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Physics 221, April 19 Key Concepts: •The first law •Thermodynamic processes •The second law •Entropy The first law of thermodynamics Energy is conserved. ΔU = ΔQ - ΔW. U = internal energy of a system (ordered and disordered energy). A system can be anything (gas, liquid, solid). The internal energy can be in the form of kinetic energy (translational or rotational) and in the form of potential energy. The internal energy U can change, if the system interact with its environment. Heat can flow in or out of the system changing U. ΔQ = heat flowing into the system ΔQ is negative if heat flows out of the system. Work can be done by the system on its environment or by external forces on the system. ΔW = work done by the system ΔW is negative if work is done on the system. The disordered internal energy U is proportional to the temperature T of the system. Extra credit: It is possible to change the temperature of a system when the system is insulated from its surroundings, so that no heat can flow into and out of the system. 1. True 2. False 0% 1 0% 2 30 Physical properties For a gas: N = # of particles that make up a system U = internal energy of the system V = volume of the system P = pressure of the system T = temperature of the system N, U, V, P, and T are examples of physical properties of the system. They depend only on the state of the system, not on the way the system was put into this state. When a system changes from state 1 to state 2, ΔQ and ΔW depend on the process. They are not physical properties of the system. Thermodynamic processes ΔU = ΔQ – ΔW (first law of thermodynamics) Which processes can change the state of a system? adiabatic process: No heat enters or leaves the system. ΔU = -ΔW isobaric process: The pressure is constant. ΔW = P(V2 - V1) for an ideal gas. isovolumetric, isometric or isochoric process: The volume is constant. ΔW = 0 for an ideal gas. isothermal process: The temperature is constant. Any sum or product of physical properties of a system is also a physical property of the system. The enthalpy H = U + PV is one such property. For an isobaric process ΔH = U + PΔV = U + W = ΔQ. So for an isobaric process ΔQ is equal to a physical property of the system and does not depend on the details of the process (or reaction). 5 moles of a monatomic gas has its pressure increased from 105 Pa to 1.5*105 Pa. This process occurs at a constant volume of 0.1 m3. Determine the change in the internal energy of the gas. Hint: PV = nRT, U = (3/2)nRT, R = 8.31 J/K 1. 2. 3. 4. 104 J 15000 J 7500 J The change in U depends on how the temperature was raised. 0% 1 0% 2 0% 3 0% 4 30 One mole of monatomic gas undergoes an adiabatic process (i.e. one where no heat enters or leaves the system). A pressure versus volume graph of the process is shown to the right. The initial and final volumes are given in the table below. Ideal gas law: PV = nRT, R = 8.31 J/K What are the initial and final temperatures? 1. 2. 3. 4. There is not enough information given to know. The same, since no heat enters the system. Initial T: 361 K; final T: 144 K Initial T: 3000 K; final T: 1200 K 30 0% 1 0% 2 0% 3 0% 4 Consider the change a gas undergoes as it transitions from point c to point a in the PV diagram. What type of process is this? 1. 2. 3. 4. 5. Adiabatic Isothermal Isobaric Isochoric none of the above 30 0% 1 0% 2 0% 3 0% 4 0% 5 Extra credit: Consider two separate processes acting on two different ideal gases. Which one involves a greater magnitude of work? The initial and final points are the same. Process 1 Process 2 1. 2. 3. 4. Process 1 Process 2 Both have the same magnitude of work. We need more information about the gases. 0% 1 0% 2 0% 3 0% 4 30 The second law of thermodynamics • Heat cannot, of itself, flow from a cold to a hot object. Can we transfer thermal energy from a cold to a hot object? Yes, but it takes ordered energy. We must do work, converting some ordered energy into thermal energy. We deliver more thermal energy to the hot object than we remove from the cold object. The minimum amount of work is Wmin = Qhot – Qcold , where Qhot/Thot = Qcold/Tcold. Examples: refrigerator, air conditioner, heat pump • Heat cannot be taken in at a certain temperature with no other change in the system and converted into work. Can we do work with heat? Yes, we can remove heat from a hot object and convert some it into ordered energy, but only if we have some colder place where we can dump the rest of it. The maximum amount of work we can get from thermal energy is Wmax = Qhot – Qcold , where Qhot/Thot = Qcold/Tcold. Examples: steam engine, internal combustion engine, fossil and nuclear power plants Extra credit: In any process, the maximum amount of heat that can be converted into mechanical energy 1. 2. 3. 4. depends on the intake and exhaust temperatures. depends on whether kinetic or potential energy is involved. depends only on the amount of friction present. is 100%. 0% 1 0% 0% 2 3 0% 4 30 The _______ the temperature difference between hot and cold, the larger the fraction of heat you can divert and transform into __________. 1. 2. 3. 4. larger; temperature larger; work smaller; work smaller; temperature 0% 1 0% 2 0% 3 0% 4 30 Entropy The total entropy of a closed system is always increasing is another way of stating the second law of thermodynamics. Change of entropy in terms of heat: ΔS = ΔQ/T Entropy is also a measure of the variety of ways in which a closed system with a fixed amount of total energy can organize itself. • The macrostate of a system is its state as viewed at a macroscopic level. • The microstate of a system is its state as viewed at the molecular level. • The multiplicity Ω is the number of ways the insides of a system can be arranged so that from the outside things looks the same. • The entropy S is S = kB lnΩ, where kB is the Boltzmann constant. For a closed system all microstate are equally likely. The macrostate with the largest multiplicity therefore is the most likely state. Equilibrium is the most likely state and the equilibrium state has the highest entropy. We flip a fair coin 10 times. Define a macrostate to be the number of heads. Some microstates for coin flips are listed below. “H” stands for heads and “T” stands for tails. Which macrostate do each of I) – IV) belong to? 1. 2. 3. I) 10 heads; II) 7 Heads; III) and IV) 5 heads I) ordered; II), III) and IV) disordered I) and II) improbable: III) and IV) probable 0% 1 0% 2 0% 3 30 We flip a fair coin 10 times. Define a macrostate to be the number of heads. Some microstates for coin flips are listed below. “H” stands for heads and “T” stands for tails. Rank the microstates from most likely to least likely to occur Most least likely 1. 2. 3. 4. I) II) III) IV) VI) III) II) I) VI) = III) II) I) VI) = III) = II) = I) 0% 1 0% 2 0% 3 0% 4 30 We flip a fair coin 10 times. Define a macrostate to be the number of heads. Which macrostate is most likely to occur. 1. 2. 3. 4. 10 heads 7 heads 5 heads 1 head 0% 1 0% 2 0% 3 0% 4 30 Why does heat, of itself, not flow from a cold to a hot object, if it is allowed by Newton's laws? 1. 2. 3. 4. Energy would not be conserved. Cold matter is more dense, and therefore this would require transfer of matter. This would require a large departure from equilibrium, which is extremely unlikely. Heat does flow from cold objects to hot objects. Every winter regions near the poles get colder while regions near the equator get even hotter. 0% 1 0% 0% 2 3 0% 4 30 A stone of mass 1.0 kg is dropped into a large lake of water from a height of 3.0 m. Both the stone and the lake have a temperature of 300 K. Find the changes in entropy of the stone – lake system. 1. 2. 3. 4. 5. 10 J/K 0 30 J/K 3 J/K 0.1 J/K 0% 1 0% 2 0% 3 0% 4 0% 5 30 Extra credit: By how much does the entropy of 100 g of water at 0 oC change if the water is slowly converted into ice at 0 oC? The latent heat of fusion is 80 kcal/kg. 1. 2. 3. 4. 5. 6. 8 kcal/K -8 kcal/K 0.029 kcal/K -0.029 kcal/K 33.5 J/K -33.5 J/K 0% 1 0% 0% 0% 2 3 4 30 0% 0% 5 6