Gaseous State

INTRODUCTION Matter is made up of atoms or molecules. The arrangement of these molecules determines the state of matter. There are three recognised state of matter : Solid, Liquid and Gas. Matter can change between states when the temperature or pressure is changed. State changes of matter are physical rather than chemical. DIFFERENCE BETWEEN STATES OF MATTER S. No. Liquid Solid Assumes the shape and volume Assumes the shape of the Retains a fixed volume and of its container. part of the container which shape. Particles can move past one another. Compressible, lots of free space Particles can move/slide past one another. Not easily compressible, little Rigid-particles locked into place. Not easily compressible, little between particles space between particles. free space between particles. Flows easily, particles can move Flows easily, particles can Does not flow easily, rigid- past one another. move/slide past one another particles cannot move/slide one past another v. Low density. Intermediate density. High density. vi. Very small intermolecular Considerable intermolecular Intermolecular forces are high, attraction but high kinetic attraction, kinetic energy vibrational motion only. energies. is less. i. Gas it occupies ii. iii. iv. GASEOUS STATE 1. Important Properties of Gases (i) Mass : Mass in gm = Moles × Molecular mass. (ii) Volume : Volume of the gas is the volume of container in which they are filled in. (iii) Temperature : Temperature of a gas is the measure of kinetic energy of gas. Kinetic energy  Temperature (iv) Pressure : Pressure of gas is defined as the force exerted by the gas on the walls of its container. It is often assumed that pressure is isotropic, i.e. it is the same in all the three directions. Force , Area v×d×g Mg P= = A A A×h×d×g = A P = hdg Pressure = Where.... h = height of the mercury column supported by the barometer. d = density of mercury. g = acceleration due to gravity. GASEOUS STATE 2. Units and Determination of Pressure of Gas (a) In SI unit the unit of pressure is the pascal (N/m2) instead, the unit bar, kPa or MPa is used. 1 bar = 105 N/m2 = 100 kN/m2 = 100 kPa (b) Pressure are also stated in mm or cm of mercury. 1 atm = 760 mm Hg = 1.01325 bar = 1.01325 × 105 Pa = 101.325 KN/m2 = 1.0332 Kgf/cm2 1 Pa = 1 Nm–2 = 1 Kgm–1 S–1 1L = 1dm3 = 10–3 m3 (SI unit) 1L atm = 101.325 J 1 KN N 1 1×103 × kg 3 = kgf/cm2 2 = 1 × 10 2 = m m 98 9.8 ×104 cm 2 101325 Pa = 133.322 Pa 760 The pressure relative to the atmosphere is called gauge pressure. The pressure relative to the perfect vacuum is called absolute pressure. Absolute pressure = Gauge pressure + Atmosphere pressure When the pressure in a system is less than atmospheric pressure, the gauge pressure becomes negative, but is frequently designated and called vacuum. Gas For example, 16 cm vacuum will be 1Torr = (d) (e) 76 – 16 × 1.013 = 0.80 bar 76 (f) Pressure is measured using a manometer which is a simple device with a horizontal arm and a U-tube carrying mercury or any other suitable liquid. Pgas h 2.4mL Hg Patm Patm+ P = Pgas (Patm+ hdg = Pgas ) An open arm manometer Ex. Calculate the mass of mercury in a uniform column 760 mm high and 1.00 cm2 in cross-sectional area. Is there any change in. (a) mass and (b) pressure of column of same height but with 2.00cm2 cross sectional area is taken ?   (a)  (density of Hg = 13.6 g/cm3) 760 mm = 76 cm V = 76 × 1 = 76 cm3 Mass = V × d = 76 × 13.6 = 1033.6 g If area of cross-section is 2cm2 (doubled) then volume is also doubled hence, Mass = 76 × 2 × 13.6 = 2067.2 g (b) This mass would rest on twice the area and so exert the same pressure. Sol. GAS LAWS The gas laws are a set of laws that describe the relationship between thermodynamic temperature (T), pressure (P) and volume (V) of gases. 1. Boyle's Law It states "at a constant temperature (T), the pressure (P) of a given mass (or moles, n) of any gas varies inversely with the volume (V)". Pressure (P) and volume (V) of gases. i.e. P 1 (for given n and T) V or, PV = K = constant For two or more gases at constant temperature P1V1 = P2V2 = ..... = K Also, density d  1 V Hence P  d or P1 P2 = = ......... = K d1 d 2 Graphical Representation T2 > T1 V V T2 1/P P P log P log P log 1/V P These plots drawn at constant temperature for a gas are called Isotherms. Ex. The diameters of a bubble at the surface of a lake is 4 mm and at the bottom of the lake is 1 mm. If atmospheric pressure is 1 atm and the temperature of the lake-water and the atmosphere are equal, what is the depth of the lake ? (The density of lake-water and Hg are 1 gm/ml and 13.6 gm/ml respectively. Also neglect the contribution of pressure due to surface tension). Sol. The pressure on the bubble = 1 atm (when it is at the surface) The pressure on the bubble = P atm (say) (when it is at the bottom) 1  (0.1)3 6 1 = 1 ×  (0.4)3 6 The volume of the bubble = 1  × (0.1)3 6 P = 64 atm. Thus the pressure due to water is 63 atm ( atmospheric pressure = 1 atm) P× or Now, consider the depth of lake is h cm  63 × 76 × 13.6 × g = h × 1 × g or h = 63 × 76 × 13.6 cm = 65116.8 cm. GASEOUS STATE Ex. A 10 cm volume of air is trapped by a column of Hg, 8 cm long in capillary tube horizontally fixed as shown below at 1 atm pressure. Calculate the length of air column when the tube is fixed at same temperature. (a) Vertically with open end up (b) Vertically with open end down 8cm 10cm (c) At 45ºC from with open end up (fig. 1) Sol. (a)  When the capillary tube is held as vertically open end up (fig.2), The pressure on the air column = atmospheric pressure + pressure of 8 cm Hg column = 76 + 8 = 84 cm of Hg. Let, at this condition the length of the air column = 2 and the length of air column when capillary is horizontally fixed = 1 = 10 cm and pressure on air column = 1atm. Let the cross section of the capillary = a cm2 76 × 10 × a = 84 × 2 × a or 2 = (b) 8cm (fig. 2) 76 ×10 = 9.04 cm 84 When the capillary tube is held as vertically open end down (fig.3), the pressure on the air column = atmospheric pressure – pressure of 8 cm Hg column = 76 – 8 = 68 cm of Hg. Let at this condition the length of air column = 3.  68 × 3 × a = 76 × 10 × a or 3 76 ×10 = 11.17 cm (fig. 3) 68 When the capillary is held at 45º with open end up, the weight of Hg is partially borne by, the gas and partially by the Hg. The pressure on the gas due to Hg column = 8 × cos 45º 8   total pressure on the gas =  76 +  cm of Hg. 2  Let length of air column at this pressure = 4. Ex. Sol.    4 × a ×  76 +  4 8   = 10 × a × 76 2 • 45  1 8 = = cm of Hg 2 2 º =8× 8c m =  4 (c) 8cm (fig. 4)  10 × 76  =  76 + 8 2  = 9.3 cm   An iron tank contains helium at a pressure of 3.0 atm at 27ºC. The tank can withstand a maximum pressure of 10 atm. The building in which tank has been placed catches fire predict whether the tank will blow up first or melt. (melting point of iron is 1535ºC) P1 = 3.0, T1 = 300 K P1 P2 3.0 P2 = =  T1 T2 300 1808 3.0 × 1808 P2 = = 18.08 atm 300 Since at melting point it is greater than the maximum pressure, therefore it will blow up. Ex. A rubber balloon contains some solid marbles each of volume 10 ml. A gas is filled in the balloon at a pressure of 2 atm and the total volume of the balloon is 1 litre in this condition. If the external pressure is increased to 4atm the volume of Balloon becomes 625 ml. Find the number of marbles present in the balloon. Sol. Let the no. of marbles be = n . volume of marble = 10 n ml. volume of balloon earlier = 1000 ml. later = 625 ml. Now for the gas inside the balloon temperature and amount of the gas is constant, hence boyles law can be applied P1V1 = P2V2 4× (625 – 10n) = 2 × (1000 – 10n) 625 × 4 = 2000 – 20n + 40n 625 × 4 – 2000 = 20n 625× 4 – 2000 = n. 20 2. 125 =n ; 5 n = 25 Charles' Law It states "at constant pressure, the volume of a given mass of a gas, increases or decreases by volume at 0ºC for every rise or fall of one degree in temperature". Vt 1 V0 = 1 + 273.15 t   Vt = V0 1 + or (at constant n and P) t   273.15  V0 (273.15 + t) 273.15 0ºC on the Celsius scale is equal to 273.15 K at the Kelvin or absolute scale. i.e. Tt (Temperature in Kelvin scale) = 273.15 + t or  Vt = From the above equation we get Vt Tt = V0 T0 Vt V0 = Tt T0 i.e. The volume of a given gas is proportional to the absolute temperature. or V1 V 2 = (at constant P) T1 T2  Graphical Representation V V log V 22.4 L mol–1 =V 0 T (K) –273.15º 0 t(ºC) 0 Graphs between V and T at constant pressure are called Isobars. 0 log T 1 th of its 273.15 GASEOUS STATE Ex. Sol. If the temp. of a particular amount of gas is increased from 27ºC to 57ºC, find final volume of the gas, if initial volume = 1 lt and assume pressure is constant. V1 V2 = T1 T2 1 V2 = (273 + 27) (273 + 57) So V2 = 1.1 lt. Ex. An open container of volume 3 litre contains air at 1 atmospheric pressure. The container is heated from initial temperature 27ºC or 300 K to tºC or (t + 273) K the amount of the gas expelled from the container measured 1.45 litre at 17ºC and 1 atm.Find temperature t. Sol.  T01 = 300 K It can be assumed that the gas in the container was first heated to (t + 273), at which a volume ‘V’ escaped from the container hence applying charles law : 3 3 + ΔV = 300 t + 273 Now, this volume ‘V’ which escapes when the container get cooled ΔV 1.45  = t + 273 290 Solve the two equations and get the value of V and t. determine  V & calculate t that will be the answer. 3. Gay-Lussac's Law : Dependence of Pressure on Temperature It states "at constant volume, the pressure of a given mass of a gas is directly proportional to the absolute temperature of the gas". P  T or P = KT or  P1 P2 = T1 T2 Graphical Representation V1 P P V2 P0 T T The plots drawn at constant volume for a gas is called as Isochore.  The Combined Gas Law It states "for a fixed mass of gas, the volume is directly proportional to absolute temperature and inversely proportional to the pressure". 1 (at constant n, T) P Charle's Law, V  T (at constant n, P) Boyle's Law, V  Therefore, V  T P or V=K T P or PV =K T or P1V1 P2 V2 = T1 T2  Volume Coefficient (v) of a Gas The ratio of increase in volume of a gas at constant pressure per degree rise of temperature to its volume at 0ºC is the volume coefficient of the gas. Vt – V0 v = V × t 0 For all gases, v =  or Vt = V0 (1 + vt) 1 273 Pressure Coefficient (p) of a Gas The ratio of increase in pressure of the gas at constant volume per degree rise of temperature to its pressure at 0ºC is the pressure coefficient of the gas. p = Pt – P0 P0 × t For all gases, p = 4. Sol. Ex. Sol. Pt = P0 (1 + pt) 1 273 Avogadro's Law It states "equal volumes of any two gases at the same temperature and pressure contain the same number of molecules". V  n (At constant P and T) or, Ex. or V1 V2 = n1 n 2 The temperature of a certain mass of a gas is doubled. If the initially the gas is at 1 atm pressure. Find the % increase in pressure ? P1 P2 T1 = T2 ; 1 P2 = T 2T % increase = 2 –1 x 100 = 100% 1 The temperature of a certain mass of a gas was increased from 27°C to 37°C at constant volume. What will be the pressure of the gas. P1 P2 = T1 T2 ; P P2 = ; 300 310 P2 = 31 P 30 THE IDEAL GAS EQUATION An ideal gas is defined to be a system in which there are no inter molecular/interatomic forces. Such a system can only exist as a gas. Any real system will approach ideal gas behaviour in the limit that the pressure is extremely low and the temperature is high enough to overcome attractive intermolecular forces. An ideal gas is a gas to go which the laws of Boyle and Charles are strictly applicable under all conditions of temperatures and pressures. From Boyle's law we get, V 1/P (at constant n and T) From charles law we get, V  T (at constant n and P) GASEOUS STATE From Avogadro's law we get, V  n (at constant T and P) Combining the above three equations we get V  nT P or V = R nT P [Where R = ideal gas constant] or PV = nRT Ideal gas equation is a relation between four variables and it describes the state of any gas. For this reason, it is also called Equation of State.  Dimension of R R  = Pressure × Volume PV = nT Mole × Temperature = (Force / Area) × (Area × Length) Mole × Degree(K) = Work or energy Force × Length = Mole × Degree (K) Mole × Degree(K) Physical Significance of R The dimentions of R are energy per mole per kelvin and hence it represents the amount of work (or energy) that can be obtained from one mole of a gas when its temperature is raised by 1K. EDUBULL KEY POINTS POINTS ETOOS KEY Units of R 1 atm × 22.4 lit = 0.0821 lit-atm mol–1K–1 273 K (i) In lit-atm R= (ii) In C.G.S system R = (iii) In M.K.S. system = 8.314 × 107 erg mole–1K–1. R = 8.314 Joule mole–1K–1. (iv) In calories R= 1× 76 ×13.6 × 980 dyne cm –2 × 22400 cm3 273 K [107 erg = 1 joule] 8.314 ×107 erg mole –1K –1 4.184 ×107 erg = 1.987  2 calorie mol–1K–1. Ex. Some spherical balloons each of volume 2 litre are to be filled with hydrogen gas at one atm & 27°C from a cylinder of volume 4 litres. The pressure of the H2 gas inside the cylinder is 20 atm at 127°C. Find number of balloons which can be filled using this cylinder. Assume that temperature of the cylinder is 27°C. Sol. No. of moles of gas taken initially = 20× 4 = 2.43 L R × 400 No. of moles of gas left in cylinder = 1× 4 = 0.162L R ×300 No. of moles of gas to be filled in balloons = 2.43 – 0.162 = 2.268 Let we have 'n' balloons that we can fill No. of moles of gas that can be filled in 1 balloon = 1× 2 = 0.081 0.082×300  0.081 × n = 2.268 n = 28 balloons. Ex. Sol. At room temperature following reaction goes to completion 2NO + O2  2NO2  N2O4 Dimer N2O4 at 262 K is solid. A 250 ml flask and a 100 ml flask are separated by a stop cock. At 300 K, the nitric oxide in the large flask exerts a pressure of 1.053 atm and the smaller on contains O2 at 0.789 atm. The gases are mixed by opening the stop cock and after the end of the reaction, the flasks are cooled to 220 K. Neglecting the vapour pressure of dimer. Find out the pressure and composition of gas remaining at 220 K (Assume gases behave ideally). For NO gas :nNO = PV 1.053× 250 = = 1.069 × 10–2 0.0821× 300 ×1000 RT For O2 gas :- n O 2 = 0.789 ×100 = 0.32 × 10–2 0.0821× 300 ×1000 Now, Mole before reaction Mole after reaction Moles of NO left PNO × V = nRT V  O2 –2 0.32 × 10 0 N2O4 0 0.32 × 10–2 = 4.29 × 10–3 Given T = 220 K 250 + 100 350 = litre 1000 1000 = PNO × 2NO + –2 1.069 × 10 1.069 × 10–2 – 2 × 0.32 × 10–2 = 0.429 × 10–2 350 = 4.29 × 10–3 × 0.821 × 220 1000 PNO (left) = 0.221 atm. DALTON'S LAW OF PARTIAL PRESSURES Dalton's law of partial pressure states "at a given temperature, the total pressure exerted by two or more non-reacting gases occupying a definite volume in equal to the sum of the partial pressures of the component gases." PTotal = p1 + p2 + p3 + ...... (At constant V and T) RT nRT  n1 n 2 n 3  + + .....  RT = (n + n + n + ......) = = + 1 2 3 V V V  V V Where n = n1 + n2 + n3 + ...... = Total moles, V = Total volume PTotal = p i = RT  ni V Dalton's law of partial pressure is applicable only to non-reacting gases. If the two non-reacting gases A and B having nA and nB number of moles respectively are filled in a vessels of volume V at temperature T, Then GASEOUS STATE PV = (nA + nB) RT ....(i) Partial pressure may be calculated as pAV = nART, pBV = nBRT Total pressure (pA + pB) V = (nA + nB)RT .....(ii) from (i) and (ii) P = pA + pB pA n A RT / V nA = = = xA (mole fraction of A) P (n A + n B )RT / V n A + n B pA = xA × P, similarly pB = xB × P Partial pressure of a component = Mole fraction × total pressure. It has been observed that gases are generally collected over water and therefore are moist. Pdry gas = P moist gas – Pwater vapour The pressure exerted by water vapour is constant with water at a particular temperature is called as aqueous tension, which varies directly with the temperature and becomes 760 mm at 100ºC. Relative Humidity (RH) = Partial pressure of water in air Vapour pressure of water Pressure of dry gas = Pressure of moist gas – aqueous tension. Moist (Gas + water vapour) Ex. The stop cock connecting the two bulbs of volume 5 litre and 10 litre containing as ideal gas at 9 atm and 6 atm respectively, is opened. What is the final pressure if the temperature remains same. Sol. After the opening of the stop cock the pressure of the each bulb will remain same. At the beginning, the no. of moles of gas in A = 10 x 6 RT At the beginning, the no. of moles of gas in B = 5x9 RT 105 RT Total no. of moles of gas before opening the stop cock  total no. of moles at the beginning = = total no. of moles of gas after opening stop cock =  pressure after the opening of the stop cock P= 105 RT 105 x = = 7 atm 10 + 5 Vtotal RT 105 RT A B 10 lit 6 atm 5 lit 9 atm Ex. A mixture of NO2 & CO having total volume of 100 ml contains 70 ml of NO2 at 1 atm, mixture is left for some time and same NO2 get dimerised to N2O4 such that final volume of the mixture become 80 ml at 1 atm, calculate the mole fraction of NO2 in final equilibrium mixture. Sol. Initial volume of NO2 = 70 ml Initial volume of CO = 100 – 70 = 30 ml Final volume of mixture = 80 ml Let the volume of NO2 in final mixture be x Let ‘v’ ml NO2 be converted to N2O4 2NO2  N2O4 V V/2 Hence final volume = volume of CO + volume of NO2 left + volume of N2O4 formed = 30 + 70 – V + V/2 = 80 V = 40 ml Hence volume of NO2 left = 70 – V = 30 ml Now as volume  moles  mole fraction = volume fraction = 30 3 = 80 8 AMAGAT'S LAW OF PARTIAL VOLUME According to this law at constant temperature and pressure, the total volume of mixture of gases which don't react is equal to the sum of partial volumes of each component present in mixture. VT = V1 + V2 + V3 +. . . . . . .  Partial Volume Partial volume of any component is defined as the volume occupied by that particular component measured under the given condition of pressure and temperature. PV = nRT V n  Partial Volume Let us consider a three non-reacting gases A, B and C are present in a container which have no. of moles nA,nB and nC respectively. For each gas partial volume is  RT  , VA = nA   P    RT   VB = nB   P  and  RT   VC = nC   P  Total Volume  RT   RT  VT =VA + VB + VC = (nA + nA +nA)   =n    P  T P  GASEOUS STATE Ex. A vessel of 2 lit capacity contains hydrogen at 380 mm pressure at 27ºC. 16gm of O2 is added to the container then find the total pressure and partial pressure where R = 0.0821 L atm mol–1 k–1. Sol. For H2 gas : Given 380 1 = atm 760 2 and p H 2 = V = 2L p H 2 V  n H 2 RT 1 = n H 2 × 0.0821 × 300 n H2 = 10000 1 = 821× 300 24.6 Moles of O2 = 16 1  32 2 Partial pressure of H2 gas (p H2 ) = 1 = 2 Ex. 1 24.6 × P 1 1 + 24.6 2 n H2 1 = ×P 2 n H 2 + n o2 or P = 6.65 atm 1 mole of N2 and 3 moles of H2 are mixed in 8.21 lit container at 300 K to form NH3. If equilibrium average molecular mass was found to be 34/3 gram then find partial pressure of each component. Sol. N2 Initially 1 + 3 at equilibrium 3H2 ‡ˆ ˆˆ ˆ† ˆˆ 2NH3 0 (1 – x) 3 (1 – x) 2x Total moles = 1 – x + 3 – 3x + 2x = (4 – 2x) moles Given Average molecular mass = 34 3 34 28(1 – x) + 2(3 – 3x) + 17(2x) = 3 (4 – 2x) 34 (4 – 2x) = 34 3  4 – 2x = 3  x = 1/2 So total no. of moles = 3 PV = nRT  P = nRT 3× 0.821× 300 = = 9 atm V 8.21 p N2 = 1– x 0.5 ×P = × 9 = 1.5 atm 3 3 p H2 = 3 – 3x 1.5 ×P = × 9 = 4.5 atm 3 3 Ex. At constant temperature of 273K, 1 Vs P are plotted for 2 ideal gases A and B as shown. Find out the number V of moles of gas A and B. Sol. PV = nRT, P= 1 nRT T V Ex. Sol. Ex. ga sB 1 1 = P V nRT Comparing by equation : y = mx + C C=0 m= 1 nRT (m = tan ) tan  = 1 nRT nA = ga 1 V 1 RTtanθ nA = 1  0.0821× 273× tan45º nB = 1 1 = RTtan60º 22.4 3 sA nA = 1 2 2 .4 45º 60º P One litre flask contains air, water vapour and a small amount of liquid water at a pressure of 200 mmHg. If this is connected to another one litre evacuated flask, what will be the final pressure of the gas mixture at equilibrium ? Assume the temperature to be 50ºC. Aqueous tension at 50ºC = 93 mmHg. The aqueous tension remains same in both the flask. Also flask are at same temperature  P 1V 1 = P 2V 2 where P1 = 200 – 93 = 107 mm V1 = 1 litre V = 1 litre V = 1 litre V2 = 2 litre  107 × 1 = P × 2 P = 53.5 mm Since aqueous tension is also present in flask, equivalent to 93 mm.  pressure of gaseous mixture = 93 + 53.5 = 146.5 mmHg. A 500 ml bulb is filled with CH 4 at a pressure of 1 atm and oxygen at a pressure of 4 atm at 27ºC. The mixture is then exploded, when the reaction CH4 + 2O2  CO2 + 2H2O () occurs. Calculate the pressure if the final temperature is 27ºC. Aqueous tension at 27ºC is 26.7 mm. Partial pressure of CH4, ( p CH 4 )= 1 atm Sol.   Partial pressure of O2, ( p O2 ) = 4 atm 1× 0.5 0.5 moles of CH4 present, ( n CH 4 ) = = RT RT 4 × 0.5 2 = RT RT CH4(g) + 2O2(g) = CO2(g) + 2H2O() 0.5 2 Initially RT RT moles of O2 present ( n O 2 ) = GASEOUS STATE Finally 1 RT – 0.5 RT After the reaction, moles of residual O2 = 2 1 1 – = RT RT RT 0.5 RT After the reaction, moles of CO2 produced = Hence total no. of moles of O2 and CO2 = Ex. 1.5 RT × = 3 atm RT 0.5  total pressure due to O2 and CO2  pressure exerted by the entire gaseous components =3+ = 1.5 RT 26.7 = 3 + 0.03513 = 3.0513 atm 760 What is the difference in the density of dry air at 1 atm and 25ºC and moist air with 50% relative humidity under the same condition. The vapour pressure of water at 25ºC is 23.7 Torr and dry air has 75.5% N2 and 24.5% O2. Sol. M (dry air) = = M1 (O 2 )X1 (% of O2 ) + M 2 (N 2 ) X 2 (% of N 2 ) X1 + X 2 32 × 24.5 + 28 × 75.5 = 28.98 g mol–1 100 d (dry air) = PM(air) 1× 28.98 = = 1.184 g L–1 = 1.184 kg m–3 RT 0.0821× 298  10 –3 kg –1 –3   Q1g L = –3 3 = 1kg m  10 m   relative humidity (50%) =  partial pressure of H 2Oin air vapour pressure of H 2O P (H2O) = 0.50 × 23.7 Torr = 11.85 torr = 11.85 atm = 0.0156 atm 760 % of H2O vapour in air = 0.0156 ×100 = 1.56% 1 % of N2 and O2 in air = 98.44% M (wet air) = d (wet air) = 28.98 × 98.44(air) + 18×1.56 (water vapour) = 28.81 g mol–1 100 1× 28.81 PM(wet air) = = 1.177 g L–1 = 1.177 kg m–3 0.0821× 298 RT difference = 1.184 – 1.177 = 0.007 kg m–3 Ex. A sample of butane gas C4H10 of unknown mass is contained in a vessel of unknown volume V at 25ºC and a pressure of 760 mmHg. To this vessel 8.6787 g of neon gas is added in such a way that no butane is lost from the vessel. The final pressure in the vessel is 1920 mm Hg at the same temperature. Calculate the volume of the vessel and the mass of butane. Sol. Partial pressure of C4H10 (p1) = 760 mm Hg By Dalton's law of partial pressure, Ptotal = p1 + p2 1920 = 760 + p2 or p2 = 1160 mm Hg p2 = x2 × ptotal where x2 = mole fraction of Ne p2 1160 8.6787 x2 = P = = 0.60 ; Moles of Ne = = 0.43 1920 20.2 total n2 0.43 x2 = n + n ; 0.60 =  n1 = 0.28 0.43 + n1 1 2 Molecular weight of C4H10 = 48 + 10 = 58 so, amount of C4H10 = n1 × 58 = 16.24 g again for C4H10 PV = n1RT ; T = 273 + 25 = 298 K, P = 760 atm = 1 atm 760 n1RT 0.28× 0.0821× 298 = = 6.85 L P 1 A mixture of CO and CO2 is found to have a density of 1.50 g/litre at 30ºC and 730 mm. Calculate the composition of mixture. V= Ex. Sol. For mixture, P = (730/760) atm, T = 303 K, d = 150 g litre–1  PV = nRT = (w/M) RT  P(w/VM) RT  P =  730 1.50 = × 0.0821 × 303 760 M  M = 38.85 d RT M Let n1 moles of CO and (100 – n1) moles of CO2 are present in 100 moles. Ex. n1 × 28 + (100 – n1 ) × 44 = 38.85  n1 = 32.19 100  Average molecular weight (M) =  mole % of CO = 32.19, and mole % of CO2 = 67.81. An open vessel at 27ºC is heated until 3/5th of the air in it has been expelled. Assuming that the volume of the vessel remains constant. Find (a) the temperature at which vessel was heated ? (b) the air escaped out if vessel is heated to 900 K ? (c) temperature at which half of the air escapes out ? Sol. One should clearly note the fact that on heating a gas in a vessel there are the number of moles of gas which go out, the volume of vessel remains constant. GASEOUS STATE Let initial moles of air at 300 K be 'n'. On heating 3/5 moles of air are escaped out at temperature T.  (a) 3  2n  Moles of air left at temperature T =  n – n  = 5   5 Under simillar conditions of P and V n1T1 = n2T2 2n ×T  T = 750 K 5 On heating vessel to 900 K, let n1 moles be left again n1T1 = n2T2 1 n1 × 900 = 300 × n  n1 = n 3 n 2 = n moles  moles escaped out = n – 3 3 Let n/2 moles are escaped out at temperature T then n1T1 = n2T2 n × 300 = (b) (c) n/2 × T = n × 300 T= 600 K GRAHAM'S LAW OF DIFFUSION AND EFFUSION Diffusion The process of mixing of gases by random motion of the molecules is called diffusion. Effusion When gas is confined to a container at high pressure than in the surrounding atmosphere it will escape from a small hole which is opened in the container until the pressure outside and inside have been equalised. This process is called effusion. The difference between diffusion and effusion is that in the former case, the gas spontaneously passes through a porous partition while in the case of effusion, the gas is passed out through a small orifice or hole by applying some external pressure. Graham's law of diffusion states "under the same conditions of temperature and pressure, the rates of diffusion of different gases are inversely proportional to the square roots of their molecular masses or their densities." r1 M2 d2 1 1 or or,, r = M = d M d 2 1 1 Where r1 and r2 are the rates of diffusion of gases 1 and 2. M1 and M2 are their molecular masses. d1 and d2 are their densities. Mathematically , r  Effect of Volume on Rate of Diffusion Volume of gas diffused Rate of diffusion = Time taken for diffusion V t Let V1 be the volume of gas 1 and V2 be the volume of gas 2, then or r= r1 =  r1 = r2 V1 t1 r2 = Vt M2 = 1 2 V2 t 1 M1 V2 t2 (i) Comparison of Times Taken for the Same Volume of Two Gases Let the times of diffusion for the same volume of two gases be t1 and t2 respectively. Then r1 V / t1 = = r2 V / t 2 (ii) M2 M1 or r1 t 2 = = r2 t1 d2 = d1 M2 M1 Comparision of the Volumes of the Gases that Diffuse in Same Time Let V1 and V2 be the volume of two gases that diffuse in same time t. Then, r1 V1 / t = = r2 V2 / t (iii) d2 = d1 d2 = d1 M2 M1 or r1 V1 = = r2 V2 d2 = d1 M2 M1 Effect of Pressure on Rate of Diffusion Rate of diffusion is proportional to the pressure of the gas in the container. r P or r P M ; again, r  1 M r1 P1 = r2 P2 M2 M1 or Application of Graham's Law of Diffusion in Enrichment of Isotopes Enrichment of Light Component If a mixture of heavier gas B and a lighter gas A is placed in contact with a porous barrier, the gas passing through will be enriched in lighter component by a factor MB M A , called enrichment factor because lighter molecules effuse more rapidly than heavier ones. The remaining gas will be enriched in the heavier component. Each passage gives an enrichment factor equal to MB M A and so thousands of such barrier in succession are necessary to provide sufficient enrichment of heavier component. Thus, enrichment factor for first barrier or operation f1 = MB MA n 'A / n 'B  overall separation or enrichment factor f = n / n A B Where nA, nB, and n 'A , n 'B are the concentration of two isotopically different components before and after processing. If required enrichment of gas A is attained in x-operation then, (f1)x = n 'A / n 'B =f nA / nB or  n 'A / n 'B  x log f1 = log    nA / nB  or  MB  x log  M  = log  A 1/2  n 'A / n 'B  n / n   A B GASEOUS STATE or x M  log  B  2  MA   n 'A / n 'B  = log    nA / nB   n' / n'  2log  A B   nA / nB  x= M  log  B   MA  or LIMITATIONS OF THIS LAW (i) The laws of diffusion and effusion are applicable only when the pressure difference is very small. (ii) Correctly speaking the rate of diffusion and effusion are not constant throughout the time interval, but these rates may be assumed constant for a small time interval. Ex. The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 47 minutes when the contained oxygen leaked through a small hole. The bulb was then completely evacuated. A mixture of oxygen and another gas of molecular weight 79 in the molar ratio of 1 : 1 at a total pressure of 4000 mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 74 minutes. Sol. The molar ratio of oxygen and the other gas in the evacuated bulb = 1 : 1 and the total pressure of the gas mixture is 4000 mm, hence the partial pressure of each gas is 2000 mm. The drop in the pressure of oxygen after 74 minutes (2000 – 1500) × 74 = 787.2 mm of Hg 47 After 74 minutes, the pressure of oxygen = 2000 – 787.2 = 1212.8 mm of Hg =  Let the rate of diffusion of other gas be rn, then rn = rO2   32 79 32 = 501.01 mm of Hg 79 Pressure of the other gas after 74 minutes = 2000 – 501.01 mm = 1498.99 mm of Hg Drop in pressure for the other gas Molar ratio = = 787.2 × Moles of unknown gas 1498.99 = = 1.236 : 1 Moles of O 2 1212.8 Ex. One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 1.6 atm takes 57 s to diffuse through the same hole. Determine the molecular formula of the compound. Sol. Rate of diffusion, r  p r  1 / M or r  1/2 r2 p 2  M1  = r1 p1  M 2  or Again, r  1 t p M 1/2 or t1  p 2  M1   t 2 =  p1   M 2  or  p2t 2  M2 =  p t  M1  11 2 2  1.6 57  ×  × 28 = 252 g mol–1 M2 =   0.8 38  Let the molecular formula of the unknown compound be XeFn. Mxe + nMF = 252 or 131 + 19 n = 252 252 – 131 = 6.36 ; 6 19 Hence, the molecular formula of the gas is XeF6. n= Ex. Sol. At 1200ºC, mixture of Cl2 and Cl atoms (both in gaseous state) effuses 1.16 times as fast as krypton effuses under identical conditions. Calculate the fraction of chlorine molecules dissociated into atoms. M (Kr) = 83.8 g mol –1. Cl ‡ˆ ˆˆ ˆ† ˆˆ 2Cl 2 r(Cl2 and Cl mix) = 1.16 r (Kr)  Mav = M(Kr) M av (Cl 2 + Cl) = 83.8 M av 83.8 = 62.28 g mol–1 (1.16) 2 Cl2 ‡ˆ ˆˆ ˆ† ˆˆ Initial mole 1 After dissociation (1 – x) (x = degree of dissociation) Total moles after dissociation = 1 – x + 2x = (1 + x)  2Cl 0 2x (1 – x) M(Cl 2 ) + 2x M (Cl) = 62.28 (1 + x) (1 – x) × 71+ 2x× 35.5 = 62.28 1+ x (1 – x + x) 71 = 62.28 1+ x 1 62.28 = = 0.88 1+ x 71 1 + x = 1.14 x = 0.14  % dissociation = 14% Alternately, following formula can also be used to determine degree of dissociation when there is equilibrium mixture. x= m(normal) – m(average) 71 – 62.28 = = 0.14 m(average) 62.28 GASEOUS STATE Ex. In a tube of length 5 m having 2 identical holes at the opposite ends. H2 & O2 are made to effuse into the tube from opposite ends under identical conditions. Find the point where gases will meet for the first time. O2 H2 5m Sol. r1 dx dt r2 = dt × dx = M2 M1 dx1 dx 2 = 32 2 dx1 dx 2 = 4  distance travelled by H 2 distance travelled by O 2 = 4 x =4 (5 – x) x = (5 – x) 4 ; x = 20 – 4x ; 5x = 20 ; x = 4 from H2 side Ex. Assume that you have a sample of hydrogen gas containing H2, HD and D2 that you want to separate into pure components (H = 1H and D = 2H). What are the relative rates of diffusion of the three molecules according to Graham’s law ? Sol. Since D2 is the heaviest of the three molecules, it will diffuse most slowly, and let we call its relative rate 1.00. We can then compare HD and H2 with D2. Comparing HD with D2, we have Rate of HD diffusion Rate of D 2 diffusion = Molecular mass of D 2 Molecular mass of HD = 4.0amu 3.0 amu = 1.15 Comparing H2 with D2 we have Mass of D 2 Rate of H 2 diffusion = Mass of H = 2 Rate of D 2 diffusion 4.0amu 2.0amu = 1.41 Thus, the relative rates of diffusion are H2(1.41) > HD (1.15) > D2(1.00). (1) Connecting Containers When we are connecting two or more containers then the movement of any component of gas occurs in other container till the final pressure of that component in all the container become equal (that means partial pressure becomes equal). (2) To calculate composition of gaseous mixture in any container, we will use law of diffusion. Ex. A 10 liter container consist of 1 mole of gas at 300 K. It is connected to another container having volume 40 liter and is initially at 300 K. The nozzle connecting two containers is opened for a long time and once the movement of gas stopped, the larger container was heated to a temperature of 600 K. Calculate (a) Moles and pressure of gas in both the containers before heating. (b) Moles and pressure in two containers after heating. (assume that initially the larger container is completely evacuated.) Sol. (a) Before heating : PV = nRT (1 – x)R × 300 x × R × 300 = 10 40 X = 0.8 moles x × R × T 0.8 × R × 300 Pressure = = 0.492 atm = V 40 (b) After heating : (1 – x1 )R × 300 x1 × R × 600 = 10 40 (1 – x) 10 lit (x) 40 lit 300K 1mole (I) 300K (II) X1 = 0.67 moles, Given T1 = 600 K Pressure = x1 × R × T1 0.67 × .0821× 600 = 0.821 atm = V 40 Ex. A mixture containing 1.12 litre D2 and 2.24 litre of H2 at NTP is taken inside a bulb connected to another bulb through a stop cock with a small opening. The second bulb is fully evacuated. The stop cock is opened for a certain time and then closed. The first bulb is now found to contain 0.10 g of D2. Determine the % by weight of the gases in second bulb. Sol. At STP D 2 = 1.12 lit.at STP = 0.2 g = 0.05 mole  Before diffusion  in I bulb H 2 = 2.24 lit.at STP = 0.2 g = 0.1mole  (I) (II) When these moles are placed in the bulb, the partial pressure of gas will be different because V and T are constant. Also P  n PD2 Thus, PH2 = 0.05 1 = 0.10 2 After diffusion D2 left in I bulb = 0.1 g or D2 diffuses from I into II (bulb) = 0.2 – 0.1 = 0.1 g Now for diffusion of D2 and H2 rH2   w D2 t D2 t H2 w H2 = M H2 × M D2 M D2 M H2 × PH2 PD2 PH2 0.1 t 4 1 × =  × t w H2 2 2 w H2 = 0.14 g   × = Wt. of gases in II bulb = wt. of D2 + wt. of H2 = 0.10 g + 0.14 g = 0.24 g  0.10 × 100 = 41.66 % 0.24 % H2 in bulb II = 58.33% % D2 by wt. = GASEOUS STATE Ex. Calculate the relative rates of diffusion for 235UF6 and 238UF6 in gaseous form. Also if naturally occurring uranium ore having U235 and U238 in the ratio 0.72 and 99.28%, and if it is desired to enrich the U235 to 10% of the sample, making use of relative rates of diffusion of UF6 having U235 and U238 isotopes, how many diffusion stages are required ? Sol. M.wt of 235U F6 = 235 + 19 × 6 = 349 M.wt of 238U F6 = 238 + 19 × 6 = 352 From Graham's law at same P and T r1 M  =  2 r2  M1  rU235F 6 rU238F 6 Also  352  =   = 1.0043  349   n' / n'  2log  A B   nA / nB  x= M  log  B   MA  Here, nA = U235 in U235F6 = 0.72 nB = U238 in U238F6 = 99.28 n = ' A U235 in U235F6 and MB = MU238F 6 and MA = MU235F 6 = 10 n 'B = U238 in U238F6 = 90  (10 / 90)  2log    (0.72 / 99.28)  = 638 steps x=  352  log    349   BAROMETRIC PRESSURE DISTRIBUTION In case of air (ideal gas) density depends on pressure. So we can not calculate pressure difference between two points directly as 'hdg'. Assume in case of air, composition of air and temperature is constant throughout. Suppose pressure at height 'h' from the ground level is 'P' and as height increase by 'dh' the pressure decreases by 'dP'. So the pressure difference between two points is : (P – dP) – P = dh × d × g – dP = dh × P  P0 PM ×g RT [PM = dRT] H H –dP Mg dh = P RT 0 l n PPP 0 –Mg H [h]0 = RT P– dP P dh { h P0 ln P –MgH P0 = RT  MgH    RT  P = P0 e– M = molar mass of air (kg/mole) g = 9.8 m/sec2 H = height in meter R = 8.314 J/mole/K Ex. Calculate the expression for density of air at any height 'H' metre from the ground level. If at the ground level atmospheric pressure is P0 atm and average molar mass of air is 'M' and temperature is 'T' kelvin. Given that average molar mass and temperature are constant.  MgH    RT  Sol. P = P0 e–  MgH  dRT   = P0 e– RT  M   P0 M  – MgH   e RT  d=   RT  Density at ground level = d0 =   MgH    RT  P0 M RT d = d0 e –   PAYLOAD OF BALLOON Payload of a balloon is defined as maximum weight which a bolloon can carry with it upward direction. V = Volume of Balloon , dout = density of autside air, w = effective weight of balloon din = density of gas inside, m = mass of balloon. Payload = dout × V × g – [ din V × g + mg] = [Buoyancy force] – [effective wt. of Balloon] Ex. Calculate payload of a balloon having volume 100 litre. It is filled with helium gas at 0.2486 atm pressure and 300 K. Density of air is 1.3 gm./litre and mass of material of bolloon is 20 gm. Sol. Buoyancy force PM = dRT din = PM He 0.2486 × 4 1 = = RT 0.0821× 300 25 1  Payload = 1.3 × 100 × 10 –  ×100 ×10 + 20 ×10   25  = 13 × 100 × 10 – [4 × 10 + 200] = 1300 – 240 = 1060 10 GASEOUS STATE (A) CALCULATION OF PAYLOAD OF BALLOON AT CONSTANT HEIGHT (1) At constant volume of balloon Payload = dout V. g – [w]  Mgh  –  RT  dout = d0 e  Payload = d0 e – Mgh RT V.g – w Density of outside air decreases with increase in height so buoyancy force decreases and payload of balloon decreases with increase in height. (2) At variable volume of balloon PV = nRT PV = constant P0V0 = Ph × Vh  Mgh  –  RT  P0V0 = P0 e × Vh  Mgh    RT  V h = V0 e  Now payload = dout × V × g – w = d0 e – Mgh RT  Mgh    RT  V0 e  xg – w = d0 V0 g – w = constant In this case buoyancy force at any height is same so payload will remain constant throughout. (B) MAXIMUM HEIGHTAT WHICH BALLOON WILLBURST Maximum volume = Vmax  Mgh max    RT  Vmax = V0 e   Mgh max   RT   Vmax = e V0  Vmax n  V  0   Mgh max  =    RT  h max = Ex. Sol. RT V  l n  max  Mg  V0  Calculate the payload of balloon of diameter 20 meter weight 100 kg. If it is filled with He at 1.0 atm and 27ºC. Density of air is 1.2 kgm–3. [R = 0.082 dm3 atm K–1 mol–1] Weight of balloon = 100 kg = 10 × 104 g Volume of balloon = 4 3 4 22  20  r = × ×  ×100  3  2  3 7 3 = 4190 × 106 cm3 = 4190 × 103 litre Weight of gas (He) in balloon = PV 1× 4190 ×103 × 4 = = 68.13 × 104 g RT 0.082 × 300 Total weight of gas and balloon = 68.13 × 104 + 10 × 104 = 78.13 × 104 g Weight of air displaced = Ex. 1.2 × 4190 ×106 = 502.8 × 104 g 103  Payload = wt. of air displaced – (wt. of balloon + wt. of gas)  Payload = 502.8 × 104 – 78.13 × 104 = 424.67 × 104 g = 4.2467 × 106 g A spherical balloon of 21 cm diameter is to be filled up with hydrogen at NTP from a cylinder containing the gas at 20 atmpsphere at 27ºC. If the cylinder can hold 2.82 litre of water at NTP, calculate the number of balloons that can be filled up. Sol. 21 cm = 10.5 cm 2 4 3 4 22 Volume of the balloon = πr = × × (10.5)3cc = 4851 cc 3 3 7 Radius of the balloon = Pressure = 1 × 76 × 13.6 × 981 = 1.014 × 106 dynes/cm2  No. of mole of H2, the balloon can contain at NTP = PV 1.014 ×106 × 4851 = = 0.2167 RT 8.314 ×107 × 273 20 × 76 ×13.6 × 981× 2820 = 2.2929 8.314 ×107 × 300 When the balloons are being filled, the pressure in the cylinder will decreases, when the pressure of the cylinder No. of mole in the cylinder = will drop to 1 atm, gas cannot be withdrawn. Now, no. of moles of H2 remaining in the cylinder unused = No. of balloons that can be filled =  = Ex. 2.2929 – 0.1146 = 10 0.2167 1.014 ×106 × 2820 = 0.1146 8.314 ×107 × 300 No.of moles of H 2 in the cylinder that can be used No.of moles of H 2 in one balloon can contain (a) How much H2 (in moles) is needed to inflate a balloon of radius 3.0 m to a pressure of 1.0 atm in an ambient temperature of 25ºC at sea level ? (b) What mass can the balloon lift at sea level, where the density of air is 1.22 kg m–3 ? (c) What would be the payload if He were used instead of H2 ? Sol. (a) PV = n H 2 RT T where P = 1 atm = 1.0132 × 105 N m–2, V= 4 ×  × (3)3 m3, 3 R = 8.314 J mol–1 K–1, T = 298 K n H2 4 22 5 3 PV 1.0132 ×10 × 3 × 7 × (3) = = = 4.62 × 103 mol RT 8.314 × 298 GASEOUS STATE (b) Density of air = 1.22 kg m–3 hence, mass of air = volume of balloon × density = 4 3 4 22 πr × d = × × (3)3 × 1.22 kg = 138.03 kg 3 3 7 mass of H2 = n H 2 × Mw of H2 = 4.62 × 103× 2 = 9.24 kg hence, the mass that the balloon can lift = mass of air – mass of H2 = 138.03 – 9.24 = 128.79 kg (c) Payload if He were used instead of H2 = 138.03 – 18.48 = 119.55 kg [Mass of He = nHe × Mw of He = 4.62 × 103 × 4 = 18.48 kg] CALCULATION OF VAPOUR DENSITY OF MIXTURE IN REACTION It is given that initially only reactant A is present with vapour density (D). It has degree of dissociation  and finally vapour density 'd' An  nB Moles before dissociation Moles after dissociation  1 1–  0 n total no. of moles after dissociation = 1 + (n – 1) Since, mol. wt.  1 no. of particles present  1  no. of particles before dissociation Normal mol. wt  1  no. of particles after dissociation Experimental mol. wt  no. of particles after dissociation Normal mol. wt.  no. of particles before dissociation Experimental mol. wt  1 + (n – 1)α M Normal = 1 M exp The entire conclusion can represented in a general way as – PNormal Pexp erimental = M Normal D Normal = = 1 + (n – 1) M exp erimental Dexp erimental KINETIC THEORY OF GASES The theoretical model for all the experimental gas laws can be analys with the help of kinetic theory of gases which is based on following assumptions : (a) All the gases consists of very small molecules or atoms whose volume is negligible compared to volume of container (in case volume of container is very small, then this assumption will be violated) (b) There is no interaction between gaseous particles (the interaction may appreciably affected under certain conditions of temperature and pressure). (c) The gaseous molecules are under a continuous state of motion which is unaffected by gravity (the random straight line motion is known as brownian motion) (d) Due to the continuous motion, collision between gaseous molecules with the wall of cantainer occurs. The collision with the wall of container are responsible for pressure exerted by the gas on the wall of container. (e) The molecule moves with different speed, however the speed of each molecule keep on changing as the collision occur. (f) All the collision occuring are considered to be perfectely elastic which implies that there is no loss of energy. (g) The average kinetic energy of gas will depends on absolute temperature only. Derivation of Equation for Kinetic Molecular Theory Let us consider a cube of side L, that has N0 molecules each of mass m moving with velocity u in all direction and thus colliding with one each other and against sides of the container. Velocity u can be resolved into three components ux, uy and uz along there axes such that u2 = u 2x + u2y + u2z (assume velocity in ms–1 and distance in meter) y x L Z Before collision After collision + ux Face B – ux Face A For a simplest case we consider motion of a molecule along x-axis only in which it moves towards face B with velocity ux. After collision against face B it moves towards face A with velocity (– ux) collissions being elastic (which results in change in direction but not velocity)  Momentum before collision on face B = mux Momentum after collision on face B = – mux Change in momentum due to one collision on face B = mux – (– mux) = 2mux To strike face B again distance travelled = 2L GASEOUS STATE Time taken to strike face B again = 2L seconds ux  Number of collisions per second on face B along x-axis =  Change in momentum due to ux 2L ux ux mu 2x collisions per second on face B along x-axis = 2mux · = 2L 2L L Change in momentum per second due to collisions of one molecule on opposite faces A and B along x-axis = 2mu 2y 2mu 2x L 2mu 2z L L Total change in momentum per second due to collisions of a single molecule on six faces along three axes Similarly for y-axis change in momentum per second = = and for z-axis = 2mu 2y 2mu 2x 2mu 2z + + L L L 2m 2 2mu 2 (u x + u 2y + u 2z ) = L L But rate of change in momentum per second = Force =   Force = 2mu 2 L 2mu 2 Force mu 2 mu 2 L Pressure = = = = 2 6L Area of six faces 3L3 3V [L3 = volume V]  Pressure due to collisions of N0 molecules on six faces of a cube = 1 mN0u2 3 1 1 mN0u2 = Mu2 3 3 mN0 = M (molar mass) N0 = Avogadro's number u = root mean square velocity (Urms) PV = Translational Kinetic Energy of n Moles 1 3 3 Mu2 = PV = nRT T 2 2 2 Average Translational Kinetic Energy Per Molecule = 3 RT 3 = KT 2 N0 2  R  Where K  =  is called Boltzmann's constant.  N0  Its numerical value is 1.38 × 10–16 erg K–1 molecule–1 Thus average K.E. is proportional to absolute temperature. If T = 0 K (i.e., – 273.15ºC), then average KE = 0 Thus, absolute zero (0 K) is the temperature at which molecular motion ceases. DIFFERENT TYPES OF MOLECULAR VELOCITIES 1 MU 2rms 3  PV = (i) Root mean square velocity (Urms) : It is defined as the square root of the mean of the squares of the velocities possessed by all the molecules present in the given sample of gas Urms =  U12 + U 22 + ...U n2 N Urms (root mean square velocity) = 3PV = M 3RT = M 3P d Where d is the density.  If N1 molecules have velocity u1 and N2 molecules have velocity u2, then Urms = (ii) N u + N u 22 N1 + N 2 Average velocity (Uav) : It is given by the arithmetic mean of square of the different velocities possessed by the molecules of the gases at a particular temperature. Uav = U12 + U 22 + ...U 2n n 8RT πM  Uav (average velocity) (iii) Most probable velocity (Ump) : It is defined as the velocity possessed by the maximum number of molecules of a gas at a given temperature.  Ump (most probable velocity) =  If P and T both are given, use equation in terms of temperature i.e. use Urms = = 3RT and not M 2RT M 3PV M  To have velocity in ms–1 (MKS) take R = 8.314 J mol–1K–1, M in kg.  If density is in kg m–3 and P in N m–2, velocity will be in ms–1. Relation between rms velocity, average velocity and most probable velocity.  Urms : Uav : Ump = 3RT 8RT 2RT : : = M πM M Urms = 1.2248 Ump Uav = 1.1284 Ump Ump = 1.0854 Uav 3 : 8 : π 2 = 1.2248 : 1.1284 : 1 GASEOUS STATE Distribution of Molecular Speeds (a) The speed of a molecule of a gas changes continuously as a result of collisions with other molecules and with the walls of the container. Thus, the net result is that we cannot speak of the speed of individual molecule; hence must consider the statistical averages of the speeds of the whole collection of gas molecules. (b) The manner in which the molecules of a gas are distributed over the possible speed ranges, from zero to very high values, was first investigated by Maxwell using the theory of probability. His results are expressed as the law of distribution of molecular speeds, one form of which is  m  dN = 4N    2πKT  3/2 e – mu 2 2πKT u2du (c) The above expression gives the no. of molecules dN having speeds between U and U + du in terms of total no. of molecules, N present in the gas m is mass of single gas molecule and T is the absolute temperature of gas. (d) The Maxwell distribution of speeds is customarily plotted with the fraction U as the abscissa. 1  dN  1  dN   =   N  dU  du  dU  1  dNu    as the ordinate and N  dU  T1 1 dN N dU T2 > T1 T2 U1 U2 U 1 dN So, the term, represents the fraction of molecules in the speed range of u to u + du per unit interval N dU of speed. Roughly speaking, this gives the probability of finding a molecule with a speed between u and (u + du). The distribution of two temperatures is shown in the fig. (e) The curve at any temperature is parabolic near the origin, since the factor u2 is dominant in this region, the exponential function being approximately equal to unity. At high values of u, however, the exponential factor dominates the behaviour of the function, causing it to decrease rapidly in value. As a consequence of the contrasting behaviour of two factors, the product function passes through a maximum at a speed known as the most probable speed (Umps). Thus, the most probable speed is the speed possessed by the maximum fraction of the molecules. (f) It can be seen from the given figure, that the fraction of molecules having either very low speeds or very high speeds are small in numbers. The total area under the curve gives the total no. of molecules in the collection i.e. the no. of molecules comprising of all speeds. The area under the curve between any two speeds, for example, U1 and U2, gives the total no. of molecules having speed between these two values. (g) The given figure illustrates the distribution of speeds at two temperatures T 1 and T2. Since the total no. of molecules is the same at both temperatures, increases the K.E. of the molecules, it follows that fraction of molecules having lower speed range decreases whereas fraction of molecules having higher speed range increases on increasing the temperature. (h) Also the curve at the higher temperature T2 has its Umps shifted to a higher value compared with that for T1, whereas corresponding fraction of molecules has decreased. But at the same time, the curve near U mps has become broader at the higher temperature indicating the more molecules possess speeds near to most probable speed. (i) The value of Umps can be calculated easily, based on the concept of maximum or minimum value of a function.  m  dN = 4N    2πRT  or 1 dN  m  = 4   N du  2πRT  3 3 2 –mU2 e 2 –mU2 e 2KT 2KT u2 du u2 Differentiating both sides with respect to U d  1 dN   m    = 4   du  N du   2πRT  3 2 –mu 2   2mu  – mu 2 2KT  2KT + u2  – e  2u e   2KT    For maximum d  1 dN   = 0 du  N du    m  4    2πRT  or 2u e 2 – mU 2KT 3 2 –mu 2   2mu  –mu 2 2KT  2KT + u2  – e  2u e  =0  2KT     mu 2  1 – 2KT  = 0 2 –mU 2KT Now at the condition of mps, U  0 and e 2 – mU  e  1– 2KT mU 2 mU 2 =0  =1 2KT 2KT or   =0 U= 2KT m 2KT = m Umps = 2RT M Maxwell Distribution Curve for Energy E= 1 MU 2 2 U= 2E M dE = 1 M × 2 UdU 2 dE = MU dU UdU = dE M By Maxwell equation : 32 2 –Mu  M  dN = 4 N   .e 2RT U. UdU  2πRT  = 0 when U = , but at the condition of Umps, U  . GASEOUS STATE 32 E  –    M   RT  dNE = 4N   .e  2πRT  32  1  dNE = 2N   .e  πRT  –E RT 1  dN E   1    = 2   dE  N   πRT  2E dE M M e - E / RT 1 dN E N dE E dE 32  E / RT E e E E E1 E2 The shaded area of this graph indicates the fraction of particle having energy between E1 and E2. From maxwell equation we are able to calculate the total no. of particles which are having energy greater than threshold energy (Activation energy) [for a chemical reaction to occur, reactant need some minimum amount of energy which is called as threshold energy or activation energy].  1  dNE = 2N    πRT  32   Ea E e  E RT dE From the above equation we calculate the total no. of particles which will participate in chemical reaction. Ex. (a) Calculate the pressure exerted by 1023 gas molecules, each of mass 10–25 kg, in a container of volume 1× 10–3 m3 and having root mean square velocity of 103 ms–1. Sol. (b) Also calculate total kinetic energy and (c) temperature of the gas. (a) By kinetic theory P= 1 mNu 2 1  1 0 2 5  1 0 23  (1 0 3 )2 = = 3.33 × 106 N m–2 3 V 3  1 0 3 1 2  Total KE =  mu  × N 2  (b) = (c) 1 1 × 10–25 × (103)2 × 1023 = × 104 = 0.5 × 104 J 2 2 Also total KE = 10 23 3 1023 nRT, where n (mole) = = NA 2 6.023×1023 3 1023 × 8.314 × T 0.5 × 104 = × 2 6.023 ×1023  T= 0.5 ×104 × 2 × 6.023 = 2415 K 3 × 8.314 Ex. A gas bulb of 1 litre capacity contains 2.0 × 10 21 molecules of nitrogen exerting a pressure of 7.57 × 103 Nm–2. Calculate the root mean square velocity of the gas molecules. Sol. No. of gas molecules = 2 × 1021 6.023 × 1023 molecules of nitrogen weight = 28 g 2 × 1021 molecules of N2 weights = Density =  Mass 0.093 0.093×10 –3 = g/L = = 0.093 kg/m3 Volume 1 10 –3 Urms = Ex. 28 × 2 ×1021 g = 0.093 g 6.023×10 23 3RT = M 3PV = M 3P = Urms = d 3 × 7.57 ×103 = 494.16 m/s 0.093 In a container of capacity 1 litre there are 1023 molecules each of mass 10–22 gms. If root mean square speed is 105 cm/ sec then calculate pressure of the gas. Sol. PV = 1 MN U2 3 P=? V = 10–3 m3 m = 10–25 kg N = 1023 5 3 U 2 = 10 cm/sec = 10 m/sec U2 = 106 m2 /sec2 P × 10–3 = 1 × 10–25 × 1023 × 106 3 1 × 10–2 × 106 × 103 3 1 P = × 107 pascals 3 P= COLLISION THEORY Collision Diameter : It is the closest distance between the centres of two molecules taking part in collision. collision diameter () = rA + rB Collision Frequency It is the number of molecular collisions taking place per second per unit volume of the gas. Determination of Number of Bimolecular Collisions (a) While considering the molecules among themselves the molecules to be rigid, non-interacting and spherical with diameter . It is also assumed that all the molecules move with the same speed. The arithmetic mean of speed is U . GASEOUS STATE (b) Two identical molecules of diameter  will just touch each other when the distance separating their centres is . Thus, a moving molecule will collide with other molecules whose centres come within a distance of  from its centre. The quantity 2 is called the collision cross-section for the rigid spherical molecule. It is obvious that this collision cross-section is an area of an imaginary sphere of radius  around the molecule within which the centre of another molecule cannot penetrate. (c) If the molecule is moving with an average speed U , then in a unit time area swept out by a single molecule is 2 U . If N* is the no. of molecules within the volume 2 U , the no. of collisions by a single molecule in a unit time will be 2 U N*. (d) So, far, we have assumed that only one molecule is moving and all other are stationary. In practice however, this is not true. In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two colliding molecules instead of average velocity of a single molecule. If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centre is 2 U as shown below.. U 2U U  The no. of collisions made by a single molecule with other molecules per unit time (collision number) are given by z1 = (e) 2 2 U N* The total number of bimolecular collision per unit time is given as z11 (collision frequency) 1 1 (z1N*) = × N* × 2 2 1 = 2 U N*2 2 2 2  U N* z11 = (f) If the collisions involve two unlike molecules, the no. of bimolecular collision is given as z 12. z12= 1 2 < u0 > N1N2 2 Where N1 and N2 are the no. of molecules per unit volume of the two types of gases, 12 is the average diameter of two molecules, that is 12= σ1 + σ 2 or < u0 > = [u 12 + u 22 7]1/2 and 2 u 12 = 8RT , πM1 u 22 = 8RT πM 2 Determination of Mean free Path and Its Dependence on Pressure and Temperature (a) The mean free path is the average distance travelled by a molecule between two successive collisions. We can express it as follows : (b) = Average distance travelled per unit time No. of collsions made bysingle molecule per unit time = U U = = z1 2πσ 2 UN * 1 2πσ 2 N * According to ideal gas equation, N1 RT T NA  N1   R   N1  or P =  (where K is known as Boltzmann's constant)   N  T =   KT V A V N1 Now = N*, V P P  P = N* KT or N* = or N*  T KT 1 = 2πσ 2 N * T 1 = or   P N* Since, according to gas law P  T at constant volume PV = nRT = (c)   = constant at constant volume. Thus there will no effect of changing T or P on , if the volume of the gas is kept constant. (d) 2 2  U Now z1 =  U = K1 T U T N*   z1 = N* P T  N*= K2 P T P 2 2  K1 K2 T  z1 = K' P T Now z1  P when T constant and z1  1 when pressure is held constant. T At constant volume P  T or P = K'' T  z1 = or z1  K'K"T T T when volume is constant. Again z1  P when volume is constant. GASEOUS STATE (e) z11 = 1 2 U N*2 2 or z11  U N*2 P T  T  or z11  or z11  Thus z11 2 P2 T 3/2  p2 when temperature is held constant and z11  T–3/2 when pressure is held constant At constant volume z11  T–1/2 and z11  P1/2. IDEAL AND REAL GAS An ideal gas is a hypothetical gas whose pressure, volume and temperature behaviour is completely described by the ideal gas equation. Actually no gas is ideal or perfect in nature. All gases are real gases. Real gases do not obey the ideal gas laws exactly under all conditions of tamperature and pressure. Real gases deviates from ideal behaviour because of mainly two assumptions of "Kinetic theory of gases".  (i) The volume of gas particle is negligible compared to volume of container (while the real gas particle have some significant volume). (ii) There is no interaction between gaseous particles (while attraction forces exist between real gas particles). Deviation from Ideal Behaviours Compressibility Factor The extent to which a real gas departs from the ideal behaviour may be expressed in terms of compressibility factor (Z), where Z is PVm [Vm = molar volume] Vm (ideal) RT When Z > 1, Vm > 22.4 lit at STP ; When Z < 1, Vm < 22.4 lit at STP Z= Vm  Plots of Compressibility Factor vs Pressure (i) For an ideal gas Z = 1 and is independent of temperature and pressure. (ii) Exceptional behaviour of H2 and He : for these gases Z > 1. For such gases repulsive forces predominants. (iii) Effect of pressure : At very low P, PV  RT i.e. Z  1 At low P, PV < RT i.e. Z < 1 At moderate P, PV  RT i.e. Z  1 At high P, PV > RT i.e. Z > 1 (iv) For the gases which are easily liquefied (e.g. CO2) Z dips sharply below the ideal line in the low pressure region. (v) Effect of temperature : An increase in temperature shows a decrease in deviation from ideal behaviours i.e. PV approaches unity or Z approaches unity with increase in temperature. Z H2 He N2 CH4 1.0 CO 2 T4 T 1 > T2 > T3 > T4 ideal gas Z = 1 100 P 200 T1 1.0 ideal gas Z Z <1 0 T3 T2 300 0 200 600 400 P Boyle's Temperature The temperature at which a real gas behaves like an ideal gas over an appreciable range of pressure is known as Boyle's temperature (TB). It means that at this temperature, the value of PV remains constant for an appreciable range of pressure, i.e. Boyle's Law is abeyed. Mathematically we may express the Boyle temperature or Boyle point as the temperature where  (PV) = 0 when P  0. P So to use ideal gas equation for real gas, we will use some correction factor for volume as well as for pressure. (A) Volume Correction : Anil ahlawat (Page 154-) In the ideal gas equation, PiVi = nRT, Vi represents the ideal volume where the molecules can move freely. In real gases, a part of the total volume is occupied by the gas molecules. Hence the free volume V i is the total volume V minus the volume occupied by the gas molecules. i.e. Real volume of gas = Actual volume of container – volume occupied by N molecules in motion. Vi = V – b for one mole of gas Vi = V – nb for n mole of gas d excluded volume Where b is termed the 'excluded volume' or 'co-volume'. It is constant and characteristic for each gas. 4 3 b =  πr  × 4N 3  [where r = radius of gas molecule, N = Avogadro's number] (B) d Illustration of excluded volume Pressure Correction In order to take account the effect of intermolecular forces of attraction, let us consider a molecule A is the midst of the vessel. This molecule is surrounded by other molecules in a symmetrical A • manner and is being attracted uniformly on all sides by the neighbouring • molecules with the result that this molecule on the whole experiences no net force of attraction. • • • • • • • • • B • • GASEOUS STATE Now, consider a molecule B near the side of the vessel, which is about to strike one of its sides, thus contributing towards the total pressure of the gas. There are molecules only on one side of the vessel, i.e. towards its centre, with the result that this molecule experiences a net force of attraction towards the centre of the vessel. This results in decreasing the velocity of the molecule, and hence its momentum. Thus, the molecule does not contribute as much force as it would have, had there been no force of attraction. Thus, the pressure of a real gas would be smaller than the corresponding pressure of an ideal gas. Van der Walls noted that the total force of attraction on any molecule about to hit a wall is proportional to the concentration of neighbouring molecules, n/V. However, the number of molecules about to hit the wall per unit wall area is also proportional to the concentration n/V. Therefore, the force per unit wall area, or pressure, is reduced from that assumed in the ideal gas wall by a factor proportional to n2/V2. Letting a be the proportionality constant, we can write P (actual) = P(ideal) – n2a/V2 or P (ideal) = P(actual) + n2a/V2 'a' is a constant which depends upon the nature of the gas, V is the volume of 1 mole of gas. Combining the two corrections, for 1 mole of gas a   T  P + 2  (V – b) = RT V    n 2a  For n mole of gas  P + 2  (V – nb) = nRT T V   The Constants 'a' and 'b' Van der Waals constant for attraction 'a' and volume 'b' are characteristic constants for a given gas. (i) The 'a' values for a given gas are measure of intermolecular forces of attraction. More are the intermolecular forces of attraction, more will be the value of a. (ii) For a given gas van der Waals constant of attraction 'a' is always greater than van der Waals constant of volume 'b'. (iii) The gas having higher value of 'a' can be liquefied easily and therefore H2 and He are not liquefied easily. (iv) Unit of 'a' is lit2 mole–2 atm or cm4 mole–2 dyne. (v) Unit of 'b' is lit mole–1 or cm3 mole–1. (vi) At low pressure, volume correction for 1 mole of gas is negligible, b=0 The gas equation can be written as a   T  P + 2  V = RT V   or Z= a PVm =1– Vm RT RT where Vm = molar volume where Z is known as the compressibility factor. (vii) At higher pressures, the pressure correction for 1 mole of gas is negligible a  2 =0 V or (P + 0) (V– b) = RT or P (Vm – b) = RT PVm Pb or PVm = RT + Pb or, Z = =1+ RT RT The Van der Walls Vonstants for Some Common Gases a (dm6 atm mol–2) b (dm3mol–1) Ammonia 4.17 0.0371 Argon 1.35 0.0322 Carbon dioxide 3.59 0.0427 Carbon monoxide 1.49 0.0399 Chlorine 6.49 0.0562 Ethane 5.49 0.0638 Ethanol 2.56 0.087 Ethylene 4.47 0.0571 Helium 0.034 0.0237 Hydrogen 0.024 0.0266 Hydrogen chloride 3.67 0.0408 Hydrogen bromide 4.45 0.0433 Methane 2.25 0.0428 Neon 0.21 0.0171 Gas Nitric oxide 1.34 0.0279 Nitrogen 1.39 0.0319 Oxygen 1.36 0.0318 Sulphur dioxide 3.71 0.0564 Water 5.44 0.0305 LIQUEFACTION OF GASES AND CRITICAL POINTS The phenomenon of converting a gas into liquid is known as liquefaction. The liquefaction of a gas takes place when the intermolecular forces of attraction become so high that exist in the liquid. A gas can be liquefied by : (a) Increasing Pressure : An increase in pressure results in an increase in attraction among molecules. (b) Decreasing Temperature : A decrease in temperature results in a decrease in kinetic energy of molecules. ANDREW'S ISOTHERMS The essential conditions for liquefaction of gases were discovered by Andrews (1869) as a result of his study of P–V–T relationship for CO2.The types of isotherms are shown in figure. Pressure F O vapour + liquid D C 48 º C 35.5ºC 31.1ºC Gas 21.5ºC E 13.1ºC B A Volume Fig : Isotherms for carbon dioxide showing critical region GASEOUS STATE Observations from Figure (a) At Low Temperatures : For the curve ABCF, as the pressure increases, volume of the curve decreases (curve A to B). At point B, at constant pressure, liquefaction commences and the volume decreases rapidly (because gas is converted to liquid with higher density). At point C, liquefaction is complete. The line CF represents the variation of V with P of the liquid state. The steepness of the line CF indicates that the liquid cannot be easily compressed. Thus AB, represent gaseous state, BC represent liquid and vapour in equilibrium. CF represent liquid state. The pressure corresponding to the line BC is there vapour pressure of the liquid. (b) At Higher Temperatures : Similar type of curve as in case (a) is obtained but the width of the horizontal portion is reduced. The pressure corresponding to this portion is higher than at lower temperatures. (c) At High Temperatures : (like 48ºC), the isotherms are like those of ideal gas. (d) At Temperature (31.1ºC) : The horizontal portion is reduced to a point. The isotherm at TC is called critical isotherm. At point O, dP = 0. dV The point O is called the point of inflection. Critical Temperature (TC) : The maximum temperature at which gas can be liquefied by the application of pressure alone i.e. the temperature above which a liquid cannot exit is called the critical temperature T C. Critical Pressure (PC) :The minimum pressure required to liquefy the gas at the temperature T C is called the critical pressure PC. Critical Volume (VC) :The volume occupied by one mole of the gas at critical temperature, TC and critical pressure, PC is called the critical volume (VC) of the gas. Determination of Value of PC, VC and TC (a) The vander waal’s equation is a   T  P + V 2  (Vm – b) = RT  m  a ab – Pb – 2 = RT T Vm Vm or PVm + or Vm + or Vm + or RT  2 a ab  V + V – V 3m –  b + =0  P  m P m P 3 a RT 1  ab  1 × – b –  2 = P Vm P  P  Vm a ab RTVm2 Vm – bV 2m – = P P P .... (1) (b) This equation has three roots in Vm for given values of a, b, P and T. It is found that either all the three roots are real or one is real and the other two are imaginary. P TC (c) The main characteristics of the figure are as follow : (i) At higher temperature than TC the isotherm exhibits a maximum and a minimum for certain values of pressures, the equation gives three roots of volume e.g., V1, V2 and V3 at pressure P1. (ii) V On increasing the temperature, the three roots become closer to each other and ultimately at critical temperature, they become identical. Thus, the cubic equation Vm can be written as (Vm – V') (Vm – V") (Vm – V"') = 0 At the critical point V' = V" = V"' = VC  the equation becomes, (Vm – VC)3 = 0 or V 3m – V 3C – 3VC V 2m + 3 V 2C Vm = 0 from eq. (1) RT  2 a ab   V m + Vm – V 3m –  b + =0  P  P P (d) By comparing the coefficients 3VC = b + RTC a ab , 3V 2C = , V 3C = PC PC PC By solving, VC = 3b, PC = (e) a 8a and TC = 27b 2 27Rb Therefore the value of critical compressibility factor in terms of vander wall’s constants is given by a a a × 3b 2 PC VC Z= = 27b = 9b = 9b 8b 8Ra RTC 8a R× 27Rb 27Rb 27b Z= (f) (g) a 27b 3 PC VC = × = = 0.375 9b 8a 8 RTC PC VC If we compart the value of RT = 0.375, with the experimental values, it has been found that the agreement C is very poor. Boyle temperatures in terms of van der Waals constant : TB = a Rb GASEOUS STATE CRITICAL CONSTANTS OF GASES Gas PC (atm) He Ne Ar Xe H2 O2 N2 CO2 H2O NH3 CH4 C2H6 2.26 26.9 48.1 58.0 12.8 50.1 33.5 72.8 218.0 111.5 45.6 48.2 Vm,c (cm3mol–1) 57.9 41.7 75.2 119.0 65.5 78.2 90.1 94.0 55.6 72.5 98.7 148.0 TC (K) 5.2 44.4 150.7 289.7 33.3 154.8 126.2 304.2 647.3 405.0 190.6 305.4 EQUATION OF CORRESPONDING STATE All the equation defined for real gases involved gas dependent constant, hence if for two real gases n,P, V are same then the fourth parameter need not to be same and can be calculated only when the gas dependent constant is known. However instead of pressure, temperature, moles and volume, new reduced parameter are defined such that P  Pr =  P   Reduced Pressure  C T Tr =  T   Reduced temperature  C  V  Vr =  V   Reduced volume  C a    P + V 2  (Vm– b) = RT T  m  a    Pr Pc + (V V ) 2  (VrVc – b) = R × TrTc r c   3    Pr + V 2  (3Vr – 1) = 8 Tr  r  From this equation the law of corresponding state can be stated two real gas having same no. of moles are at same reduced temperature and same reduced volume then they have same reduced pressure. Ex. For 1 mole gas reduced temperature is 3 and reduced volume is 10. If critical pressure is 42 atm then calculate the pressure exerted by 1 mole of real gas. 3    Pr + V 2  (3Vr – 1) = 8 Tr  r  Sol.  3    Pr + (10) 2  (3× 10 – 1) = 8 × 3    3   24 – P = PrPc =   42 = 33.49 atm  29 100  Pr = 24 3 – 29 100 Law of Equiparition of Energy This law states that is energy of the molecule (due to translational, rotational, vibrational motion) is represented as a sum of independent significant terms (which are proportional to v2 (velocity), 2 (angular velocity) and x2 RT (displacement) then each of the significant term contribute to in average K.E. per mole of the system). 2 DEGREE OF FREEDOM Number of independent co-ordinates or axis required to represent position or motion of the molecules is called as degree of freedom (for a system comprising of N atoms, total 3N degree of freedom are available, out of which some of translational, some are rotational and the other are vibrational). Translation Degree of Freedom Motion of any molecule can be represented along three independent coordinates or axis so translational degree of freedom is always three ( these are significant at all temperatures for the gas). 1 1 1 2 2 mVx2 + mVy + mVZ 2 2 2 RT RT RT 3RT T.E. (per mole) = + + = 2 2 2 2 RT Each degree of freedom contributes to in average K.E. per mole. 2 Rotational Degree of Freedom T.E. = (i) For monoatomic molecule rotational energy is (ii) (a) 1 2 I will be insignificant (if rotation is about its own axis.) 2 For polyatomic gases : Non-linear complete rotation can be represented along three perpendicular axis (this is significant at moderate temperature). (b) Linear : only two independent axis are required to represent total energy due to rotation. For non Linear (NL) : 1 2 1 2 1 2 Iωx + Iω y + Iωz 2 2 2 RT RT RT 3RT R.E. per mole = + + = 2 2 2 2 For Linear : R.E. = (Here I = moment of Inertia) 1 2 1 2 1 2 1 2 R.E. =  Iω x + Iω y  or  Iω y + Iω z  or 2 2  2 2  1 2 1 2  Iω x + Iω z  2 2  RT RT + = RT T 2 2 These are influencial only at moderate temperature. Vibrational Degree of Freedom (a) For linear molecule = 3N – 5. (b) For non-linear molecule = 3N – 6. where N is no. of atoms present in a molecule. Each vibrational degree of freedom comprises of two energy terms. R.E. per mole = 1 1 mv2  Kx2 2 2 vibrational degree of freedom contributes RT to average K.E. per mole. These are significant only at a very high temperature. V.E. = GASEOUS STATE HEAT CAPACITY It is the amount of heat required to raise the temperature of 1 mole or 1 gram of a substance through 1ºC or 1 K. If it is measured per mole of substance then it is called as molar heat capacity. If it is measured in per gram of substance then it is called as specific heat. q n × ΔT C= (i) Heat capacity at constant pressure CP = (ii) (C = molar heat capacity, q = Heat given) qp n × ΔT = ΔH nΔT qp = H = Enthalpy change. Heat capacity at constant volume CV = qp nΔT = ΔU n × ΔT qV = U = change in internal energy molar heat capacity is a function of temperature. C = f(T) C = (a + bT + CT2 + .....) (where a, b, c are constant) T2  C dT q= n T1 qV = CV (T2 – T1) = K.E./ mole K.E./ mole = CV (T) CV = K.E / mole ΔT at T temperature : CV = K.E / mole T For ideal gas CP – CV = R (here R = ideal gas constant) CV CP 3 5 Monoatmic gas Diatomic gas 2 7 2 R R 2 9 2 CP/CV =  (Poission's Ratio) R R 5 3 9 7 = 1.66 = 1.28 For 'N' atomic molecules the degree of dissociation is 3N. (i) If the gas is monoatomic : (a) It will have three degree of freedom for K.E. since each degree of freedom provides therefore total K.E. is 1 KT amount of energy 2 3 KT (It will not have any rotational or vibrational degree of freedom). 2 (ii) If gas is diatomic : Total degree of freedom is six. (a) Out of which 3 is translational degree of freedom, therefore total energy (b) Out of which 2 will be rotational degree of freedom - 3 KT.. 2 1 2 1 2 1 1 Iω x + Iω y = KT + KT = KT 2 2 2 2 Remaining is vibrational degree of freedom is one. R.E. = (c) Energy concerned with the vibration motion corresponds to energy because of relative motion + change in position from equilibrium r0 position. 2 V.E. = 1  dr  1 m   + K (ro – r) 2  dt  2 2 = 1  dr  1 m   + Kx 2 (Here x is displacement from equilibium position) 2  dt  2 1 1 KT + KT  KT 2 2 So each vibrational degree of freedom will provide KT amount of energy. = We consider the vibrational degree of freedom at very high temperature. (iii) If gas is triatomic : Total degree of freedom is nine. (a) Out of which (b) 3 KT is translational. 2 If linear then it will have two degree of freedom of rotational. (c) Vibrational will have four degree of freedom T.E. = 3 13 KT KT + KT + 4KT = 2 2 CV CP CP/CV =  (Poission's Ratio) Triatomic (linear) 13 R 2 15 R 2 15 = 1.15 3 Non linear 6R 7R 7 = 1.16 6 CV CP CP/CV = Diatomic 5 R 2 7 R 2 7 = 1.4 5 Triatomic (linear) 5 R 2 7 R 2 7 = 1.4 5 Triatomic (Non linear) 3R 4R 4 = 1.33 3 If vibration is ignored at lower temperature. GASEOUS STATE Temperature above which gases gives heating effect and below this cooling effect in process of adiabatic expansion. Texp. > Ti Texp. < Ti Texp. = Ti Ti =  heating  Cooling  No cooling, No heating 2a = 2 × Tb Rb Ti for H2 gas = 193 K Ti for He gas = 33 K Ex. Arrange following in decreasing ’a’ factor (H2O, CO2, Ar) H2O > CO2 > Ar polar  For Non Polar Molecules Greater the size or surface area, greater will be vander waals' forces, so greater will be ’a’ constant. Ex. Arrange following gases according to ‘a’ He , Ar, Ne, Kr. aKr > aAr > aNe > aHe  More ‘a’ factor means higher will be boiling point. liquification Pressure Is the pressure required to convert gas into liquid. for easy liquefaction When Ex. a and P Z < 1, Vm < Vm, ideal  easily liquifiable Z > 1, Vm > Vm, ideal  more difficult to compress. Arrange the following according to liquification pressure. n-pentane ; iso-pentane , neo pentane. an-pentene > aiso-pentane > aneo-pentane liquification pressure = LP L Pn  pen tan e < L P iso pen tan e < L Pneo pen tan e  b is roughly related with size of the molecule. (Thumb rule) 4 3 b = NA 4  π r  3  Ex. Two vander waals gases have same value of b but different a values. Which of these would occupy greater volume under identical conditions ? Sol. If two gases have same value of 'b' but different values of 'a', then the gas having a larger value of 'a' will occupy lesser volume. This is because the gas with a larger value of 'a' will have larger force of attraction and hence lesser distance between its molecules. Ex. Calculate the pressure exerted by 5 mole of CO2 in one litre vessel at 47ºC using van der waal's equation. Also report the pressure of gas if it behaves ideal in nature. Given that a = 3.592 atm lt2 mol–2, b = 0.0427 L/mol. Also, if the volume occupied by CO2 molecules is negligible, Sol. then calculate the pressure exerted by one mole of CO2 gas at 273 K. Vander waal's equation  n 2a  T  p + V 2  [V – nb] = nRT   n CO 2 = 5, V = 1 litre, T = 320 K, a = 3.592, b = 0.0427  3.592   P + 25 × [1 – 5 × 0.0427] = 5 × 0.0821 × 320  1   P = 77.218 atm For ideal behaviour of gas, PV = nRT  P × 1 = 5 × 0.0821 × 320  P = 131.36 atm a   For one mole  P + 2  [V – b] = RT T V    P= a RT – 2 V V  P= 3.592 0.0821× 273 – (22.4) 2 22.4  P = 0.9922 atm The volume occupied by 1 mole at 273 K is 22.4 litre if b is negligible. Ex. Two moles of ammonia gas are enclosed in a vessel of 5 litre capacity at 27ºC. Calculate the pressure exerted by the gas, assuming that (i) the gas behaves like an ideal gas (using ideal gas equation). (ii) the gas behaves like a real gas (using van der Waal’s equation) Given that for ammonia, a = 4.17 atm litre2 mol–2 and b = 0.037 litre mol–1. Sol. Given, n = 2 moles V = 5 litres T = 27ºC = (27 + 273) K = 300 K a = 4.17 atm litre2 mol–2 b = 0.037 litre mol–1 Also, we know that R = 0.0821 litre atm K–1 mol–1 (i) If the gas behaves like an ideal gas, we have PV = nRT  P= nRT 2 × 0.0821× 300 = = 9.85 atm V 5 GASEOUS STATE (ii) If the gas behaves like a real gas, we apply van der Waal's equation i.e.  an 2  T  P + 2  (V – nb) = nRT V   or Ex. P= 2 nRT an 2 2 × 0.0821× 300 4.17 × (2) – = – = 9.33 atm. 2 V – nb V (5) 2 5 – 2 × 0.037 1 mole of CCl 4 vapours at 77ºC occupies a volume of 35.0 L. If vander waal's constants are a = 20.39 L2 atm mol–2 and b = 0.1383 L mol–1, calculate compressibility factor Z under, (a) low pressure region. Sol. (a) Under low pressure region, V is high  (V – b)  V (b) high pressure region. a    P + 2  V = RT T V   PV + a = RT T V PV a + =1 RT RTV Z= (b) a  PV  20.39 = 1 –  =1– = 0.98 RT  RTV  0.0821× 350× 35 Under high pressure region, P is high, a   P + 2  P V    P (V – b) = RT PV – Pb = RT PV Pb – =1 RT RT  Z= PV Pb =1+ RT RT Z=1+ Ex. Sol.  PV = 1, or Q  RT P 1 =  RT V  b 0.1383 =1+ = 1 + 0.004 = 1.004 V 35  B  One way of writing the equation of state for real gases is PV = RT 1 + + ... where B is a constant. Derive  V  an approximate expression for B in terms of van der Waal’s constants a and b. According to van der Waal's equation a   P + 2  [V – b] = RT T V   or P= RT a – (V – b) V 2 Multiply by V, then PV = a RTV – (V – b) V or a   V – PV = RT   V – b VRT  or –1  b a  PV = RT 1 –  –  V VRT   –1 2  b b b  1 – V  = 1 + V +  V  + ....  b a   PV = RT 1 + + ... – VRT   V a  1     . + ... PV = RT 1 +  b – RT  V     B=b– a RT Ex. The critical temperature and pressure of CO2 gas are 304.2 K and 72.9 atm respectively. What is the radius of CO2 molecule assuming it to behave as vander Waal’s gas ? Sol. TC = 304.2 K PC = 72.9 atm TC = 8a 27Rb PC = a 27b 2  8a TC 27Rb 8a 8b 27b 2 = = × = PC a 27Rb R a 2 27b or b= RTC 1 0.082 × 304.2 = × = 0.04277 lit mol–1 8PC 8 72.9 b = 4 NA × 4 3 πr = 42.77 cm3 3 4 3 πr = 42.77 3  4 × NA × or r3 = or r3 = 0.424 × 10–23 = 4.24 × 10–24 or r = (4.24)1/3 × 10–8 cm = 1.62 × 10–8 cm  radius of CO2 molecule = 1.62 Å 3 × 42.77 × 10 –23 16 × 6.023 × 3.14 GASEOUS STATE Ex. The vander waals constant for HCI are a = 371.843 KPa.dm6 mol–2 and b = 40.8 cm3 mol–1 find the critical constant of this substance. Sol. The critical pressure, PC = 371.843 ×103 371.843 ×109 a = = = 8.273 x 106 2 –6 27 × (40.8) ×10 27 × (40.8) 2 27b 2 PC = 8.273 MPa The critical pressure, T C = 8a 27Rb R = 8.314 KPa dm3 K–1 mol–1 TC = 8×371.843 8a = = 324.79 = 324.8 K 8.314× 27 × 40.8×10 –3 27Rb The critical volume, VC = 3b = 3 x 40.8 = 122.4 cm3 Ex. The vander waals constant for gases A, B and C are as follows : Gas a/dm6 KPa mol–2 b/dm3 mol–1 A 405.3 0.027 B 1215.9 0.030 C 607.95 0.032 Which gas has (i) Highest critical temperature (ii) The largest molecular volume (iii) Most ideal behaviour around STP ? Sol. 8a Since, R is constant, higher the value of a/b, higher will be critical temperature. 27Rb VC = 3b and VC  Vm (for a particular gas) therefore higher the value of VC , higher will be molar volume of the gas. TC = If the critical temperature is close to 273 K, gas will behave ideally around the STP. Let us illustrate the result in a tabular form. Gas a/dm6 KPa mol–2 b/dm3 mol–1 TC VC a/b A 405.3 0.027 534.97 K 0.081 1.501 x 104 B 1215.9 0.030 1444.42 K 0.09 4.053 x 104 C 607.95 0.032 677.07 K 0.096 1.89 x 104 (i) B gas has the largest critical temperature. (ii) C gas has the largest molecular volume. (iii) A gas has the most ideal behaviour around STP Ex. Sol. In a container of capacity 1 litre, air and some liquid water is present in equilibrium at total pressure of 200 mm of Hg. This container is connected to another one litre evacuated container. Find total pressure inside the container when equilibrium is again stablised (aqueous tension or vapour pressure at this temp. is 96 mm Hg). Total pressure = 200 mm of Hg = Pgas + Pvapour water  Pgas + 96 = 200 Pgas = 104 mm of Hg Initially when second container is connected P1 = 104 mm of Hg P2 = ? V1 = 1 V2 = 2 litre P1 V1 = P2 V2 104 × 1 = P2 × 2 52 = P2 After equilibrium is established Ptotal = 52 + 96 = Pgas + Pwater = 148 mm of Hg at equilibrium. Ex. Carbon dioxide gas (CO2) measuring 1 litre is passed over heated coke the total volume of the gases coming out becomes 1.6 litre. Find % conversion of CO2 into carbon monoxide. Sol. CO2 + C  2CO 1 0 at time t CO2 CO 1–x 2x Initial volume = 1 litre final volume = 1.6 litre Final volume = (1 + x) litres 1 + x = 1.6 x = 0.6 x = 0.6 0.6 × 10 = 60% of CO2 will be converted into CO 1 Ex. Sol. 100 ml of hydrocarbon is mixed with excess of oxygen and exploded. On cooling, the mixture was reported to have a contraction of 250 ml. The remaining gas when passed through a solution of aqueous KOH, the mixture shows a further contraction of 300 ml. Find molecular formula of the hydrocarbon. Y  Cx Hy +  x +  O2  x CO2 + 4  100 ml x 100 y H2 O 2 y 100. 2 mixture contains CO2 , H2O & excess O2 on cooling, H2O is separated, volume of H2O = 250 ml y × 100 = 250 ; y = 5, 2 as KOH absorbs CO2, hence 300 ml contraction is because of CO2 that has been absorbed. Volume of CO2 = 100 x = 300 ; x =3 Empirical formula = C3 H5 ; molecular formula = C6 H10. Note : If water is already condensed out then total contraction in reaction mixture = {volume of reactants} – {volume of products + volume of unused species excluding volume of H2O}. GASEOUS STATE Ex. 100 ml of an hydrocarbon is burnt in excess of oxygen in conditions so that water formed gets condensed out the total contraction in volume of reaction mixture was found to be 250 ml when the reaction mixture is further exposed to aqueous KOH a further contraction of 300 ml is observed find molecular formula of hydrocarbon. Sol. Cx Hy + O2 CO2  100 ml excess By POAC on ‘C’ atoms x × 100 = 300 x=3 POAC on ‘H’ atoms y × 100 = 2 × moles of H2O 600 + 50y 2 {V = volume of O2 consumed} volume of O2 consumed The total volume contraction is 250 ml. Hence, 100 + V – 300 = 250 – 200 + V = 250 2 × 450 – 600 = 50 y 300 =y=6 50 Hydro carbon will be C3 H6 H 2O 300 ml POAC on O atoms 2 × v = 2 × 300 + 1 × H2O 2 × v = 600 + 50 y v= + EDUBULL TIPS & FORMULAS 1. GAS LAW (I) Boyle’s Law 1 (n,T  co nst) P P1V1 = P2V2 V (II) Charle’s Law V  T (n, P = const) V2 T2  V1 T1 (III) Gay lussac’s Law P  T (n, V = const) P2 T2  P1 T1 (IV) 2. Avogadro’s Law V  moles  number of molecules (P, T = cosnt) Ideal gas equation PV = nRT R = 0.0821 lit atm mol–1 K–1 R = 8.314 J K–1 mol–1 or 8.314 N × K–1 mol–1 –1 –1 R = 2 cal K mol R = 8.314 × 107 erg K–1 mol–1 GRAHAM’S DIFFUSION LAW It is applicable for non reacting gases r VD  1 d r 1 VD r 1 Mw (P, T = const) dgas Mw  dH 2 2 rate of diffusion r  l diffused gas t time taken (Where,  = distance travelled by differed gas) r r 3. Vdiffused gas t time taken n diffused gas t time taken DALTON’S LAW OF PARTIAL PRESSURE GASEOUS STATE Pmixture = P1 + P2 + P3 ........ (T & V const) Pt = Pmoist gas = Pdry gas + P water vapours It is applicable for non racting gases. Methods of determination of partial pressure (PA & PB are partial pressure) (I) from ideal gas equation PAV = nA RT and PBV = nBRT (II) In the form of mole fraction. nA PA = XA PT = n PT t nB PB = XB PT = n PT t PT = sum of partial pressure of al gases (III) In the form of volume fraction PA = (IV) VA PT V and PB = VA PT V In individual pressure and individual volume are given PA = VA P1 V and PB = VB P2 V P1, P2 = pressure of gases before mixing PA, PB = pressure of gases after mixing 4. Kinetic gtas equation PV = 5. 1 mN V2rms 3 Average KE (KEav) KEav = 3 nRT T 2 (m moles) KEav = 3 RT T 2 ( 1 mol or NA molecules) KEav = 3 KT 2 (1 molecules) K = 1.38 × 10–23 JK–1 molecule–1 6. Vrms  V12  V22  ...Vn2 N Vav  Vrms  3RT Mw Vav  8RT  Mw Vmp  2RT Mw Vrms  3PV Mw Vav  8PV  Mw Vmp  2PV Mw Vrms  3P d Vav  8P d Vmp  2P d Vrms : Vav : Vmp Vmp : Vav : Vrms = 3 : 8  : 2 =1 : 0.92 : 0.82 : 8  : 3 : 1.128 : 1.224 = 2 =1 7. Compressibility factor (z) = If z = 1, If z > 1, If z < 1, 8. V1  V2  V3 ...Vn N (Vm )obs P(Vm )obs  Vi RT the gas show ideal gas behaviour. the gas show positive deviatioin. the gas show negative deviation. Vanderwaal’s Sequation  an 2  P    (V – nb) = nRT T V2   Pi = PR + an 2 V2  Pi > P R a  force of attraction  , liquification  b  , effective size of molecule  incompressible vol  , compressible vol  (I) At high pressure, Vanderwaal’s equation is PVm – Pb = RT (II) At low pressure Vanderwaal’s equation is PVm + a  RT Vm