Thermodynamics PDF

Thermodynamics Thermodynamics y A branch of chemistry which gives information about the flow of heat. Limitations y Doesn’t give information about the rate of reaction. y Thermodynamics process are not applicable for micro system such as e–, p+, n etc...

Thermodynamics Thermodynamics y A branch of chemistry which gives information about the flow of heat. Limitations y Doesn’t give information about the rate of reaction. y Thermodynamics process are not applicable for micro system such as e–, p+, n etc Thermodynamics 1.  y For a system the change in energy is identical in magnitude but opposite in sign Rack your Brain to the change in energy of its surrounding. An open container and open system. Are they same ? Types of system 1. Open system : The type of system where both mass and Concept Ladder heat transfer takes place. with surrounding. Example. Boiled water in an open vessel In lab experiments, the 2. Closed system : room is considered as surrounding. The type of system where only heat transfer Uuniv  Usys  Usurr  0 takes place with surrounding but there is no Usys   Usurr mass transfer. Example. Boiled water in a closed vessel. 3. Isolated system : Rack your Brain The type of system where neither heat transfer nor mass transfer takes place. Can you compare the relative Thermodynamics Example. Boiled water in thermo flask. magnitude of the change in energy of system and surrounding? 2. y The universe is isolated, because it contains everything by definition, and thus there can Concept Ladder be no exchange of energy with anything. Reactants undergo reaction to decrease Universe is considered as an their energy and will proceed until they isolated system. So, all the reach a state of low energy and will remain laws applicable for universe in this state unless disturbed. This state is are applicable for isolated called equilibrium. system. Q.1 Write down system of following: 1. Helium filled balloon 2. Coffee in a thermos flask 3. The earth 4. Satellite in an orbit 5. Human being 6. Refrigeration cycle A.1 1. Helium filled balloon Closed 2. Coffee in a thermos flask Isolated 3. The earth Open 4. Satellite in an orbit Closed 5. Human being Open 6. Refrigeration cycle Closed Thermodynamic properties 1. Intensive properties : (i) Those properties which are independent of mass. Thermodynamics (ii) Their values remain uniform throughout system. (iii) They are non-additive. 3.  2. Extensive properties : (i) Those properties which depend on Rack your Brain mass. (ii) Ratio of two extensive properties Is Pressure an intensive property? become an intensive property. (iii) They are additive. Extensive properties Intensive Properties Volume Molar volume Number of Moles Density Mass Refractive index Free Energy (G) Surface tension Entropy Viscosity Enthalpy Free energy per mole Internal Energy (E & U) Specific heat Thermodynamics Pressure, Temperature, Boiling Heat Capacity point, freezing Point. Etc. 4. State Functions Those thermodynamic properties which Concept Ladder depend on initial and final state. Both q and w are not state Eg: Pressure, volume, temperature, Gibb’s free functon sice their values energy, internal energy, entropy etc. depend upon the path Path function by which the change is Those properties which depend on path. carried, but the quality q + w is a state function, this is Eg: Heat, Work, Loss of energy due to friction. because q + w = DV and DU is a state function. Thermodynamic Process Rack your Brain The change of thermodynamic state from Thermodynamics one condition to another condition is called What are the condition for the thermodynamic process. various process can a occur? 5.  Rack your Brain When a system undergoes a change at constant pressure then the process will be? Initial and5 final molestate of O is is same Q.2 2 expanded from 1L to 10L at 300K then which relation is correct? (i) DE = 0 (ii) DH = 0 (iii) W = 0 (iv) DS = 0 (1) i,ii,iii (2) ii,iii (3) i,ii (4) ii,iii,iv A.2 (3) As moles and temperature are constant DH = DE + DngRT DH = 0 Reversible process Irreversible process Driving force is infinitesimally Driving force is large and finite. small. PV work is done across PV work is done across pressure pressure difference dP. difference DP. A reversible heat transfer takes Irreversible heat transfer take place across temperature place across difference DT. difference dT. It is an ideal process. It is a real process. It takes infinite time for It takes finite time for completion of process. completion of process. It is an imaginary process and It is a natural process and Thermodynamics can not be realised in actual occurs in particular direction practice. under given set of conditions. 6.  The system is far away from Throughout the process, the state of equilibrium and exact system remain infinitesimally path of process can not be closer to state of equilibrium defined as different part of and exact path of process the system are under different can be drawn. conditions. Thermodynamics 7.  1. State Functions (i) Those thermodynamic properties Rack your Brain which depend on initial and final state (ii) e.g DE or DU Why the energy of system at DH equilibrium is minimum? DS DG etc. 2. Path function (i) Those properties which depend on path (ii) e.g Work and heat General Terms : Those properties which depends on path ; Concept Ladder e.g Work, Heat 1. Work : The negative work sign (i) Thermodynamic work represents decrease in W = + PDV energy content of system. 2. Sign Convention During compression, the (i) Compression : (Positive) → Work done sign of dV is negative which on the system gives positive value of W (ii) Expansion : (Negative) → Work done representing the increase by the system, means expansions in energy content of system 3. Unit of work : atm × litre during compression. 1 atm × lit = 24.23 cal 1 atm × lit = 101.3 J 1 cal = 4.18 J 1 J = 107 erg 1 J = 0.24 calorie 1 atm × litre > 1 cal > 1 joule > 1 erg Thermodynamics 8. y For a small displacement dx due to force F, work done on the system. Concept Ladder dW = F.dx Also F = PA If the system is at lower dW = PA.dx temperature than the [Here P = pressure, A = Area, V = volume] surroundings the energy is V = (l–x)A gained by the system from ⇒ dV = –A. dx the surrounding causing a ⇒ dW = –Pext. dV rise in the temperature of v2 the system. ⇒ WPV = - ∫ Pext.dV v1 4. Heat (Q) (i) By difference in temperature the total amount of energy transferred from one body to another is known as Heat. (ii). Sign convention : Heat absorb by the system (+) Heat evolved by the system (–) Definitions Internal Energy (E or U) Or Hidden Energy It is the amount of heat 1. Sum of various type of energy related with a evolved or absorbed when a system is known as internal energy chemical reaction is carried E or U = P.E + K.E + T.E +...... out at constant volume and 2. Energy due to gravitational pull is not temperature. considered in internal energy 3. It is impossible to calculate the absolute value of internal energy because it is not possible to calculate the exact value of all type of energy at a time 4. Internal energy is an extensive property 5. It is a state function 6. Relation between Internal Energy & Pressure Rack your Brain for 1 mole of ideal gas and per unit volume Why there is no change in internal 3 In ideal gas internal energy = K.E = RT energy in a cyclic process? 2 Thermodynamics (1 mole) PV = nRT P × 1 = 1 × RT 9.  P = RT 3 U= P 2 3 Concept Ladder I.E = K.E = P 2 7. The properties which arise out of collective The macroscopic energy behavior of large number of chemical changes with velocity and entities. elevation of the system are Example. Pressure, volume temperature, not considered in internal composition, colour refractive index etc. energy change of system. Zeroth Law of Thermodynamics When two different system are in thermal equilibrium with 3rd system separately then will also in thermal equilibrium with each other Thermodynamics 10. Types of Thermodyanmic Processes 1. Isobaric process (i) Pressure → constant DP = 0 (ii) Thermodynamic work W   PV Rack your Brain W = F × dl = P × A ×dl When a system undergoes a = Pressure × Area × change in length change at constant pressure = P × V = D(PV) then the process will be? W = P (DV) because P constant (iii) If in question process is not given then we will consider Isobaric because maximum processes are carried out in open vessel where pressure is constant. 2. Isochoric Process : (i) Volume → Constant DV = 0 (ii) Thermodynamics work W = 0 Thermodynamics 11.  Note : If not given then we may consider. 3. Isothermal Process (i) Temperature → constant DT = 0 (ii) Ideal gas equation, PV = nRT Concept Ladder So, PV = Constant P1V1 = P2V2 Generally all non reacting (iii) For ideal gas and isothermal process, gases such as H2, O2, N2, Ne then Change in internal energy DU = 0 etc are considered as an Internal Energy = Kinetic Energy = 3/2 ideal gas. RT Internal Energy ∝ Temperature = DI.E. ∝ DT = DU = 0 (iv) Ideal gas Isothermal & moles are constant : Change in enthalpy DH = 0 DH = DU + DngRT Rack your Brain DH = 0 (vi) Graph log P v/s log V Why work done during the For isotherm = PV = K isothermal reversible process is log P + log V = log K greater than adiabatic process? log P = –log V + log K y = mx + c Thermodynamics 12.  Concept Ladder log P v/s log V log PVg = log K 4. Adiabatic process [log P + g log V] = log K log P = – g log V + log K (i) No transformation of heat with surrounding means Q =0 (ii) (a) PVg = constant (b) TVg–1 = constant (c) TP(1–g)/g = constant g = Poison Ratio C p  Cv PV   K Rack your Brain In adiabatic expansion, final volume is less than that of an isothermal expansion. Why? Thermodynamics 13.  Rack your Brain Why the energy of system at equilibrium is minimum? (iv) Slope of PV graph of adiabatic process is more than isothermal. 5. Equilibrium : It is process for no change in thermodynamic property (P,V,T etc) of system with time. 6. Cyclic process : Previous Year’s Question (i) If a system goes through different changes and finally obtains its initial An ideal gas expands isothermally position then this process is known as from 10–3 m3 to 10–2 m3 at 300 K cyclic process. against a constant pressure of (ii) DU = 0 ; DH = 0 105 N m-2. The work done on the gas is [NEET] (1) +270 kJ (2) –900 J (3) + 900 kJ (4) –900 kJ Work done in Reversible process 1. Isobaric Process Rack your Brain Work done = + [VH – VL] 2. Isochoric Process (DV =0) At which condition in So, Work done = 0 Thermodynamics thermodynamics a process is called reversible ? 14. 3. Isothermal process T  W   2.303 nRT log  H   TL  Concept Ladder R = 8.31 Joule/ K/ mole R = 2 cal / K/ mole For an isothermal isobaric R = 0.082 atm × litre /K/mole expansion, At constant T 4. Adiabatic Process and P nR q = -W W TH  TL  DH = 0, DU = 0 1 y In adiabatic process increment in temperature indicates compression of gas. Work Done In Irreversible Process : y Generally these processes are carried out in open vessel in which gas show expansion It is two types. 1. Free Expansion Rack your Brain Expansion of gas in vacuum is known as free expansion. In this process, Why maximum work is done is P=0 case of reversible isothermal So, W = 0 expansion process? 2. Intermediate Expansion Gas do work against the external pressure to expand i.e known as intermediate work. W  Pext  V2  V1  [Pext = External Pressure] PV = nRT PDV = DngRT W = –DngRT Note: These formulae are applicable for all Concept Ladder irreversible processes : Relationship between q, (-w) Total Moles  Total Moles  DU and DH in intermediate ng   of gaseous    of gaseous  expansion  Pr oduct   rec tant  0 < Pex (V2-V1) 0 y Rapid change or sudden change indicate (3) q < 0 , DT = 0 and w = 0 that the process is adiabatic (4) q > 0 , DT > 0 and w > 0 y In P–V graph Area under the curve shows the work done. Thermodynamics 16. Q.4 2 lit of a gas is expanded against an external pressure of 2 atm upto 12 lit. then calculate the work done in joule (1) –2206 J (2) –2026 J (3) –1996 J (4) –2006 J A.4 (2) W = –Pext (V2–V1) W= –2 (12–2) = –20 atm lit = – 20× 101.3 J = –2026 J Q.5 2 mole of an ideal gas is expanded in reversible and isothermal process from 2.24 lit. to 22.4 litre. Then find out the amount of work done for expansion in cal. at 300 K. (1)–2763 cal (2) 3276 cal (3) –3276 cal (4) 2763 cal A.5 (4) V W = –2303 nRT log H VL 22.40  = 2.303×2×2×300 log   2.24  = –2.303×1200×1 = 2763.600 = –2763.6 cal W = 2763 cal Q.6 260 g Zn reacts with HCl Zn (s) + 2HCl (l) → ZnCl2 (s) + H2 (g) Then calculate the work done in cal (Zn = 65) (1) –3200 cal (2) +3200 cal (3) –3100 cal (4) +3100 cal A.6 (1) 4 [Zn (s) + 2HCl (l) → ZnCl2 (s) + H2 (g)] 4 Zn + 8 HCl → 4 ZnCl2 + 4 H2 260 =n = 4 65 Dng = 4–0 = 4 Thermodynamics W = DngRT = –4 × 2 × 400 = –3200 cal 17.  Q.7 22 gm CO2 changes from 500 ml, 300 K to 4l reversible and adiabaticaly then find out (i) final temperature, (ii) work done in cal (g = 4/3) (1) 150 K, –450 cal (2) 250 K, –450 cal (3) 150 K, –250 cal (4) 250 K, –250 cal A.7 (1) (i) T1V1  1  T2 V2  1  1 4 1 T1  V2   4000  3     T2  V1   500  = 1/3 = 1/3 300 ⇒ =2 T2 ⇒ T2 = 150 K nR (ii) W TH  TL  1 0.5  2 = 4 300  150 1 3 = –1×3 = –450 cal Q.8 2 lit. N2 is at 0° C and 5 atm pressure it is expanded isothermally against an external pressure of 1 atm until the pressure of gas becomes 1 atm. Calculate the amount of work in expansion in Joule. (1) –810.4 J (2) –635.5 J (3) 635.5 J (4) 810.4 J A.8 (4) P1V1 = P2V2 5 × 2 = 1 × V2 V2 = 10 lit W = – Pext (V2–V1) = – 1(10–2) = –8 atm × 1 Thermodynamics = –8 × 101.3 = –810.4 J → Amount work = 810.4 J 18.  Q.9 Indicate the sign of work in following process : 1. PCl5 → PCl3 + Cl2 2. N2 + 3H2 → 2NH3 A.9 1. Dng = 2 – 1 = 1 (+ve) 2. Dng = 2–4 = –2 W = –ngRT W = –ve First Law of Thermodynamics (FLOT) Robert Mayer & Helmholtz 1. Total energy of universe always remains constant. Therefore, energy can neither be Concept Ladder created nor be destroyed. 2. According to FLOT one form of energy can DE = q + w is invalid for be completely converted into another form. open system. 3. Mathematically. Let work done on the system = W Heat absorbed by the system =q Then DE = q + W Thermodynamics 19.  Application of First law of Thermodynamics : Hess’s law of constant heat summation: According to this law the net amount of heat change in the complete process is the same Previous Year’s Question regardless of the method employed when the chemical change can be made to take place in An ideal gas expands isothermally two or more ways which involves one or more from 10–3 m3 to 10–2 m3 at 300 K steps. against a constant pressure of 105 N m3. The work done on the gas is [NEET] (1) +270 kJ (2) –900 J (3) + 900 kJ (4) –900 kJ According to FLOT different thermodynamic processes 1. Isothermal Process U  q  w    DU = 0 ; q = –W  0  q  w   q  w  Concept Ladder   According to FLOT in isothermal process Transfer of heat at constant heat absorb by the system is equal to work volume brings about a done by the system. change in the internal 2. Adiabatic process (q = 0) energy of the system DU = W ; whereas that at constant pressure brings about a  3  IE  T .  RT  IE IE.  T change in the enthalpy of  2  the system. In adiabatic process work done on the system (compression) indicates the increment in internal energy so temperature of system increases. 3. Isochoric Process : (W =0) Thermodynamics DU = qv Means heat at constant volume is known as 20.  internal energy change. 4. Isobaric Process : (P → constant) Concept Ladder Generally these process are carried out in open vessel where gas shows expansion so Unit of pressure FLOT. 1 Pascal = 1 kg m-1s-2 DU = q – W [q = DU + PDV] 1 bar 1 × 105 Pa DU = q – PDV qp = DH 1 atmosphere (atm) = 101,325 Pa y Heat at constant pressure is known as 1 torr = 1/760 atm Note : 1 L-atm = 101.3 J Q.10 In the compression of air 5KJ work is done and amount of heat evolved is 3 KJ then find out DU (1) 2 KJ (2) 4 KJ (3) 5KJ (4) 6 KJ A.10 (1) DU = q + w w = + 5KJ q = –3 KJ DU = q + w = –3 + 5 = 2 KJ Q.11 At 300K temperature 1 mole of ideal gas expanded from 1 litre to 10 litre then calculate change in internal energy (DU in cal). (1) 1381 (2) –1381 (3) Zero (4) –690 A.11 (3) Isothermal DU = 0 Q.12 1 mole of ideal gas is expanded at 400 K temperature from 10 litre to 100 litre reversibly then calculate the heat in cal : (1) –1842 (2) 1842.4 (3) –2418 (4) 2418 A.12 (2) 1842.4 V W = –2.303 nRT log H VL = –2.303 × 1× 2 × 400 × 1 Thermodynamics = – 2.303 × 800 = –1842.4 cal DU = 0 q = –w = – (–1842.4)cal = 1842.4 cal. 21.  Q.13 An ideal gas is expanded reversibly and adiabatically then which relation is correct : (1) Tf>Ti (2) Tf = Ti (3) Tf N2 > O2 Like U and H, S is also a e.g : C2H4 > N2 state function. 9. a. Unit of Entropy – J cal or K K b. Unit of molar Entropy- J cal or K - mole K - mole 10. a. Entropy is an extensive property b. Molar entropy is an intensive property Previous Year’s Question c. It is a state function 11. At equilibrium position or a reversible process at constant P,T For the reaction 2Cl (g)   Cl2 (g) The correct option is H S  [NEET] T (1) DrH > 0 and DrS > 0 G  H  TS (2) DrH > 0 and DrS < 0 At equilibrium G  0 , so 0 = DH – TDS (3) DrH < 0 and DrS > 0 H  TS (4) DrH < 0 and DrS < 0 H  S Thermodynamics T 12. Total entropy change for the universe is always (+ve) 32.  STotal  0 (DS)sys +(DS)surr > 0 13. At equilibrium position total entropy change Rack your Brain is 0 STotal    0 Ssys   Ssurr   0 A real crystal has higher entropy than the ideal crystal? 14. An ideal crystal has a perfect order of its constituent particles while a real crystal has less order because of some defects. Therefore a real crystal has more entropy than an ideal crystal. Second law of Thermodynamics 1. Total entropy change is always positive for universe. 2. For a spontaneous or irreversible process total entropy change is always positive.  Ssystem  Ssurrounds  0  Q.24 Indicate the sign of Entropy change for following : 1. Fe (300K) → Fe (500 K) 2. N2 + 3H2 → 2NH3 3. Boiling an egg 4. Stretching a rubber 5. Rusting of Iron A.24 1. Fe (300K) → Fe (500 K)   Positive 2. N2 + 3H2 → 2NH3   Negative 3. Boiling an egg → Positive In the liquid of egg peptide bonding is present between protein molecule. It will dissociate when egg will boil so, randomness increases 4. Stretching a rubber → Negative On stretching a rubber its molecules indicate an orderly arrangement so randomness decreases 5. Rusting of Iron → Positive In iron, metallic bond is present between Fe atoms due to rusting there bonds dissociate so DS = Positive Thermodynamics 33.  Q.25 Latent heat of fusion of ice at 1 atm pressure & 0°C is 6016 J/mol then find out the entropy change: (i) When solid convert into liq. (ii) Liq. Convert into solid J J (1) (i) 22 (ii)—22 mol. K mol. K J J (2) (i)–22 (ii)22 mol. K mol. K (3) (i) & (ii) both are same J J (4) (i)11 (ii)—11 mol. K mol. K A.25 (1) For solid to liquid H 6016 J S    22 T 273 mol. K For liquid to solid J S  22 mol. K Q.26 Cu block is at 130°C, amount of heat evolved by Cu block to surrounding is 340 J. Temperature of surrounding is 32°C. Then calculate 1. Entropy change for the reaction 2. Entropy change for the surrounding 3. Total entropy change Assuming that temperature of system and surround remain same A.26 1.  SSys.  q 340   0.85J / cal T 403 q 340 2.  SSurr.    1.1J / cal T 305 3.  STotal.  0.85  1.1 = 1.95 J/cal Thermodynamics 34. y Entropy change for different Thermodynamic process : y Maximum process are carried out in open Previous Year’s Question vessel where gas show expansion So FLOT In which case change in entropy DU = q-W is negative [NEET] q U  W (1) 2H (g)    H2 (g)  T T (2) Evaporation of water (3) Expansion of a gas at constant nCv dT PdV temperature dS   T T (4) Sublimation of solid to gas nCv dT nRdV dS   T T y Integrate both side from stage (1) → (2) 2 T2 V2 dT dV 1 dS  nCV  T T  nR  V V 1 1 Rack your Brain S  nCV  nT T  nR  nV V T2 V2 y 1 1 Heat exchanged in a chemical reaction at constant temperature  T V  y S   nCV loge 2  nR loge 2  and pressure is called ?  T1 V1   T V  y S   2.303 nCV log 2  2.303 nR log 2   T1 V1  Process Constants Value of entropy Isothermal V2 T1 = T2 S  2.303 nR log Process V1 Isochoric T2 V1 = V2 S  2.303 nCV log Process T1 Isobaric T2 P1 = P2 S  2.303 nCP log Process Thermodynamics T1 Adiabatic q =0 S  0 35.  Q.27 1 Mole of ideal gas expanded reversibly and isothermally from 1 lit to 10 lit. then find out the entropy change in cal/Kelvin. (1) 2.303 (2) 6.604 (3) 7.606 (4) 4.606 A.27 S  2.303 nR log V2 V1 10 = 2.303 × 1 × 2 × log 1 DS = 4.606 cal/K Carnot cycle The Carnot cycle consists of the following 4 processes: y This process has a reversible isothermal gas expansion. In it the ideal gas in the system absorbs qin amount of heat from a heat source at a high temperature Thigh, expands Concept Ladder and does work on surroundings. y This process has a reversible adiabatic gas Carnot engine is used as expansion. In it the system is thermally standard performans of all the heat engines operating insulated. The gas continuously expands between two body of and do work on surroundings, which causes different temerpatures i.e., the system to cool to a lower temperature, high temeprature body and law temperature body. Tlow. y This process has a reversible isothermal gas compression. In it surroundings do work to the gas at Tlow, and causes a loss of heat, qout. y This process has a reversible adiabatic gas compression. In it the system is thermally Thermodynamics insulated. Surroundings continuously do work to the gas, which causes the temperature to rise back to Thigh. 36. 37. Thermodynamics  Process W q DU DH V  V  1 nRThigh ln  2  nRThigh ln  2  0 0  V1   V1  2 nCv  Tlow  Thigh  0 nC v  Tlow  Thigh  nCp  Tlow  Thigh  V  V  3 nRTlow ln  4  nRTlow ln  4  0 0  V3   V3  4 nCv  Thigh  Tlow  0 nCv  Thigh  Tlow  nCp  Thigh  Tlow  V  V  V  V  Full Cycle nRThigh ln  2   nRTlow ln  4   V3  nRThigh ln  2   nRTlow ln  4  0 0 V  1    V1   V3  Spontaneous Process : i. The process in which physical and chemical change occur due to its own means without any external help. ii. Spontaneous process are irreversible. Thermodynamics 38. 1. Gibbs free energy (G) (i) Energy obtained from a system which can be put into useful work mean DG = (W)useful According to FLOT DU = q + W Here (W) includes 2 types of work A. Thermodynamic work = PDV B. Useful work = Wuseful Example of useful work : Previous Year’s Question 1 y H2O (l)  H2 (g)  O2 (g) 2 For a sample of perfect gas 3 when its pressure is changed y KClO3 (s)  KCl (s)  O2 (g) 2 isothermally from Pi to Pf the CaCO3 (s)  CaO (s)  CO2 (g) entropy change is given by y [NEET] In above reactions 2 types of work are present (A) Thermodynamic work due to expansion of  Pf  (1) S  nR ln   gases  Pi  (B) Useful work to dissociate the compound P  (2) S  nR l n  i  2. Galvanic cell reaction is spontaneous. So  Pf  obtained free energy is converted into  Pf  useful work (3) S  nRT ln    Pi  G  Wuseful  Welect  nFE 3. Cdiamond → Cgraphite  Pi  (4) S  RT ln   DG =Wusrful = PDV  Pf  To convert one allotropic form to another free energy change is equal to useful work Mathematically, free energy is the difference of enthalpy and the product of temperature and entropy. G = H – TS G = (U + PV) –TS 4. It is impossible to calculate the absolute value of free energy because it is not Thermodynamics possible to calculate the absolute value of enthalpy or internal energy. DG = DU + D(PV) – D(TS) 39.  At constant P,T DG = (DU + PDV) – TDS DG = DH – TDS for system y In standard condition[P = 1 atm] [T = 25°C] DG° = DH° – TDS° y Free energy is an extensive property y Free energy is a state function y Free energy change for a reversible reaction G  G0f  2.303 RT log Q DH DS DG Process Negative Positive Negative Spontaneous Non-spontaneous Positive Negative Positive Spontaneous Non-spontaneous Negative Negative High temp then DG = (+) Spontaneous Spontaneous Positive Positive Low temp then DG= (–) Non-Spontaneous y If temperature is more than the equilibrium temperature then it is known as high temperature and if temperature is less than equilibrium temperature then it’s known as low temperature. Q.28 At 1 atm pressure heat of vapourisation of H2O is 44.3 KJ/mol and entropy J change is 120 then find out the equilibrium temperature. mol. K (1) 396 K (2) 369 K (3) 693 K (4) 639 K Thermodynamics A.28 DG = 0 H 44.3  1000 T   369 K S 120 40.  Q.29 2 mole of ideal gas is expanded reversibly and isothermally from 1 lit to 10 lit at 127°C. Then find out the free energy change in cal (1) –230.3 cal (2) –3386.5 cal (3) –3366.8 cal (4) –3683.8 cal A.29 DG = –TDS because DH = 0 V2 DG = –T (2.303 nR log ) V1 10 DS = –2.303 × 2 × 2 × 400 log 1 DS = –2.303 × 1600 = –3683.8 cal y Coupled reaction A reaction whose DG is positive is non spontaneous another reaction whose DG is negative is spontaneous when both reaction are coupled then non spontaneous reaction may become spontaneous. Third law of Thermodynamics Rack your Brain y Entropy of pure crystalline substance is zero at absolute zero temperature. What is the condition for a y Information about the entropy is given by process to be spontaneous SLOT while it is calculated by TLOT. according to second law of thermodynamics ? Limitation y Mixture of isotopes do not show zero entropy at absolute zero temperature. Thermodynamics y Glassy solid such as CO, CO2, NO, N2O, NO2, H2O etc doesn’t show zero entropy at absolute zero temperature. 41.  Q30 Find the work done, when one mole of ideal gas in 10 litre container at 1 atm. is allowed to enter evacuated bulb of capacity 100 litre. Sol. W = – PDV Sol. But since gas enters the vacuum bulb and pressure in vacuum is zero. W=0 Q31 If 1 mole of gas expands from 1 litre to 5 litre against constant atmospheric pressure than calculate the work done. Sol. W = –PDV = –1 (5 – 1) = –4 litre-atm. Q32 A system expands from 5 L to 10 L against a constant external pressure of 2 atm. If it absorbs 800 J of energy in the process. Calculate the change in its internal energy. Sol. DU = q + w w = – P(V2 – V1) = –2 (10 – 5) = –10 atm -L × 101.3 J = –1013 J U = 800 – 1013 J DU = 800 – 1013 = –213 J Q33 FeCO3(s) decomposes at constant pressure as FeO(s) + CO2(g) FeCO (s) ∆ → FeO(s) + CO (g) 3 2 at 25°C, the heat absorbed during the reaction is 80 kJ. Calculate ∆H & DU for the reaction Sol. FeCO3(s) → FeO(s) + CO2(g) ∆H = qp= 80 kJ ∆H = ∆U = ∆ngRT Thermodynamics [1 × 8.314 × 298] ⇒ 80 kJ = kJ ⇒ DU = 77.522 kJ 1000 42. Q34 Gases ∆G°f (Cal/mole) CO –32.80 H2O –54.69 CO2 –94.26 H2 0 Estimate the standard free energy change in the chemical reaction. CO + H2O CO2 + H2 Sol. Sol. Using the necessary data from the table. CO H2O CO2 H2 ∆G° –32.8 –54.69 –94.26 0 kcal ∴ ∆G° –94.26 + 0 – (–32.8) – (–54.69) = –6.8 kcal/mol Q35 A liquid of volume of 100 L and at the external pressure of 10 atm–litre the liquid is confined inside an adiabatic bath. External pressure of the liquid is suddenly increased to 100 atm and the liquid gets compressed by 1 L against this pressure then find, (i) work (ii) ∆U (iii) ∆H Sol. Sol. Dq = 0 Work done =– 100 × –1 = 100 L. Atm Dw = DU ⇒ 100 = DU ⇒ ∆H = DU + (P1V2 – P1V1) = 100 + (100 × 99 – 100 × 10) = 100 + 100 × 89 = 9000 lit atm. ∴ 1 L. Atm = 101.3 Joule Q36 For the combustion of 1 mole of liquid benzene at 25°C, the heat of reaction at constant pressure is given by , 1 C6H6 (  ) + 7 O2 (g) → 6 CO2 (g) + 3H2O(  ); ∆H =–780980 cal. 2 What would be the heat of reaction at constant volume ? Sol. We have, ∆H = ∆U + DngRT Here, Dng = 6 – 7.5 = –1.5. Thus, ∆U = ∆H + DngRT = – 780980 – (–1.5.) × 2 × 298 = –780090 calories. Q37 At 25°C, a 0.01 mole sample of a gas is compressed in volume from 4.0 L to Thermodynamics 1.0 L at constant temperature. What is work done for this process if the ex- ternal pressure is 4.0 bar ? 43.  Sol. 1.2 × 103 J W = –Pext (V2 – V1) = –4 (1 – 4) bar. L = 1.2 × 103 J. Q38 Calculate q, W, ∆U and ∆H when 100 gm of CaCO3 is converted into its arago- nite form given density of calcite = 2g/cc and density of aragonite = 2.5 g/cc. Sol.  CaCO3 CaCO3  Calcite Aragonite ∆H = 2kJ/mole Generally for solid Solid solid Liquid solid Liquid Transitions W 0 For process at equilibrium DG = 0 Thermodynamics 46.