1 / 54

THERMODYNAMIC

THERMODYNAMIC. Conducted by Efa Mai Inanignsih. Objective. to be able to formulate work, heat and internal energy based on the principal law of thermodynamics and apply it in problem solving to be able to apply general equation of ideal gases in isothermal, isochoric and isobaric processes

guilford
Download Presentation

THERMODYNAMIC

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. THERMODYNAMIC Conducted by Efa Mai Inanignsih

  2. Objective • to be able to formulate work, heat and internal energy based on the principal law of thermodynamics and apply it in problem solving • to be able to apply general equation of ideal gases in isothermal, isochoric and isobaric processes • to be able to analyze ideal gases process based on pressure against volume graph • to be able to describe the principle of Carnot engine

  3. The phenomenonWhat do you think of This is it!! If a balloon is inflated and kept in a hot place, it will finally explode, because the particles of gas in the balloon continue to expand and press the balloon wall so that when the wall is unable to stand the gas pressure, the balloon will explode Balloon explosion??

  4. Introduction • Thermodynamics discuss the relationship between heat and mechanical work • also the basic knowledge about temperature and heat, the influence of temperature and heat to the characteristics of substances, and kinetic theory of gases • System and Environment is the everything becoming our observation object is the everything which is outside the system

  5. ∆ s External Work The figure is a tube containing ideal gas closed with a piston If the gas in is heated at a constant pressure, then the gas will expand and push the piston with force F so that the piston will shift as far as s

  6. External Work (continued) W = F.s Because F = pA ThenW = p.A.s Where W = mechanical work of gas (J) p = gas pressure A = section area of tube (m2) s = displacement of piston (m) Because A s = V, then the work equation above becomes W = p.V Where V is change of volume (m3) Work done by the gas is directly proportional to pressure (p) and change of volume (V) of the gas.

  7. Sample Problem Solution V1 = 5.0 L V2= 3.5 L p = 1.0 x 105 Pa because the gas is compressed, then W = p V = p (V2 – V1) = 1.0 x 105 Pa (3.5 L – 0.5 L) = 1.0 x 105 Pa (-1.5 x 10-3 m3) = -150 J Thus, the external work applied to the gas is -150 J • A gas is compressed so that the volume decreases from 5.0 L to 3.5 L at a constant pressure of 1.0 x 105 Pa. Calculate the external work applied to the gas!

  8. Exercises • A number of ideal gases are heated at a constant pressure of 2 x 105 N/m2 so that the volume changes from 20 liters to 30 liters. Calculate the external work done by the gas during expansion! • A gas is compressed at constant pressure 2.00 x 105 Pa from a volume of 2.00 m3 to a volume of 0.500 m3. What is the work done to the gas? If the temperature initially was 40oC, what is the final temperature of the gas?

  9. Thermodynamics Processes • Isothermal Process • Isobaric Process • Isochoric Process • Adiabatic Process

  10. Isothermal Process • Isothermal process is the change process of gas state at constant temperature • From the ideal gas state equation, pV = nRT obtained pV = constant, because nRT has a constant value. So pV = constant then p1V1 = p2V2 Where p1 = initial pressure V1 = initial volume p2 = final pressure V2 = final volume complies with the Boyle’s law

  11. External work in Isothermal Process • The external work done by gas in the isothermal process can be determined from the equation pV = nRT and W = p V.

  12. Isobaric Process • Isobaric process is the change process of gas state at constant pressure because p is constant and nR is always constant, then the Gay-Lussac’s law

  13. Isobaric Process(continued) • Graph of isobaric process

  14. External work in Isobaric Process • The external work done by gas in isobaric process can be determined by equation W = p V = p (V1 – V2) it can also be determined by the area under the p – V graph (shaded region)

  15. Isochoric Process • Isochoric process is the change process of gas state at constant volume • because V is constant and nR is always constant, then • This process complies the Gay-Lussac’s law. graph of isochoric process

  16. External work in Isochoric Process • The external work done by gas in isochoric process: • In other words, in isochoric process (constant volume), the gas does not do any external work. So the area under the p – V graph only forms a point. The area under the p – V graph for isochoric process is zero

  17. Adiabatic Process • Adiabatic process is the process change of a gas state which does not experience any transfer of heat or there is no heat entering or coming out of the system (gas) • This process complies with the Poisson’s formula Where  = Laplace constant = Cp/Cv Cp = specific heat of gas at constant pressure CV= specific heat of gas at constant volume

  18. Adiabatic Process(contd.) • The equation above can also be expressed in another equation as follows Curvature of p – V adiabatic graph is steeper than isothermal

  19. External work in Adiabatic Process • The external work done by gas in adiabatic process is expressed as follows. The external work done by gas is equal to the area of shaded region under the p – V graph

  20. Sample Problem Solution Given n = 2 mol R = 8.31 J/mol K T = (-23 + 273) K = 250 K V2 = ½ V1 so V2/V1 = 1/2 The external work done in isothermal process W = = (2 mol) (8.31 J/mol K) (250 K) (ln ) = 4155 J (ln 1 – ln 2) = 4155 J (0 – 0.69) = -2866.95 J Thus, the external work done is -2866.95 J (-) sign indicates that at the gas is applied work. • Two moles of gas is compressed at a constant temperature of -23oC so that its volume becomes half of the initial. Calculate the external work done by the gas! (R = 8.31 J/mol K, ln 1 = 0, ln 2 = 0.69)

  21. Sample Problem • A monoatomic ideal gas ( = 5/3) is compressed adiabatically and the volume decreases to its half. Determine the ratio of the final pressure to the initial pressure! Solution Because  = 5/3 and V2/V1 = ½ then p1V1= p2V2 p2 /p1 = V1 /V2 = 2 (5/3) = Thus, the ratio of the final to the initial pressure is

  22. Exercise • A gas occupying a room of 40 cm3 is heated at a constant pressure so that the volume becomes twice the initial. The gas pressure is 105 Pa. Calculate the external work done by the gas! • Two moles of ideal gas initially has a temperature of 27oC, a volume V1 and pressure p1 = 6.0 atm. The gas expands isothermally to V2 volume and pressure p2 = 3.0 atm. Calculate the external work done by the gas! (R = 8.31 J/mole K)

  23. The First Law of Thermodynamic • If an external work is applied on a system, then the temperature of the system will increase. This happens because the system receives energy from the environment. This increase of temperature relates to the increase of the internal energy. • In adiabatic process, the external work applied on the ideal gas will be equal to the change of internal energy of the ideal gas. Where W = external work (J) U = the change of internal energy (J)

  24. The First Law • in non-adiabatic process, the gas will not only receive external work but also heat. Where Q = heat (J) This is it!!.....The first law of thermodynamics That states “Though heat energy has turned into the change of internal energy and external work, the amount of all energy is always constant”.

  25. The First Law • If a system receives (absorbs) heat from the environment Q = U + (+W) = U + W • If a system receives heat from the environment Q = U + (-W) = U – W The rules of W and Q values

  26. The First Law in isothermal process • In isothermal process (constant temperature), the change of internal energy U = 0, because the change of temperature T = 0. So that the first law of thermodynamics becomes

  27. The First Law in isochoric process • In isochoric process (constant volume), the work applied by gas W = 0 because the change of volume U = 0. So that the first law of thermodynamics becomes

  28. The First Law in isobaric process • In isobaric process (constant pressure), the work done by gas W = p V = p (V2 – V1). So, the first law of thermodynamics becomes

  29. The First Law in adiabatic process • In adiabatic process, the system does not receive heat or release heat, so Q = 0. Therefore, the first law of thermodynamics becomes

  30. Heat Capacity • Heat capacity of gas is the amount of heat energy needed to increase gas temperature by one Kelvin (1 K) or one degree Celsius (1oC). Where C = heat capacity (J/K) Q = absorbed heat (J) T = the change of temperature (K)

  31. Derivation • The first law of thermodynamics can be derived as

  32. For isobaric process (p = constant) Monatomic gas: Diatomic gas:

  33. For isochoric process (V = constant). Monatomic gas: Diatomic gas: • Relationship between Cp and CV :

  34. Other parameters • Molar heat capacity of gas

  35. Specific heat of gas

  36. Sample Problem Solution Because n = 1 mol T = (-23 + 273) K = 250 K V2 = ½ V1 R = 8.31 J/mol K Then W = nRT ln V2/V1 = 1 x 8.31 x 250 ln ½ = -1.4 x 103 joule Thus, the work done by the gas is -1.4 x 103 joule. • One mole of gas is compressed at a constant temperature of -23oC so that its volume decreases to half of its initial volume. Calculate the work done by the gas! (R = 8.31 J/mole K; ln 1 = 0, ln 2 = 0.69)

  37. Exercise • Two moles of ideal gas initially has temperature of 27oC, volume V1 and pressure p1 = 6.0 atm. The gas expands in isothermic process and reaches volume of V2 and pressure p2 = 3.0 atm. Calculate the external work done by the gas! (R = 8.3 J/mole K) (Answer : 11.5 J) • 2.5 m3 of neon gas with temperature 52oC is heated in isobaric process to 91oC. If the pressure of the gas is 4.0 x 105 N/m2, determine the work done by the gas! • 56 x 10-3 kg of nitrogen is heated from -3oC to 27oC. If it is heated in a free expanding vessel, then required heat of 2.33 kJ. If the nitrogen is heated in a stiff vessel (cannot expand), then the heat required is 1.66 kJ. If the relative mass of nitrogen molecules is 28 g/mole, calculated (a) the heat capacity of nitrogen, (b) the general gas constant!

  38. Thermodynamic Cycle • Cycle means the process which runs from the initial state and returns to that initial state after gas does work Random cycle in p – V diagram • In the process a – b, the gas expands in adiabatic process and the work done by the gas is the area of plane abV2V1, its value is negative. In process b – c the compressed gas in isothermal process is the area of plane bcV1V2, its value is positive. In process c – a the gas does not do any work because its volume is constant. The process c – a is an isochoric process which is done to make the gas returns to its initial state.

  39. The total external work done by the gas in one cycle a – b – c – a is the area of abca. • A thermodynamics cycle can occurs in a heat engine, such as otto engine (Otto cycle), diesel engine (diesel cycle), steam engine (Rankine cycle), and carnot engine Wabca = Wab + Wbc + Wca area of abV2V1 + (-area of bcV1V2) + 0 Wabca = area of abca

  40. Thermodynamic Cycle: Carnot engine • Carnot engine is assumed as an ideal heat engine which works cyclically and reversible between two temperatures without any loss of energy. Within one cycle, the gas returns to its initial state, so there is no change of internal energy (U = 0). Q = U + W Q1 – Q2 = 0 + W W = Q1 – Q2 imaginary Carnot engine

  41. Carnot cycle Work process of Carnot engine to produce Carnot cycle Entire of process in the Carnot can be represented in pressure (P) against volume (V) graph:

  42. Efficiency • Efficiency of engine • In the Carnot engine, holds W = Q1 – Q2

  43. Sample Problem • The figure below indicates the thermodynamics change of system from iitial state A to B and C and back to A. If VA = 0, VB = 30 joule and heat given to the system in process B  C = 50 J • Determine • the system internal energy in state C, • the heat given to the system in A  B process, and • the heat given to the system or taken from in C  A process.

  44. Solution a. The heat which is given away to the system in B  C process, is QBC = +50 joule. QBC = +50 joule WBC = 0, because B  C process is isochoric Use the first law of thermodynamics QBC = UBC + WBC UBC = QBC – WBC UBC= 50 – 0 = 50 J UBC= UC – UBUC = UBC + UB UC= 50 + 30 = 80 J   Thus, the system internal energy in state C is 80 J.

  45. b. Process from A  B WAB = area of ABED = AB x BE = 2 x 30 = 60 J UAB = UB – UA = 30 – 0 = 30 J QAB is calculated by using the first law of thermodynamics QAB = UAB + WAB QAB = 30 + 60 = 90 J Thus, the heat given in A  B process is 90 J.

  46. c. Process from C  A WCA= -area of ACED = -(area of ABED + area ABC) = -120 J UCA = UA – UC= 0 – 80 = 80 J QCA is calculated by using the first law of thermodynamics. QCA = UCA + WCA= -80 + (-120) = -200 J Thus, the energy taken in C  A process is -200 J.

  47. The Second Law of Thermodynamics • is a restriction of the first law of thermodynamics which expresses the energy conservation • It is states: “energy cannot be created or destroyed but can only change from one form to another” • Rudolf Clausius:Heat flows spontaneously from an object of high temperature to an object of lower temperature and it does not flow spontaneously in the opposite direction without external work.

  48. Entropy • The total entropy of the universe does not change when a reversible process occurs (Suniverse = 0) and increase when the irreversible process occurs (Suniverse > 0). • entropy is a measurement of the amount of energy or heat which cannot changed into work • the total change of entropy of the Carnot engine is with S = the change of entropy (J/K)

  49. Kelvin and Planck formulate the second law of thermodynamics about heat engine that it is impossible to make an engine with 100% efficiency • Principle of cooler engines: is flowing heat from the cool reservoir T2 to the hot reservoir T1 by exerting external effort on the system. • The magnitude of external work needed in a cooler engine is formulated as Where : Q1 = heat absorbed from low temperature Q2 = heat given at high temperature

  50. Sample Problem • A motor operates a cooler engine for producing ice. Q2 heat is taken from a cooling room which contains an amount of water at 0oC and Q1 heat is given away to the air around it at 15oC. Suppose the cooler engine has a coefficient of performance of 20% of the coefficient of performance of an ideal cooler engine. • Calculate the work done by the motor to make 1 kg of ice? (ice latent heat is 3.4 x 105 J/kg) • What is time required to make 1 kg of ice if the power of the motor is 50 W?

More Related