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Page 1 THERMODYNAMICS Thermodynamic processes : 1. Isothermal process : T = constant dT = 0 ?T = 0 2. Isochoric process : V = constant dV = 0 ?V = 0 3. Isobaric process : P = constant dP = 0 ?P = 0 4. Adiabatic process : q = 0 or heat exchange with the surrounding = 0(zero) IUPAC Sign convention about Heat and Work : Work done on the system = Positive Work done by the system = Negative 1 st Law of Thermodynamics ?U = (U 2 ? U 1 ) = q + w Law of equipartion of energy : U = 2 f nRT (only for ideal gas) ?E = 2 f nR (?T) where f = degrees of freedom for that gas. (Translational + Rotational) f = 3 for monoatomic = 5 for diatomic or linear polyatmic = 6 for non  linear polyatmic Page 2 THERMODYNAMICS Thermodynamic processes : 1. Isothermal process : T = constant dT = 0 ?T = 0 2. Isochoric process : V = constant dV = 0 ?V = 0 3. Isobaric process : P = constant dP = 0 ?P = 0 4. Adiabatic process : q = 0 or heat exchange with the surrounding = 0(zero) IUPAC Sign convention about Heat and Work : Work done on the system = Positive Work done by the system = Negative 1 st Law of Thermodynamics ?U = (U 2 ? U 1 ) = q + w Law of equipartion of energy : U = 2 f nRT (only for ideal gas) ?E = 2 f nR (?T) where f = degrees of freedom for that gas. (Translational + Rotational) f = 3 for monoatomic = 5 for diatomic or linear polyatmic = 6 for non  linear polyatmic Calculation of heat (q) : Total heat capacity : C T = dT dq T q ? ? ? = J/ºC Molar heat capacity : C = ndT dq T n q ? ? ? = J mole ?1 K ?1 C P = 1 ? R ? ? C V = 1 ? R ? Specific heat capacity (s) : S = mdT dq T m q ? ? ? = J gm ?1 K ?1 WORK DONE (w) : Isothermal Reversible expansion/compression of an ideal gas : W = ? nRT ln (V f /V i ) Reversible and irreversible isochoric processes. Since dV = 0 So dW = ? P ext . dV = 0. Reversible isobaric process : W = P (V f ? V i ) Adiabatic reversible expansion : ? 1 2 2 V T ? ? = 1 1 1 V T ? ? Reversible Work : W = 1 V P V P 1 1 2 2 ? ? ? = 1 ) T T ( nR 1 2 ? ? ? Irreversible Work : W = 1 V P V P 1 1 2 2 ? ? ? = 1 ) T T ( nR 1 2 ? ? ? = nC v (T 2 ? T 1 ) = ? P ext (V 2 ? V 1 ) and use 2 2 2 1 1 1 T V P T V P ? Free expansion?Always going to be irrerversible and since P ext = 0 so dW = ? P ext . dV = 0 If no. heat is supplied q = 0 then ?E = 0 so ?T = 0. Page 3 THERMODYNAMICS Thermodynamic processes : 1. Isothermal process : T = constant dT = 0 ?T = 0 2. Isochoric process : V = constant dV = 0 ?V = 0 3. Isobaric process : P = constant dP = 0 ?P = 0 4. Adiabatic process : q = 0 or heat exchange with the surrounding = 0(zero) IUPAC Sign convention about Heat and Work : Work done on the system = Positive Work done by the system = Negative 1 st Law of Thermodynamics ?U = (U 2 ? U 1 ) = q + w Law of equipartion of energy : U = 2 f nRT (only for ideal gas) ?E = 2 f nR (?T) where f = degrees of freedom for that gas. (Translational + Rotational) f = 3 for monoatomic = 5 for diatomic or linear polyatmic = 6 for non  linear polyatmic Calculation of heat (q) : Total heat capacity : C T = dT dq T q ? ? ? = J/ºC Molar heat capacity : C = ndT dq T n q ? ? ? = J mole ?1 K ?1 C P = 1 ? R ? ? C V = 1 ? R ? Specific heat capacity (s) : S = mdT dq T m q ? ? ? = J gm ?1 K ?1 WORK DONE (w) : Isothermal Reversible expansion/compression of an ideal gas : W = ? nRT ln (V f /V i ) Reversible and irreversible isochoric processes. Since dV = 0 So dW = ? P ext . dV = 0. Reversible isobaric process : W = P (V f ? V i ) Adiabatic reversible expansion : ? 1 2 2 V T ? ? = 1 1 1 V T ? ? Reversible Work : W = 1 V P V P 1 1 2 2 ? ? ? = 1 ) T T ( nR 1 2 ? ? ? Irreversible Work : W = 1 V P V P 1 1 2 2 ? ? ? = 1 ) T T ( nR 1 2 ? ? ? = nC v (T 2 ? T 1 ) = ? P ext (V 2 ? V 1 ) and use 2 2 2 1 1 1 T V P T V P ? Free expansion?Always going to be irrerversible and since P ext = 0 so dW = ? P ext . dV = 0 If no. heat is supplied q = 0 then ?E = 0 so ?T = 0. Application of Ist Law : ?U = ?Q + ?W ? ?W = ?P ?V ? ?U = ?Q ?P?V Constant volume process Heat given at constant volume = change in internal energy ? du = (dq) v du = nC v dT C v = dT du . n 1 = 2 f R Constant pressure process : H ? Enthalpy (state function and extensive property) H = U + PV ? C p ? C v = R (only for ideal gas) Second Law Of Thermodynamics : ?S universe = ?S system + ??S surrounding > 0 for a spontaneous process. Entropy (S) : ?S system = ? B A rev T dq Entropy calculation for an ideal gas undergoing a process : State A irr S irr ? ?? ? State B P 1 , V 1 , T 1 P 2 , V 2 , T 2 ?S system = nc v ln 1 2 T T + nR ln 1 2 V V (only for an ideal gas) Third Law Of Thermodynamics : The entropy of perfect crystals of all pure elements & compounds is zero at the absolute zero of temperature. Gibb?s free energy ?G) : (State function and an extensive property) G system = H system ? TS system Criteria of spontaneity : (i) If ?G system is (?ve) < 0 ? process is spontaneous (ii) If ?G system is > 0 ? process is non spontaneous (iii) If ?G system = 0 ? system is at equilibrium. Page 4 THERMODYNAMICS Thermodynamic processes : 1. Isothermal process : T = constant dT = 0 ?T = 0 2. Isochoric process : V = constant dV = 0 ?V = 0 3. Isobaric process : P = constant dP = 0 ?P = 0 4. Adiabatic process : q = 0 or heat exchange with the surrounding = 0(zero) IUPAC Sign convention about Heat and Work : Work done on the system = Positive Work done by the system = Negative 1 st Law of Thermodynamics ?U = (U 2 ? U 1 ) = q + w Law of equipartion of energy : U = 2 f nRT (only for ideal gas) ?E = 2 f nR (?T) where f = degrees of freedom for that gas. (Translational + Rotational) f = 3 for monoatomic = 5 for diatomic or linear polyatmic = 6 for non  linear polyatmic Calculation of heat (q) : Total heat capacity : C T = dT dq T q ? ? ? = J/ºC Molar heat capacity : C = ndT dq T n q ? ? ? = J mole ?1 K ?1 C P = 1 ? R ? ? C V = 1 ? R ? Specific heat capacity (s) : S = mdT dq T m q ? ? ? = J gm ?1 K ?1 WORK DONE (w) : Isothermal Reversible expansion/compression of an ideal gas : W = ? nRT ln (V f /V i ) Reversible and irreversible isochoric processes. Since dV = 0 So dW = ? P ext . dV = 0. Reversible isobaric process : W = P (V f ? V i ) Adiabatic reversible expansion : ? 1 2 2 V T ? ? = 1 1 1 V T ? ? Reversible Work : W = 1 V P V P 1 1 2 2 ? ? ? = 1 ) T T ( nR 1 2 ? ? ? Irreversible Work : W = 1 V P V P 1 1 2 2 ? ? ? = 1 ) T T ( nR 1 2 ? ? ? = nC v (T 2 ? T 1 ) = ? P ext (V 2 ? V 1 ) and use 2 2 2 1 1 1 T V P T V P ? Free expansion?Always going to be irrerversible and since P ext = 0 so dW = ? P ext . dV = 0 If no. heat is supplied q = 0 then ?E = 0 so ?T = 0. Application of Ist Law : ?U = ?Q + ?W ? ?W = ?P ?V ? ?U = ?Q ?P?V Constant volume process Heat given at constant volume = change in internal energy ? du = (dq) v du = nC v dT C v = dT du . n 1 = 2 f R Constant pressure process : H ? Enthalpy (state function and extensive property) H = U + PV ? C p ? C v = R (only for ideal gas) Second Law Of Thermodynamics : ?S universe = ?S system + ??S surrounding > 0 for a spontaneous process. Entropy (S) : ?S system = ? B A rev T dq Entropy calculation for an ideal gas undergoing a process : State A irr S irr ? ?? ? State B P 1 , V 1 , T 1 P 2 , V 2 , T 2 ?S system = nc v ln 1 2 T T + nR ln 1 2 V V (only for an ideal gas) Third Law Of Thermodynamics : The entropy of perfect crystals of all pure elements & compounds is zero at the absolute zero of temperature. Gibb?s free energy ?G) : (State function and an extensive property) G system = H system ? TS system Criteria of spontaneity : (i) If ?G system is (?ve) < 0 ? process is spontaneous (ii) If ?G system is > 0 ? process is non spontaneous (iii) If ?G system = 0 ? system is at equilibrium. Physical interpretation of ?G : ??The maximum amount of nonexpansional (compression) work which can be performed. ?G = dw nonexp = dH ? TdS. Standard Free Energy Change (?Gº) : 1. ?Gº = ?2.303 RT log 10 K 2. At equilibrium ??G = 0. 3. The decrease in free energy (??G) is given as : ??G = W net = 2.303 nRT log 10 1 2 V V 4. ? º f G for elemental state = 0 5. ? º f G = º products G ? º ts tan ac Re G Thermochemistry : Change in standard enthalpy ?H° = 0 2 , m H ? 0 1 , m H = heat added at constant pressure. = C P ?T. If H products > H reactants ? Reaction should be endothermic as we have to give extra heat to reactants to get these converted into products and if H products < H reactants ? Reaction will be exothermic as extra heat content of reactants will be released during the reaction. Enthalpy change of a reaction : ?H reaction = H products ? H reactants ?H° reactions = H° products ? H° reactants = positive ? endothermic = negative ? exothermic Temperature Dependence Of ?H : (Kirchoff's equation) : For a constant pressure reaction ?H 2 ° = ?H 1 ° + ?C P (T 2 ? T 1 ) where ?C P = C P (products) ? C P (reactants). For a constant volume reaction ? ? ? ? ? ? dT . C E E V 0 1 0 2 Page 5 THERMODYNAMICS Thermodynamic processes : 1. Isothermal process : T = constant dT = 0 ?T = 0 2. Isochoric process : V = constant dV = 0 ?V = 0 3. Isobaric process : P = constant dP = 0 ?P = 0 4. Adiabatic process : q = 0 or heat exchange with the surrounding = 0(zero) IUPAC Sign convention about Heat and Work : Work done on the system = Positive Work done by the system = Negative 1 st Law of Thermodynamics ?U = (U 2 ? U 1 ) = q + w Law of equipartion of energy : U = 2 f nRT (only for ideal gas) ?E = 2 f nR (?T) where f = degrees of freedom for that gas. (Translational + Rotational) f = 3 for monoatomic = 5 for diatomic or linear polyatmic = 6 for non  linear polyatmic Calculation of heat (q) : Total heat capacity : C T = dT dq T q ? ? ? = J/ºC Molar heat capacity : C = ndT dq T n q ? ? ? = J mole ?1 K ?1 C P = 1 ? R ? ? C V = 1 ? R ? Specific heat capacity (s) : S = mdT dq T m q ? ? ? = J gm ?1 K ?1 WORK DONE (w) : Isothermal Reversible expansion/compression of an ideal gas : W = ? nRT ln (V f /V i ) Reversible and irreversible isochoric processes. Since dV = 0 So dW = ? P ext . dV = 0. Reversible isobaric process : W = P (V f ? V i ) Adiabatic reversible expansion : ? 1 2 2 V T ? ? = 1 1 1 V T ? ? Reversible Work : W = 1 V P V P 1 1 2 2 ? ? ? = 1 ) T T ( nR 1 2 ? ? ? Irreversible Work : W = 1 V P V P 1 1 2 2 ? ? ? = 1 ) T T ( nR 1 2 ? ? ? = nC v (T 2 ? T 1 ) = ? P ext (V 2 ? V 1 ) and use 2 2 2 1 1 1 T V P T V P ? Free expansion?Always going to be irrerversible and since P ext = 0 so dW = ? P ext . dV = 0 If no. heat is supplied q = 0 then ?E = 0 so ?T = 0. Application of Ist Law : ?U = ?Q + ?W ? ?W = ?P ?V ? ?U = ?Q ?P?V Constant volume process Heat given at constant volume = change in internal energy ? du = (dq) v du = nC v dT C v = dT du . n 1 = 2 f R Constant pressure process : H ? Enthalpy (state function and extensive property) H = U + PV ? C p ? C v = R (only for ideal gas) Second Law Of Thermodynamics : ?S universe = ?S system + ??S surrounding > 0 for a spontaneous process. Entropy (S) : ?S system = ? B A rev T dq Entropy calculation for an ideal gas undergoing a process : State A irr S irr ? ?? ? State B P 1 , V 1 , T 1 P 2 , V 2 , T 2 ?S system = nc v ln 1 2 T T + nR ln 1 2 V V (only for an ideal gas) Third Law Of Thermodynamics : The entropy of perfect crystals of all pure elements & compounds is zero at the absolute zero of temperature. Gibb?s free energy ?G) : (State function and an extensive property) G system = H system ? TS system Criteria of spontaneity : (i) If ?G system is (?ve) < 0 ? process is spontaneous (ii) If ?G system is > 0 ? process is non spontaneous (iii) If ?G system = 0 ? system is at equilibrium. Physical interpretation of ?G : ??The maximum amount of nonexpansional (compression) work which can be performed. ?G = dw nonexp = dH ? TdS. Standard Free Energy Change (?Gº) : 1. ?Gº = ?2.303 RT log 10 K 2. At equilibrium ??G = 0. 3. The decrease in free energy (??G) is given as : ??G = W net = 2.303 nRT log 10 1 2 V V 4. ? º f G for elemental state = 0 5. ? º f G = º products G ? º ts tan ac Re G Thermochemistry : Change in standard enthalpy ?H° = 0 2 , m H ? 0 1 , m H = heat added at constant pressure. = C P ?T. If H products > H reactants ? Reaction should be endothermic as we have to give extra heat to reactants to get these converted into products and if H products < H reactants ? Reaction will be exothermic as extra heat content of reactants will be released during the reaction. Enthalpy change of a reaction : ?H reaction = H products ? H reactants ?H° reactions = H° products ? H° reactants = positive ? endothermic = negative ? exothermic Temperature Dependence Of ?H : (Kirchoff's equation) : For a constant pressure reaction ?H 2 ° = ?H 1 ° + ?C P (T 2 ? T 1 ) where ?C P = C P (products) ? C P (reactants). For a constant volume reaction ? ? ? ? ? ? dT . C E E V 0 1 0 2 Enthalpy of Reaction from Enthalpies of Formation : The enthalpy of reaction can be calculated by ?H r ° = ? ? B ?H f °, products ? ? ? B ?H f °, reactants ? B is the stoichiometric coefficient. Estimation of Enthalpy of a reaction from bond Enthalpies : ?H = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? atoms gasesous the from products form to released Enthalpy atoms gasesous into reactants break to required Enthalpy Resonance Energy : ?H° resonance = ?H° f, experimental ? ?H° f, calclulated = ?H° c, calclulated ??? ?H° c, experimentalRead More
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