Chemistry Notes for NEET Chapter 6 PDF

60 E3 Chapter 6 Gaseous state 1 m 3  10 3 dm 3  10 6 cm 3  10 6 mL  10 3 L ID The state of matter in which the molecular forces of attraction between the particles of matter are minimum, is known as gaseous state. It is the simplest state and shows great uniformity in behaviour. Characteristics...

60 E3 Chapter 6 Gaseous state 1 m 3  10 3 dm 3  10 6 cm 3  10 6 mL  10 3 L ID The state of matter in which the molecular forces of attraction between the particles of matter are minimum, is known as gaseous state. It is the simplest state and shows great uniformity in behaviour. Characteristics of gases U (1) Gases or their mixtures are homogeneous in composition. (2) Gases have very low density due to negligible intermolecular (3) Mass : (i) The mass of a gas can be determined by weighing the container in which the gas is enclosed and again weighing the container after removing the gas. The difference between the two weights gives the mass of the gas. (ii) The mass of the gas is related to the number of moles of the gas i.e. forces. U D YG (3) Gases have infinite expansibility and high compressibility. (4) Gases exert pressure. (5) Gases possess high diffusibility. (6) Gases do not have definite shape and volume like liquids. (7) Gaseous molecules move very rapidly in all directions in a random manner i.e., gases have highest kinetic energy. (8) Gaseous molecules collide with one another and also with the walls of container with perfectly elastic collisions. (9) Gases can be liquified, if subjected to low temperatures (below critical) or high pressures. (10) Thermal energy of gases >> molecular attraction. (11) Gases undergo similar change with the change of temperature and pressure. In other words, gases obey certain laws known as gas laws. Measurable properties of gases ST (1) The characteristics of gases are described fully in terms of four parameters or measurable properties : (i) The volume, V, of the gas. (ii) Its pressure, P (iii) Its temperature, T (iv) The amount of the gas (i.e., mass or number of moles). (2) Volume : (i) Since gases occupy the entire space available to them, the measurement of volume of a gas only requires a measurement of the container confining the gas. (ii) Volume is expressed in litres (L), millilitres (mL) or cubic centimetres (cm 3 ) or cubic metres (m 3 ). (iii) 1L  1000 mL ; 1 mL  10 3 L ; 1 L  1 dm 3  10 3 m 3 moles of gas (n)  Mass in grams m  Molar mass M (4) Temperature : (i) Gases expand on increasing the temperature. If temperature is increased twice, the square of the velocity (v 2 ) also increases two times. (ii) Temperature is measured in centigrade degree ( o C) or celsius degree with the help of thermometers. Temperature is also measured in Fahrenheit (F ). o (iii) S.I. unit of temperature is kelvin (K) or absolute degree. K  o C  273 C F o  32  5 9 (5) Pressure : (i) Pressure of the gas is the force exerted by the gas per unit area of the walls of the container in all directions. Thus, Pressure Force( F) Mass(m )  Acceleration(a)  (P)  Area( A) Area(a) o (iv) Relation between F and o C is (ii) Pressure exerted by a gas is due to kinetic energy 1 (KE  mv 2 ) of the molecules. Kinetic energy of the gas molecules 2 increases, as the temperature is increased. Thus, Pressure of a gas  Temperature (T). (iii) Pressure of a pure gas is measured by manometer while that of a mixture of gases by barometer. (iv) Commonly two types of manometers are used, (a) Open end manometer; (b) Closed end manometer (v) The S.I. unit of pressure, the pascal (Pa), is defined as newton per metre square. It is very small unit. 1 1Pa  1 Nm 2  1 kg m 1 s 2 (vi) C.G.S. unit of pressure is dynes cm 2. (vii) M.K.S. unit of pressure is kgf / m 2. The unit kgf / cm 2 sometime called ata (atmosphere technical absolute). (viii) Higher unit of pressure bar, is KPa T1< T2< T3 PV T3 T2 T1 MPa. or (ix) Several other units used for pressure are, Symbol Value O O (4) At constantP mass and temperature densitylogof1/Va gas is directly bar bar 1bar  10 Pa atmosphere atm 1 atm  1.01325  10 5 Pa Torr Torr 101325 1 Torr  Pa  133.322 Pa 760 millimetre of mercury mm Hg 1 mm Hg  133.322 Pa 5 proportional to its pressure and inversely proportional to its volume. Thus, d  P  or mass    V  d    d1 P V  1  2 .......  K d 2 P2 V1 (5) At altitudes, as P is low d of air is less. That is why mountaineers carry oxygen cylinders. Charle's law ID (x) 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. (xi) When the pressure in a system is less than atmospheric pressure, the gauge pressure becomes negative, but is frequently designated and called vacuum. For example, 16 cm vacuum will be 1 V E3 Name log P 60 1 bar  10 5 Pa  10 5 Nm 2  100 KNm 2  100 KPa proportional to the absolute temperature ( o C  273) at constant pressure”. Thus, V  T at constant pressure and mass D YG U 76  16  1.013  0.80 bar. 76 (xii) If ‘h’ is the height of the fluid in a column or the difference in the heights of the fluid columns in the two limbs of the manometer, d is the density of the fluid (1) French chemist, Jacques Charles first studied variation of volume with temperature, in 1787. (2) It states that, “The volume of a given mass of a gas is directly or V  KT  K(t( o C)  273.15) , (where k is constant), (Hg  13.6  10 3 Kg / m 3  13.6 g / cm 3 ) and g is the gravity, then K pressure is given by, Pgas  Patm  h dg (xiii) Two sets of conditions are widely used as 'standard' values for reporting data. Condition S.T.P./N.T.P. S.A.T.P. * T 273.15 K 298.15 K V (Molar volume) 22.414 L 24.800 L P V V V or 1  2  K (For two or more gases) T T1 T2 (3) If t  0 o C , then V  V0 V0  K  273.15 hence, m 1 atm 1 bar U * Standard ambient temperature and pressure. Boyle's law ST (1) In 1662, Robert Boyle discovered the first of several relationships among gas variables (P, T, V). (2) It states that, “For a fixed amount of a gas at constant temperature, the gas volume is inversely proportional to the gas pressure.” Thus, P  1 / V at constant temperature and mass or P  K / V (where K is constant) or PV  K or P1 V1  P2 V2  K (For two or more gases) (3) Graphical representation of Boyle's law : Graph between P and V at constant temperature is called isotherm and is an equilateral (or rectangular) hyperbola. By plotting P versus 1 / V , this hyperbola can be converted to a straight line. Other types of isotherms are also shown below,  K V V0 273.15 V0 t   [t  273.15]  V0 1    V0 [1   v t] 273.15 273. 15   where  v is the volume coefficient, v  V  V0 1   3.661  10  3 o C 1 tV0 273.15 Thus, for every 1 o change in temperature, the volume of a gas changes by 1 1     of the volume at 0 o C. 273.15  273  (4) Graphical representation of Charle's law : Graph between V and T at constant pressure is called isobar or isoplestics and is always a straight line. A plot of V versus t( o C) at constant pressure is a straight line cutting T3 P O T1 P T3 T2 T1 V or 1/d the temperature axis at temperature. T2 T1< T2< T3 T1< T2< T3 O 1/V or d  273.15 o C. It is the lowest possible 1/d or V 1/d or V Avogadro's law 22.4 L mol–1 = V0 T(k) C C 0o (1) According to this law, “Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules.” t C) (o (5) At constant mass and pressure density of a gas is inversely proportional to it absolute temperature. Thus, d  1 1  T V mass    V  d    Gay-Lussac's law (Amonton's law) proportional to the absolute temperature ( o C  273) at constant volume.” (Avogadro's number  6.02  10 23 ) and by this law must occupy the same volume at a given temperature and pressure. The volume of one mole of a gas is called molar volume, V which is 22.4 L mol 1 at S.T.P. or N.T.P. m (3) This law can also express as, “The molar gas volume at a given temperature and pressure is a specific constant independent of the nature of the gas”. U (where K is constant) D YG P P P or 1  2  K (For two or more gases) T T1 T2 (3) If t  0 C , then P  P0 Hence, P0  K  273.15 K P0 273.15 Ideal gas equation (1) The simple gas laws relating gas volume to pressure, temperature and amount of gas, respectively, are stated below : Boyle's law : U where  P is the pressure coefficient, ST P  P0 1 P    3.661  10  3 o C 1 tP0 273.15 Thus, for every 1 o change in temperature, the pressure of a gas 1 1   changes by   of the pressure at 0 o C. 273.15  273  (4) This law fails at low temperatures, because the volume of the gas molecules be come significant. (5) Graphical representation of Gay-Lussac's law : A graph between P and T at constant V is called isochore. V1< V2< V3< V4 V1 O T(k) O T 1 1 or V  V P (n and T constant) (n and P constant) (T and P constant) If all the above law's combines, then V or V nT P nRT ( R  Ideal gas constant) P PV  nRT or This is called ideal gas equation. R is called ideal gas constant. This equation is obeyed by isothermal and adiabatic processes. (2) Nature and values of R : From the ideal gas equation, PV Pressure  Volume R  nT mole  Temperatur e Force  Volume Force  Length Area   mole  Temperatur e mole  Temperatur e  Work or energy. mole  Temperatur e R is expressed in the unit of work or energy mol 1 K 1. V2 V3 V4 P P Charle's law : VT Avogadro's law : V  n P0 t   P [t  273.15]  P0 1   P0 [1  t] 273.15 273.15   P 2 moles 2 volumes 2 litres 1 litre 1n litre Thus, Vm  specific constant  22.4 L mol 1 at S.T.P. or N.T.P. Thus, P  T at constant volume and mass  1 mole 1 volume 1 litre 1 / 2 litre 1 / 2 n litre ID (2) It states that, “The pressure of a given mass of a gas is directly or P  KT  K(t(o C)  273.15) 2 moles 2 volumes 2 litres 1 litre 1n litre (2) One mole of any gas contains the same number of molecules (1) In 1802, French chemist Joseph Gay-Lussac studied the variation of pressure with temperature and extende the Charle’s law so, this law is also called Charle’s-Gay Lussac’s law. o V1 V2  .......  K n1 n 2 Example, 2 H 2 (g) O 2 (g)  2 H 2 O(g) (6) Use of hot air balloons in sports and meteorological observations is an application of Charle's law. K or V  Kn (where K is constant) or d1 T2 V2   ......  K d 2 T1 V1 or Thus, V  n (at constant T and P) 60 –273.15o E3 O Since different values of R are summarised below : R  0.0821 L atm mol 1 K 1 Thus, Ptotal  P1  P2  P3 .........  8.3143 joulemol 1 K 1 (S.I. unit) Where P1 , P2 , P3 ,...... are partial pressures of gas number 1, 2, 3  8.3143 Nm mol 1 K 1.........  8.3143 KPa dm 3 mol 1 K 1 K (2) Partial pressure is the pressure exerted by a gas when it is present alone in the same container and at the same temperature. Partial pressure of a gas 1  5.189  10 19 eV mol 1 K 1  1.99 cal mol 1 K (P1 )  1 (3) Gas constant, R for a single molecule is called Boltzmann constant (k) R 8.314  10 7  ergs mole 1 degree 1 N 6.023  10 23 k or 1.38  10 joulemol 1 degree 1 m RT M d d T d T dT Thus, or 1 1  2 2 = Constant P1 T2 P (For two or more different temperature and pressure) (5) Gas densities differ from those of solids and liquids as, (i) Gas densities are generally stated in g/L instead of g / cm 3. ST U (ii) Gas densities are strongly dependent on pressure and temperature as, d  P  1 / T Densities of liquids and solids, do depend somewhat on temperature, but they are far less dependent on pressure. (iii) The density of a gas is directly proportional to its molar mass. No simple relationship exists between the density and molar mass for liquid and solids. (iv) Density of a gas at STP  d ( N 2 ) at STP  (X 1 ) in a mixture of gas Partial pressure of a gas (P1 ) PTotal ID D YG dT M M  Constant  , P R R (  M and R are constant for a particular gas) or RT ( n  n1  n 2  n3.....) V (4) Applications : This law is used in the calculation of following relationships, (ii) % of a gas in mixture  m   d   V  ( PV  nRT ) or  n  mRT PV PM RT RT V Partial pressure of a gas (P1 )  100 PTotal (iii) Pressure of dry gas collected over water : When a gas is collected over water, it becomes moist due to water vapour which exerts its own partial pressure at the same temperature of the gas. This partial perssure of water vapours is called aqueous tension. Thus, Pdry gas  Pmoist gas or PTotal  Pwater vapour U M P1 V1  P2 V2  P3 V3..... V (i) Mole fraction of a gas   mass of the gas (m )  n   Molecular weight of the gas (M )    P1 , P2 , P3........ are mixed together in container of volume V, then, or  (n1  n 2  n 3.....) (4) Calculation of mass, molecular weight and density of the gas by gas equation PV  nRT  (3) If a number of gases having volume V1 , V2 , V3...... at pressure PTotal   1.38  10 16 ergsmol 1 degree 1 23 Number of moles of the gas (n1 )  PTotal  Mole fraction ( X 1 )  PTotal Total number of moles (n) in the mixture 60 1 E3  8.3143 MPacm mol 3 molar mass 22.4 28  1.25 g L1 , 22.4 32  1.43 g L1 d (O 2 ) at STP  22.4 Dalton's law of partial pressures (1) According to this law, “When two or more gases, which do not react chemically are kept in a closed vessel, the total pressure exerted by the mixture is equal to the sum of the partial pressures of individual gases.” or Pdry gas  Pmoist gas  Aqueous tension (Aqueous tension is directly proportional to absolute temperature) (iv) Relative humidity (RH) at a given temperature is given by, RH  Partial pressure of water in air. Vapour pre ssure of water (5) Limitations : This law is applicable only when the component gases in the mixture do not react with each other. For example, N 2 and O 2 , CO and CO 2 , N 2 and Cl 2 , CO and N 2 etc. But this law is not applicable to gases which combine chemically. For example, H 2 and Cl 2 , CO and Cl 2 , NH 3 , HBr and HCl, NO and O 2 etc. (6) Another law, which is really equivalent to the law of partial pressures and related to the partial volumes of gases is known as Law of partial volumes given by Amagat. According to this law, “When two or more gases, which do not react chemically are kept in a closed vessel, the total volume exerted by the mixture is equal to the sum of the partial volumes of individual gases.” Thus, VTotal  V1  V2  V3 ...... Where V1 , V2 , V3 ,...... are partial volumes of gas number 1, 2, 3..... Graham's law of diffusion and Effusion (1) Diffusion is the process of spontaneous spreading and intermixing of gases to form homogenous mixture irrespective of force of gravity. While Effusion is the escape of gas molecules through a tiny hole such as pinhole in a balloon.  Diffusion into a vacuum will take place much more rapidly than diffusion into another place.  Both the rate of diffusion of a gas and its rate of effusion depend on its molar mass. Lighter gases diffuses faster than heavier gases. The gas with highest rate of diffusion is hydrogen. (2) According to this law, “At constant pressure and temperature, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its vapour density.” 1 Thus, rate of diffusion (r)  (T and P constant) d For two or more gases at constant pressure and temperature, d2  2  d1  2 (vi) The pressure of a gas is due to the continuous bombardment on the walls of the containing vessel. M2 M1 (vii) At constant temperature the average K.E. of all gases is same. r1 V /t w /t  1 1  1 1  r2 V2 / t 2 w 2 / t 2 Volume of a gas diffused (r)  Time taken for diffusion d2 d1 PV  (a) When equal volume of the two gases diffuse, i.e. V1  V2 r1 t 2   r2 t1 d2 d1 D YG then, (b) When volumes of the two gases diffuse in the same time, i.e. t1  t 2 then, r1 V  1  r2 V2 d2 d1 (iii) Since, r  p (when p is not constant)  P   r    M   (4) Rate of diffusion and effusion can be determined as, (i) Rate of diffusion is equal to distance travelled by gas per unit time through a tube of uniform cross-section. (ii) Number of moles effusing per unit time is also called rate of diffusion. (iii) Decrease in pressure of a cylinder per unit time is called rate of effusion of gas. (iv) The volume of gas effused through a given surface per unit time is also called rate of effusion. (5) Applications : Graham's law has been used as follows, (i) To determine vapour densities and molecular weights of gases. (ii) To prepare Ausell’s marsh gas indicator, used in mines. (iii) Atmolysis : The process of separation of two gases on the basis of their different rates of diffusion due to difference in their densities is called atmolysis. It has been applied with success for the separation of isotopes and other gaseous mixtures. r1 P  1 r2 P2 M2 M1 ST U then, Kinetic theory of gases (3) Kinetic gas equation : On the basis of above postulates, the following gas equation was derived, U then, (viii) The average K.E. of the gas molecules is directly proportional to the absolute temperature. ID where, M 1 and M 2 are the molecular weights of the two gases. (ii) Since, rate of diffusion (iii) Gaseous molecules are perfectly elastic so that there is no net loss of kinetic energy due to their collisions. (v) Gaseous molecules are considered as point masses because they do not posses potential energy. So the attractive and repulsive forces between the gas molecules are negligible. (3) Graham's law can be modified in a number of ways as, (i) Since, 2  vapour density (V.D.) = Molecular weight d2  d1 (ii) The volume of the individual molecule is negligible as compared to the total volume of the gas. (iv) The effect of gravity on the motion of the molecules is negligible. d2 d1 r then, 1  r2 (i) Every gas consists of a large number of small particles called molecules moving with very high velocities in all possible directions. E3 r1  r2 (1) Kinetic theory was developed by Bernoulli, Joule, Clausius, Maxwell and Boltzmann etc. and represents dynamic particle or microscopic model for different gases since it throws light on the behaviour of the particles (atoms and molecules) which constitute the gases and cannot be seen. Properties of gases which we studied earlier are part of macroscopic model. (2) Postulates 60  All gases spontaneously diffuse into one another when they are brought into contact. 1 2 mnu rms 3 where, P = pressure exerted by the gas V = volume of the gas m = average mass of each molecule n = number of molecules u = root mean square (RMS) velocity of the gas. (4) Calculation of kinetic energy We know that, K.E. of one molecule  1 mu 2 2 K.E. of n molecules  1 3 1 mnu 2  PV ( PV  mnu 2 ) 2 2 3 n = 1, Then K.E. of 1 mole gas  3 RT 2 ( PV  RT )  3  8.314  T  12.47 T Joules. 2  Average K.E.per mole 3 RT 3   KT N (Avogadro number ) 2 N 2 R    Boltzmann constant  K  N   This equation shows that K.E. of translation of a gas depends only on the absolute temperature. This is known as Maxwell generalisation. Thus average K.E.  T. If T  0 K (i.e.,  273.15 o C) then, average K.E. = 0. Thus, absolute zero (0K) is the temperature at which molecular motion ceases. Molecular collisions (1) The closest distance between the centres of two molecules taking part in a collision is called molecular or collision diameter (). The molecular diameter of all the gases is nearly same lying in the order of 10 8 cm. (2) According to Maxwell, at a particular temperature the distribution of speeds remains constant and this distribution is referred to as the Maxwell-Boltzmann distribution and given by the following expression, dn0  M   4   n  2RT  where n is the number of molecules per unit molar volume, Avogadro number( N 0 ) 6.02  10 23 3  m 0.0224 Vm (ii) The total number of bimolecular collision per unit time are given 1 by, Z AA   2 u av. n 2 2 (iii) If the collisions involve two unlike molecules, the number of bimolecular collision are given by,  A  B (c) At particular volume; Z  T 1 / 2 (3) During molecular collisions a molecule covers a small distance before it gets deflected. The average distance travelled by the gas molecules between two successive collision is called mean free path (). Average distance travelled per unit time( u av ).  No. of collisionsmade by singlemolecule per unit time (Z A )  1 2n  2 T (4) Based on kinetic theory of gases mean free path,  . Thus, P (i) Larger the size of the molecules, smaller the mean free path, i.e., 1 U  2 2 u avr.n 300 K (T1) T1 SO > PCl is order of rate of diffusion.  Vapour density is independent of temperature and has no unit 2 2 3 3 while absolute density is dependent of temperature and has unit of gm –1  The isotherms of CO were first studied by Andrews.  1 Cal = 4.2 Joule, 1 Kcal = 4200 Joule 2  The gas which has least mean free path has maximum value of a, is easily liquefied and has maximum value of T. b  T