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“Bioenergetics” Prof. Dr. Metin TULGAR

“Bioenergetics” Prof. Dr. Metin TULGAR. Energy Concept In Living Creature s. Energy is the ability of an object to do work . Some examples: Transmittion of water from roots to the leaves, A ny movement of an animal, Pump of blood to the vessels .

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“Bioenergetics” Prof. Dr. Metin TULGAR

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  1. “Bioenergetics” Prof. Dr. Metin TULGAR

  2. Energy Concept In Living Creatures Energy is the ability of an object to do work. Some examples: • Transmittion of water from roots tothe leaves, • Any movement of an animal, • Pumpof blood to the vessels. Energy source of livings: Nutrients.

  3. Kinds of Energies for Livings A mechanical work: in muscle contraction, An electrical work: ions, crossing across themembrane, A chemical work: during the synthesis of a substance. If livings can’t transform energy to other forms, they can not survive.

  4. Typical events proving that the cells use energy: • Cells keep substances in high concentration. • That the cells move. • Ability to synthesize macromolecules from micromolecules.

  5. Bioenergetics is; The science that deals with howthe energy is producedand transformedin living creatures.

  6. BIOTHERMODYNAMICS Bio(living) +thermo(heat) + dynamics(power): Biothermodynamics is the science that deals with theenergy and its transformations in livings.

  7. Here are some concepts of thermodynamics: W: Work that the System does. Q: Energy amount of the System E: Internal Energy of the System Thermodynamics deal with the relationships of these concepts. An example System: Amip (Single-celled organism). • Amip consumes energy when it moves. • Heat is transfered between the invironment and Amip. • Internal energy source of Amip is food.

  8. Laws of Thermodynamics In order to determine the lastsituation of this System. Physical charasteristics of a System: • Mass, • Volume, • And Temperature.

  9. There are three laws of thermodynamics: 1. Zeroth law of thermodynamics 2. First law of thermodynamics 3. Second law of thermodynamics

  10. Zeroth Law of Thermodynamics • The zeroth law of thermodynamics state that: “If two badies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other.”

  11. Figure #1: Three bodies in thermalequilibrium. B TB C TC A TA • Consider three bodies A, B, and C with absolute temperatures TA, TB, and TC in thermal equilibrium. • According to the zeroth law of thermodynamics; if TA= TB, and TA =TC; then TB =TC .

  12. First Law of Thermodynamics • Also known asconservation of energy principle. • energy change of(heat+work)= internal energychange. • Internal Energy: Total energy of all the microscobic forms of energy.( atoms,molecules,ions )

  13. Scientific definition of First Law: ΔEi = Q + W . • ΔEi:internal energychange, Q:heat, W:wrok . • Internal Energy Function:Ei = u( V, T ) . • V: volume , T: absolute temperature . • Differentiation of Internal Energy Function: • dEi = (u/T)VdT+(u/V)TdV

  14. Consider conservation of energy principle; • Eu = Ei + Eo • Eu: energy of the universe, • Ei: internal energy of system, • Eo: energy of environment. • If “energy can be neither created nor destroyed”, then Eu is constant. • Change in internal energy of the system: Ei to Ei’ • Change in energy of environment: Eo to Eo’ • Ei + Eo = Ei’ + Eo’ • Ei – Ei’ = Eo’ – Eo • Then, ΔEi = ΔEo

  15. First Law could be concluded as this: “Energy can be neither created nor destroyed, it can only change forms”

  16. Ideal Gas Equation of State • Ideal gas is an imaginary gas which has • no volume, • no attractive and repulsive forces between its molecules, and • no loss of kinetic energy when its molecules collided with each other.

  17. Ideal gas equation of state: P.V = n.R.T • P: pressure, • V: volume, • n: # of moles, • R: gas constant(8,314j/Kmol), • T: temperature(K)

  18. Isochoric Process • Process in which volume is kept constant.

  19. “Mathematical analysis of isochoric process” • dEi = (u/T)VdT + (u/V)TdV = dQ + P.dV • For isochoric processdV = 0 => (u/T)VdT = dQ = CV.dT CV : specific heat of System in isochoric process • For n moles of gases, n. Ei = qV = n.CV.dT • Thus, for an isochoric process heat energy is absorbed by the system to increase internal energy.

  20. Isobaric Process • process in which pressure is kept constant

  21. Isobaric Process (Continuing) • dEi = dQ - dW • dQ = dEi + dW = dEi + P.dV • Isobaric situations: dQP=dEi + d(P.V) = d(Ei + P.V) = dH • => Enthalpy: H = Ei + P.V dQP = dH QP = H • These equations result in a conclusion that: “Change in heat energy amount is equal to the enthalpy”

  22. Isobaric Process (Continuing) • dH = CP.dT CP: specific heat at isobaric process • For n moles of gases: dH = n.CP.dT

  23. Enthalpy Consider the Enthalpy with the Ideal Gas Law: H = Ei + P.V = Ei + R.T Take the derivative of above equation: dH = dEi + R.dT The resultant equation indicates that: “Internal Energy of ideal gas is only dependent on T”

  24. Enthalpy (Continuing) Consider the other equations together: CP.dT = CV.dT + R.dT CP = CV + R => R = CP – CV

  25. Isothermal Process • Examples: Ideal gases & ideal paramagnetic crystals • process in which temperature is held constant

  26. Adiabatic Process (Continuing) • For an adiabatic system: ΔQ = 0 ΔEi = ΔQ + ΔW => ΔEi = ΔW • If the system is compressed Ei = - W(internal energy ↑, Work is -) • If the system is expanded - Ei = W(internal energy ↓, Work is +)

  27. Adiabatic Process • Process in which no heat transfer takes place • Cork, mineral wool, isolated glass must be used to isolate the system Or; The process should be completed immediately.

  28. Second Law of Thermodynamics

  29. Second Law of Thermodynamics i. states in which direction a process can take place • heat does not flow spontaneously from a cold to a hot body • heat cannot be transformed completely into mechanical work • it is impossible to construct an operational perpetual motion machine ii. introduces concept of entropy

  30. Second Law of Thermodynamics (Continuing) • “It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.” 2nd Law Statement By Kelvin-Planck • To have such a cycle another reservoir with different temperature is needed.

  31. Reversible and Irreversible Processes • Irreversible Process: “Once having taken place , these processes can not reverse themselves spontaneously and restore the system to its initial state.” • For Example, Once a cup of hot coffee cools, it will not heat up by retrieving the heat it lost from the surroundings.

  32. Reversible Process: “A process that can be reversed without leaving any trace on the surroundings.” This is possible only if the net heat and net work exchange between system and surroundings is zero for the combined (original and reverse) process.

  33. Entropy • property that indicates the direction of a process Entropy , • is a measure of disorder • is a measure of a system’s ability to do useful work • determines direction of time

  34. Entropy (S) equation: S = Q/T or ΔS = ΔQ / Tsystem Q: heat T: Abs. Temp. Unit Analysis: [Cal]/[˚K] [Cal/mol]/[˚K](for molar entropy)

  35. For example: An ice (crystalline) molecules are strongly order So, the Entropy is small for ice. Ice →(heat) → Water (liquid) That is, the strong (crystalline) structure is broken. So, the Entropy increases. Water (liquid) →(heat)→ Vapor (gas) Gas molecules are more unordered. That is to say “the Entropy increases more”.

  36. The example can be concluded as this; “Entropy is increased by the energy given to the system” • Relationship bw. the change in Energy & Entropy of System: ΔS > ΔQ/T

  37. Some examples for entropy change • If ordered system →an independent system, Entropy ↑. The reverse cause Entropy ↓. • If solids and liquids reacts a gases Entropy ↑. The reverse cause Entropy ↓. • If a Volume ↑, Entropy ↑. If a Volume ↓, Entropy ↓.

  38. Standard Absolute Entropy: Change in Entropy, caused by the increase in temperature of pure crystals, in isobaric conditions, from 0K to 298K.

  39. Entropy of Reaction: Change in Entropy of system in isobaric and isothermal conditions, for chemical reactions in which maximum of reaction parameter is 1 mole.

  40. Entropy is dependent of three parameters: • Temperature • Pressure • Reaction variable • S = f(T,P,ξ) • T: absolute temperature, • P: pressure, • ξ: reaction variable.

  41. Free Energy • (also known as Gibbs’ Energy oruseful energy) • In biological systems Free Energy conceptis the best method for explaining the energyconversions. • Explains the usable component of system’s total energy under isobaric and isothermic conditions.

  42. The Concept of Free Energy Combine the 1st and 2nd laws of thermodynamics: ΔEi = ΔQ – P. ΔV, “ΔQ = T. ΔS” => ΔEi = T.ΔS – P. ΔV ΔEi + P. ΔV – T. ΔS = ΔG, “ΔEi + P. ΔV = ΔH” => ΔH – T. ΔS = ΔG • H: enthalpy • ΔH: change in enthalpy

  43. For the reaction entropy conditions; • P & T areconstants. Then: G = Ei + P.V – T.S By this formula the free energy can be calculated.

  44. In a reaction, the change in free energy can be zero, positive or negative value. • The sign of free energy change determines the direction of reaction.

  45. The condition that free energy change is; • Zero (ΔG = 0), indicates the equilibrium state. • Negative (ΔG < 0), explains that the reaction has occurred by giving energy to the environment. • Positive (ΔG > 0), indicatesthe system has taken energy from the environment during the reaction.

  46. SUN Radiation Energy (photons) GREEN PLANTS Chemical energy of carbohydrates and other products of cells carbon dioxide water oxygen Thermal and other energies ANIMALS Chemical energy Biological energy

  47. Three main steps of Biologic Energy Flow: • Solar energy → Organic Materials (by ototrof organisms) • Organic Materials → ATP (energy) (by respiration) • ATP→ other forms of energy (by biological processes)

  48. Photosynthesis & Respiration • Photosynthesis: Conversion of solar energy (photons)into chemical energywith chlorophyllby the green plants. (Chlorophyll) • 6CO2 + 6H2O + n.h.f.  C6H12O6 + 6O2 • n: Planck constant (6,62.10-34 j.s) • h: photon frequency • f: number of photons in the reaction

  49. During photosynthesis; • Free energychange: Gs = 2867 kj/mol “glucose formation needs energy” “2867 kJ solar energy is used for each glucose mole” • Enthalpychange: Hs = 2810 kj/mol • Entropy change: Ss = - 182 j/(mol)K “reaction is endothermic”

  50. The inverse mechanism of photosynthesis is called respiration. • C6H12O6 + 6O26CO2 + 6H2O (Glucose)

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