay-Lussac's law and Avogadro's law. Gaseous reactions for investigation purposes are studied in a closed graduated tube open at one end and the other closed end of which is provided with platinum terminals for the passage of electricity through the mixture of gases. Such a tube is known as Eudiometer tube and hence the name Eudiometry also used for Gas analysis. 25 ml of a gaseous hydrocarbon (which has same rate of diffusion as nitrogen) and methane effuse through a hole in 173 min. while the same volume of nitrogen effuse in 200 minutes under the same conditions. Calculate the contraction in volume if 25 ml of this mixture is exploded with excess of 02 and cooled to room temperature. Example 9 : Solution: By the Graham's law of diffusion M2 173 M2 Ml - 200 28 • M2 = 20.95 Let the volume of hydrocarbon be y ml in 25 ml of the mixture. y x 28 + (25 - y) x 16 Combined molecular weight 20.95 = 25 Since the gaseous hydrocarbon has same rate of diffusion as N2, the molecular weight of gaseous hydrocarbon = 28, i.e., it must be C2H4. C2H4 +302 + 21420 i.e., for 1 volume of hydrocarbon, there is contraction of 2 volumes of water, Contraction for 10.31 ml = 20.62 ml Similarly, for methane CH4 + 202 —.>C02 + 21420 Contraction for 14.69 ml of CH4 = 29.38 ml Total contraction = 50 ml

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