Gaseous State PDF

10 C H A P T E R Gaseous State C O N T E N T S THE GASEOUS STATE GENERAL CHARCTERISTICS OF GASES PARAMETERS OF A GAS GAS LAWS BOYLE’S LAW CHARLES’S LAW THE COMBINED GAS LAW GAY LUSSAC’S LAW AVOGADRO’S LAW THE IDEAL-GAS EQUATION KINETIC MOLECULAR THEORY OF GASES DE...

10 C H A P T E R Gaseous State C O N T E N T S THE GASEOUS STATE GENERAL CHARCTERISTICS OF GASES PARAMETERS OF A GAS GAS LAWS BOYLE’S LAW CHARLES’S LAW THE COMBINED GAS LAW GAY LUSSAC’S LAW AVOGADRO’S LAW THE IDEAL-GAS EQUATION KINETIC MOLECULAR THEORY OF GASES DERIVATION OF KINETIC GAS A ll matter exists in three states: gas, liquid and solid. A EQUATION molecular level representation of gaseous, liquid and DISTRIBUTION OF MOLECULAR solid states is shown in Fig. 10.1 VELOCITIES A gas consists of molecules separated wide apart in empty DIFFERENT KINDS OF space. The molecules are free to move about throughout the VELOCITIES container. CALCULATION OF MOLECULAR A liquid has molecules touching each other. However, the VELOCITIES intermolecular space, permit the movement of molecules throughout the liquid. COLLISION PROPERTIES A solid has molecules, atoms or ions arranged in a certain EXPLANATION OF DEVIATIONS : order in fixed positions in the crystal lattice. The particles in a VAN DER WAALS EQUATION solid are not free to move about but vibrate in their fixed positions. VAN DER WAALS EQUATION Of the three states of matter, the gaseous state is the one most studied and best understood. We shall consider it first. LIQUEFACTION OF GASES : CRITICAL PHENOMENON GENERAL CHARACTERISTICS OF GASES LAW OF CORRESPONDING 1. Expansibility STATES Gases have limitless expansibility. They expand to fill the entire vessel they are placed in. METHODS OF LIQUEFACTION OF GASES 2. Compressibility Gases are easily compressed by application of pressure to a movable piston fitted in the container. 355 356 10 PHYSICAL CHEMISTRY Figure 10.1 Molecular representation of the gaseous, liquid and solid states. 3. Diffusibility Gases can diffuse rapidly through each other to form a homogeneous mixture. 4. Pressure Gases exert pressure on the walls of the container in all directions. 5. Effect of Heat When a gas, confined in a vessel is heated, its pressure increases. Upon heating in a vessel fitted with a piston, volume of the gas increases. The above properties of gases can be easily explained by the Kinetic Molecular Theory which will be considered later in the chapter. PARAMETERS OF A GAS A gas sample can be described in terms of four parameters (measurable properties): (1) the volume, V of the gas (2) its pressure, P (3) its temperature, T (4) the number of moles, n, of gas in the container  GASEOUS STATE 357 The Volume, V The volume of the container is the volume of the gas sample. It is usually given in litre (l or L) or millilitres (ml or mL). 1 1itre(l) = 1000 ml and 1 ml = 10–3 l One millilitre is practically equal to one cubic centimetre (cc). Actually 1 litre(l) = 1000.028 cc The SI unit for volume is cubic metre (m3) and the smaller unit is decimeter3 (dm3). The pressure of a gas is defined as the force exerted by the impacts of its molecules per unit surface area in contact. The pressure of a gas sample can be measured with the help of a mercury manometer (Fig. 10.2) Similarly, the atmospheric pressure can be determined with a mercury barometer (Fig. 10.3). Vacuum Vacuum h mm 760 mm Hg 1 Atm pressure GAS Pressure = h mm Hg Figure 10.2 Figure 10.3 A mercury manometer. A mercury barometer. The gas container is connected with a U-tube A long tube (80 × 1 cm) filled with Hg inverted into containing Hg having vacuum in the closed end. dish containing Hg. The atmospheric pressure is The difference in Hg height in two limbs gives the equal to 760 mm Hg column supported by it at sea gas pressure in mm Hg. level. The pressure of air that can support 760 mm Hg column at sea level, is called one atmosphere (1 atm). The unit of pressure, millimetre of mercury, is also called torr. Thus, 1 atm = 760 mm Hg = 760 torr The SI unit of pressure is the Pascal (Pa). The relation between atmosphere, torr and pascal is : 1 atm = 760 torr = 1.013 × 105 Pa The unit of pressure ‘Pascal” is not in common use. Temperature, T The temperature of a gas may be measured in Centigrade degrees (°C) or Celsius degrees. The SI unit of temperature is Kelvin (K) or Absolute degree. The centigrade degrees can be converted to kelvins by using the equation. K = °C + 273 The Kelvin temperature (or absolute temperature) is always used in calculations of other parameters of gases. Remember that the degree sign (°) is not used with K. 358 10 PHYSICAL CHEMISTRY The Moles of a Gas Sample, n The number of moles, n, of a sample of a gas in a container can be found by dividing the mass, m, of the sample by the molar mass, M (molecular mass). mass of gas sample (m) moles of gas (n) = molecular mass of gas (M ) THE GAS LAWS The volume of a given sample of gas depends on the temperature and pressure applied to it. Any change in temperature or pressure will affect the volume of the gas. As results of experimental studies from 17th to 19th century, scientists derived the relationships among the pressure, temperature and volume of a given mass of gas. These relationships, which describe the general behaviour of gases, are called the gas laws. BOYLE’S LAW In 1660 Robert Boyle found out experimentally the change in volume of a given sample of gas with pressure at room temperature. From his observations he formulated a generalisation known as Boyle’s Law. It states that : at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. If the pressure is doubled, the volume is halved. Figure 10.4 Boyle’s law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. If the pressure is doubled, the volume is halved. The Boyle’s Law may be expressed mathematically as V ∝ 1/P (T, n are constant) or V = k × 1/P where k is a proportionality constant. PV = k If P1, V1 are the initial pressure and volume of a given sample of gas and P2, V2 the changed  GASEOUS STATE 359 pressure and volume, we can write P1 V1 = k = P2 V2 or P1 V1 = P2 V2 This relationship is useful for the determination of the volume of a gas at any pressure, if its volume at any other pressure is known. V (litre) V (litre) P (mm Hg) 1/P (mm Hg) Figure 10.5 Graphical representation of Boyle's law. (a) a plot of V versus P for a gas sample is hyperbola; (b) a plot of V versus 1/P is a straight line. The Boyle’s law can be demonstrated by adding liquid mercury to the open end of a J-tube. As the pressure is increased by addition of mercury, the volume of the sample of trapped gas decreases. Gas pressure and volume are inversely related; one increases when the other decreases. Figure 10.6 Demonstration of Boyle’s law. CHARLES’S LAW In 1787 Jacques Charles investigated the effect of change of temperature on the volume of a fixed amount of gas at constant pressure. He established a generalisation which is called the Charles’ Law. It states that : at constant pressure, the volume of a fixed mass of gas is directly proportional to the Kelvin temperature of absolute temperature. If the absolute temperature is doubled, the volume is doubled. Charles’ Law may be expressed mathematically as V ∝T (P, n are constant) or V =kT 360 10 PHYSICAL CHEMISTRY where k is a constant. V or =k T If V1, T1 are the initial volume and temperature of a given mass of gas at constant pressure and V2, T2 be the new values, we can write V1 V =k = 2 T1 T2 V1 V2 or = T1 T2 Using this expression, the new volume V2, can be found from the experimental values of V1, T1 and T2. Figure 10.7 Charles law state that at constant pressure, the volume of a fixed mass of gas is directly proportional to the absolute temperature. 500 400 Volume (ml) 300 200 100 0 0 100 200 300 400 500 In Kelvin (K) o 273 173 –73 27 127 227 In Celsius ( C) Figure 10.8 Graph showing that at constant pressure, volume of a given mass of gas is directly proportional to the Kelvin temperature.  GASEOUS STATE 361 THE COMBINED GAS LAW Boyle’s Law and Charles’ Law can be combined into a single relationship called the Combined Gas Law. 1 Boyle’s Law V ∝ (T, n constant) P Charles’ Law V ∝T (P, n constant) T Therefore, V ∝ (n constant) P The combined law can be stated as : for a fixed mass of gas, the volume is directly proportional to kelvin temperature and inversely proportional to the pressure. If k be the proportionality constant, kT V = (n constant) P PV or =k (n constant) T If the pressure, volume and temperature of a gas be changed from P1, V1 and T1 to P2, T2 and V2, then P1 V1 P2 V2 =k =k T1 T2 P1 V1 P2 V2 or = T1 T2 This is the form of combined law for two sets of conditions. It can be used to solve problems involving a change in the three variables P, V and T for a fixed mass of gas. SOLVED PROBLEM. 25.8 litre of a gas has a pressure of 690 torr and a temperature of 17°C. What will be the volume if the pressure is changed to 1.85 atm and the temperature to 345 K. SOLUTION Initial conditions : Final Conditions : V1 = 25.8 litres V2 = ? 690 P1 = = 0.908 atm P2 = 1.85 atm 760 T1 = 17 + 273 = 290 K T2 = 345 K Substituting values in the equation P1 V1 P2 V2 = T1 T2 0.908 atm × 25.8 litre 1.85 atm × V2 = 290 K 345 K 0.908 × 25.8 × 345 Hence, V2 = = 15.1 litres 290 × 1.85 GAY LUSSAC’S LAW In 1802 Joseph Gay Lussac as a result of his experiments established a general relation between the pressure and temperature of a gas. This is known as Gay Lussac’s Law or Pressure-Temperature Law. It states that : at constant volume, the pressure of a fixed mass of gas is directly proportional to the Kelvin temperature or absolute temperature. 362 10 PHYSICAL CHEMISTRY The law may be expressed mathematically as P ∝T (Volume, n are constant) or P = kT P or =k T For different conditions of pressure and temperature P1 P =k = 2 T1 T2 P1 P2 or = T1 T2 Knowing P1, T1, and T2, P2 can be calculated. AVOGADRO’S LAW Let us take a balloon containing a certain mass of gas. If we add to it more mass of gas, holding the temperature (T) and pressure (P) constant, the volume of gas (V) will increase. It was found experimentally that the amount of gas in moles is proportional to the volume. That is, V ∝n (T and P constant) or V = An where A is constant of proportionality. V or = A n For any two gases with volumes V1, V2 and moles n1, n2 at constant T and P, V1 V = A= 2 n1 n2 If V1 = V2, n1 = n2 Thus for equal volumes of the two gases at fixed T and P, number of moles is also equal. This is the basis of Avogadro’s Law which may be stated as : equal volumes of gases at the same temperature and pressure contain equal number of moles or molecules. If the molar amount is doubled, the volume is doubled. Figure 10.9 Avogadro’s law states that under equal conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules.  GASEOUS STATE 363 The Molar Gas Volume. It follows as a corollary of Avogadro’s Law that one mole of any gas at a given temperature (T) and pressure (P) has the same fixed volume. It is called the molar gas volume or molar volume. In order to compare the molar volumes of gases, chemists use a fixed reference temperature and pressure. This is called standard temperature and pressure (abbreviated, STP). The standard temperature used is 273 K (0°C) and the standard pressure is 1 atm (760 mm Hg). At STP we find experimentally that one mole of any gas occupies a volume of 22.4 litres. To put it in the form of an equation, we have 1 mole of a gas at STP = 22.4 litres THE IDEAL GAS EQUATION We have studied three simple gas laws : 1 Boyle’s Law V ∝ (T, n constant) P Charles’ Law V ∝T (n, P constant) Avogadro’s Law V ∝n (P, T constant) These three laws can be combined into a single more general gas law : nT V ∝...(1) P This is called the Universal Gas Law. It is also called Ideal Gas Law as it applies to all gases which exhibit ideal behaviour i.e., obey the gas laws perfectly. The ideal gas law may be stated as : the volume of a given amount of gas is directly proportional to the number of moles of gas, directly proportional to the temperature, and inversely proportional to the pressure. Introducing the proportionality constant R in the expression (1) we can write nT V =R P or P V = nRT...(2) The equation (2) is called the Ideal-gas Equation or simply the general Gas Equation. The constant R is called the Gas constant. The ideal gas equation holds fairly accurately for all gases at low pressures. For one mole (n = 1) of a gas, the ideal-gas equation is reduced to PV = RT...(3) The ideal-gas equation is called an Equation of State for a gas because it contains all the variables (T, P, V and n) which describe completely the condition or state of any gas sample. If we know the three of these variables, it is enough to specify the system completely because the fourth variable can be calculated from the ideal-gas equation. The Numerical Value of R. From the ideal-gas equation, we can write PV R=...(1) nT We know that one mole of any gas at STP occupies a volume of 22.4 litres. Substituting the values in the expression (1), we have 1 atm × 22.4 litres R= 1 mole × 273 K = 0.0821 atm. litre mol–1 K–1 It may be noted that the unit for R is complex; it is a composite of all the units used in calculating the constant. 364 10 PHYSICAL CHEMISTRY If the pressure is written as force per unit area and volume as area times length, from (1) (force/area) × area × length force × length R= = n×T n×T work = nT Hence R can be expressed in units of work or energy per degree per mole. The actual value of R depends on the units of P and V used in calculating it. The more important values of R are listed in Table 10.1. TABLE 10.1. VALUE OF R IN DIFFERENT UNITS 0.0821 litre-atm K–1 mol–1 8.314 × 107 erg K–1 mol–1 82.1 ml-atm K–1 mol–1 8.314 Joule K–1 mol–1 62.3 litre-mm Hg K–1 mol–1 1.987 cal K–1 mol–1 DALTON’S LAW OF PARTIAL PRESSURES John Dalton visualised that in a mixture of gases, each component gas exerted a pressure as if it were alone in the container. The individual pressure of each gas in the mixture is defined as its Partial Pressure. Based on experimental evidence, in 1807, Dalton enunciated what is commonly known as the Dalton’s Law of Partial Pressures. It states that : the total pressure of a mixture of gases is equal to the sum of the partial pressures of all the gases present (Fig. 10.10). Figure 10.10 Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures exerted by each gas. The pressure of the mixture of O2 and N2 (Tanks) is the sum of the pressures in O2 and N2 tanks. 364 10 PHYSICAL CHEMISTRY If the pressure is written as force per unit area and volume as area times length, from (1) (force/area) × area × length force × length R= = n×T n×T work = nT Hence R can be expressed in units of work or energy per degree per mole. The actual value of R depends on the units of P and V used in calculating it. The more important values of R are listed in Table 10.1. TABLE 10.1. VALUE OF R IN DIFFERENT UNITS 0.0821 litre-atm K–1 mol–1 8.314 × 107 erg K–1 mol–1 82.1 ml-atm K–1 mol–1 8.314 Joule K–1 mol–1 62.3 litre-mm Hg K–1 mol–1 1.987 cal K–1 mol–1 DALTON’S LAW OF PARTIAL PRESSURES John Dalton visualised that in a mixture of gases, each component gas exerted a pressure as if it were alone in the container. The individual pressure of each gas in the mixture is defined as its Partial Pressure. Based on experimental evidence, in 1807, Dalton enunciated what is commonly known as the Dalton’s Law of Partial Pressures. It states that : the total pressure of a mixture of gases is equal to the sum of the partial pressures of all the gases present (Fig. 10.10). Figure 10.10 Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures exerted by each gas. The pressure of the mixture of O2 and N2 (Tanks) is the sum of the pressures in O2 and N2 tanks.  GASEOUS STATE 365 Mathematically the law can be expressed as Ptotal = P1 + P2 + P3... (V and T are constant) where P1, P2 and P3 are partial pressures of the three gases 1, 2 and 3; and so on. Dalton’s Law of Partial Pressures follows by application of the ideal-gas equation PV = n RT separately to each gas of the mixture. Thus we can write the partial pressures P1, P2 and P3 of the three gases ⎛ RT ⎞ ⎛ RT ⎞ ⎛ RT ⎞ P1 = n1 ⎜ ⎟ P2 = n2 ⎜ ⎟ P3 = n3 ⎜ ⎟ ⎝ V ⎠ ⎝ V ⎠ ⎝ V ⎠ where n1, n2 and n3 are moles of gases 1, 2 and 3. The total pressure, Pt, of the mixture is RT Pt = (n1 + n2 + n3 ) V RT or Pt = nt V In the words, the total pressure of the mixture is determined by the total number of moles present whether of just one gas or a mixture of gases. SOLVED PROBLEM 1. What pressure is exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10 litre vessel ? SOLUTION Applying the ideal-gas equation RT P=n V we can find the partial pressure of H2 and N2 2.00 Moles of H2 = = 0.990 2.02 8.00 Moles of N2 = = 0.286 28 0.990 mole × 0.0821 atm. litre K –1 mol –1 × 273 K ∴ PH 2 = 10.0 litre = 2.20 atm. 0.286 mole × 0.0821 atm. litre K –1 mol –1 × 273 K and PN 2 = 10.0 litre = 0.64 atm Thus Ptotal = PH 2 + PN 2 = 2.20 atm + 0.64 atm = 2.84 atm Thus the pressure exerted by the mixture of H2 and N2 is 2.84 atm. SOLVED PROBLEM 2. A sample of oxygen is collected by the downward displacement of water from an inverted bottle. The water level inside the bottle is equalised with that in the trough. Barometeric pressure is found to be 757 mm Hg, and the temperature of water is 23.0°C. What is the partial pressure of O2 ? Vapour pressure of H2O at 23°C = 19.8 mm Hg. SOLUTION The total pressure inside the bottle is 366 10 PHYSICAL CHEMISTRY Ptotal = PO2 + PH 2O Since the water levels inside and outside the bottle were equalised, the total gas pressure inside the bottles must be equal to Patm. ∴ Ptotal = Patm = PO2 + PH 2O But Patm is given as 757 mm Hg ∴ PO2 = 757 mm Hg – PH 2O = 757 mm Hg – 19.8 mm Hg = 737.2 mm Hg Thus the partial pressure of O2 is 737.2 mm Hg. GRAHAM’S LAW OF DIFFUSION When two gases are placed in contact, they mix spontaneously. This is due to the movement of molecules of one gas into the other gas. This process of mixing of gases by random motion of the molecules is called Diffusion. Thomas Graham observed that molecules with smaller masses diffused faster than heavy molecules. Figure 10.11 A light molecule diffuses quicker than a heavy molecule. In 1829 Graham formulated what is now known as Graham’s Law of Diffusion. It states that : 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. Mathematically the law can be expressed as r1 M2 = r2 M1 where r1 and r2 are the rates of diffusion of gases 1 and 2, while M1 and M2 are their molecular masses. When a gas escapes through a pin-hole into a region of low pressure of vacuum, the process is called Effusion. The rate of effusion of a gas also depends, on the molecular mass of the gas. (a) Diffusion (b) Effusion Figure 10.12 (a) Diffusion is mixing of gas molecules by random motion under conditions where molecular collisions occur. (b) Effusion is escape of a gas through a pinhole without molecular collisions.  GASEOUS STATE 367 Dalton’s law when applied to effusion of a gas is called the Dalton’s Law of Effusion. It may be expressed mathematically as Effusion rate of Gas 1 M2 = (P, T constant) Effusion rate of Gas 2 M1 The determination of rate of effusion is much easier compared to the rate of diffusion. Therefore, Dalton’s law of effusion is often used to find the molecular mass of a given gas. SOLVED PROBLEM 1. If a gas diffuses at a rate of one-half as fast as O2, find the molecular mass of the gas. SOLUTION Applying Graham’s Law of Diffusion, r1 M2 = r2 M1 1 32 2 = 1 M1 Squaring both sides of the equation. 2 ⎛1⎞ 32 1 32 ⎜ ⎟ = or = ⎝2⎠ M1 4 M1 Hence, M1 = 128 Thus the molecular mass of the unknown gas is 128. SOLVED PROBLEM 2. 50 ml of gas A effuse through a pin-hole in 146 seconds. The same volume of CO2 under identical conditions effuses in 115 seconds. Calculate the molecular mass of A. SOLUTION Effusion rate of CO 2 MA = Effusion rate of A M CO2 50 /115 MA = 50 /146 44 MA or (1.27) 2 = 44 Hence MA = 71 ∴ Molecular mass of A is 71. KINETIC MOLECULAR THEORY OF GASES Maxwell and Boltzmann (1859) developed a mathematical theory to explain the behaviour of gases and the gas laws. It is based on the fundamental concept that a gas is made of a large number of molecules in perpetual motion. Hence the theory is called the kinetic molecular theory or simply the kinetic theory of gases (The word kinetic implies motion). The kinetic theory makes the following assumptions. Assumptions of the Kinetic Molecular Theory (1) A gas consists of extremely small discrete particles called molecules dispersed throughout 368 10 PHYSICAL CHEMISTRY the container. The actual volume of the molecules is negligible compared to the total volume of the gas. The molecules of a given gas are identical and have the same mass (m). Actual volume of gas molecules Figure 10.13 Figure 10.14 A gas is made of molecules dispersed Actual volume of the gas in space in the container. molecules is negligible. Molecular collision Collision with wall Figure 10.15 Figure 10.16 Gas molecules are in constant Molecules move in straight line and motion in all possible directions. change direction on collision with another molecule or wall of container. (2) Gas molecules are in constant random motion with high velocities. They move in straight lines with uniform velocity and change direction on collision with other molecules or the walls of the container. Pool table analogy is shown in Fig.10.17. Figure 10.17 Gas molecules can be compared to billiard balls in random motion, bouncing off each other and off the sides of the pool table.  GASEOUS STATE 369 (3) The distance between the molecules are very large and it is assumed that van der Waals attractive forces between them do not exist. Thus the gas molecules can move freely, independent of each other. (4) All collisions are perfectly elastic. Hence, there is no loss of the kinetic energy of a molecule during a collision. (5) The pressure of a gas is caused by the hits recorded by molecules on the walls of the container. ⎛1 2⎞ (6) The average kinetic energy ⎜ mv ⎟ of the gas molecules is directly proportional to absolute ⎝2 ⎠ temperature (Kelvin temperature). This implies that the average kinetic energy of molecules is the same at a given temperature. How Does an Ideal Gas Differ from Real Gases ? A gas that confirms to the assumptions of the kinetic theory of gases is called an ideal gas. It obeys the basic laws strictly under all conditions of temperature and pressure. The real gases as hydrogen, oxygen, nitrogen etc., are opposed to the assumptions (1), (2) and (3) stated above. Thus : (a) The actual volume of molecules in an ideal gas is negligible, while in a real gas it is appreciable. (b) There are no attractive forces between molecules in an ideal gas while these exist in a real gas. (c) Molecular collisions in an ideal gas are perfectly elastic while it is not so in a real gas. For the reasons listed above, real gases obey the gas laws under moderate conditions of temperature and pressure. At very low temperature and very high pressure, the clauses (1), (2) and (3) of kinetic theory do not hold. Therefore, under these conditions the real gases show considerable deviations from the ideal gas behaviour. DERIVATION OF KINETIC GAS EQUATION Starting from the postulates of the kinetic molecular theory of gases we can develop an important equation. This equation expresses PV of a gas in terms of the number of molecules, molecular mass and molecular velocity. This equation which we shall name as the Kinetic Gas Equation may be derived by the following clauses. Let us consider a certain mass of gas enclosed in a cubic box (Fig. 10.18) at a fixed temperature. Suppose that : the length of each side of the box = l cm the total number of gas molecules =n the mass of one molecule =m the velocity of a molecule =v The kinetic gas equation may be derived by the following steps : (1) Resolution of Velocity v of a Single Molecule Along X, Y and Z Axes According to the kinetic theory, a molecule of a gas can move with velocity v in any direction. Velocity is a vector quantity and can be resolved into the components vx, vy, vz along the X, Y and Z axes. These components are related to the velocity v by the following expression. v 2 = v x2 + v 2y + v z2 Now we can consider the motion of a single molecule moving with the component velocities independently in each direction. 370 10 PHYSICAL CHEMISTRY (2) The Number of Collisions Per Second on Face A Due to One Molecule Consider a molecule moving in OX direction between opposite faces A and B. It will strike the face A with velocity vx and rebound with velocity – vx. To hit the same face again, the molecule must travel l cm to collide with the opposite face B and then again l cm to return to face A. Therefore, 2l the time between two collisions of face Av = seconds vx vx the number of collisions per second on face A = 2l 1 cm is Ax Y- X-Axis Vz Vx V B A Z-Axis Vx Vx Vy Figure 10.18 Figure 10.19 Resolution of velocity v into Cubic box showing molecular components Vx , Vy and Vz. collisions along X axis. (3) The Total Change of Momentum on All Faces of the Box Due to One Molecule Only Each impact of the molecule on the face A causes a change of momentum (mass × velocity) : the momentum before the impact = mvx the momentum after the impact = m (– vx) ∴ the change of momentum = mvx – (– mvx) = 2 mvx vx But the number of collisions per second on face A due to one molecule = 2l Therefore, the total change of momentum per second on face A caused by one molecule 2 ⎛ v ⎞ m vx = 2m v x × ⎜ x ⎟ = ⎝ 2l ⎠ l The change of momentum on both the opposite faces A and B along X-axis would be double i.e., 2 2mvx2 / l similarly, the change of momentum along Y-axis and Z-axis will be 2mv y / l and 2mvz2 / l respectively. Hence, the overall change of momentum per second on all faces of the box will be 2 2mvx2 2 mv y 2mvz2 = + + l l l 2m 2 = (vx + v y2 + vz2 ) l 2m v 2 = ( v 2 = v x2 + v y2 + v z2 ) l  GASEOUS STATE 371 (4) Total Change of Momentum Due to Impacts of All the Molecules on All Faces of the Box Suppose there are N molecules in the box each of which is moving with a different velocity v1, v2, v3, etc. The total change of momentum due to impacts of all the molecules on all faces of the box 2m 2 = (v1 + v22 + v32 +...) l Multiplying and dividing by n, we have 2mN ⎛ v12 + v22 + v32 +... ⎞ = ⎜ ⎟ l ⎝ n ⎠ 2mN u 2 = l where u2 is the mean square velocity. (5) Calculation of Pressure from Change of Momentum; Derivation of Kinetic Gas Equation Since force may be defined as the change in momentum per second, we can write 2mN u 2 Force = l Total Force But Pressure = Total Area 2mNu 2 1 1 mNu 2 P= × 2 = l 6l 3 l3 Since l3 is the volume of the cube, V, we have 1 mNu 2 P= 3 V 1 or P V = mNu 2 3 This is the fundamental equation of the kinetic molecular theory of gases. It is called the Kinetic Gas equation. This equation although derived for a cubical vessel, is equally valid for a vessel of any shape. The available volume in the vessel could well be considered as made up of a large number of infinitesimally small cubes for each of which the equation holds. Significance of the term u. As stated in clause (4) u2 is the mean of the squares of the individual velocities of all the N molecules of the gas. But u = u 2. Therefore u is called the Root Mean Square (or RMS) Velocity. KINETIC GAS EQUATION IN TERMS OF KINETIC ENERGY If N be the number of molecules in a given mass of gas, 1 P V = mNu 2 (Kinetic Gas equation) 3 2 1 = N × mu 2 3 2 2 = N ×e 3 where e is the average kinetic energy of a single molecule. 2 2 ∴ PV = Ne = E 3 3 372 10 PHYSICAL CHEMISTRY 2 or PV = E...(1) 3 where E is the total kinetic energy of all the N molecules. The expression (1) may be called the kinetic gas equation in terms of kinetic energy. We know that the General ideal gas equation is PV = nRT...(2) From (1) and (2) 2 E = nRT...(3) 3 For one mole of gas, the kinetic energy of N molecules is, 3RT E=...(4) 2 Since the number of gas molecules in one mole of gas in N0 (Avogadro number), E 3RT e= = N0 2 N0 3RT or e=...(5) 2 N0 substituting the values of R, T, N0, in the equation (5), the average kinetic energy of a gas molecule can be calculated. SOLVED PROBLEM 1. Calculate the average kinetic energy of a hydrogen molecule at 0°C. SOLUTION 3 RT e= 2 N0 Here R = 8.314 × 107 erg K–1 mol–1 T = 273 K ; N0 = 6.02 × 1023 3 8.314 × 107 × 273 ∴ e= × = 5.66 × 10 –14 erg 2 6.02 × 1023 Thus the average kinetic energy of H2 at 0°C is 5.66 × 10–14 erg SOLVED PROBLEM 2. Calculate the kinetic energy of two moles of N2 at 27°C. (R = 8.314 JK–1 mol–1) SOLUTION 3 We know E= nRT 2 Here, T = 27 + 273 = 300 K ; n = 2; R = 8.314 JK–1 mol–1 Substituting these values, we have 3 E= × 2 × 8.314 × 300 = 7482.6 J 2 Therefore the kinetic energy of two moles of N2 is 7482.6 J.  GASEOUS STATE 373 DEDUCTION OF GAS LAWS FROM THE KINETIC GAS EQUATION (a) Boyle’s Law According to the Kinetic Theory, there is a direct proportionality between absolute temperature and average kinetic energy of the molecules i.e., 1 mNu 2 ∝T 2 1 or mNu 2 = kT 2 3 1 or × mNu 2 = kT 2 3 1 2 or mNu 2 = kT 3 3 1 Substituting the above value in the kinetic gas equation PV = mNu 2 , we have 3 2 PV = kT 3 The product PV, therefore, will have a constant value at a constant temperature. This is Boyle’s Law. (b) Charles’ Law As derived above, 2 PV = kT 3 2 k or V = × T 3 P At constant pressure, ⎛ 2 k⎞ V = k' T where ⎜ k ′ = × ⎟ ⎝ 3 P⎠ or V ∝T That is, at constant pressure, volume of a gas is proportional to Kelvin temperature and this is Charles’ Law. (c) Avogadro’s Law If equal volume of two gases be considered at the same pressure, 1 PV = m1 N1u12...Kinetic equation as applied to one gas 3 1 PV = m2 N 2 u22...Kinetic equation as applied to 2nd gas 3 1 1 ∴ m1 N1u12 = m2 N 2 u22...(1) 3 3 When the temperature (T) of both the gases is the same, their mean kinetic energy per molecule will also be the same. 1 1 i.e., m1u12 = m2u22...(2) 3 3 Dividing (1) by (2), we have  GASEOUS STATE 373 DEDUCTION OF GAS LAWS FROM THE KINETIC GAS EQUATION (a) Boyle’s Law According to the Kinetic Theory, there is a direct proportionality between absolute temperature and average kinetic energy of the molecules i.e., 1 mNu 2 ∝T 2 1 or mNu 2 = kT 2 3 1 or × mNu 2 = kT 2 3 1 2 or mNu 2 = kT 3 3 1 Substituting the above value in the kinetic gas equation PV = mNu 2 , we have 3 2 PV = kT 3 The product PV, therefore, will have a constant value at a constant temperature. This is Boyle’s Law. (b) Charles’ Law As derived above, 2 PV = kT 3 2 k or V = × T 3 P At constant pressure, ⎛ 2 k⎞ V = k' T where ⎜ k ′ = × ⎟ ⎝ 3 P⎠ or V ∝T That is, at constant pressure, volume of a gas is proportional to Kelvin temperature and this is Charles’ Law. (c) Avogadro’s Law If equal volume of two gases be considered at the same pressure, 1 PV = m1 N1u12...Kinetic equation as applied to one gas 3 1 PV = m2 N 2 u22...Kinetic equation as applied to 2nd gas 3 1 1 ∴ m1 N1u12 = m2 N 2 u22...(1) 3 3 When the temperature (T) of both the gases is the same, their mean kinetic energy per molecule will also be the same. 1 1 i.e., m1u12 = m2u22...(2) 3 3 Dividing (1) by (2), we have 374 10 PHYSICAL CHEMISTRY N1 = N2 Or, under the same conditions of temperature and pressure, equal volumes of the two gases contain the same number of molecules. This is Avogadro’s Law. (d) Graham’s Law of Diffusion If m1 and m2 are the masses and u1 and u2 the velocities of the molecules of gases 1 and 2, then at the same pressure and volume 1 1 m1 N1u12 = m2 N 2 u22 3 3 By Avogadro’s Law N1 = N2 ∴ m1u12 = m2u22 2 ⎛ u1 ⎞ m2 or ⎜u ⎟ = m ⎝ 2⎠ 1 If M1 and M2 represent the molecular masses of gases 1 and 2, 2 ⎛ u1 ⎞ M2 ⎜u ⎟ = M ⎝ 2⎠ 1 u1 M2 = u2 M1 The rate of diffusion (r) is proportional to the velocity of molecules (u), Therefore, Rate of diffusion of gas 1 r1 M2 = = Rate of diffusion of gas 2 r2 M1 This is Graham’s Law of Diffusion. DISTRIBUTION OF MOLECULAR VELOCITIES While deriving Kinetic Gas Equation, it was assumed that all molecules in a gas have the same velocity. But it is not so. When any two molecules collide, one molecule transfers kinetic energy ( 12 mv 2 ) to the other molecule. The velocity of the molecule which gains energy increases and that of the other decreases. Millions of such molecular collisions are taking place per second. Therefore, the velocities of molecules are changing constantly. Since the number of molecules is very large, a fraction of molecules will have the same particular velocity. In this way there is a broad distribution of velocities over different fractions of molecules. In 1860 James Clark Maxwell calculated the distribution of velocities from the laws of probability. He derived the following equation for the distribution of molecular velocities. 3/ 2 – MC 2 dN c ⎛ M ⎞ = 4π ⎜ ⎟ e 2 RT C dc 2 N ⎝ 2πRT ⎠ where dNc = number of molecules having velocities between C and (C + dc) N = total number of molecules M = molecular mass T = temperature on absolute scale (K) The relation stated above is called Maxwell’s law of distribution of velocities. The ratio dnc/n gives the fraction of the total number of molecules having velocities between C and (C + dc). Maxwell plotted such fractions against velocity possessed by the molecules. The curves so obtained illustrate the salient features of Maxwell distribution of velocities.  GASEOUS STATE 375 Fig. 10.20. Shows the distribution of velocities in nitrogen gas, N2, at 300 K and 600 K. It will be noticed that : (1) A very small fraction of molecules has either very low (close to zero) or very high velocities. (2) Most intermediate fractions of molecules have velocities close to an average velocity represented by the peak of the curve. This velocity is called the most probable velocity. It may be defined as the velocity possessed by the largest fraction of molecules corresponding to the highest point on the Maxvellian curve. (3) At higher temperature, the whole curve shifts to the right (dotted curve at 600 K). This shows that at higher temperature more molecules have higher velocities and fewer molecules have lower velocities. Fraction of molecules Most probable velocity 300 K 600 K V Molecular velocity Figure 10.20 Distribution of molecular velocities in nitrogen gas, N2 , at 300 K and 600 K. DIFFERENT KINDS OF VELOCITIES In our study of kinetic theory we come across three different kinds of molecular velocities : (1) the Average velocity (V) (2) the Root Mean Square velocity (μ) (3) the Most Probable velocity (vmn) Average Velocity Let there be n molecules of a gas having individual velocities v1, v2, v3..... vn. The ordinary average velocity is the arithmetic mean of the various velocities of the molecules. v + v2 + v3..... + vn v = 1 n From Maxwell equation it has been established that the average velocity vav is given by the expression 8RT v = πM Substituting the values of R, T, π and M in this expression, the average value can be calculated. Root Mean Square Velocity If v1, v2, v3..... vn are the velocities of n molecules in a gas, μ2, the mean of the squares of all the velocities is 376 10 PHYSICAL CHEMISTRY v12 + v22 + v32..... + vn2 μ2 = n Taking the root v12 + v22 + v32..... + vn2 μ= n μ is thus the Root Mean Square velocity or RMS velocity. It is denoted by u. The value of the RMS of velocity u, at a given temperature can be calculated from the Kinetic Gas Equation. 1 PV = mNu 2...Kinetic Equation 3 3PV u2 = mN For one mole of gas PV = RT 3RT Therefore, u2 =...M is molar mass M 3RT u= M By substituting the values of R, T and M, the value of u (RMS velocity) can be determined. RMS velocity is superior to the average velocity considered earlier. With the help of u, the total Kinetic energy of a gas sample can be calculated. Most Probable Velocity As already stated the most probable velocity is possessed by the largest number of molecules in a gas. According to the calculations made by Maxwell, the most probably velocity, vmp, is given by the expression. 2 RT vmps M Substituting the values of R, T and M in this expression, the most probably velocity can be calculated. Relation between Average Velocity, RMS Velocity and Most Probable Velocity We know that the average velocity, v , is given by the expression 8RT v = πM 3RT and μ= M v 8 RT M 8 ∴ = × = μ πM 3RT 3π = 0.9213 or v = μ × 0.9213...(1) That is, Average Velocity = 0.9213 × RMS Velocity The expression for the most probably velocity, vmp, is  GASEOUS STATE 377 2 RT vmp = M 3RT and μ= M vmp 2 RT M 2 ∴ = × = = 0.8165 μ M 3RT 3 or vmp = μ × 0.8165...(2) That is, Most Probable Velocity = 0.8165 × RMS Velocity RMS can be easily calculated by the application of Kinetic Gas equation. Knowing the value of RMS, we can find the average velocity and the most probable velocity from expressions (1) and (2). CALCULATION OF MOLECULAR VELOCITIES The velocities of gas molecules are exceptionally high. Thus velocity of hydrogen molecule is 1,838 metres sec–1. While it may appear impossible to measure so high velocities, these can be easily calculated from the Kinetic Gas equation. Several cases may arise according to the available data. While calculating different types of velocities, we can also make use of the following expressions stated already. 3RT RMS velocity, μ= M 8RT Average velocity, v = M 2 RT Most Probable velocity, vmp = M Case 1. Calculation of Molecular Velocity when temperature alone is given 1 PV = mNu 2 (Kinetic Gas equation) 3 where N = N0 (Avogadro’s number) Thus we have, M = m × N0 = molecular mass of the gas 3PV 3RT ∴ u= = (∵ PV = RT for 1 mole) M M But R = 8.314 × 107 ergs deg–1 mol–1 = 0.8314 × 108 ergs deg–1 mol–1 3 × 0.8314 × 108 × T ∴ u= M T = 1.58 × 10 × 4 cm sec –1 M where T is Kelvin temperature and M the molar mass. 378 10 PHYSICAL CHEMISTRY SOLVED PROBLEM. Calculate the root mean square velocity of CO2 molecule at 1000°C. SOLUTION T = 273 + 1000 = 1273 K; M = 44 Applying the equation T u = 1.58 × 104 × M 1273 we have u = 1.58 × 104 × 44 –1 u = 84985 cm sec or 849.85 m sec–1 Case 2. Calculation of Molecular Velocity when temperature and pressure both are given. In such cases we make use of the following relation based on Kinetic Gas equation. 3PV u= M We know that 1 mole of a gas at STP occupies a volume of 22400 ml (known as molar volume). But before applying this relation the molar volume is reduced to the given conditions of temperature and pressure. SOLVED PROBLEM. Calculate the RMS velocity of chlorine molecules at 12°C and 78 cm pressure. SOLUTION At STP : At given conditions : V1 = 22400 ml V2 = ? T1 = 273 K T2 = 12 + 273 = 285 K P1 = 76 cm P2 = 78 cm P1 V1 P2 V2 Applying = T1 T2 P1 V1 T2 76 × 22400 × 285 we have V2 = = = 22785 ml T1 P2 273 × 78 3PV we know that u= M P = hdg = 78 × 13.6 × 981 dynes cm–2 V = 22785 ml; M = 71 3 × 78 × 13.6 × 981 × 22785 ∴ u= 71 u = 31652 cm sec or 316.52 m sec–1 –1 Case 3. Calculation of Molecular Velocity at STP Here we use the relation  GASEOUS STATE 379 3PV u= M where P = 1 atm = 76 × 13.6 × 981 dynes cm–2 V = 22,400 ml M = Molar mass of the gas SOLVED PROBLEM. Calculate the average velocity of nitrogen molecule at STP. SOLUTION In this example we have, P = 1 atm = 76 × 13.6 × 981 dynes cm–2 V = 22,400 ml M = 28 Substituting these values in the equation 3PV u= M 3 × 76 × 13.6 × 981 × 22400 we have = 28 = 49,330 cm sec–1 ∴ Average velocity = 0.9213 × 49330 cm sec–1 = 45,447 cm sec–1 Case 4. Calculation of Molecular Velocity when pressure and density are given In this case we have 3PV 3P ⎡ M ⎤ u= or u = ⎢ = D⎥ M D ⎣ V ⎦ where P is expressed in dynes cm–2 and D in gm ml–1. SOLVED PROBLEM. Oxygen at 1 atmosphere pressure and 0°C has a density of 1.4290 grams per litre. Find the RMS velocity of oxygen molecules. SOLUTION We have P = 1 atm = 76 × 13.6 × 981 dynes cm–2 1.4290 D = 1.4290 g l–1 = g ml–1 1000 = 0.001429 g ml–1 3P Applying u= D 3 × 76 × 13.6 × 981 we get u= = 46138 cm sec –1 0.001429 Case 5. Calculation of most probable velocity In this case we have T vmp = 1.29 × 104 M where T expressed in Kelvin and M to mass. 380 10 PHYSICAL CHEMISTRY SOLVED PROBLEM. Calculate the most probable velocity of nitrogen molecules, N2, at 15°C. SOLUTION T = 273 + 15 = 288 K We know that T vmp = 1.29 × 104 M 288 = 1.29 × 104 28 = 4.137 × 104 cm sec–1 COLLISION PROPERTIES In the derivation of Kinetic gas equation we did not take into account collisions between molecules. The molecules in a gas are constantly colliding with one another. The transport properties of gases such as diffusion, viscosity and mean free path depend on molecular collisions. We will now discuss some properties of gases which determine the frequency of collisions. The Mean Free Path At a given temperature, a molecule travels in a straight line before collision with another molecule. The distance travelled by the molecule before collision is termed free path. The free path for a molecule varies from time to time. The mean distance travelled by a molecule between two successive collisions is called the Mean Free Path. It is denoted by λ. If l1, l2, l3 are the free paths for a molecule of a gas, its free path l + l2 + l3 +..... + ln λ= 1 n where n is the number of molecules with which the molecule collides. Evidently, the number of molecular collisions will be less at a lower pressure or lower density and longer will be the mean free path. The mean free path is also related with the viscosity of the gas. Collision l1 l2 l3 Free path Figure 10.21 The mean free path illustrated. The mean free path, λ, is given by the expression 3 λ=η Pd where P = pressure of the gas d = density of the gas η = coefficient of viscosity of the gas  GASEOUS STATE 381 By a determination of the viscosity of the gas, the mean free path can be readily calculated. At STP, the mean free path for hydrogen is 1.78 × 10–5 cm and for oxygen it is 1.0 × 10–5 cm. Effect of Temperature and Pressure on Mean Free Path (a) Temperature The ideal gas equation for n moles of a gas is PV = n R T...(i) where n is the number of moles given by Number of molecules N n= = Avogadro's Number N0 Substituting this in equation (i) we get N PV = RT No N PN 0 or = V RT At constant pressure 1 N ∝...(ii) T The mean free path is given by Distance travelled by the molecule per second λ= Number of collisions per c.c. v = 2 πσ 2 vN 1 =...(iii) 2 πσ2 N combining equations (ii) and (iii), we get λ ∝T Thus, the mean free path is directly proportional to the absolute temperature. (b) Pressure We know that the pressure of a gas at certain temperature is directly proportional to the number of molecules per c.c. i.e. P∝N and mean free path is given by 1 λ= 2πσ2 N Combining these two equations, we get 1 λ∝ P Thus, the mean free path of a gas is directly proportional to the pressure of a gas at constant temperature. 382 10 PHYSICAL CHEMISTRY SOLVED PROBLEM 1. At 0°C and 1 atmospheric pressure the molecular diameter of a gas is 4Å. Calculate the mean free path of its molecule. SOLUTION. The mean free path is given by 1 λ= 2 π σ2 N where σ is the molecular diameter and N is the no. of molecules per c.c. Here σ = 4Å = 4 × 10–8 cm. We know 22400 ml of a gas 0°C and 1 atm. pressure contains 6.02 × 1023 molecules. 6.02 × 1023 ∴ No. of molecules per c.c., N = 22400 = 2.689 × 1019 molecules Substituting the values, we get 1 σ= 1.414 × 3.14 × (4 × 10 –8 ) 2 × 2.689 × 1019 1 = 1.414 × 3.14 × 16 × 2.689 × 103 = 0.524 × 10–5 cm SOLVED PROBLEM 2. The root mean square velocity of hydrogen at STP is 1.83 × 105 cm sec–1 and its mean free path is 1.78 × 10–5 cm. Calculate the collision number at STP. SOLUTION. Here root mean square velocity μ = 1.831 × 105 cm sec–1 We know average velocity v = 0.9213 × RMS velocity = 0.9213 × 1.831 × 105 cm sec–1 = 1.6869 × 105 cm sec–1 Average velocity The mean free path = Collision Number Average velocity ∴ Collision Number = Mean free path 1.6869 × 105 cm sec –1 = 1.78 × 10 –5 cm. = 9.4769 × 109 sec–1 The Collison Diameter When two gas molecules approach one another, they cannot come closer beyond a certain distance. The closest distance between the centres of the two molecules taking part in a collision is called the Collision Diameter. It is denoted by σ. Whenever the distance between the centres of two molecules is σ, a collison occurs. The collision diameter is obviously related to the mean free path of molecules. The smaller the collision or molecular diameter, the larger is the mean free path. 382 10 PHYSICAL CHEMISTRY SOLVED PROBLEM 1. At 0°C and 1 atmospheric pressure the molecular diameter of a gas is 4Å. Calculate the mean free path of its molecule. SOLUTION. The mean free path is given by 1 λ= 2 π σ2 N where σ is the molecular diameter and N is the no. of molecules per c.c. Here σ = 4Å = 4 × 10–8 cm. We know 22400 ml of a gas 0°C and 1 atm. pressure contains 6.02 × 1023 molecules. 6.02 × 1023 ∴ No. of molecules per c.c., N = 22400 = 2.689 × 1019 molecules Substituting the values, we get 1 σ= 1.414 × 3.14 × (4 × 10 –8 ) 2 × 2.689 × 1019 1 = 1.414 × 3.14 × 16 × 2.689 × 103 = 0.524 × 10–5 cm SOLVED PROBLEM 2. The root mean square velocity of hydrogen at STP is 1.83 × 105 cm sec–1 and its mean free path is 1.78 × 10–5 cm. Calculate the collision number at STP. SOLUTION. Here root mean square velocity μ = 1.831 × 105 cm sec–1 We know average velocity v = 0.9213 × RMS velocity = 0.9213 × 1.831 × 105 cm sec–1 = 1.6869 × 105 cm sec–1 Average velocity The mean free path = Collision Number Average velocity ∴ Collision Number = Mean free path 1.6869 × 105 cm sec –1 = 1.78 × 10 –5 cm. = 9.4769 × 109 sec–1 The Collison Diameter When two gas molecules approach one another, they cannot come closer beyond a certain distance. The closest distance between the centres of the two molecules taking part in a collision is called the Collision Diameter. It is denoted by σ. Whenever the distance between the centres of two molecules is σ, a collison occurs. The collision diameter is obviously related to the mean free path of molecules. The smaller the collision or molecular diameter, the larger is the mean free path.  GASEOUS STATE 383 σ Figure 10.22 Collision diameter of molecules. The collision diameter can be determined from viscosity measurements. The collision diameter of hydrogen is 2.74 Å and that of oxygen is 3.61Å. The Collision Frequency The collision frequency of a gas is defined as : the number of molecular collisions taking place per second per unit volume (c.c.) of the gas. Let a gas contain N molecules per c.c. From kinetic consideration it has been established that the number of molecules, n, with which a single molecule will collide per second, is given by the relation n = 2 π v σ2 N where v = average velocity; σ = collision diameter. If the total number of collisions taking place per second is denoted by Z, we have Z = 2 π v σ2 N × N = 2 π v σ2 N 2 Since each collision involves two molecules, the number of collision per second per c.c. of the gas will be Z/2. 2 π v σ2 N 2 Hence the collision frequency = 2 π v σ2 N 2 = 2 Evidently, the collision frequency of a gas increases with increase in temperature, molecular size and the number of molecules per c.c. Effect of Temperature and Pressure on Collision Frequency (i) Effect of Temperature We know collision frequency is given by π v σ2 N 2 Z =...(i) 2 From this equation it is clear that Z ∝v But μ∝ T or Z∝ T Hence collision frequency is directly proportional to the square root of absolute temperature. 384 10 PHYSICAL CHEMISTRY (ii) Effect of Pressure From equation (i), we have Z ∝ N2...(ii) where N is the number of molecules per c.c. But we know that the pressure of the gas at a certain temperature i.e. P∝N...(iii) combining equation (ii) and (iii) we get Z = P2 Thus the collision frequency is directly proportional to the square of the pressure of the gas. SPECIFIC HEAT RATIO OF GASES The Specific heat is defined as the amount of heat required to raise the temperature of one gram of a substance through 1°C. It may be measured at constant volume or at a constant pressure and though the difference in the two values is negligible in case of solids and liquids, it is appreciable in case of gases and a ratio of the two values gives us valuable information about the atomicity of a gas molecule. Specific Heat at Constant Volume It is the amount of heat required to raise the temperature of one gas through 1°C while the volume is kept constant and the pressure allowed to increase. It is denoted by the symbol Cv. In Physical Chemistry it is more common, however, to deal with one gram mole of the gas and the heat required in such case is called Molecular Heat and is represented at constant volume by Cv. It is possible to calculate its value by making use of the Kinetic theory. 1 2 Consider one mole of a gas at the temperature T. Its kinetic energy is mnu. From the kinetic 2 gas equation 1 PV = mnu 2 2 2 1 = × mnu 2 = RT 3 2 1 3 or mnu 2 (= KE) = RT 2 2 3 If the temperature is raised by 1°C to (T + 1)K kinetic energy becomes R (T + 1). 2 3 3 ∴ Increase in kinetic energy = R (T + 1) – RT 2 2 3 = R 2 If, therefore, it be assumed that the heat supplied to a gas at constant volume is used up entirely in increasing the kinetic energy of the moving molecules, and consequently increasing the temperature, 3 the value of Cv should be equal R. It is actually so for monoatomic gases and vapours because 2 such molecules can execute only translatory motion along the three co-ordinate axes. Motion of monoatomic gas molecules is the simplest and can be resolved into three perpendicular components  GASEOUS STATE 385 along the co-ordinate axes. Thus the energy of such a molecule can be considered to be composed of three parts as 1 2 1 2 1 2 1 2 mv = mvx + mv y + mvz 2 2 2 2 The number of square terms involved in determining the total kinetic energy of a molecule is often referred to as the Degrees of freedom of motion. Such molecules have three degrees of freedom of motion. According to the principle of equipartition of energy, total energy of the molecule is equally distributed among all its degrees of freedom. But in the case of diatomic and polyatomic molecules, the heat supplied may not only increase this kinetic energy of translation of the molecules as a whole but also cause an increase in the energy in the inside of the molecules which we may call as intramolecular energy. This intramolecular energy may be the vibration energy i.e., energy of the atoms executing vibrations with respect to each other along their line of centres or rotational energy which manifests itself in the rotation of the molecules about axes perpendicular to the line of centres. There will be other degrees of freedom for rotational and vibrational modes of motion also. For such cases the heat needs will be complex and are denoted by ‘x’ – a factor which depends upon vibrational and rotational degrees of freedom. Vibrational degrees of freedom rapidly increase with the increase in the total number of atoms in a molecule but the degrees of freedom are two for linear diatomic and three for non- linear diatomic molecules in case of rotational motion. 3 Consequently in such cases the molecular heat will be greater than R by the factor x. 2 3 or Cv = R + x 2 The value of x varies from gas to gas and is zero for monoatomic molecules. Specific Heat at Constant Pressure It may be defined as the amount of heat required to raise the temperature of one gram of gas through 1°C, the pressure remaining constant while the volume is allowed to increase. It is written as cp and the Molecular heat in this gas is represented as Cp. Now, whenever a gas expands it has to do work against external pressure. It means that when a gas is heated under constant pressure, the heat supplied is utilised in two ways : (1) in increasing the kinetic energy of the moving molecules and this has already been shown to be equal to 3/2 R + x cal. (2) in performing external work done by the expanding gas. The work done by the gas is equivalent to the product of the pressure and the change in volume. Let this change in volume be ΔV when the constant pressure is P and the initial volume is V. For 1 g mole of the gas at temperature T, PV = RT...(i) At temperature (T + 1) K P (V + ΔV) = R (T + 1)...(ii) Subtracting (i) from (ii) P × ΔV = R 3 Hence R cal must be added to the value of R cal in order to get the thermal equivalent of the 2 energy supplied to one gram mole of the gas in the form of heat when its temperature is raised by 1°C. 3 5 ∴ Cp = R+R= R (for monoatomic molecules) 2 2 386 10 PHYSICAL CHEMISTRY 3 For di– and polyatomic molecules, it will be R + x. 2 Specific Heat Ratio The ratio of the molecular heats will be the same as the ratio of the specific heats. It is represented by the symbol γ. Cp 5 R+x γ= = 2 Cv 3 2 R+x For monoatomic molecules, x = 0 Cp 5 R 5 γ= = 2 3 = = 1.667 Cv R 2 3 For diatomic molecules in most cases, S = R Cp 7 R 7 γ= = 2 5 = = 1.40 Cv 2 R 5 3 For polyatomic molecules, very often x = R 2 Cp 5 R + 23 R 8 γ= = 2 = = 1.33 Cv 3 2 R+ 3 2 R 6 These results are found to be in accord with experimental observations at 15°C given in the Table that follows and thus specific heat ratio helps us to determine the atomicity of gas molecules. The theoretical difference between Cp and Cv as calculated above is R and its observed value also shown in the table below comes out to about 2 calories. Gas Cp Cv Cp–Cv = R g = Cp/Cv Atomicity Helium 5.00 3.01 1.99 1.661 1 Argon 4.97 2.98 1.90 1.667 1 Mercury vapour 6.93 4.94 1.99 1.40 2 Nitrogen 6.95 4.96 1.99 1.40 2 Oxygen 6.82 4.83 1.49 1.41 2 Carbon dioxide 8.75 6.71 2.04 1.30 3 Hydrogen sulphide 8.62 6.53 2.09 1.32 3 DEVIATIONS FROM IDEAL BEHAVIOUR An ideal gas is one which obeys the gas laws or the gas equation PV = RT at all pressures and temperatures. However no gas is ideal. Almost all gases show significant deviations from the ideal behaviour. Thus the gases H2, N2 and CO2 which fail to obey the ideal-gas equation are termed nonideal or real gases. Compressibility Factor The extent to which a real gas departs from the ideal behaviour may be depicted in terms of a new function called the Compressibility factor, denoted by Z. It is defined as PV Z = RT The deviations from ideality may be shown by a plot of the compressibility factor, Z, against P.  GASEOUS STATE 387 For an ideal gas, Z = 1 and it is independent of temperature and pressure. The deviations from ideal behaviour of a real gas will be determined by the value of Z being greater or less than 1. The difference between unity and the value of the compressibility factor of a gas is a measure of the degree of nonideality of the gas. For a real gas, the deviations from ideal behaviour depend on (i) pressure; and temperature. This will be illustrated by examining the compressibility curves of some gases discussed below with the variation of pressure and temperature. Effect of Pressure Variation on Deviations Fig. 10.23 shows the compressibility factor, Z, plotted against pressure for H2, N2 and CO2 at a constant temperature. 2.0 N2 H2 1.5 CO2 Ideal Gas PV RT 1.0 Z= 0.5 0 0 200 400 600 800 1000 P (atm) Figure 10.23 Z versus P plots for H2, N2 and CO2 at 300 K. At very low pressure, for all these gases Z is approximately equal to one. This indicates that at low pressures (upto 10 atm), real gases exhibit nearly ideal behaviour. As the pressure is increased, H2 shows a continuous increase in Z (from Z = 1). Thus the H2 curve lies above the ideal gas curve at all pressures. For N2 and CO2, Z first decreases (Z < 1). It passes through a minimum and then increases continuously with pressure (Z > 1). For a gas like CO2 the dip in the curve is greatest as it is most easily liquefied. Effect of Temperature on Deviations Fig 10.24 shows plots of Z or PV/RT against P for N2 at different temperatures. It is clear from the shape of the curves that the deviations from the ideal gas behaviour become less and less with increase of temperature. At lower temperature, the dip in the curve is large and the slope of the curve is negative. That is, Z < 1. As the temperature is raised, the dip in the curve decreases. At a certain temperature, the minimum in the curve vanishes and the curve remains horizontal for an appreciable range of pressures. At this temperature, PV/RT is almost unity and the Boyle’s law is obeyed. Hence this temperature for the gas is called Boyle’s temperature. The Boyle temperature of each gas is characteristic e.g., for N2 it is 332 K. Conclusions From the above discussions we conclude that : (1) At low pressures and fairly high temperatures, real gases show nearly ideal behaviour and the ideal-gas equation is obeyed. (2) At low temperatures and sufficiently high pressures, a real gas deviates significantly from ideality and the ideal-gas equation is no longer valid. 388 10 PHYSICAL CHEMISTRY 3 200K 500K 2 PV RT 1000K Z= 1 Ideal Gas 0 0 300 600 900 1200 P (atm) Figure 10.24 Z versus P plots for N2 at different temperatures. (3) The closer the gas is to the liquefaction point, the larger will be the deviation from the ideal behaviour. EXPLANATION OF DEVIATIONS – VAN DER WAALS EQUATION van der Waals (1873) attributed the deviations of real gases from ideal behaviour to two erroneous postulates of the kinetic theory. These are : (1) the molecules in a gas are point masses and possesses no volume. (2) there are no intermolecular attractions in a gas. Therefore, the ideal gas equation PV = nRT derived from kinetic theory could not hold for real gases. van der Waals pointed out that both the pressure (P) and volume (V) factors in the ideal gas equation needed correction in order to make it applicable to real gases. Volume Correction The volume of a gas is the free space in the container in which molecules move about. Volume V of an ideal gas is the same as the volume of the container. The dot molecules of ideal gas have zero-volume and the entire space in the container is available for their movement. However, van der Waals assumed that molecules of a real gas are rigid spherical particles which possess a definite volume. Ideal volume = V Volume = V – b Excluded volume (b) V–b Ideal Gas Real Gas Figure 10.25 Volume of a Real gas. The volume of a real gas is, therefore, ideal volume minus the volume occupied by gas molecules (Fig. 10.25). If b is the effective volume of molecules per mole of the gas, the volume in the ideal gas  GASEOUS STATE 389 equation is corrected as : (V – b) For n moles of the gas, the corrected volume is : (V – nb) where b is termed the excluded volume which is constant and characteristic for each gas. Excluded volume 2r Figure 10.26 Excluded volume for a pair of gas molecules. Excluded volume is four times the actual volume of molecules. The excluded volume is not equal to the actual volume of the gas molecules. In fact, it is four times the actual volume of molecules and can be calculated as follows. Let us consider two molecules of radius r colliding with each other (Fig. 10.26). Obviously, they cannot approach each other closer than a distance (2r) apart. Therefore, the space indicated by the dotted sphere having radius (2r) will not be available to all other molecules of the gas. In other words the dotted spherical space is excluded volume per pair of molecules. Thus, 4 excluded volume for two molecules = π(2r )3 3 ⎛4 ⎞ = 8 ⎜ πr 3 ⎟ ⎝3 ⎠ 1 ⎛4 ⎞ excluded volume per molecule (Ve) = × 8 ⎜ πr 3 ⎟ 2 ⎝