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310 9 PHYSICAL CHEMISTRY STANDARD ENTROPY OF FORMATION It is the entropy of formation of 1 mole of a compound from the elements under standard conditions. It is denoted ΔSºf. We can calculate the value of entropy of a given compound from the values of Sº of elements....

310 9 PHYSICAL CHEMISTRY STANDARD ENTROPY OF FORMATION It is the entropy of formation of 1 mole of a compound from the elements under standard conditions. It is denoted ΔSºf. We can calculate the value of entropy of a given compound from the values of Sº of elements. Sºf = Sº (compound) – ∑Sº (elements) SOLVED PROBLEM 1. Urea hydrolyses in the presence of water to produce ammonia and carbon dioxide. CO(NH2)2 (aq) + H2O (l) ⎯⎯ → CO2(g) + 2NH3(g) What is the standard entropy change for this reaction when 1 mole of urea reacts with water ? The standard entropies of reactants and products are listed below : Substance Sº(cal/mole K) CO(NH2)2 41.55 H2O (l) 16.72 CO2 (g) 51.06 NH3 (g) 46.01 SOLUTION We know that ΔSº = ∑Sº (products) – ∑Sº (reactants) or ( ΔS º = ( S º CO2 + 2S º NH3 ) – S º CO(NH2 ) + S º H2O 2 ) = [51.06 + 2 × 46.01] – [41.55 + 16.72] cal K–1 = 84.81 cal K–1 SOLVED PROBLEM 2. Calculate the standard entropy of formation, ΔSºf , of CO2(g). Given the standard entropies of CO2(g), C(s), O2(g), which are 213.6, 5.740, and 205.0 JK–1 respectively. SOLUTION We know that Sºf = Sºcompound – ∑Sºelements or S º f = S º CO2( g ) – ⎡ S º C( s ) + S º O2 ( g ) ⎤ ⎣ ⎦ Substituting the values Sºf = 213.6 – [5.740 + 205.0] JK–1 = (213.6 – 210.74) JK–1 = 2.86 JK–1 SOME USEFUL DEFINITIONS (1) Cyclic Process When a system undergoes a series of changes and in the end returns to its original state, it is said to have completed as cycle. The whole process comprising the various changes is termed a cyclic process. Since the internal energy of a system depends upon its state, it stands to reason that in cyclic process the net change of energy is zero. Or, we can say that the work done by the system during all these changes should be equal to the heat absorbed by the system. ΔE = 0 = q – w or q = w  SECOND LAW OF THERMODYNAMICS 311 (2) Heat Engines Heat source The flow of heat from a hotter body to a colder body is T2 spontaneous process. The heat that flows out spontaneously can be used to do work with the help of a suitable device. q2 A machine which can do work by using heat that flows out spontaneously from a high-temperature source to a low-temperature sink, is called a heat engine. Heat A heat engine takes heat energy from a high-temperature reservoir engine and converts some of it into work, returning the unconverted heat to Work = q2 – q1 a low-temperature sink. A basic heat engine is illustrated in Fig. 9.6. A steam engine is a typical heat engine. It takes heat from the boiler q1 (high-temperature source), converts some heat to work and returns the unused heat to the surroundings (low-temperature sink). A heat engine running on a periodic cyclic process can yield Heat sink T1 work continuously. (3) Efficiency of a Heat Engine Figure 9.6 The ratio of the work obtained in a cyclic process (w) to the Principle of heat engine heat taken from the high-temperature reservoir (q) is referred to (illustration). as the efficiency of a heat engine. No heat engine, no matter how well constructed, can convert all the heat from the high-temperature reservoir into work. Such an engine would be 100% efficient. Sadi Carnot was the first scientist to realise this and deduce an expression showing the limitations of heat engines. THE CARNOT CYCLE In 1824 Sadi Carnot proposed a theoretical heat engine to show that the efficiency was based upon the temperatures between which it operated. Carnot’s imaginary engine could perform a series of operations between temperatures T1 and T2, so that at the end of these operations the system was restored to the original state. This cycle of processes which occurred under reversible conditions is referred to as the Carnot cycle. The medium employed in operating Carnot’s engine was one mole of an ideal gas which could be imagined to be contained in a cylinder fitted with a frictionless piston. The Carnot cycle comprises four operations or processes. (1) Isothermal reversible expansion (2) Adiabatic reversible expansion (3) Isothermal reversible compression (4) Adiabatic reversible compression The above four processes are shown in the indicator diagram of Carnot cycle (Fig. 9.7). A (P1, V1 ) Iso th erm al ( T 2) B (P2 ,V2 ) Pressure Adiabatic Adiabatic D Isot he rma (P4 ,V4 ) l (T1 ) C (P3 ,V3 ) Volume Figure 9.7 Indicator diagram of the Carnot cycle. 312 9 PHYSICAL CHEMISTRY First Operation – Isothermal Reversible Expansion Let T2, P1 and V1 be the temperature, pressure and volume respectively of the gas enclosed in the cylinder initially. The cylinder is placed in the heat reservoir at the higher temperature (T2). Now the gas is allowed to expand isothermally and reversibly so that the volume increases from V1 to V2. AB represents the path of the process in the diagram. Work done. Since the process in operation 1 is isothermal, ΔE = 0. If q2 be the heat absorbed by the system and w1 the work done by it, according to the first law equation (ΔE = q – w), q 2 = w1 V2 But w1 = RT2 In V1 V2 Therefore, q 2 = RT2 In...(1) V1 Second Operation – Adiabatic Reversible Expansion The gas at B is at a temperature T2 and has volume V2 under the new pressure P2. The gas is now allowed to expand reversibly from volume V2 to V3 when the temperature drops from T2 to T3 (along BC). Work done. Since this step is adiabatic, q = 0. If w2 be the work done, according to the first law equation (Δ E = q – w), ΔE = – w2 or w2 = – ΔE But ΔE = Cv (T1 – T2) Therefore, w2 = Cv (T2 – T1)...(2) Third Operation – Isothermal Reversible Compression Now the cylinder is placed in contact with a heat reservoir at a lower temperature, T1. The volume of the gas is then compressed isothermally and reversibly from V3 to V4 (represented by CD in diagram). Work done. During compression, the gas produces heat which is transferred to the low temperature reservoir. Since the process takes place isothermally, ΔE = 0. If q1 is the heat given to the reservoir and w3 the work done on the gas, using proper signs for q and w, we have V – q1 = – w3 = RT1 1n 4...(3) V3 Fourth Operation – Adiabatic Reversible Compression The gas with volume V4 and temperature T1 at D is compressed adiabatically (along DA) until it regains the original state. That is, the volume of the system becomes V1 and its temperature T2. Work done. In this step work is done on the system and, therefore, bears the negative (–) sign. If it is denoted by w4, we can write – w4 = – Cv (T2 – T1)...(4) Net Work Done in One Cycle Adding up the work done (w) in all the four operations of the cycle as shown in equations (1), (2), (3) and (4), we have w = w1 + w2 + (– w3) + (– w4) V V = RT2 1n 2 + Cv (T2 – T1 ) + RT1 1n 4 – Cv (T2 – T1 ) V1 V3  SECOND LAW OF THERMODYNAMICS 313 V2 V = RT2 1n + RT1 1n 4 V1 V3 Net Heat Absorbed in One Cycle If q is the net heat absorbed in the whole cycle. q = q2 – q1 where q2 is heat absorbed by the system in operation 1 and q1 is the heat transferred to the sink reservoir. From (1) and (3) V2 V q = q2 – q1 = RT2 1n + RT1 1n 4 V1 V3 V2 V or q = RT2 1n – RT1 1n 3...(5) V1 V4 According to the expression governing adiabatic changes, γ –1 T2 ⎛ V3 ⎞ = T1 ⎜⎝ V2 ⎟⎠...for adiabatic expansion γ –1 T1 ⎛ V1 ⎞ = T2 ⎜⎝ V2 ⎟⎠...for adiabatic compression V3 V4 or = V2 V1 V3 V2 or = V4 V1 Therefore, substituting the value of V3/V4 in equation (5), the value of net heat may be given as V2 V q = RT2 1n – RT1 1n 2 V1 V1 V2 = R (T2 – T1) 1n V...(6) 1 Calculation of Thermodynamic Efficiency Since the total work done in a cycle is equal to net heat absorbed, from (6) we can write V2 w = R (T2 – T1) 1n...(7) V1 The heat absorbed, q2, at higher temperature T2 is given by equation (1), V2 q 2 = RT2 1n...(8) V1 Dividing (7) by (8) w R (T2 – T1 ) 1n V2 / V1 T2 – T1 = = q2 RT2 1n V2 / V1 T2 w T2 – T1 or =...(9) q2 T2 The factor w/q2 is called thermodynamical efficiency. It is denoted by η and gives the fraction of the heat taken from the high-temperature reservoir which it is possible to convert into work by a heat 314 9 PHYSICAL CHEMISTRY engine. Therefore, the efficiency of a Carnot engine, the most ideal of all engines, is limited by the operating temperatures of the engine. The larger the temperature difference (T2 – T1) between the high and the low temperature reservoirs, the more the heat converted to work by the heat engine. For a given temperature of the high-temperature reservoir, the lower the temperature of the sink, the greater will be the efficiency of the machine. Similarly, for a given temperature of the sink, the efficiency will be increased by using a high temperature of the source reservoir. Carnot Theorem We have shown above that w T2 – T1 = q2 T2 This result deduced for a perfect gas depends upon the temperature limits between which the cycle operates. It is independent of all other factors. Thus Carnot stated an important relation known as the Carnot theorem. It states that : every perfect engine working reversibly between the same temperature limits has the same efficiency, whatever be the working substance. MORE STATEMENTS OF THE SECOND LAW From equation (9) w T2 – T1 T = =1– 1 q2 T2 T2 Evidently w/q2 is less than 1, or q2 is greater than w. This means that heat transferred by a spontaneous process is never completely converted into work (If so, w/q2 would be 1). This leads to another statement of the Second law (Lord Kelvin). It is impossible to take heat from a hotter reservoir and convert it completely into work by a cyclic process without transferring a part of heat to a cooler reservoir. This statement recognises the fact that heat engines could never be 100% efficient, since some heat must be returned to a low-temperature reservoir. Another statement of the Second law was given by Clausius. It is impossible for a cyclic process to transfer heat from a body at a lower temperature to one at higher temperature without at the same time converting some work to heat. This statement recognises that heat flows spontaneously from hot objects to cold objects and to get it flow in the opposite direction, we have to expend some work. SOLVED PROBLEM 1. An engine operating between 150ºC and 25ºC takes 500 J heat from a high temperature reservoir. Assuming that there are no frictional losses, calculate the work that can be done by this engine. SOLUTION From equation (9) w T2 – T1 Efficiency = = q2 T2 Here T2 = 423 K; T1 = 298 K w 423 – 298 Substituting the values = 500 423 ∴ w = 147.75 J  SECOND LAW OF THERMODYNAMICS 315 WONDERFUL ACHIEVEMENTS IN SCIENCE AND ENGINEERING The Second Law of Thermodynamics "A spontaneous change is accompanied by an increase in the total entropy of the system and its surroundings." ΔSSys + ΔSSurr > 0 ΔSSys is the entropy change of the system. ΔSSurr is the entropy change of the surroundings. Scientists responsible for the formulation and development of the Second Law Include Rudolph Clausius (1822-1888), Lord Kelvin (1824-1907), Josiah Willard Gibbs (1839-1903) and Ludwig Boltzmann (1844-1906). The Second Law explains why the rusting of Iron is inevitable, why car engines can never be 100% efficient and why batteries eventually run down. It governs the direction of all biochemical reactions including those responsible for glucose catabolism, DNA replication and protein synthesis - even life obeys the Second Law of Thermodynamics. SOLVED PROBLEM 2. The boiling point of water at a pressure of 50 atmosphere is 265ºC. Compare the theoretical efficiencies of a steam engine operating between the boiling point of water at (i) 1 atmosphere (ii) 50 atmosphere, assuming the temperature of the sink to be 35ºC in each case. SOLUTION (i) At 1 atmosphere : T2 (boiling point of water) = 100ºC or 373 K T1 (temperature of sink) = 35ºC or 308 K T2 – T1 373 – 308 Efficiency = = = 0.174 T2 373 (ii) At 50 atmosphere : T2 (boiling point of water) = 265ºC or 538 K T1 (temperature of sink) = 35ºC or 308 K T2 – T1 538 – 308 Efficiency = = = 0.428 T2 538 The possible increase of efficiency is very marked. SOLVED PROBLEM 3. If a Carnot engine operating between two heat reservoirs at 227ºC and 27ºC absorbs 1000 calories from the 227ºC reservoir per cycle, how much heat is discharged into the 27ºC reservoir and how much work is done per cycle ? What is the efficiency of the cycle ? SOLUTION (a) We know that : w T2 – T1 Efficiency = = q2 T2 316 9 PHYSICAL CHEMISTRY q2 (T2 – T1 ) or w= T2 1000 (500 – 300) = 500 = 400 cal ∴ The work done per cycle is 400 cal (b) The heat from the high-temperature reservoir (q2) minus the heat discharged into the low- temperature reservoir (q1) is converted into work (w). Thus, q2 – q1 = w ∴ 1000 – q1 = 400 or q 1 = 600 cal. (c) Efficiency : w T2 – T1 = q2 T2 500 – 300 200 = = 500 500 = 0.4 Therefore, efficiency of the engine is 0.4 DERIVATION OF ENTROPY FROM CARNOT CYCLE Because processes cannot be 100% efficient, a term to describe the energy available for doing useful work becomes necessary. Although we have discussed the concept of entropy already, its classical derivation deserves attention. In a Carnot cycle, q2 has a positive value and q1 has a negative value, since former is taken up at a higher temperature and the latter is given out at the lower temperature. Thus thermodynamic efficiency may be expressed as q – q1 T2 – T1 η= 2 = q2 T2 q T or 1– 1 =1– 1 q2 T2 q1 T1 or = q2 T2 q1 q2 or =...(i) T1 T2 using sign convention, heat absorbed (i.e., q2) is given the +ve sign and heat lost (i.e. q1) is given the – ve sign. Equation (i) becomes q q + 2 =– 1 T2 T1 q2 q1 or + =0 T2 T1 q or ∑ =0 T Any reversible cycle may be regarded as made up of a number of Carnot cycles. Consider, for example, the cycle represented in Fig. 9.8 by the closed curve ABA. Imagine a series of isothermal  SECOND LAW OF THERMODYNAMICS 317 A Pressure B Volume Figure 9.8 Carnot cycle. and adiabatic curves drawn across the diagram so that a number of Carnot cycles are indicated. Starting at A and going through all the cycles successively from A to B, it can be shown that all paths inside the diagram cancel each other leaving only zigzag outer path. The larger the number of cycles taken in this manner, the closer will the resultant path correspond to ABA which represents the reversible cycle under consideration. The reversible cycle can, therefore, be regarded as being made up of an infinite number of Carnot cycles, for each of which the sum of the two q/T terms involved is zero i.e., q1 q2 + =0 T1 T2 For the reversible cycle ABA comprising a series of Carnot cycles, therefore, the above expression takes the form q ∑ =0 T and for an infinite number of Carnot cycles dq =0∫ T Since the cycle is performed in two steps, viz., from A to B and then back from B to A, we have: ∫ dq dq B A dq T = T ∫A (Path I) + ∫ B T (Path II) = 0 B dq A dq B dq ∫A T (Path I) = – ∫B T (Path II) = ∫A T (Path II) It is evident, therefore, that both these integrals are independent of the path taken from A to B. Both depend upon the value of some function at A and the same function at B. This function is called entropy (S). Let SB be the entropy at the state B and SA in the state A. Then, the increase in entropy, ΔS, is given by the expression B dq ΔS = SB – SA = ∫A T 318 9 PHYSICAL CHEMISTRY and for each infinitesimally small change dq dS = T Like ΔE and ΔH, ΔS is dependent only on the state of the system and can be calculated if the substance can be brought reversibly from one state to the other. It is independent of the path taken. SOLVED PROBLEM 1. Calculate the entropy change in the evaporation of 1 mole of water at 100ºC. Latent heat of evaporation of water is 9,650 cals per mole. SOLUTION Entropy change on the evaporation of 1 mole of water is obtained by dividing the latent heat of evaporation of 1 mole of water by the absolute temperature 9, 650 ∴ ΔS = = 25.87 cal K–1 mol–1 373 SOLVED PROBLEM 2. Calculate the increase in entropy when one gram molecular weight of ice at 0ºC melts to form water. Latent heat of fusion of ice = 80 calories. SOLUTION q ΔS = T q for one mole of ice = 80 × 18 calories and T = (0 + 273) = 273 K 80 × 18 ∴ ΔS = = 5.274 cal K–1 mol–1 273 ENTROPY CHANGE IN AN IRREVERSIBLE PROCESS The efficiency of an irreversible Carnot cycle is always less than that of a reversible one operating between the same two temperatures. q2 – q1 T2 – T1 < q2 T2 where q2 is the heat absorbed at temperature T2 and q1 is the heat returned at temperature T1. q2 q1 – V1 or P1 > P2 hence ΔST is positive whereas in isothermal contraction V2 < V1 or P1 < P2 , ΔST is negative Case 2 : At constant pressure (Isobaric process) In this case P1 = P2 The equation (iv) reduces to T ΔS P = 2.303 n CP log 2 T1 Case 3 : At constant volume for an isobaric process In this case V1 = V2 The equation (iv) reduces to T ΔS v = 2.303 × n × Cv log 2 T1 SOLVED PROBLEM 1. Calculate the entropy change involved in thermodynamic expansion of 2 moles of a gas from a volume of 5 litres to a volume of 50 litres at 303 K. SOLUTION. Here n = 2; V1 = 5 litres; V2 = 50 litres V using the relation ΔST = 2.303 × n × R log 2 V1 on substituting the values we get 50 ΔST = 2.303 × 2 × 8.314 × log 5 = 38.29 JK–1  SECOND LAW OF THERMODYNAMICS 321 SOLVED PROBLEM 2. Calculate the entropy change when 2 moles of an ideal gas are allowed to expand isothermally at 293 K from a pressure of 10 atmosphere to a pressure of 2 atmosphere. SOLUTION. We know P ΔST = 2.303 × n × R × log 1 P2 Here n = 2; R = 8.314 J P1 = 10 atm; P2 = 2 atm. Substituting the values we get 10 ΔST = 2.303 × 2 × 8.314 × log 2 = 2.303 × 2 × 8.314 × 0.6990 = 26.76 JK–1 ENTROPY CHANGE ACCOMPANYING CHANGE OF PHASE When there is a change of state from solid to liquid or liquid to vapours or solid to vapours (melting, evaporation and sublimation respectively), there is a change in entropy. This change may be carried out at constant temperature reversibly as two phases are in equilibrium during the change. Let us consider the process of melting of 1 mole of the substance being carried out reversibly. It would absorb molar heat of fusion at temperature equal to its melting point. The entropy change is given by ΔH f ΔS f = Tf where ΔHf is the Molar heat of fusion at its melting point, Tf at constant pressure. Similarly, when one mole of liquid is boiled reversibly it would absorb molar heat of vaporisation at a temperature equal to its boiling point. In this case entropy change is given by ΔH v ΔSv = Tb where ΔHv is Molar heat of vaporisation at its boiling point at constant pressure. On similar lines we can calculate the change in entropy when one mole of a solid changes reversibly from one allotropic form to another at its transition temperature. We can write ΔH t ΔSt = Tt Where ΔHt is the Molar heat of transition at its transition temperature Tt. SOLVED PROBLEM 1. Calculate the entropy change when 1 mole of ethanol is evaporated at 351 K. The molar heat of vaporisation of ethanol is 39.84 kJ mol–1. SOLUTION. We know ΔH v ΔSv = Tb Here ΔHv = 39.84 kJ mol–1 = 39840 J mol–1 Tb = 351 K 39840 ∴ ΔS v = 351 = 113.5 JK–1 mol–1 322 9 PHYSICAL CHEMISTRY SOLVED PROBLEM 2. 30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 J mol–1 K–1. Calculate the melting point of sodium chloride. SOLUTION We know ΔH f ΔS f = Tf Here ΔS f = 28.4 JK–1 mol–1 and ΔHf = 30.4 kJ K–1 mol–1 = 30400 JK–1 mol–1 On substitution, we get ΔH f 30400 Tf = = ΔS f 28.4 = 1070.4 K Free Energy Function (G) and Work Function (A) The free energy function (G) is defined as G = H – TS where H is the heat content or enthalpy of the system, T is its temperature and S its entropy. It is a single valued function of thermodynamic state of the system and is an extensive property. Let us consider a system which undergoes a change of state from (1) to (2) at constant temperature. We have G2 – G1 = (H2 – H1) – T(S2 – S1) or ΔG = ΔH – T Δ S We know ΔH = ΔE + PΔV ∴ ΔG = ΔE + PΔV – TΔS Also ΔA = ΔE – TΔS ∴ ΔG = ΔA + PΔV (At constant P & T) But PΔV represents the work done due to expansion against a constant external pressure P. Therefore, it is clear that the decrease in free energy (– ΔG) accompanying a process taking place at constant temperature and pressure is equal to the maximum work obtainable from the system other than the work of expansion. This quantity is referred to as the net work, so that Net work = w – PΔV = – ΔG This quantity is of great importance in thermodynamics because the change in free energy is a measure of net work which may be electrical, chemical or surface work. The work function (A) is defined as A = E – TS Where E is the energy content of the system, T its absolute temperature and S its entropy. Since E, T and S depend upon the thermodynamic state of the system only and not on its previous history, it is evident that the function A is also a single valued function of the state of the system. Consider an isothermal change at temperature T from the initial state indicated by subscript (1) to the final state indicated by subscript (2) so that A1 = E1 – TS1...(i)  SECOND LAW OF THERMODYNAMICS 323 and A2 = E2 – TS2...(ii) Subtracting (i) from (ii), we have A2 – A1 = (E2 – E1) – T (S2 – S1) ΔA = ΔE – TΔS...(iii) Where ΔA is the change in work function A, ΔE is the corresponding change in internal energy and ΔS as change in the entropy of the system. q Since ΔS = rev where qrev is the heat taken up when the change is carried out in a reversible T manner and constant temperature, we have ΔA = ΔE – qrev...(iv) According to first law of thermodynamics dE = qrev – wrev or – w = ΔE – qrev substituting the value in equation (iv), we get – ΔA = wrev i.e., decrease in the work function A in any process at constant temperature gives the maximum work that can be obtained from the system during any change. Variation of Free Energy with Temperature and Pressure By definition G = H – TS and H = E + PV ∴ G = E + PV – TS Differentiating, we get dG = dE + PdV + VdP – TdS – SdT...(i) For an infinitesimal stage of a reversible process dq = dE + dw dq and dS = T If the work done is restricted to work of expansion, then dq = dE + PdV and dS = dE + PdV...(ii) comparing equation (i) and (ii) we have dG = dE + PdV + VdP – dE – PdV – SdT = VdP – SdT If the pressure remains constant dGp = – SdTp ⎛ dG ⎞ or ⎜ ⎟ = –S...(iii) ⎝ dT ⎠ p But at constant temperature dT = 0 and we have (dG)T = VdP ⎛ dG ⎞ or ⎜ ⎟ =V...(iv) ⎝ dP ⎠T Equations (iii) and (iv) give the variation of free energy with temperature and pressure respectively. 324 9 PHYSICAL CHEMISTRY Isothermal change in Free Energy By definition G = H – TS and H = E + PV ∴ G = E + PV – TS on differentiating we get dG = dE + PdV + VdP – TdS – SdT... (i) and from first law of thermodynamics we have dq = dE + PdV...(ii) From equation (i) and (ii) we get dG = dq + VdP – TdS – SdT...(iii) For a reversible process dq dS = or TdS = dq T Substituting this in equation (iii) we get dG = TdS + VdP – TdS – SdT = VdP – SdT...(iv) Since the process is isothermal (there is no change in temperature) dT = 0, the equation (iv) reduces to dG = VdP Integrating with in the limits G1 and G2; P1 and P2, we get P2 Δ G = G2 – G1 = ∫ VdP P1 For 1 mole of the gas PV = RT RT or V = P On substitution we get P2 dP dG = RT ∫ P P1 P2 = RT 1n...(v) P1 For n moles of the gas we have P2 ΔG = nRT 1n...(vi) P1 We know that P1 V1 = P2 V2 P2 V1 or = P1 V2 From equation (v) we get V1 ΔG = RT 1n V2  SECOND LAW OF THERMODYNAMICS 325 and from equation (vi) we get V1 ΔG = n RT 1n V2 Changing to natural logarithms, we get P2 ΔG = 2.303 × nRT log P1 V1 or ΔG = 2.303 × nRT log V2 With the help of these equations we can calculate the change in free energy in isothermal process having an ideal gas. SOLVED PROBLEM 1. Four moles of an ideal gas expand isothermally from 1 litre to 10 litres at 300 K. Calculate the change in free energy of the gas. (R = 8.314 JK–1 mol–1) SOLUTION For an isothermal process V1 ΔG = 2.303 × nRT log V2 Here V1 = 1 litre ; V2 = 10 litres T = 300 K; R = 8.314 JK–1 mol–1 n =4 Substituting the values in the formula we get 1 ΔG = 2.303 × 300 × 4 × 8.314 × log 10 = 22976.5 × [– 1.0] = – 22976.5 J = – 22.9765 kJ SOLVED PROBLEM 2. Two moles of an ideal gas are allowed to expand reversibly and isothermally at 300 K from a pressure of 1 atm to a pressure of 0.1 atm. What is the change in Gibbs free energy ? SOLUTION. We know for an isothermal process P2 ΔG = 2.303 × nRT log P1 Here P1 = 1.0 atm; P2 = 0.1 atm. n = 2 moles R = 8.314 JK–1 mol–1 T = 300 K On substitution we get 0.1 ΔG = 2.303 × 2 × 8.314 × 300 × log 1 = 11488.2 × [– 1.0] = – 11488.2 J = – 11.4882 kJ 326 9 PHYSICAL CHEMISTRY GIBB’S HELMHOLTZ EQUATIONS These are two equations derived by J.W. Gibbs and H.Von Helmholtz and are known as Gibbs Helmholtz equations. One equation can be expressed in terms of changes in free energy (ΔG) and enthalpy (ΔH), while the other can be expressed in terms of changes in internal energy (ΔE) and work function (ΔA). The former is generally employed and is applicable to all processes, chemical or physical, but in a closed system. (a) In Terms of Free Energy and Enthalpy We know that dG = VdP – SdT At constant pressure dP = 0, then dG = – SdT Let G1 be the initial free energy at temperature T, G1 + dG1, be the free energy at temperature T + dT. Then dG1 = – S1 dT...(i) Where S1 is the entropy of the system in the initial state. Now suppose the free energy of the system in the final state is G2 at temperature T, and G2 + dG2 is the free energy at temperature T + dT in the final state. Then dG2 = – S2 dT...(ii) where S2 is the entropy of the system in the final state. Subtracting (i) from (ii) we get dG2 – dG1 = –(S2 – S1) dT or d (ΔG) = – ΔS dT At constant pressure ⎛ ΔG ⎞ d ⎜ ⎟ = – ΔS...(iii) ⎝ dT ⎠ p We know ΔG = ΔH – T ΔS ΔG – ΔH or – ΔS =...(iv) T Comparing equations (iii) and (iv) ΔG – ΔH ⎛ ΔG ⎞ =d ⎜ ⎟ T ⎝ dT ⎠p ⎛ ΔG ⎞ or ΔG = ΔH + Td ⎜ ⎟ ⎝ dT ⎠p This equation is called Gibb’s Helmholtz equation in terms of free energy and enthalpy change at constant pressure. (b) In terms of Internal Energy and Work Function By definition the work function A = E – TS...(i) or ΔA = ΔE – TΔS ΔA – ΔE or – ΔS =...(ii) T  SECOND LAW OF THERMODYNAMICS 327 Differentiating equation (i) we get dA = dE – TdS – SdT...(iii) From the first law of thermodynamics dq = dE + PdV and at constant volume dV = 0 dq = dE For a reversible change dq dS = T or dq = TdS – dE...(iv) Comparing equations (iii) and (iv) we get dA = – SdT Let A1 be the work function in its initial state at temperature T and A1 + dA1 be the work function in its initial state at T + dT. And A2 be the work function in its final state at temperature T and A2 + dA2 be the work function in its final state at T + dT. Then dA1 = – S1 dT...(v) and dA2 = – S2 dT...(vi) where S1 and S2 are the entropies of the system in initial and final states of the system respectively. Subtracting equation (v) from equation (vi) we get dA2 – dA1 = – (S2 – S1) dT or d (ΔA) = – ΔS dT At constant volume ⎛ ΔA ⎞ d⎜ ⎟ = – ΔS ⎝ dT ⎠v From equation (ii) we have ΔA – ΔE = – ΔS T On comparison we have ΔA – ΔE ⎛ ΔA ⎞ = d⎜ ⎟ T ⎝ dT ⎠v ⎛ ΔA ⎞ or ΔA = ΔE + T d ⎜ ⎟ ⎝ dT ⎠v This equation is called Gibbs Helmholtz equation in terms of internal energy and work function at constant volume. Importance of Gibb’s Helmholtz Equations Gibb’s Helmholtz equations are used to calculate the heats of reaction (ΔH or ΔE) when ΔG or ΔA at two temperatures are given. This point is made clear in the following examples. SOLVED PROBLEM 1. For the reaction 1 H2 (g) + O2 ( g ) ⎯⎯ → H 2O (l ) 2 The values of enthalpy change and free energy change are – 68.32 and – 56.69 kcals respectively. Calculate the value of free energy change at 25ºC. 328 9 PHYSICAL CHEMISTRY SOLUTION. Using the Gibb’s Helmholtz equation ⎛ ΔG ⎞ ΔG = ΔH + Td ⎜ ⎟ ⎝ dT ⎠ p Here ΔG = – 56.69 Kcal ΔH = – 68.32 kcals and T = 273 + 25 = 298 K On substitution we get ⎡ ⎛ ΔG ⎞ ⎤ – 56.69 = – 68.32 + 298 ⎢ d ⎜ ⎟ ⎥ ⎣ ⎝ dT ⎠ p ⎦ ⎛ ΔG ⎞ – 56.69 + 68.32 or d⎜ ⎟ = ⎝ dT ⎠ p 298 = 0.0390 kcal ΔG ⎞ Assuming that d ⎛⎜ ⎟ remains constant over this temperature range. At 30ºC we can write ⎝ dT ⎠ p ΔG = 68.32 + 303 × 0.039 = – 68.32 + 11.817 = – 56.503 kcals SOLVED PROBLEM 2. For the following reaction N2 (g) + 3H2 (g) = 2NH3 (g) The free energy changes at 25ºC and 35ºC are – 33.089 and – 28.018 kJ respectively. Calculate the heat of reaction. SOLUTION. We know ⎛ ΔG ⎞ ΔG = ΔH + Td ⎜ ⎟ ⎝ dT ⎠ p Here G1 = – 33.089 kJ T1 = 273 + 25 = 298 K G2 = – 28.018 kJ T2 = 273 + 35 = 308 K ⎛ ΔG ⎞ – 28.018 – (– 33.089) ∴ d⎜ ⎟ = ⎝ dT ⎠p 308 – 298 = 0.507 kJ At 35ºC ΔG = – 28.018 kJ T = 273 + 35 = 308 K ⎛ ΔG ⎞ ∴ ΔG = ΔH + Td ⎜ ⎟ ⎝ dT ⎠p – 28.018 = ΔH + 308 × 0.507 or ΔH = 28.018 + 156.156 = 184.174 kJ Conditions of Equilibrium and Criterion for a Spontaneous Process (a) In Terms of Entropy Change The entropy of a system remains unchanged in a reversible change while it increases in an irreversible change i.e.  SECOND LAW OF THERMODYNAMICS 329 ∑ dS = 0 For Reversible change and ∑ dS > 0 For Irreversible change For a system with its surroundings we can write dSsystem + dSsurroundings = 0 and dSsystem + dSsurroundings > 0 combining the two relations, we have dSsystem + dSsurroundings ≥ 0...(i) If we assume the change in surroundings as reversible and surroundings evolve dq heat reversibly, then dq dSsurrounding = – T From the first law of Thermodynamics we know dq dE + dw dS = – = T T where dE is the increase in internal energy of the system and dw be the work done by the system. From equation (i) we get dE + dw dSsystem – ≥0 T or TdSsystem – dE – dw ≥ 0 or TdSsystem – dE – PdV ≥ 0...(ii) [∵ dw = PdV] Case I. When E and V are constant In this case dE = 0 and dV = 0 the equation (i) reduces to dSE.V ≥ 0 or dSE.V > 0 for an irreversible change (spontaneous) and dSE.V = 0 for a reversible change (equilibrium) Case II. When S and V are constant Here dS = 0 and dV = 0 The equation (i) reduces to – dE ≥ 0 or – dE > 0 for an irreversible change (spontaneous) and – dE = 0 for a reversible change (equilibrium) (b) In Terms of Enthalpy Change We know H = E + PV on differentiating dH = dE + PdV + VdP or – dE – PdV = – dH + VdP Putting in equation (ii) TdS – dH + VdP ≥0 330 9 PHYSICAL CHEMISTRY or dH – VdP – TdS ≤ 0 At constant S and P Here dS = 0 and dP = 0 ∴ we have dH ≤ 0 or dH < 0 for an irreversible change (spontaneous) and dH = 0 for a reversible change (equilibrium) (c) In Terms of Free Energy Change We know G = H + TS or G = E + PV + TS [∵ H = E + PV] on differentiating we get dG = dE + PdV + VdP + TdS + SdT TdS – dE – PdV = – dG + VdP – SdT Substituting in equations (ii) we get – dG + VdP – SdT ≥ 0 or dG – VdP + SdT ≤ 0 At constant pressure in an isothermal process (T is also constant) this equation reduces to dG ≤ 0 or dG < 0 for an irreversible change (spontaneous) and dG = 0 for a reversible change (equilibrium) Thus the conditions for spontaneity and equilibrium may be summed up in the Table 9.2. TABLE 9.2. CONDITIONS FOR SPONTANEITY AND EQUILIBRIUM Conditions Irreversible Process Reversible Process (Spontaneous) (Equilibrium) At Constant E, V dS > 0 dS = 0 At Constant S, V – dE > 0 – dE = 0 At Constant S, P dH < 0 dH = 0 At Constant P, T dG < 0 dG = 0 THE CLAPEYRON EQUATION A useful thermodynamic relation which gives us important information about a system consisting of any two phases of a single substance in chemical equilibrium is the Clapeyron equation. It is derived from the Gibbs-Helmholtz equation mentioned above. Let the system studied be Liquid U Vapour Consider one gram mole of a liquid confined in a cylinder by a frictionless piston. Let the volume of the liquid be V1 and its vapour pressure equal to p. Now allow the liquid to evaporate reversibly at a constant temperature T and when the whole of it has vaporised, let the volume of the vapour be V2. ∴ Work done during evaporation w = p (V2 – V1)...(i) Differentiating equation (i) with respect to temperature at constant (V2 – V1), we get,  SECOND LAW OF THERMODYNAMICS 331 dw dp = (V2 – V1 )...(ii) dT dT Heat absorbed from the surroundings is the latent heat of vaporisation L which on substitution in the first law equation gives us ΔE = w – L...(iii) On substitution of expressions (ii) and (iii) in the Gibbs-Helmholtz equation, we have dw w + ΔE = T dT dp w + (L – w) = T (V2 – V1) dT dp or L=T (V2 – V1 ) dT dp L or = dT T (V2 – V1 ) This is the Clapeyron equation which in its general form may be written as dp ΔH =...(1) dT T (V2 – V1 ) where ΔH is the heat of transition when a volume V1 of a definite weight of one form changes to a volume V2 of the same weight of other form at the temperature T. CLAUSIUS–CLAPEYRON EQUATION The above equation can be simplified by neglecting the small volume of the liquid in comparison with the volume of the vapour. Equation (1) given above in such a case becomes dp ΔH = dT TV Supposing the vapour obeys the ideal gas laws RT V2 = p dp ΔH p = dT RT 2 1 dp ΔH or × = p dT RT 2 1 dp d log e p But × = p dT dT p ΔH ∴ d log e =...(2) dT RT 2 Equation (2) above is known as the Clausius-Clapeyron equation and though approximate is of very great value. If ΔH is regarded as constant, we may integrate the above equation dT d log e P = ΔH RT 2 ΔH dT ∫ d log e p = R ∫ T 2 332 9 PHYSICAL CHEMISTRY – ΔH or log e p = + Constant RT – ΔH log10 p = +C 2.303 RT If p1 is the vapour pressure at T1 and p2 the vapour pressure at T2 we have – ΔH log10 p1 = +C...(i) 2.303 RT1 – ΔH and log10 p2 = +C...(ii) 2.303 RT2 Subtracting (i) from (ii) p2 ΔH ⎛ 1 1 ⎞ log10 = – 2.303 R ⎜⎝ T1 T2 ⎟⎠...(3) p1 APPLICATIONS OF CLAPEYRON-CLAUSIUS EQUATION (1) Calculation of Latent Heat of Vaporisation If the vapour pressure of a liquid at two temperatures T1 and T2 be p1 and p2 respectively, the molar heat of vaporisation ΔHv can be calculated by substituting these values in Clapeyron-Clausius equation. SOLVED PROBLEM. If the vapour pressures of water at 95ºC and 100ºC are 634 and 760 mm respectively, calculate the latent heat of vaporisation per mole. SOLUTION The Clapeyron Clausius equation states that p2 ΔH v ⎡1 1⎤ log10 = ⎢T – T ⎥ p1 2.303 R ⎣ 1 2⎦ In this case, we have : p 1 = 634 mm p 2 = 760 mm T1 = 273 + 95 = 368 K T2 = 273 + 100 = 373 K R = 1.987 cal ΔHv = ? Substituting the above values in the Clapeyron-Clausius equation, we have : 760 ΔH v ⎡ 1 1 ⎤ log = – 634 2.303 R ⎢⎣ 368 373 ⎥⎦ ∴ ΔHv = 9886 cal mol–1 In a similar manner, if the vapour pressures at two different temperatures of a solid in equilibrium with its liquid phase are known, the latent heat of fusion can be calculated. (2) Calculation of Boiling Point or Freezing Point If the freezing point or the boiling point of a liquid at one pressure is known, it is possible to calculate it at another pressure by the use of the Clapeyron-Clausius equation.  SECOND LAW OF THERMODYNAMICS 333 SOLVED PROBLEM. At what temperature will water boil under a pressure of 787 mm? The latent heat of vaporisation is 536 cals per gram. SOLUTION The data is follows p 1 = 760 mm T1 = 373 K p 2 = 787 mm T2 = ? ΔHv = 536 × 18 cal mol –1 According to Clapeyron-Clausius equation, we have : p2 ΔH v ⎡ 1 1⎤ log10 = ⎢ – ⎥ p1 4.576 ⎣ T1 T2 ⎦ Substituting the above values, we have : 787 536 × 18 ⎡ T2 – 373 ⎤ log10 = 760 4.576 ⎢⎣ 373 T2 ⎥⎦ T2 = 374 K ∴ Water will boil at 101ºC under a pressure of 787 mm. (3) Calculation of Vapour Pressure at Another Temperature If the mean heat of vaporisation is available, it is possible to calculate the vapour pressure of a liquid at given temperature if the vapour pressure at another temperature is known. SOLVED PROBLEM. Calculate the vapour pressure of water at 90.0ºC if its value at 100.0ºC is 76.0cm. The mean heat of vaporisation of water in the temperature range 90º and 100ºC is 542 calories per gram. SOLUTION In this problem, we have : ΔHv = 542 × 18 cal per mole p2 = ? p 1 = 76.0 cm T2 = 90 + 273 = 363 K T1 = 100 + 273 = 373 K Using the Clapeyron-Clausius equation we have : p2 ΔH v ⎡ T2 – T1 ⎤ log10 = p1 4.576 ⎢⎣ T1 T2 ⎥⎦ p2 542 × 18 ⎡ 363 – 373 ⎤ log10 = or 76 4.576 ⎢⎣ 363 × 373 ⎥⎦ ∴ p 2 = 52.88 cm or 528.8 mm (4) Calculation of Molal Elevation Constant The molal elevation constant (Kb) of a solvent is defined as the elevation in boiling point which may theoretically be produced when one mole of any solute is dissolved in 1000 g of the solvent. Accordingly, if w g of a solute of molecular weight M is dissolved in 1000 g of the solvent and ΔT is the elevation produced, the molal elevation Kb is given by the equation 334 9 PHYSICAL CHEMISTRY M ΔT Kb = w Let the boiling point of the pure solvent be T and that of the solvent (T + ΔT) when the atmospheric pressure is p1. While p1 is the vapour pressure of the solution at (T + ΔT) and is also the vapour pressure of the solvent at T, p2 the vapour pressure of the solvent at (T + ΔT) can be calculated by the application of Clapeyron-Clausius equation. p2 L ⎡1 1 ⎤ log e = ⎢ – p1 R ⎣ T (T + ΔT ) ⎥⎦ L ⎡ ΔT ⎤ = R ⎣ T (T + ΔT ) ⎥⎦ ⎢ L ΔT = × when ΔT is very small. R T2 p2 ⎛ p – p1 ⎞ p2 – p1 Now log e = log e ⎜ 1 + 2 = p1 ⎟⎠ since (p2 – p1) is very small and the remainder p1 ⎝ p1 of the terms can be neglected. According to Raoult’s law, the relative lowering of vapour pressure in a dilute solute is equal to the molar fraction of the solute in solution p2 – p1 n ∴ = p2 N p2 – p1 When the difference between p2 and p1 is very small as has just been supposed, may p2 p2 – p1 be taken equal to. p1 p2 – p1 n L ΔT ∴ = = p2 N RT 2 RT 2 n or ΔT = × L N w W But n= and N = m M RT 2 wM ∴ ΔT = × L mW If 1 mole of the solute is dissolved in 1000 g of the solvent, the above equation becomes : RT 2 ΔT = L × 1000 M ΔT But Kb = w M RT 2 ∴ Kb = L × 1000 L But = l, the latent heat of vaporisation per gram of the solvent. M  SECOND LAW OF THERMODYNAMICS 341 If there is a small change in the temperature and pressure of the system as well as the amounts of its constituents, the change in the property X is given by ⎛ δX ⎞ ⎛ δX ⎞ ⎛ dX ⎞ dX = ⎜ ⎟ dT +⎜ ⎟ dp +⎜ ⎟ dn1 ⎝ δT ⎠ P n1 , n2..... n j ⎝ δ p ⎠T n1 , n2..... n j ⎝ δ n ⎠T , P n2.....n j ⎛ δX ⎞ ⎛ δX ⎞ +⎜ ⎟ dn2 +⎜ ⎟ dn3 +..... + ⎝ δn2 ⎠T , P n1 , n3.....n j ⎝ δn3 ⎠T , P n1 , n2.....n j ⎛ δX ⎞ ⎜ δn ⎟ dn j...(ii) ⎝ j ⎠T , P n1 , n2..... ⎛ δX ⎞ The quantity ⎜ δn ⎟ is called partial molar property for the constituent 1. It is represented ⎝ 1 ⎠T , P , n2.....n j by writing a bar over its symbol for the particular property i.e. X so that ⎛ δX ⎞ ⎛ δX ⎞ X1 = ⎜ ⎟ ; X2 = ⎜ ⎟ ⎝ δn1 ⎠T , P , n2....n j ⎝ δn2 ⎠T , P , n1 , n3....n j The equation (ii) may be written as : ⎛ δX ⎞ ⎛ δX ⎞ dX = ⎜ ⎟ dT + X =⎜ ⎟ dP ⎝ δT ⎠ P, n1 , n2.....n j ⎝ δP ⎠T , n1 , n2.....n j + X 1 dn1 + X 2 dn2 + X 3 dn3 +..... X j dn j If the temperature and the pressure of the system are kept constant dT and dP are zero so that dX = X 1 dn1 + X 2 dn2 +..... X j dn j...(iii) and this on integration for a system of definite composition represented by the number of moles n1, n2..... nj gives X = n1 X1 + n2 X + n3 X 3 +.....n j X j...(iv) i.e., the partial molal property X of any constituent may be regarded as the contribution of 1 mole of that constituent to the total value of the property of the system under specified conditions. Partial Molar Free Energy : Chemical Potential If the extensive property under study is free energy (G), G will represent the partial molar free energy so that ⎛ δG ⎞ ⎛ δG ⎞ G1 = ⎜ ⎟ and G j = ⎜ ⎟ δ ⎝ 1 ⎠T , P , n n.....n n ⎝ δn j ⎠ 2 3 j T , P , n1 n2.....n j –1 This quantity is, for most purposes, identical with the function known as chemical potential represented by the symbol μ. Accordingly we have ⎛ δG ⎞ μ1 = G1 = ⎜ ⎟ ⎝ δn1 ⎠T , P , n2.....n j Thus, it is the partial derivative of the free energy with ni when all other variables are kept constant. Physical Significance of Chemical Potential By definition the chemical potential of a given substance is the change in free energy of the system produced on addition of one mole of the substance at constant temperature and pressure to a 342 9 PHYSICAL CHEMISTRY large bulk of the mixture so that its composition does not undergo any change. It is an intensive property and it may be regarded as the force which drives the chemical system to equilibrium. At equilibrium the chemical potential of the substance in the system must have the same value through the system. In other words, the matter flows spontaneously from a region of high chemical potential to low chemical potential. The chemical potential may also be regarded as the escaping tendency of that system. Greater the chemical potential of a system greater will be its escaping tendency. Gibbs Duhem Equation It has already been discussed that free energy G is an intensive thermodynamic property. It can be determined by fixing the variables T, P and number of moles of various constituents (composition of the mixture under study). Mathematically, we can write. G = f (T, P, n1, n2..... nj)...(i) where n1, n2....... nj are the number of moles of various constituents. Differentiating equation (i), we get ⎛ δG ⎞ ⎛ δG ⎞ dG = ⎜ ⎟ dT +⎜ ⎟ dP + ⎝ δT ⎠ P , n1 , n2.....n j ⎝ δP ⎠T , n1 , n2.....n j ⎛ δG ⎞ ⎛ δG ⎞ ⎛ δG ⎞ ⎜ δn ⎟ dn1 + ⎜ ⎟ dn2 + ⎜ ⎟ dn j ⎝ 1 ⎠T , P , n2.....n j ⎝ 2 ⎠T , P , n1.....n j ⎝ δn j ⎠T1 , P , n1.....n j –1 δn We know the chemical potential is given by ⎛ δG ⎞ μi = ⎜ ⎟ = G1 ⎝ δni ⎠T , P , n1 , n2..... Substituting in equation (ii) we get ⎛ δG ⎞ ⎛ δG ⎞ dG = ⎜ ⎟ dT + ⎜ ⎟ dP + μ1 dn1 , + μ 2 dn2 , +.... + μ j dn j...(iii) ⎝ δT ⎠ P , n1 , n2.....n j ⎝ δP ⎠T , n1 , n2.....n j For a closed system there is no change in the composition and equation (iii) reduces to ⎛ δG ⎞ ⎛ δG ⎞ dG = ⎜ ⎟ dT + ⎜ ⎟ dP...(iv) ⎝ δT ⎠ P, n1 , n2.....n j ⎝ δP ⎠T , n1 , n2.....n j But we know dG = – SdT + VdP...(v) [∵ G = H – TS H = E + PV and G = dE + PdV + VdP – TdS – SdT] comparing equation (iv) and (v) ⎛ δG ⎞ ⎜ ⎟ =–S ⎝ δT ⎠ P ,n1 n2....n j ⎛ δG ⎞ and ⎜ ⎟ = –V ⎝ δT ⎠T ,n1 n2....n j Putting these values in equation (iii) we get dG = – SdT + VdP + μ1 dn1 + μ2 dn2 +..... μj dnj...(vi) At constant temperature and pressure equation (vi) reduces to (dG)T.P = μ1 dn1 + μ2 dn2 +..... μj dnj...(vii) 346 9 PHYSICAL CHEMISTRY 3. (a) Derive an expression for entropy change for ideal gas associated with temperature and pressure changes. (b) Calculate the total entropy change when 5 grams of ice at 0°C is converted into steam at 100°C. (Latent heat of evaporation of water = 540 cals/g; Cp for water = 18 cals/moles; Latent heat of water = 80 cal/mole) Answer. (b) 10.265 cal/degree 4. Calculate the work performed when two gram of hydrogen gas is expanded isothermally and reversibly at 27°C from 10 to 100 litres. What is the amount of heat absorbed? What is the change in the internal energy? Answer. 1372.81 cal; Zero 5. Two moles of an ideal gas undergo isothermal reversible expansion from 15 lit to 30 lit at 300 K. Calculate the work done and the change in entropy. Answer. 826.5 cal; 2.755 cal K–1 6. Water boils at 373 K at one atm pressure. At what temperature will it boil at a hill station where the atmospheric pressure is 500 mm Hg? (Latent heat of vaporisation of water is 2.3 kJ g–1 and R = 8.314 JK–1 mol–1) Answer. 361.65 K 7. A Carnot’s engine works between the temperature 27° and 127°C. Calculate the efficiency of the engine. Answer. 25% 8. (a) Explain the term Thermodynamic efficiency. (b) Calculate the work done on the system if one mole of an ideal gas at 300 K is compressed isothermally and reversibly to one-fifth of its original volume. Answer. (b) 4014.98 × 107 ergs 9. What do you understand by the term enthalpy? Calculate the work done in the expansion of 3 moles of hydrogen reversibly and isothermally at 27°C from 21.0 litres to 70.3 litres. Answer. 9042.47 × 107 ergs 10. (a) “It is not profitable to carry out a process reversibly although maximum work can be obtained by doing so”. Comment. (b) Two moles of hydrogen are compressed adiabatically from NTP conditions to occupy a volume of 4.48 litres. Calculate the final pressure and temperature (γ = 1.41) (c) Derive a relation between pressure and volume for an adiabatic reversible expansion of an ideal gas. Answer. (b) 25.7 atm; 429.1°C 11. (a) Derive the Clapeyron-Clausius equation giving the temperature dependence of water pressure indicating clearly the assumption involved. (b) An engine operates between 100°C and 0°C and another engine operates between 100°C and 0 K (absolute zero). Find the efficiencies in two cases. Answer. (b) 26.8%; 100 % 12. (a) Explain giving reason “The net entropy of the universe tends to increase”. (b) For the reaction H2(g) + ½O2(g) → H2O( ) the values of enthalpy change and free energy change are –68.32 and –56.69 kcal respectively at 25°C. Calculate the value of free energy change at 30°C. (c) Write down the applications of Gibb’s Helmholtz equation. Answer. (b) –56.495 kcal 13. (a) Bring about clearly the criteria for reversibility and irreversibility in terms of S, E, H and G. (b) 1.0 mole of steam is condensed at 100°C and water is cooled to 0°C and frozen to ice. Calculate the  SECOND LAW OF THERMODYNAMICS 347 entropy change for the process. Latent heat of fusion of ice and evaporation of water are 80 and 540 cals/g respectively. Answer. (c) 25.716 cal deg–1 mol–1 14. (a) Write a note on “Carnot’s Cycle”. (b) How are work function and free energy related? Discuss the criteria of spontaneity of a chemical reaction. (c) Calculate the entropy increase in the evaporation of a mole of water at 100°C (Heat of vaporization = 540 cal g–1) 15. (a) Give the expression for Gibb’s free energy change (ΔG) for the reaction nA + mB pC + qD (b) How is ΔG of a chemical reactions is related to ΔS and ΔH? (c) The enthalpy change for the transition of liquid water to steam is 40.8 kJ mol–1 at 373 K. Calculate ΔG for the process. Answer. (b) 109.38 3 JK–1 mol–1 16. Calculate ΔS when 28 gm of N2 gas expands reversibly from 2 litres to 20 litres at 27°C. Answer. 38.294 JK–1 17. (a) Prove that Enthalpy remains constant when a real gas passes through a porous plug in adiabatic expansion. (b) Explain the relationship between entropy and probability. (c) Derive an expression for the efficiency of a Carnot’s engine working between the two temperatures T1 and T2. 18. (a) Define standard heat of formation and standard entropy change of a reaction. (b) Explain why in case of non-polar solvents, the ΔS is nearly equal to 88 JK–1 mol–1. (c) Calculate the entropy change for the reaction : N2(g) + 3H2(g) → 2NH3(g) Given : S° of N2, H2 and NH3 as 191.5, 130.6 and 192.2 JK–1 mol–1 respectively. Answer. –198.9 J K–1 mol–1 (Delhi BSc, 2001) 19. (a) Determine the entropy change for an ideal gas when temperature and volume are varied. (b) Calculate the entropy change involved in isothermal expansion of 2 moles of the gas from a volume of 5 litres to a volume of 50 litres at 30°C. Answer. 38.29 JK–1 (Guru Nanak Dev BSc, 2002) 20. Calculate entropy change if 2 moles of water at 373 K are evaporated to vapours at 373 K. Give its units also. (Given molar heat of vaporisation of water is 9650 cal) Answer. 216.49 JK–1 (Andhra BSc, 2002) 21. 0.5 g of nitrogen is enclosed in a cylinder fitted with a piston at 25°C. Calculate the change in entropy, if the gas is expanded adiabatically to double its volume. Answer. zero (Guru Nanak Dev BSc, 2002) 22. (a) Starting from appropriate definition of chemical potential of a component in an ideal gas solution, derive an expression for ΔGmixing for the formation of ideal binary solution and show further that ΔHmixing=0 for the solution. (b) A solution is prepared by mixing 2 moles of CS2 and 3 moles of CCl4 at 298 K and 1 atm pressure. Assuming ideal behaviour, calculate ΔGmixing for the solution. Answer. –3620.22 J (Guru Nanak Dev BSc, 2002) 23. Calculate entropy change for the fusion of one mole of a solid which melts at 300 K. The latent heat of fusion is 2.51 k J mol–1. Answer. 8.366 J (Arunachal BSc, 2003) 24. Calculate the change in entropy when 3 moles of an ideal gas is heated from 323 K to 423 K at a constant volume. (Cv = 32.94 JK–1 mol–1) Answer. 26.658 JK–1 (Nagpur BSc, 2003) 348 9 PHYSICAL CHEMISTRY 25. Calculate the value of dT/dP for water ice system at 273 K. ΔHf for ice is 6007.8 J mol–1; Molar Volume of water and ice are 0.018 dm3 mol–1 and 0.1963 dm3 mol–1 respectively. (Given 1 J = 9.87 × 103 dm3 atm) Answer. 0.0100654 K atm–1 (Nagpur BSc, 2003) 26. Calculate standard Gibb’s Free energy change for the combustion of methane : CH4(g) + 2O2(g) → CO(g) + 2H2O(g) at 25°C. ΔH° = –191.8 kcal and ΔS° = 1.2 cal K–1 Answer. –549.4 cal (Sambalpur BSc, 2003) 27. ΔG and ΔH values for a reaction at 300 K are : –66.944 kJ and –41.84 kJ respectively. Calculate the free energy change at 330 K, assuming that ΔH and ΔS remain constant over this temperature range. Answer. –69.454 kJ (Nagpur BSc, 2003) 28. (a) Derive the integral Clausius-Clapeyron equation in the form P2 ΔH ⎡ T2 − T1 ⎤ log = ⎢ ⎥ P1 2.303 R ⎣ T1 T2 ⎦ for an ideal gas. (b) At 373.6 K and 372.6 K the vapour pressure of H2O( ) are 1.018 and 0.982 atm respectively. What is the heat of vaporization of water? (R = 1.987 cal) Answer. (b) 41675.8 J (Jamia Millia BSc, 2003) 29. Explain the term fugacity. How is fugacity of a gas determined? (Panjab BSc, 2003) 30. (a) State Carnot’s theorem and second law of thermodynamics. (b) Define Chemical potential. Derive effect of temperature and pressure on chemical potential. (Indore BSc, 2004) 31. Explain the following : (a) Under what conditions can an isothermal expansion of a gas become a free expansion process. (b) Increase in volume of a gas for a given decrease in pressure is less in an adiabatic expansion than in isothermal expansion. (c) All spontaneous processes lead to increase the entropy of the universe. (d) Free Energy of formation of an element at 1 atm and 298 K is assumed to be zero but entropy is not zero under the same conditions. (Gulbarga BSc, 2004) 32. (a) Describe Carnot’s cycle for establishing the maximum convertibility of heat into work. How does it lead to the definition of Second Law of Thermodynamics? (b) Heat supplied to a Carnot engine is 1897.86 kJ. How much useful work can be done by the engine which works between 0°C and 100°C? Answer. (b) 508.80 kJ (Madurai BSc, 2004) 33. What do you understand by the term enthalpy? Calculate the work done in the expansion of 3 moles of hydrogen reversibly and isothermally at 27°C from 21.0 litres to 70.3 litres. Answer. 9042.47 × 107 ergs (Kakatiya BSc, 2004) 34. The heat of vaporisation, ΔHvap of carbon tetrachloride, CCl4, at 25 ºC is 43 kJ mol–1. If 1 mole of liquid carbon tetrachloride at 25 ºC has an entropy of 214 JK–1, what is the entropy of 1 mole of the vapour in equilibrium with the liquid at this temperature ? Answer. 358 JK–1 mol–1 (Kalyani BSc, 2005) 35. Calculate the amount of the heat supplied to Carnot cycle working between 105 ºC and 20 ºC if the maximum work obtained is 200 cal ? Answer. 889.4 cal (Panjab BSc, 2005)  SECOND LAW OF THERMODYNAMICS 349 36. The enthalpy change involved in oxidation of glucose is – 2880 kJ mol–1. Twenty five percent of this energy is available for muscular work. If 100 kJ of muscular work is needed to walk one kilometer, what is the maximum distance a person will be able to walk after eating 120 g of glucose ? Answer. 4.8 km (Delhi BSc, 2005) 37. What is the entropy change for conversion of one mole of ice to water at 273 K and 1 atm pressure (Given ΔH of ice = 6.025 kJ mol–1). Answer. 22.069 J mol–1 (Bundelkhand BSc, 2005) 38. Calculate the efficiency of steam engine operating between 100 ºC and 25 ºC. What would be the efficiency of the engine if the boiler temperature is raised to 150 ºC, the temperature of the sink remaining same ? Answer. 22.1% ; 29.55% (Agra BSc, 2006) 39. At 373 K the entropy change for the transition of liquid water to steam is 109 JK mol–1. Calculate the –1 enthalpy change ΔHvap for the process. Answer. 40.657 kJ mol–1 (Madurai BSc, 2006) 40. Ethanol boils at 78.4 ºC and standard enthalpy of vaporisation of ethanol is 42. kJ mol–1. Calculate the entropy of vaporisation of ethanol. Answer. 120.66 JK–1 mol–1 (Barodra BSc, 2006) MULTIPLE CHOICE QUESTIONS 1. A process which proceeds of its own accord, without any outside assistance, is called (a) non-spontaneous process (b) spontaneous process (c) reversible process (d) irreversible process Answer. (b) 2. The tendency of a process to occur naturally is called (a) momentum of the reaction (b) spontaneity of the reaction (c) equilibrium of the reaction (d) equilibrium of the reaction Answer. (b) 3. Which of the following is true about the criteria of spontaneity? (a) a spontaneous change is unidirectional (b) a spontaneous change to occur, time is no factor (c) once a system is in equilibrium, a spontaneous change is inevitable (d) all of the above Answer. (d) 4. A spontaneous change is accompanied by _______ of internal energy or enthalpy. (a) increase (b) decrease (c) neither increase nor decrease (d) none of these Answer. (b) 5. Mixing of two or more gases is a (a) spontaneous process (b) non-spontaneous process (c) reversible process (d) none of these Answer. (a)