11/15/2017. F GonP. F PonG THERMAL ENERGY OF IDEAL GAS HIGH PRESSURE GAS IN A CYLINDER REMEMBER HIGH PRESSURE GAS IN A CYLINDER

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1 UNIT Thermodynamics: Laws of thermodynamics, ideal gases, and kinetic theory A HYSICS THERMAL ENERGY OF IDEAL GAS IDEAL GAS ASSUMTION Ideal gas particles do not interact at a distance; thus the system has no potential energy due to particle interactions. In an ideal gas, the total internal energy of the gas particles equals its thermal energy. The thermal energy depends only on its absolute temperature and on the number of moles of gas. CHATER 1 FIRST LAW OF THERMODYNAMICS U = 3Nk BT U = 3nRT REMEMBER HIGH RESSURE GAS IN A CYLINDER U THERMAL = N 3k BT KE avg = 3k BT Assume you have a gas at a high pressure in a cylinder with a movable piston. The gas exerts a force on the piston to the left. The piston exerts a force on the gas to the right (Newton s 3 rd Law) F Gon F ong HIGH RESSURE GAS IN A CYLINDER HIGH RESSURE GAS IN A CYLINDER Now assume that you pull the piston a small distance to the left. Distance = x olume changes from v i to v f. Change in volume = A ( X) System: gas (F ong ) ( x) = THE ISTON IS DOING WORK ON THE GAS! W = F ong X 1

2 IDENTIFY THE SYSTEM WORK DONE ON THE GAS W = F ong X = A X W = F ong A System: piston System: gas W = cos 180 F Gon x = 0 Gas expands ositive work F ong x = 180 Gas expands negative work W = f f f f Work 1 i 1 i Work 1 Work i 1 1 Work i W = Work = Area diagram Work depends on the process Gas expands is ositive W = Gas contracts is negative W = MATHEMATICAL MODEL FOR WORK If you push piston, gas compresses. Collisions between molecules become more frequent. Work done is positive (particles gain energy). If you pull piston away, gas expands. Collisions between molecules become less frequent. Work done is negative (particles lose energy). WORK DONE BY/ON A GAS (Constant ressure rocess) Work done by the environment (piston) on a gas when a gas changes from volume i to f W ENonG = Area under vs graph Negative if olume of the gas increases ositive if olume of the gas decreases. Work done by the gas on the environment (piston) when a gas changes from volume i to f W GonEN = Area under vs graph ositive if olume of the gas increases Negative if olume of the gas decreases.

3 HOW TO CHANGE THE ENERGY ( U ) OF A SYSTEM Increase, Decrease, or stays the same? rocess: Heat is added to a gas. (HYSICS 1) Temperature: increases By doing work on the system U i + W = U f Thermal energy: olume: Work done? increases stays the same No, because the volume does not change. Increase, Decrease, or stays the same? rocess: Heat is added to a gas. Temperature: increases Thermal energy: increases olume: increases Work done? Yes, negative work is done. The gas is expanding. HOW TO CHANGE THE ENERGY ( U ) OF A SYSTEM (HYSICS 1) By doing work on the system (HYSICS ) By transferring heat to the system U i + W + Q = U f 3

4 ARTICLE MOTION EXLAINS THE CHANGE IN THERMAL ENERGY OF THE GAS The energy of a hot flame is transferred to a gas inside a cooler cylinder without any work being done. This transfer of energy from an object at one temperature (the hot flame) to an object at a different temperature (the cool gas) is called heating. HEATING ( Q ) hysical quantity that characterizes a process of transferring energy from the environment to a system, which is at a different temperature. The environment does not do work on the system. HEATING ( Q ) The SI unit of heating is the calorie. One calorie is the amount of energy that must be transferred to 1 g of water to increase its temperature by 1 C. THE FIRST LAW OF THERMODYNAMICS THE THE FIRST FIRST LAW LAW OF OF THERMODYNAMICS THERMODYNAMICS W + Q = U DRUM ROLL LEASE! work done on the system Heat energy transferred to the system Change in the internal energy of the system

5 THE MEANING OF THE FIRST LAW OF THERMO ΔU = W + Q Energy can be added or removed from a gas system in two ways: Work (compression or expansion of the gas) Heating/cooling (thermal energy from/to the environment) Most often, you will have to consider both W and Q at the same time in order to determine if ΔU and ΔT are positive or Negative! WORK HEATING ENERGY RINCILE We can rewrite the first law of thermodynamics by including all types of energy in a system: ossible Confusion Alert!! Heating the gas doesn t necessarily means that its temperature will go up! All it means is that we have some positive Q. We must also consider W in order to get a complete picture of the process. WORK-HEATING-ENERGY vs WORK-ENERGY THINK OF A COLD WINTER DAY First law of thermodynamics The work-heating-energy equation explains many phenomena that the work-energy principle could not: The change in the temperature of water placed on a hot electric stove The change in the temperature of a hot hard-boiled egg placed in a bowl of cold water In both cases, the system contacts a part of the environment that is at a different temperature than the system. The first law of Thermodynamics now explains... Heat is added to a gas. iston moves up as gas warms: Work-Energy principle: U thermal > 0 W iston on gas < 0 #EpicFail Work-heating-energy principle: W iston on gas < 0 U thermal > 0 W + Q Flame to gas = U thermal The first law of Thermodynamics now explains... Heat is added to a gas: Work-Energy principle: W = 0 U thermal > 0 #EpicFail Work-heating-energy principle: W = 0 U thermal > 0 Q Flame to gas = U thermal 5

6 TWO IMORTANT OINTS ABOUT THE QUANTITATIE ANALYSIS OF HEATING roviding the same amount of energy to two equal amounts of a gas through heating might not lead to the same rise in the gases' temperatures if work was done on one gas but not the other. The energy that needs to be transferred through heating to change the temperature of 1 kg of air by 1 C is different when the volume of the gas is constant versus when the pressure is constant. A NOTE ABOUT TEMERATURE, THERMAL ENERGY, AND HEATING Temperature is the physical quantity that measures the average random kinetic energy of the individual particles that make up the object. Thermal energy is the physical quantity that measures the total random kinetic energy of all the particles. Heating is the physical quantity that measures the process through which some amount of thermal energy is transferred. JOULE vs CALORIE The SI unit of heating is the calorie. One calorie is the amount of energy that must be transferred to 1 g of water to increase its temperature by 1 C. W + Q = U JOULE S EXERIMENT Joule placed a paddle wheel in water. The wheel was connected with a string passing over pulleys tied to a heavy block. When the block went down, the paddle wheel rotated in the water. Temperature in water increased. U G = Q Joules Calories Joules m B g h = 1 cal kg K m w T JOULE EXERIMENT JOULE EXERIMENT Joule lifted the block multiple times to get a bigger temperature (greater work done on the block). m B g h = 1 cal kg K m w T Change in temperature of water agreed with predicted work done on the block. Joules calories T = m B g h 1 cal g K m w m B g h = c m w T 6

7 JOULE vs CALORIE SECIFIC HEAT ( c ) 1.0 calorie =.18 Joules. 1.0 Joule = 0.39 calorie The amount of energy needed to transfer to 1 kg of water to change its temperature by 1 C is 1000 calories or 180 joules. W + Q = U Joules Joules Joules It is the physical quantity equal to the amount of energy that needs to be added to 1 kg of a substance to increase its temperature by 1 C. The symbol for specific heat is c and the units [J/kg C]. This energy is added through heating or work or both. SECIFIC HEAT The energy ΔU must be added to a substance of mass m and specific heat c to cause its temperature to change by ΔT : m U = m c T Q = U W + Q = U W = 0 m Q = U = mc T SECIFIC HEATS OF ARIOUS SOLIDS AND LIQUIDS SECIFIC HEAT The specific heats of sand, bricks, and concrete are about one-fifth that of water. These materials exhibit much greater temperature changes than water when equal masses of these materials absorb the same amount of energy. This is one reason why the sand on a beach or the concrete beside a swimming pool feels so much hotter than the adjacent water on a sunny day. 7

8 Tip m A cup is made of an experimental material that can hold hot liquids without significantly increasing its own temperature. The cup s mass is 0.75 kg, and its specific heat is 1860 J/kg C. If the temperature of the cup increases from 0.0 C to 36.5 C, what is amount of energy that has been transferred by heat into the cup? U = 0.75 kg solution U = m c T 1860 kg J U = J m 50 J of energy are required to increase the temperature of a kg block of copper from C to 80 C. What is the specific heat of copper? If the same amount of energy is transferred to water ( kg at C) by how much would its temperature change? (c water = 180 J/kg C) solution U = m c T c = c = c = U m T 50 J kg J kg U = m c T T = T = U m c 50 J kg 180J kg T = 5.1 A car has a 16-kg, mostly aluminum (c aluminum = 900 J/kg C) engine that, when operating at highway speeds, converts chemical potential energy into thermal energy at a rate of = 1.8 x 10 5 J/s (0 hp). Suppose the cooling system shuts down and all of the thermal energy continually generated by the engine remains in the engine. How long will it take for the engine's temperature to change from T i = 0 C to T f = 100 C, the boiling temperature of water? 8

9 = U T T = U T = solution m c T 900 kg 16 kg J T = 1.8x10 5 J S T = 65.6 s SHARING ENERGY THROUGH THE ROCESS OF HEATING WHEN OBJECTS ARE IN CONTACT (CALORIMETRY) A hot object loses thermal energy to a cold object through the process of heating, and a cold object gains energy through the process of heating from a hot object. We can summarize this relation as follows: m m Tip kg of milk at 5ºC are added to 0. kg of coffee at 90ºC. After thermal equilibrium is established, what is the temperature of the liquid in the cup? (Assume the specific heat of water for both liquids) solution solution Q c = Q H T F T im = T F T ic T F T im = T F + T ic m m c m T F T im = m c c c T F T ic T F + T F = T ic + T im c m = c c 0.05 kg T F T im = 0.kg T F T ic 5T F = T F = 365 T F = 73 9

10 solution The kayak of a 70-kg man tips on a spring day and he falls into a cold stream. When he is rescued from the cold water, the kayaker's body temperature is 33 C (91. F). You place him in 50 kg of warm bath water at a temperature of 1 C (105.8 F). What is the final temperature of the man and the water? Assume (c person = 370 J/kg C) Q c = Q H m p c p T F T ip = m w c w T F T iw T F T im = T F T ic,900 T F T im = 0900 T F T ic solution T F T ip = 0.86 T F T iw T F T ip = 0.86T F T iw T F T F = 0.86T iw + T ip WORK DONE ON A GAS.86T F = T F = 68.8 T F = 36.7 Effect of a moving piston on the temperature of an enclosed gas When you move the piston, you do work on the gas and the particles collide with the moving piston. When the piston is expanding, the particles move more slowly after the collision. When the piston is compressing, the particles move more quickly after the collision. [ka] B 0 WORK DONE ON A GAS (isobaric process) Work = Area under the graph W = A [m 3 ] Negative work if the olume of the gas increases (expansion) ositive work if the olume of the gas decreases (compression) 10

11 WORK DONE ON A GAS (isochoric process) WORK DONE ON THE GAS [ k a ] B No area under the graph To determine the work done by the piston on the gas, we could use calculus A [ m 3 ] W = 0 Work = Area diagram We can also apply some hysics knowledge. [ k a ] WORK DONE ON A GAS (isothermal process) A You need calculus to find the area under the graph! 1. 5 ALYING THE FIRST LAW OF THERMODYNAMICS TO GAS ROCESSES B W = nrt ln [ m 3 ] F i Same rules from isobaric process apply! W + Q = U Isothermal process (constant temperature) The gas is in a non-insulated container within an environment of constant temperature, which is the same temperature as the gas. Isothermal The container has a piston or some form of movable walls. If the piston is pulled out, the gas molecules rebound at a slower speed when they collide elastically with the piston, which cause the temperature to decrease momentarily. But the walls of the container are always at the same temperature as the environment, heating the system to immediately bring it back to the same temperature. 11

12 higher temperature 11/15/017 Isothermal rocess Temperature is constant (ΔT = 0) Internal energy is constant (ΔU int = 0) Work is cancelled by heating (W = -Q) oints of equal temperature Isotherms higher temperature Isochoric process (constant volume) The gas is in a noninsulated container with a fixed size. Since volume is constant, no work can be done on or by the gas. Any energy transfer between the gas and environment occurs by heating. Isochoric rocess Isochoric If the environment is warmer than the gas, the particles will rebound at faster speeds when they hit the walls of the container. This will cause an increase the average speed of the particles, increasing temperature and pressure. olume is constant (Δ = 0) No work is done on or by the gas (W = 0) Internal energy is changed by heating (ΔU int = Q) 1

13 Isobaric process (constant pressure) The gas is in a noninsulated container that has a piston that can move freely (up or down), keeping the gas at constant pressure. Isobaric The gas may exchange energy with the environment by work and heating. In the scenario shown, the higher-temperature environment warms the gas by heating. But the gas warms less because the piston also moves outward, slowing the particles that hit it and causing negative work to be done on the gas by the environment. Isobaric rocess ressure is constant (Δ = 0) Work is done on or by the gas (W = -Δ) Internal energy is changed by work and heating (ΔU int = W + Q) Diagram for Isobaric rocess This would be an isobaric compression. What happens to the temperature of the gas during this process? (Think of the isotherms!) The temperature goes down! (Because the product of goes down) Adiabatic process (no energy transferred through heating) The gas is in a thermally insulated container, or a compression/expansion occurs very quickly. Adiabatic curves are steeper than isotherms, and pass through points of different temperature (different product of ). Any energy transfer between the gas and environment occurs by work. 13

14 An adiabatic curve is defined by the mathematical function for which the negative of the area under the line (work done on the gas) is completely accountable for the change in internal energy of the gas. > 1 1 Adiabatic In the scenario shown, the gas is compressed very quickly. As the particles collide with the piston moving inward, the speed of the rebounding particles is greater than before. The gas s temperature increases. Adiabatic rocess No heating (Q = 0) Work is done on or by the gas (W = -Area) Internal energy is changed by work only (ΔU int = W ) Adiabatic Compression: Diagram The gas is compressed with zero Q, so its temperature must always be going up. Must constantly go to higher isotherms, while having the volume of the gas decrease This line is called an adiabat. It is much steeper than an isotherm, and is not a hyperbola! 1

15 [ka] 100 A 0.01 B [m 3 ] 1.60 moles of a gas are compressed isothermally. a) What is the final pressure of the gas? b) What is the temperature of the gas? c) How much work is done compressing the gas. d) How much heat is removed from the gas? e) Draw an energy bar chart [ka] A solution B = nrt [m 3 ] T = nr 1 1 = = 1 1 = 00,000 a T = 300 K 0.01 W + Q = U J + Q = 0 J Q = J - solution [ka] 500 W = nrt ln F A i B W = ln [m 3 ] ositive work (compressing) W = 553.5J 8,000 6,000,000, ,000 -,000-6,000-8,000 W Q DU A sample of helium is taken through the cycle shown in the diagram. The temperature of state A is 00 K. For each process in the cycle, indicate whether the quantities W, Q and U are positive, negative, or zero. W Q U A B B C C A

16 A burner heats 1.0 m 3 of air inside a small hot air balloon. The air is at atmospheric pressure, and initially at 37 C. W = W = 1x W = 0,000 J 1. Draw a diagram (assume the gas expands slowly so that it remains at constant atm ). Determine the amount of energy that needs to be transferred to the air through heating (in joules) to make it expand from 1.0 m 3 to 1. m Draw an energy bar chart. Gas is expanding, negative work - SOLUTION - solution 1 T 1 = T T = T m3 T = K 1.0 m 3 T = K U = 3 U 1 = 150,000 J U = 180,000 J U = U U 1 U = 30,000 J W + Q = U Q = U W Q = 30,000 J ( 0,000 J) Q = 50,000 J - solution An ideal gas is heated in a container of fixed volume, so that its absolute temperature doubles. final A fixed volume process must correspond to a vertical line on the diagram. A If the initial state of the gas corresponds to point A shown, draw the process as well as the final state of the gas on the diagram. initial In order for the temperature to double with a fixed volume, the pressure of the gas must double. 16

17 - solution Explosive Start! Notice, this will also bring the gas to a higher isotherm! This makes sense, since it goes to a higher temperature. While testing the strength of a steel-reinforced cargo crate made for shipping fireworks, the engineers at BOOM Co light the fuse to a pile of fireworks while they are sealed inside one of the cargo crates. The massive transformation of chemical energy to thermal energy cases the temperature within the container to skyrocket. Explosive Start! The fixed-volume explosion involves 1,00 moles of gas, and undergoes the following process. a) The melting point of steel is 1,370 C. Will the container survive? b) How much is the gas heated during the explosion? c) Draw an energy bar chart When working with gas processes, you will need to use all three of these to solve the problems: = nrt U = 3 Remember: Celsius to Kelvin add W + Q = U Also remember: nr and Nk B are interchangeable based on given info = nrt = Nk B T T 1 = K T 1 = 7.7 T = 150. K T = Sadly, someone forgot to lock the container and the process was no longer isochoric. Finally, since the process was isochoric, we know that no work could be done. All of that energy increase had to come from heating when the fireworks exploded. So, W + Q = U 0 + Q = U Q = 18,000,000 J Tank will survive melting temperature U 1 =,500,000 J U =,500,000 J U = 18,000,000 J 0,000,000 15,000,000 10,000,000 5,000,000 0 W Q DU 17

18 - solution [a] [m 3 ] How much heating is transferred to this isobaric expansion? Q = How do you know is an isobaric expansion? olume is increasing. Gas is expanding at constant pressure [a] WORK (negative is expanding) [m 3 ] T 1 = nr U 1 = T = nr U = U = W = W + Q = U Q = Q = Build your own adiabat! Draw the diagram for an adiabatic expansion. State whether the temperature of the gas increases or decreases, and draw in the isotherms to support your answer. - solution [ka] moles of an ideal gas undergo the adiabatic expansion shown below [m 3 ] a) Find the temperature at each vertex. b) Find the change in internal energy for the process. c) How much work is done by the gas during this expansion? d) Draw an energy bar chart [ka] [m 3 ] T 1 = 75.1 K U 1 = 7500 J ,000-1,500 -,000 U = 1500 J W + Q = U W + 0 = J W = J T = K U = 6000 J W Q DU 18

19 A QUESTION One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and eventually the state of the gas returns to its initial state with a - diagram as shown below. Answer the following in terms of 0, 0, and R. a) Find the temperature at each vertex. b) Find the change in internal energy for each process. c) Find the work done on the gas for each process. d) Find the heat transferred to the gas in each process e) Find total W, Q and U. A B C T U W Q U A B R B C R C A R CYCLE U AB = U BC = U CA = solution T A = 0 0 R U A = W AB = W BC = 0 W CA = U NET = 0 W NET = T B = 0 0 R T C = 0 0 R U B = U C = Q AB = Q BC = Q CA = Q NET = A QUESTION One mole of ideal gas is at pressure 0 and volume 0. The gas then undergoes three processes: i. The gas expands isothermally to 0 while heat Q flows into the gas. ii. The gas is compressed at constant pressure back to the original volume. iii. The pressure is increased while holding the volume constant until the gas returns to its initial state. a) Draw a - diagram that depicts this cycle. Label relevant points on the axes. In terms of 0, 0, and R b) Find the temperature at each vertex. c) Find the change in thermal energy (internal energy) for each leg of the cycle. d) Find the work done on the gas on each leg of the cycle. e) Find the heat transferred on each leg of the cycle. - solution [ka] 50/ 0 A [ka] 50/ 0 A 30/ 30/ 0/ 0/ 0/ C B 0 30/ 0 50/ [m 3 ] 0/ 0/ 0/ C B 0 30/ 0 50/ [m 3 ] 19

20 A B C T U W Q U 0 0 R 0 0 R 0 0 R A B ln() 0 0 ln () 0 0 B C C A CYC LE ln() 0 0 ln [ka] 50/ 0 30/ 0/ 0/ 0/ A C U AB = 0 U BC = U CA = B [m / 0 50/ ] - solution T A = 0 0 R U A = W AB = ln() 0 0 W BC = 0 0 W CA = 0 T B = 0 0 R U B = T C = 0 0 R U C = Q AB = ln () 0 0 Q BC = Q CA = Skills for solving gas problems using the first law of thermodynamics In addition to the standard problem-solving strategy, when doing the "simplify and diagram" step: Decide whether you can model the system as an ideal gas. Decide whether the gas undergoes one of the isoprocesses. When doing the "mathematical representation" step: Use a work-heating-energy bar chart to help apply the first law of thermodynamics. 1.6 CHANGING STATE Q [J] kg of water at 0 C is heated up to 100 C Graph Q vs. T (shape) Write an equation for the graph Write an expression for the slope of the graph Q = m c T slope = m c T [ C] 0

21 Assume 1 kg of ice at -0 C is heated up to 150 C Find the slopes for sections A, C, E. Sections A 100 Sections C 180 Sections E 1960 slope = m c In this specific case, since the mass of the sample is 1 kg, the slope reflects the specific heat of the substance. Sections A c ice = 100 kg/j C Sections C c water = 100 kg/j C Sections E c vapor = 1960 kg/j C SECTIONS B & D hase change graph becomes vertical. Section B The ice starts to melt. Section D water starts to boil. The thermometer reading does not change, even though energy is being added to the system. Slope = 0 Melting and freezing Boiling and condensation When we transfer energy to the solid material at the melting temperature, all of this energy goes into changing the potential energy of particle interactions, not the kinetic energy thus the temperature does not change. For a molecule to leave the surface of a liquid, it must have enough kinetic energy to break away from the neighboring molecules, which are exerting attractive forces on it. The energy transferred to the liquid leads to a change in the potential energy component of the internal energy. 1

22 How much energy is added between sections A & C (Section B)? Energy to melt or freeze: Latent heat of fusion [L F ] Sections B Q = 33,000 J How much energy is added between sections C & E (Section D)? Sections D Q =,56,000 J The energy in joules needed to melt a mass m of a solid at its melting temperature, or the energy released when a mass m of the liquid freezes at that same temperature. U int = ± m L F L F is the heat of fusion of the substance. The plus sign is used when the substance melts and the minus when the substance freezes. Energy to boil or condense: Heat of vaporization [L ] Heats of fusion and vaporization The energy in joules needed to vaporize a mass m of a liquid at its boiling temperature, or the energy released when a mass m of a gas condenses at that same temperature. U int = ± m L L is the heat of vaporization of the substance. The plus sign is used when the substance vaporizes and the minus when the substance condenses. L f = 3.35x105 J kg WATER L =.56x106 J kg Things to notice about melting and boiling The values for the heats of fusion and vaporization are much larger than the specific heat. Much more energy is needed to change the state of a substance than to change its temperature. The values for heat of vaporization are significantly larger than the values for heat of fusion. More energy is required to boil the same mass of the same substance than to melt it. How much energy is required to vaporize 3 kg of ice at - 0 C? 1. Heat to increase the temperature of ice to 0 C Q = m c T Q = [0 C (-0 C)] Q = 16,000 J

23 . Heat to melt the ice Q = +m LF Q = 3 335,000 Q = 1,005,000 J. Heat to vaporize water Q = +m Lv Q = 3,56,000 Q = 6,768,000 J 3. Heat to increase the temperature of water to 100 C Q = m c T Q = [100 C 0 C] Q = 1,5,000 J 5. Total heat Q total = Qice + QF + Q water + Q v Q = 16, ,005, ,5, ,768,000 Q total = 9,153,000 J solution You add 10 g of ice at temperature 0 C to 00 g of coffee at 90 C. Once the ice and coffee reach equilibrium, what is their temperature? m i L Fi + m i c w T F T iw Q Fusion + Q w = Q c Energy to melt the ice (Heat of fusion) = m c c c T F T ic x T F 0 = T F 90 solution T F 0 = 836 T F T F = 836T F T F = T F = 81.9 HEATING MECHANISMS Conduction Convection Radiation Evaporation 3

24 HEATING MECHANISMS HEATING MECHANISMS evaporation CONDUCTION (H CONDUCTION ) Energy transfer from particle to particle via contact. convection radiation HEATING MECHANISMS CONECTION (H CONECTION ) Energy transfer by particles moving from one place to another. HEATING MECHANISMS RADIATION (H RADIATION ) Energy transfer via absorption or emission of radiation. HEATING MECHANISMS EAORATION (H EAORATION ) Energy transfer due to evaporation/condensation on surface.

25 HEATING MECHANISMS Evaporation When gaseous water vapor is converted to liquid water (condenses), energy is released and returned to the object on which the vapor condenses, raising its temperature. To stay cool, you want the rate of evaporation to be somewhat greater than the rate of condensation. 5