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Thermal Physics

Thermal Physics. Topic 10.2 Processes. The First Law of Thermodynamics. Is a statement of the Law of Conservation of Energy in which the equivalence of work and thermal energy flow is taken into account. It can be stated as

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Thermal Physics

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  1. Thermal Physics Topic 10.2 Processes

  2. The First Law of Thermodynamics Is a statement of the Law of Conservation of Energy in which the equivalence of work and thermal energy flow is taken into account. It can be stated as the heat added to a closed system equals the change in the internalenergy of the system plus the work done by the system.

  3. That is,  Q = U +  W = U + pV or U = Q - W

  4. Explanation Where `+ Q' is the thermal energy added to the system and `+ W' is the work done by the system. If thermal energy leaves the system, then Q is negative. If work is done on the system, then W is negative. For an isolated system, then W = Q = 0 and U = 0.

  5. Example If 22 J of work is done on a system and 3.4 x 102 J of heat is added, what is the change in internal energy of the system?

  6. Solution Using the formula, Q = U + W We have that 340 J = U + (-22) J So that U = 340 J + 22 J = 362 J That is, the change in internal energy of the system is 3.6 x 102 J.

  7. Isobaric A graph of pressure as a function of volume change when the pressure is kept constant is shown below

  8. p p V V p -V diagram Isobaric process V1 V2 Area = work done = p (V1 - V2)

  9. Work Done Such a process is said to be isobaric. Note that the work done by the gas is equal to the area under the curve An isobaric transformation requires a volume change at constant pressure For this to occur, the temperature needs to change to keep the pressure constant

  10. Example • 6.0 dm3 of an ideal gas is at a pressure of 202.6 kPa. It is heated so that it expands at constant pressure until its volume is 12 dm3. Find the work done by the gas.

  11. Solution • Using the formula W = pV • we have that • W = 202.6 kPa x(12 - 6.0) dm3 • = 202.6 x 103 Pa x (12 - 6.0) x 10-3m3 • = 1.216 x 103 J • That is, the work done by the gas in the expansion is 1.2 x 103 J.

  12. p V Isochoric / Isovolumetric A graph of pressure as a function of volume change when the volume is kept constant is shown below Isochoric / Isovolumetric

  13. Work Done Such a process is said to be isochoric. When the volume is kept fixed, the curve of the transformation is said to be an isochore.Note that the work done by the gas is equal to zero as V = 0. There is zero area under the curve on a p-V diagram.

  14. p V p - V diagram Isochoric / Isovolumetric

  15. However The temperature and pressure can both change and so such a transformation will be accompanied by a thermal energy change.

  16. Example A thermal system containing a gas is taken around the cycle as in the next slide (Cyclic processes such as this one have important applications in heat engines that convert internal energy into useful mechanical energy).

  17. p V B C 6 2 A D 4 10 Starting at point A on the diagram, describe the cycle, and, calculate the work done by the system in the completion of one cycle.

  18. Solution From A to B, the volume is kept constant (isovolumetric) as the pressure increases. This can be achieved by heating the gas. Since V = 0, then no work is done by the gas, W = 0.

  19. From B to C, the gas expands (volume increases) (isobaric expansion) while the pressure is kept constant. The amount of work done by the gas is given by the area under the 6 kPa isobar. Now, we have that W = p V So that W = 6 kPa. (10 - 4) m3 = 36,000J

  20. From C to D, the gas is cooled to keep the volume constant as the pressure is decreased (isovolumetric) Again V = 0 and no work is done by the gas, W = 0

  21. From D to A, the gas is compressed (volume decreases) (isobaric compression) while the pressure is kept constant. The amount of work done on the gas is given by the area under the 2 kPa isobar. Again, using the fact that W = p V we have that W = 2 kPa. (4 -10) m3 = - 12,000 J

  22. That is, the net work done by the gas is therefore 36,000 J – 12,000 J =24,000 J. Because the cycle is traced in a clockwise direction, the net work done on the surroundings is positive. If the cycle was traced in a counter-clockwise direction, the work done would be negative

  23. Isothermal Processes A thermodynamic process in which the pressure and the volume are varied while the temperature is kept constant is called an isothermal process

  24. In other words, when an ideal gas expands or is compressed at constant temperature, then the gas is said to undergo an isothermal expansion or compression.

  25. p T1 T2 T3 V The figure below shows three isotherms for an ideal gas at different temperatures where Tl < T2 < T3 Isothermal process

  26. Isothermal Processes The curve of an isothermal process represents a Boyle's Law relation pV =constant = nRT • the moles of gas n, • the molar gas constant R, • the absolute temperature T are constant

  27. How? In order to keep the temperature constant during an isothermal process • the gas is assumed to be held in a thin container with a high thermal conductivity that is in contact with a heat reservoir - an ideal body of large mass whosetemperature remains constant when heat is exchanged with it. e.g.. a constant-temperature water bath.

  28. And • the expansion or compression should be done slowly so that no eddies are produced to create hot spots that would disrupt the energy equilibrium of the gas.

  29. An Ideal Gas Consider an ideal gas enclosed in a thin conducting vessel that is in contact with a heat reservoir, and is fitted with a light, frictionless, movable piston If an amount of heat Q is added to the system which is at point A of an isotherm, then the system will move to another point on the graph, B.

  30. The heat taken in will cause the gas to expand isothermally and willbe equivalent to the mechanical work done by the gas Because the temperature is constant, there is no change in internal energy of the gas

  31. That is, T = 0 and U = 0 Q = W The work done by gas is equal to the Area under the curve between A and B

  32. Isothermally If the gas expands isothermally from A to B and then returns from B to A following exactly the same path during compression Then the isothermal change is said to be reversible.

  33. Adiabatic Processes An adiabatic expansion or contraction is one in which no heat Q is allowed to flow into or out of the system For the entire adiabatic process, Q = 0

  34. How To ensure that no heat enters or leaves the system during an adiabatic process it is important to: • make sure that the system is extremely well insulated • carry out the process rapidly so that the heat does not have the time to leave the system

  35. p T1 T2 Adiabatic compression V Adiabatic Expansion

  36. Example The compression stroke of an automobile engine is essentially an adiabatic compression of the air-fuel mixture The compression occurs too rapidly for appreciable heat transfer to take place

  37. What happens In an adiabatic compression the work done on the gas will lead to an increase in the internal energy resulting in an increase in temperature. U=Q - W but Q=O therefore U= - W

  38. In an adiabatic expansion the work done by the gas will lead to a decrease in the internal energy Resulting in a decrease in temperature.

  39. p p p V V V Work done Area between curves = Net work done by the gas Area = work done on the gas as it is compressed Area = work done by the gas expanding

  40. Example For the compression stroke of an experimental diesel engine, the air is rapidly decreased in volume by a factor of 15, the compression ratio. The work done on the air-fuel mixture for this compression is measured to be 550 J (a) What type of thermodynamic process is likely to have occurred? (b) What is the change in internal energy of the air-fuel mixture? (c) Is the temperature likely to increase or decrease?

  41. Solution (a) Because the compression occurs rapidly appreciable heat transfer does not take place, and the process can be considered to be adiabatic, Q = 0

  42.  U = Q - W = 0 - (-550) J Therefore, the change in internal energy is 550J

  43. (c) The temperature rise will be very large resulting in the spontaneous ignition of the air-fuel mix

  44. Thermodynamic Cycle and Engines A thermodynamic engine is a device that transforms thermal energy to mechanical energy Cars, Steam trains, jets and rockets have engines that transform fuel energy (chemical energy) into kinetic energy of their motion

  45. Efficiency In all engines the conversion is accompanied by the emission of exhaust gases that waste some of the thermal energy These engines are not very efficient as only part of the thermal energy is converted to mechanical energy

  46. Presenters The EngineBy Vija and Bombo Internal Combustion By Adil and Suhayb Schematic of a Heat Engine By Ham and Heisei Refrigerators By Aleem and Bhavik Heat Pumps By Alex and Said Schematic for a Refrigerator by Karim and Chilla Carnot´s Theorem By Martin and Mitul

  47. The Engine By Vija and Bombo Has two crucial features: • It must work in cycles to be useful • the cyclic engine must have more than one heat reservoir

  48. The Thermodynamic Cycle Is the process in which the system is returned to the same state from which it started That is, the initial and final states are the same in the cyclic process

  49. Combustion Engines The steam engine and the steam turbine are examples of external combustion engines The fuel is burnt outside the engine and the thermal energy is transferred to the piston or a turbine chamber by means of steam

  50. The Steam Engine Water is heated in a boiler heat reservoir (high temperature) to produce steam As the piston returns to its original position, the intake valve is closed and the exhaust valve opens

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