General Physics 1, Q2 Week 8 Thermodynamics PDF
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This document is a learning kit for learners to understand thermodynamics, a branch of physics related to heat, work, and temperature. It will guide learners through the key concepts of thermodynamics, and their application, including the first and second laws of thermodynamics, PV diagrams. This kit covers ideal gas laws within the context of the thermodynamics concepts.
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THERMODYNAMICS for GENERAL PHYSICS 1/ Grade 12 Quarter 2/ Module 8 1 FOREWORD This learning kit will serve as guide for learners to understand thermodynamics. Thermodynamics is a branch of physics that deals with heat, work, and temper...
THERMODYNAMICS for GENERAL PHYSICS 1/ Grade 12 Quarter 2/ Module 8 1 FOREWORD This learning kit will serve as guide for learners to understand thermodynamics. Thermodynamics is a branch of physics that deals with heat, work, and temperature, and their relation to energy and other physical properties of matter. This relationship between absolute pressure, volume and temperature is explained in the Ideal Gas Law. Central to the concept of thermodynamics are its laws. The first law speaks that there is a constant amount of energy in the universe (conservation of energy). The second law states that the energy of the universe is not conserved but rather more energy is wasted as heat. Learners will understand the relationship between changes in internal energy, work done, and thermal energy, explain entropy, calculate entropy in different systems and realize that disorder increases until it reaches its equilibrium and what happens after that. Further, they will also learn entropy as a measure of disorder and identify if it is a reversible or irreversible process. Moreover, learners will understand PV diagrams of a thermodynamic system. In doing so, learners are expected to learn how to interpret PV diagrams. This learning kit is carefully prepared with a set of activities guided with contextualized discussions and illustrations that meet the standards of the K12 curriculum. In using this learning kit, learners will realize that physics is a boundless discipline because it covers almost everything man can imagine. The activities included herein are simple, readily understandable, and easy to do. In doing so, learners will be given opportunity to broaden their knowledge and enhance their resourcefulness and creativity in performing activities provided to them. This will enable them to develop their critical thinking skills. The mathematics involved is simple and does not require students to be math wizards to fit into analyses. It is hoped that their understanding of the basic concepts will benefit them in many ways and the skills they acquired in using this kit may help them in dealing with practical problems. 2 OBJECTIVES At the end of this Self-Learning Kit, you should be able to: K: differentiate the thermodynamic processes; : interpret PV diagrams of a thermodynamic process; S : solve problems involving ideal gas equations and thermodynamic processes; : derive the expression for the work done by a gas; and A: discuss the application of the first and second laws of thermodynamics in everyday life. LEARNING COMPETENCIES Enumerate the properties of an ideal gas (STEM_GP12GLTIIh- 57). Solve problems involving ideal gas equations in contexts such as, but not limited to, the design of metal containers for compressed gases (STEM_GP12GLTIIh- 58). Interpret PV diagrams of a thermodynamic process (STEM_GP12GLTIIh- 60). Compute the work done by a gas using dW=PdV (STEM_GP12GLTIIh- 61). State the relationship between changes internal energy, work done, and thermal energy supplied through the First Law of Thermodynamics (STEM_GP12GLTIIh- 62). Differentiate the following thermodynamic processes and show them on a PV diagram: isochoric, isobaric, isothermal, adiabatic, and cyclic (STEM_GP12GLTIIh- 63). Calculate the efficiency of a heat engine (STEM_GP12GLTIIi-67). Describe reversible and irreversible processes (STEM_GP12GLTIIi-68). Explain how entropy is a measure of disorder (STEM_GP12GLTIIi-69). State the 2nd Law of Thermodynamics (STEM_GP12GLTIIi-70). Calculate entropy changes for various processes e.g., isothermal process, free expansion, constant pressure process, etc. (STEM_GP12GLTIIi-71). 3 I. WHAT HAPPENED Hello classmate! Hi! I‘m Lovely. We are It‘s me Drei. It‘s nice to here to help you learn about meet you. Today is a thermodynamic systems. wonderful day to learn. Allow us to help you in completing different activities we will meet along the way. Can we expect a full blast of energy and active That‘s good to participation from you? hear. Come and let us join hands in learning PV diagrams. Let‘s begin this with an activity. Are you ready? PRE-TEST: Let us test your stock knowledge! MULTIPLE CHOICE: Read each statement carefully. Choose the letter of the BEST answer and write it down on your General Physics 1 notebook. 1. Which mathematical formula summarizes the Ideal Gas Law? A. PV = Dt/r B. PV = ma C. PV = nRT D. PV = mgh 2. What is the definition of thermodynamics? A. The energy available to do work. B. The study of the relationship between heat, work, and energy. C. The amount of heat it takes to move an engine. D. The movement of heat. 4 3. When heat is added to a system, all of the following may happen except A. increase in internal energy B. decrease in the system‘s temperature C. external work is done by the system D. increase in the pressure in the system 4. The average molecular kinetic energy of a gas depends on: A. pressure B. volume C. temperature D. number of moles 5. The temperature of an ideal gas increases from 20 C to 40 C while the pressure stays the same. What happens to the volume of the gas? A. It doubles B. It quadruples C. It is cut to one-half D. It slightly increases 6. Internal energy of an ideal gas depends on: i. the volume of the ideal gas ii. the pressure of the ideal gas iii. the absolute temperature of the ideal gas A. I B. II C. III D. I and II 7. What quantities appear in the first law of thermodynamics? A. Heat, work, and thermal energy B. Work, heat, and entropy C. Inertia, torque, angular momentum D. Force, mass, acceleration 8. When heat is added to a system, all of the following may happen except A. increase in internal energy B. decrease in the system‘s temperature C. external work is done by the system D. increase in the pressure in the system 9. Heat is the _____. A. average amount of energy per molecule contained in an object. B. total amount of energy contained in an object. C. energy transferred between objects because of a temperature difference. D. amount of energy all the molecules have. 10. Which of the following summarizes the first law of thermodynamics? A. Two systems that are in equilibrium with a third system are in equilibrium with one another. B. The total entropy of a system increases over time. C. Energy cannot be created or destroyed. D. Energy cannot be transferred or transformed. 5 11. All the following statements define the second law of thermodynamics except, A. Processes that involve the transfer or conversion of heat energy are irreversible. B. The total entropy of an isolated system can never decrease over time. C. Energy can be changed from one form to another, but it cannot be created or destroyed. D. Heat is a form of energy, and thermodynamic processes are Therefore subject to the principle of conservation of energy. 12. What is relation between thermodynamic probability and entropy? A. thermodynamic probability increases with entropy B. thermodynamic probability decreases with entropy C. thermodynamic probability does not change with change of entropy D. none of the above 13. At the most probable state of a system the entropy of the system is A. minimum B. maximum C. constant D. none of the above 14. Which of the statements below shows the connection of entropy in our daily lives? A. melting of ice B. cooking rice C. getting old D. all of the above 15. Why do engines cannot reach 100% efficiency? A. because it needs additional pressure B. because friction generates heat C. because some parts are not functional D. all of the above 6 II. WHAT I NEED TO KNOW DISCUSSION: IDEAL GAS LAW Since it's hard to exactly describe a real gas, people created the concept of an ideal gas that helps us model and predict the behavior of real gases. Real gases often exhibit behavior very close to ideal. The properties of an ideal gas are: An ideal gas consists of a large number of identical molecules. The volume occupied by the molecules themselves is negligible compared to the volume occupied by the gas. The molecules obey Newton's laws of motion, and they move in random motion. The molecules experience forces only during collisions; any collisions are completely elastic, and take a negligible amount of time. An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly flexible and in which there are no intermolecular attractive forces. In such a gas, all the internal energy is in the form of kinetic energy and any change in internal energy is accompanied by a change in temperature. An ideal gas can be characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them may be deduced from kinetic theory and is called the Ideal Gas Law. The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation of the behavior of many gases under many conditions, although it has several limitations. It was first stated by Benoît Paul Émile Clapeyron in 1834 as a combination of the empirical Boyle's law, Charles's law, Avogadro's law, and Gay-Lussac's law. The behavior of an ideal gas, that is, the relationship of pressure (P), volume (V), and temperature (T), can be summarized in the ideal gas law: PV = nRT where n is the number of moles of gas, and R = 8.31 J / (mol K) is known as the universal gas constant. 7 An alternate way to express the ideal gas law is in terms of N, the number of molecules, rather than n, the number of moles. N is simply n multiplied by Avogadro's number, so the ideal gas law can be written as: PV = NkT (Boltzmann’s constant k= R/NA = 1.38 x 10 -23 J/K) LAWS OF THERMODYNAMICS Thermodynamics is a branch of physics which deals with study of systems involving energy in the form of heat and work. It was born in the 19th century as scientists were first discovering how to build and operate steam engines. It deals with the relationship between heat and other properties (such as pressure, density, temperature, etc.) in a substance. A thermodynamic system includes anything whose thermodynamic properties are of interest. It is embedded in its surroundings or environment; it can exchange heat with, and do work on, its environment through a boundary, which is the imagined wall that separates the system and the environment. A good example of a thermodynamic system is gas confined by a piston in a cylinder. If the gas is heated, energy is added to the gas molecules. We can observe the increase in average kinetic energy of the molecules by measuring how the temperature of the gas increases. As the gas molecules move faster, they also collide with the piston more often. These increasingly frequent collisions transfer energy to the piston and move it against an external pressure, increasing the overall volume of the gas. In this example, the gas has done work on the surroundings, which includes the piston and the rest of the universe. This is one example of how a thermodynamic system can do work. FIRST LAW OF THERMODYNAMICS The First Law of Thermodynamics states that energy cannot be created or destroyed. The first law is a restatement of the principle of conservation of energy which states that energy can be transferred and transformed, but it cannot be created or destroyed. The first law is all about the relationship between internal energy, heat, and work. Let us see how these three quantities are related. 8 Imagine yourself cooking a popcorn. When the popcorn inside the pot is placed over a flame, heat is being added. Energy is added to the popcorn by conduction of heat. Each kernel begins to pop and expand. In this process, the changes that took place in the popcorn were first, the temperature of the popcorn increased; second, as the popcorn expands, its volume also increased and consequently pressure is also raised. Because of these changes in the state of the popcorn, its internal energy is also increased. Now, when almost all the popcorn kernels have popped, the lid begins to move up as the popcorn pushes their way out of the pot to give space for other popcorns inside to expand. The popcorn has done work by pushing the lid up to a certain distance. This is the first law of thermodynamics, the heat added to a system is equal to the sum of the increase in internal energy plus the external work done by the system. In thermodynamics, internal energy refers to the total energy (in the form of heat) of the system. Heat is energy transferred between substances or systems due to a temperature difference between them. The relationship between internal energy and work is given by the following equation: Q = ΔU + W where Q is the heat added to the system, ΔU is the change in the internal energy of the system, and W is the work done by the system on its surroundings. The first law of thermodynamics may seem kind of abstract, but if we start to look around us, we will find that transfers and transformations of energy take place all the time. For example, light bulbs in our homes transform electrical energy into light energy (radiant energy); in an electric fan the electrical energy is converted into mechanical work which moves the blade. Plants convert the energy of sunlight (radiant energy) into chemical energy stored in organic molecules. You are transforming this chemical energy from your last snack into kinetic energy as you walk, breathe, and move. The human body also obeys the first law of thermodynamics. How? When you are in a crowded room with other people you start to feel warm and you start to sweat. This is the body‘s way to cool itself. The heat from the body is transferred to the sweat. As the sweat absorbs more heat, it evaporates from your body, becoming more 9 disordered and transferring heat to the air, which heats up the air temperature of the room. See! Thermodynamics is working around us. What are PV diagrams? Pressure and volume exhibit a causal relationship, meaning that the change of one variable will cause the change in the other. To understand how pressure directly affects volume and vice versa, let‘s consider a gas sealed in a container with a tightly fitting yet movable piston as example. If a force is applied, the piston moves down, and the gas would compress— decreasing the volume in the system and causing an increase in pressure. Moreover, if the piston moves up, the volume of the system would increase, decreasing the pressure of the system. Work can be done on the gas by pressing the piston downward. However, if an increase (or decrease) in pressure and/or volume is desired, an external heat source (or a cooling source) from its surroundings must be added. A convenient way to visualize these changes in the pressure and volume is by using a Pressure Volume diagram or PV diagram. Thus, a PV diagram is a graph of pressure as a function of volume. Each point on a PV diagram corresponds to a different state of the gas. The pressure is given on the vertical axis and the volume is given on the horizontal axis. Retrieved from https://www.khanacademy.org/science/physics/thermodynamics/laws-of- thermodynamics/a/what-are-pv-diagrams 10 Every point on a PV diagram represents a different state for the gas (one for every possible volume and pressure). As a gas goes through a thermodynamics process, the state of the gas will shift around in the PV diagram, tracing out a path as it moves (as shown in the diagram below). Retrieved from https://www.khanacademy.org/science/physics/thermodynamics/laws-of- thermodynamics/a/what-are-pv-diagrams If we press the piston downward, the volume of the gas will decrease, so the state must shift to the left toward smaller volumes. Since the gas is being compressed, we can also say that positive work W is being done on the gas. Similarly, if we let the gas expand, pushing the piston upward, the volume of the gas will increase, so the state must shift to the right toward larger volumes. Since the gas is expanding, we can also say that negative work W is being done on the gas. Being able to decode the information shown in a PV diagram allows us to make statements about the change in internal energy ΔU, heat transferred Q, and work done W on a gas. There are four different situations that you can expect to see in PV diagrams: 1. Isobaric: the gas is held at a constant pressure 2. Isochoric: the gas is held at a constant volume 3. Isothermal: the gas is held at a constant temperature 4. Adiabatic: No heat flows in or out of the gas 1. Isobaric Process Imagine a cylinder filled with a gas. At the top is a movable piston that has a mass. At the start the pressure is a certain amount for the volume of gas involved. A heat source is added under the cylinder 11 and slowly heat it up. In an isobaric process, the pressure must stay the same as it was at the start. We expect that with the additional internal energy, the molecules will be bouncing around a bit more. The gas will do work against the piston as it expands, where W = PdV. The PV diagram in an Isobaric process shows a gas going from a smaller to a bigger volume, while the pressure stays constant. Since the volume of the gas increased, work was done by the gas (this is represented by the area under the line on a PV diagram). Retrieved from https://commons.wikimedia.org/wiki/File:P-V_diagram,_Isobaric_Process.jpg 2. Isochoric Process In an isochoric process, the gas must keep a constant volume. Imagine a gas filled in a sealed container that cannot change size or shape. As heat is added, the gas will want to expand even though it can't. This will build up pressure inside the cylinder. The increase in pressure will also create more force against the sides of the cylinder, but since nothing is moving, no work is done. W = Fd = F (0) = 0 J The first law of thermodynamics shows that since no work is being done (W = 0 J), all that heat being added just increases the internal energy. ΔU = Q + W ΔU = Q 12 For the PV diagram of the isochoric process, the volume stays the same while the pressure builds. Since it is a vertical line, there is no area under the line, which means no work is done. Retrieved from https://commons.wikimedia.org/wiki/File:Isochoric_process.png 3. Isothermal Process During this process heat can enter and leave the gas but the total energy in the gas will not change. Imagine a cylinder with a piston on top that is filled with a gas. The cylinder is sitting inside a large container filled with water. If anything happens to the cylinder that would normally cause a temperature change in the gas, the water will either absorb or release heat to keep the gas in the cylinder at a constant temperature. We have been holding down on the piston a bit, but now we release it so that the piston rises. The volume of the gas is expanding, and the pressure inside will be dropping. Work is being done by the gas (work is negative), so to keep the internal energy (and temperature) constant, the gas must be absorbing heat from the water. As long as the energy lost (by the gas doing work) equals the heat gained from the water, the internal energy will not change and the temperature stays the same. ∆U=Q+W ∆ U=−W+W ∆ U=0 The PV diagram we get shows a curved line as the pressure and volume both changed. The line is called an isotherm because the temperature stayed the same. The isotherm is a curve. This happens because when the piston was being released there was a sudden 13 drop in pressure as the gas expanded. As the process continued the pressure dropped more slowly as the volume increased. The area under the curve is still the work done. Retrieved from https://commons.wikimedia.org/wiki/File:Isothermal-process-in-p-V-diagram.svg 4. Adiabatic Process An adiabatic process is a process during which no heat enters or leaves the system. In an adiabatic process, heat flow (Q) is zero. Applying the first law of thermodynamics, if ΔU=Q+W, and Q is 0, the change in internal energy of the gas must be equal to the work done on the gas (ΔU=W). If the gas is compressed (work done on the gas is positive) the internal energy increases. If the gas expands (work done by the gas is negative) the internal energy decreases. The PV diagram of an adiabatic process looks very much like an isothermal process, but it drops off to a lower point. This means it is on a different isotherm. In the adiabatic process where a gas expands, the work done by the gas causes the internal energy to decrease, so the temperature must decrease as well. Therefore, there is a drop to a lower isotherm. 14 Retrieved from https://commons.wikimedia.org/wiki/File:Adiabatic.png Work Done by a Gas When a force acts upon an object to cause a displacement of the object, it is said that work was done upon the object. In thermodynamics, however, work is defined as the energy it takes to move an object against a force. Work (W) is one of the fundamental ways energy enters or leaves a system, and it has units of Joules (J). Gases can do work through expansion or compression against a constant external pressure. Work done by gases is also sometimes called pressure-volume. Let's consider gas contained in a piston. This is a good example of a thermodynamic system. If the gas is heated, the gas will exert more pressure on the container. These increasingly frequent collisions of gas molecules transfer energy to the piston and move it against an external pressure, increasing the overall volume of the gas. When the gas expands, it will push the piston up thereby doing work on the piston; this is one example of how a thermodynamic system can do work (Figure 1). Retrieved from https://commons.wikimedia.org/wiki/File:Gax_expanding_doing_work_on_a_piston_in_a_cylinder.jpg Figure 1. Gas expanding doing work on a piston in a cylinder 15 When the gas expands against an external pressure, the gas has to transfer some energy to the surroundings. Thus, the negative work decreases the overall energy of the gas. When the gas is compressed, energy is transferred to the gas so the energy of the gas increases due to positive work. The force of an expanding gas on the piston is equal to the pressure exerted by the gas multiplied by the area of the piston. The work the gas does to move the piston to a certain distance x is equal to: W = PA (∆x); where P is exerted by the gas, and A is the area of the piston. But, A (∆ x) is equal to the change in the volume ∆V of the gas. Hence: W = P∆V Therefore, the work done by a system can be calculated by considering transfer of energy by gas molecules when the piston is moving. Example 1: When 1g of water (V= 1 x 10 -6 m3) is heated at one atmospheric pressure and allowed to vaporize, its volume increases to 1.671 x 10 -3 m3. Compute the work done by the water in its surroundings as it turns into steam. Given: m = 1 g = 1 x 10 -3 kg V1 = 1 g = 1 x 10 -6 m3 V2 = 1.671 x 10 -3 m3 P = 1 atm = 1.013 x 10 5 N/m2 Find: W Solution: W = PdV = (1.013 x 10 5 N/m2)( 1.671 x 10 -3 m3 - 1 x 10 -6 m3) = 169. 171 J = 169 J 16 Example 2: What is the work done when a sample of a gas expands from 12.5L to 17.2L against a pressure of 1.29 atm? Solution: When a gas expands at constant pressure then for a small change in volume 'dv' work done is dW=PdV. If the volume changes from 'V1' to 'V2' at constant pressure 'p', the work done is dW=P(V2−V1). When work is done by the system against external pressure then dW=PdV ⇒W=∫V2V1 PdV. So, by calculating with given numerical, we get, dW=1.29atm (17.2L−12.5L) = 1.29atm × (4.7L) = 6.063 atm. L To have the answer, we use the units below: 1atm=101325 Pa 1Pa=1J/m3 (because J=Pa.m3) Substitute '1Pa' value to get '1atm' value, ⇒1atm=101325J/m3 Now as we know 1 Liter=10−3m3. Converting the work done 'dW' value into Joules using the above required units we get, dW= 6.063 × 101325 J.m3 ×10−3m3 = 614.3334 J. Therefore, the work done is 614.33 J. What are Reversible and Irreversible Processes? A thermodynamic process is said to be reversible if the process can be turned back such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. If the system and the surroundings do not return to their original condition once the process is initiated, it is an irreversible process. 17 Taking an example of an automobile engine, that has travelled a distance with the aid of fuel equal to an amount ‗x‘. During the process, the fuel burns to provide energy to the engine, converting itself into smoke and heat energy. We cannot retrieve the energy lost by the fuel and cannot get back the original form. In reality, no such processes as reversible processes can exist. SECOND LAW OF THERMODYNAMICS AND ENTROPY The Second Law of Thermodynamics is about the quality of energy. It states that as energy is transferred or transformed, more and more of it is wasted. The Second Law also states that there is a natural tendency of any isolated system to degenerate into a more disordered state. The state of disorder is called entropy. Other examples of this entropy are the following: No matter how you take care of your clothes, as time goes by, it will become old and the colors will fade away, losing its quality when you first bought it. Your room, even if you have cleaned it already, you can see that after a few hours, it will become messy again. Have you observed that there are people who used whitening and rejuvenating products? If yes, then, these people are trying to reverse the effects of entropy, (though it cannot be). As we grow older, our skin, our body and our youthful looks will fade, and disorder will take its natural place. In keeping with our everyday experience, the second law says that whenever something happens, the universe (system + surroundings) gets messier (entropy increases). For engineers, the second law sets an upper limit on the efficiency of a heat engine. A heat engine is a device used for converting heat into work. There are two types of heat engine, the steam heat engine and the heat pump. Depressingly, it also shows that the amount of energy available to do work is constantly decreasing. Useful energy in the universe (potential energy) is slowly but surely being dissipated as heat. Figure 2. A schematic diagram of a refrigerator heat engine 18 Retrieved from https://www.qsstudy.com One thing, the Second Law explains is that it is impossible to convert heat energy to mechanical energy with 100 percent efficiency. After the process of heating a gas to increase its pressure to drive a piston, there is always some leftover heat in the gas that cannot be used to do any additional work. This waste heat must be discarded by transferring it to a heat sink. In the case of a car engine, this is done by exhausting the spent fuel and air mixture to the atmosphere. Additionally, any device with movable parts produces friction that converts mechanical energy to heat that is generally unusable and must be removed from the system by transferring it to a heat sink. Retrieved from energy.gov Figure 3. Schematic diagram of energy transfer for a heat engine Entropy can also be defined as the measure of a system‘s thermal energy per unit temperature that is unavailable for doing useful work. Because work is obtained from ordered molecular motion, the amount of entropy is also a measure of the molecular disorder, or randomness, of a system. The concept of entropy provides deep insight into the direction of spontaneous change for many everyday phenomena. When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder. The greater the number of atoms or molecules in the gas, the greater the disorder. The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it (in this case, the number of atoms or molecules); that is, the greater the number of microstates, the greater the entropy. 19 A disordered system has a greater number of possible microstates than does an ordered system, so it has a higher entropy. This is most clearly seen in the entropy changes that accompany phase transitions, such as solid to liquid or liquid to gas. As you know, a crystalline solid is composed of an ordered array of molecules, ions, or atoms that occupy fixed positions in a lattice, whereas the molecules in a liquid are free to move and tumble within the volume of the liquid; molecules in a gas have even more freedom to move than those in a liquid. Each degree of motion increases the number of available microstates, resulting in a higher entropy. Thus, the entropy of a system must increase during melting (ΔSfus > 0). Similarly, when a liquid is converted to a vapor, the greater freedom of motion of the molecules in the gas phase means that ΔSvap > 0. Conversely, the reverse processes (condensing a vapor to form a liquid or freezing a liquid to form a solid) must be accompanied by a decrease in the entropy of the system: ΔS < 0. Entropy can be calculated by using the ratio of Q/T, and this ratio can be defined as the change in entropy ΔS for a reversible process, Q ΔS = — T where Q is the heat transfer, which is positive for heat transfer into and negative for heat transfer out, and T is the absolute temperature at which the reversible process takes place. The SI unit for entropy is Joules/Kelvin (J/K). If temperature changes during the process, then it is usually a good approximation to take T as the average temperature, avoiding the need to use the integral calculus to find ΔS. If we are going to calculate the change of entropy in a reversible process, the total would be zero. However, for an irreversible process, change of entropy is always greater than zero. For example, heat transfer from hot to cold. Example 3: Calculate the total change in entropy if 4000 J of heat transfer occurs from a hot reservoir at T = 600 K (327 °C) to a cold reservoir at T = 250 K (−23 °C). Assuming there is no temperature change in either reservoir. 20 Solution: We calculate the two changes in entropy using ΔStot = ΔSh + ΔSc. First, for the heat transfer from the hot reservoir, ΔSh = −Qh / Th = −4000 J / 600 K = −6.67 J/K. And for the cold reservoir, ΔSc = Qc / Tc = 4000 J / 250 K = 16.0 J/K. Thus, the total is ΔStot = ΔSh + ΔSc = (−6.67 + 16.0) J/K = 9.33 J/K. Discussion: There is an increase in entropy for the system of two heat reservoirs undergoing this irreversible heat transfer. This means there is a loss of ability to do work with this transferred energy. Entropy has increased, and energy has become unavailable to do work. It is reasonable that entropy increases for heat transfer from hot to cold. Since the change in entropy is Q/T, there is a larger change at lower temperatures. The decrease in entropy of the hot object is therefore less than the increase in entropy of the cold object, producing an overall increase, just as in the previous example. This result is very general: There is an increase in entropy for any system undergoing an irreversible process. With respect to entropy, there are only two possibilities: entropy is constant for a reversible process, and it increases for an irreversible process. There is a fourth version of the second law of thermodynamics stated in terms of entropy: The total entropy of a system either increases or remains constant in any process; it never decreases. For example, heat transfer cannot occur spontaneously from cold to hot, because entropy would decrease. 21 Entropy is very different from energy. Entropy is not conserved but increases in all real processes. Reversible processes (such as in Carnot engines) are the processes in which the most heat transfer to work takes place and are also the ones that keep entropy constant. Thus, we are led to make a connection between entropy and the availability of energy to do work. What does a change in entropy mean, and why should we be interested in it? One reason is that entropy is directly related to the fact that not all heat transfer can be converted into work. The next example gives some indication of how an increase in entropy results in less heat transfer into work. Example 4: a. Calculate the work output of a Carnot engine operating between temperatures of 600 K and 100 K for 4000 J of heat transfer to the engine. b. (b) Now suppose that the 4000 J of heat transfer occurs first from the 600 K reservoir to a 250 K reservoir (without doing any work, and this produces the increase in entropy calculated above) before transferring into a Carnot engine operating between 250 K and 100 K. What work output is produced? Solution (a): The Carnot efficiency is given by Effc = 1 – Tc / Th. Substituting the given temperatures yields Effc = 1− 100 K / 600 K = 0.833 = 0.833% Now, the work output can be calculated using the definition of efficiency for any heat engine as given by Eff = W / Qh Solving for W and substituting known terms gives W = EffcQh = (0.833)(4000 J) = 3333 J. 22 Solution (b): Similarly, Effc = 1 − Tc / T′c =1 − 100 K / 250 K = 0.600% so that W = EffcQh = (0.600)(4000 J) = 2400 J Discussion: There is 933 J less work from the same heat transfer in the second process. This result is important. The same heat transfer into two perfect engines produces different work outputs, because the entropy change differs in the two cases. In the second case, entropy is greater and less work is produced. Entropy is associated with the unavailability of energy to do work. Figure 4. (a) Carnot engine working between 600 K and 100 K has 4000 J of heat transfer and performs 3333 J of work. (b) The 4000 J of heat transfer occurs first irreversibly to a 250 K reservoir and then goes into a Carnot engine. The increase in entropy caused by the heat transfer to a colder reservoir results in a smaller work output of 2400 J. There is a permanent loss of 933 J of energy for the purpose of doing work. 23 When entropy increases, a certain amount of energy becomes permanently unavailable to do work. The energy is not lost, but its character is changed, so that some of it can never be converted to doing work—that is, to an organized force acting through a distance. For instance, in the previous example, 933 J less work was done after an increase in entropy of 9.33 J/K occurred in the 4000 J heat transfer from the 600 K reservoir to the 250 K reservoir. It can be shown that the amount of energy that becomes unavailable for work is Wunavail = ΔS⋅T0 where T0 is the lowest temperature utilized. In the previous example, Wunavail = (9.33 J/K)(100 K) = 933 J Entropy is related not only to the unavailability of energy to do work—it is also a measure of disorder. This notion was initially postulated by Ludwig Boltzmann in the 1800s. For example, melting a block of ice means taking a highly structured and orderly system of water molecules and converting it into a disorderly liquid in which molecules have no fixed positions. There is a large increase in entropy in the process, as seen in the following example. Example 5: Find the increase in entropy of 1.00 kg of ice originally at 0 °C that is melted to form water at 0 °C. Solution: The change in entropy is defined as: ΔS = Q/T. Here, Q is the heat transfer necessary to melt 1.00 kg of ice and is given by Q = mLf, where m is the mass and Lf is the latent heat of fusion. Lf = 334 kJ/kg for water, so that Q = (1.00 kg)(334 kJ/kg) = 3.34 × 105 J 24 Now, the change in entropy is positive, since heat transfer occurs into the ice to cause the phase change; thus, ΔS = Q / T = 3.34 × 105 J / T. T is the melting temperature of ice. That is T = 0 °C = 273 K. So, the change in entropy is ΔS = 3.34 × 105 J / 273 K =1.22 × 103 J/K. This is a significant increase in entropy accompanying an increase in disorder. When ice melts, it becomes more disordered and less structured. The systematic arrangement of molecules in a crystal structure is replaced by a more random and less orderly movement of molecules without fixed locations or orientations. Its entropy increases because heat transfer occurs into it. Performance Task: Activity 1 Understanding the Laws of Thermodynamics Directions: Predict what will happen after a few weeks/months given the following situations. Write your answers on your notebook/Activity Sheet. SITUATION PREDICTION 1.A well-folded clothes in a cabinet 2. A clean but abandoned classroom 3.Your clean, shiny floor in your house 4. Your new, and nice bag 5. Your beautiful garden Questions: 1. What usually happens to the situations above? 2. Why do you think that a disorder usually occurs even ifyou are not doing anything about it? 3. How could we reverse the situation from disorder to order? 25 II. PICTURE ANALYSIS: Analyze the picture below and explain how is it related to the second law of thermodynamics (5 pts.) (Image © Hayati Kayhan | Shutterstock) III. INTERPRETING PV DIAGRAMS The PV diagram below shows the state of an ideal gas is being changed from position 1 to position 2. What is the change in the internal energy of the gas during this process? (A) ΔU = W (B) ΔU = Q (C) ΔU > 0 (D) ΔU = 0 (E) ΔU < 0 Explain your answer. Retrieved from http://content.njctl.org/courses/science/ap-physics b/thermodynamics/thermodynamics-practice-problems/thermodynamics-practice-problems-2012-05- 07.pdf 26 Activity 2 Problem Solving Directions: Solve the given problems below. Write your answers on your General Physics 1 notebook/Activity Sheet. Show your solutions (5 points each). 1. When 1.00 g of water at 100°C changes from the liquid to the gas phase at atmospheric pressure, its change in volume is 1.67×10 −3m3. How much work is done by the water against the atmosphere in its expansion? 2. The P-V diagram on the next page show an expansion of a gas from state 1 to state 2 at constant pressure, followed by another expansion that takes the system to state 3 along the path indicated. Find the work done by the gas in the two processes shown in the figure above. Given: P = 80 kPA V1= 4 L V2 = 8 L 3. What is the decrease in entropy of 25.0 g of water that condenses on a bathroom mirror at a temperature of 35.0°C, assuming no change in temperature and given the latent heat of vaporization to be 2450 kJ/kg?. 27 III. WHAT I HAVE LEARNED Great job! Now that we already learned Are you ready the concepts, I am for our last confident that we can challenge? Be sure apply this learning in our to apply what you daily activities. What have learned. Here‘s should we do next, your final test. Good Lovely? luck! EVALUATION/POST TEST: A. MULTIPLE CHOICE: Read each statement carefully. Choose the letter of the BEST answer and write it down on your General Physics 1 notebook/Activity Sheet. 1. It is known that curves A, B, C are thermodynamic processes. Which of the following associations is correct? A. A - Adiabatic, B - Isothermal,, C - Isobaric B. A - Isobaric, B - Isothermal C – Adiabatic Retrieved from C. A- Isothermal, B - Adiabatic, C - Isobaric ttps://physicscatalyst.com/ D. None of these heat/thermodynamics- multiple-choice- questions.php 2. The process in which the internal energy of the system remains constant is: A. Adiabatic B. Isobaric C. Isochoric D. Isothermal 3. Lines on a PV diagram describing any process held at constant temperature are therefore called _______. A. adiabats B. curves C. isotherms d. none of the choices 28 4. When heat is added to a system, all of the following may happen except A. decrease in the system‘s temperature B. external work is done by the system C. increase in internal energy D. increase in the pressure in the system 5. Internal energy of an ideal gas depends on: i. the volume of the ideal gas ii. the pressure of the ideal gas iii. the absolute temperature of the ideal gas A. I B. II C. III D. I and II 6. Heat is the _____. A. Average amount of energy per molecule contained in an object B. Total amount of energy contained in an object C. Energy transferred between objects because of a temperature difference D. Amount of energy all the molecules have 7. Which of the following summarizes the first law of thermodynamics? A. Two systems that are in equilibrium with a third system are in equilibrium with one another. B. The total entropy of a system increases over time. C.Energy cannot be created or destroyed. D. Energy cannot be transferred or transformed 8. All the following statements define the second law of thermodynamics except, A. Processes that involve the transfer or conversion of heat energy are irreversible. B. The total entropy of an isolated system can never decrease over time. C. Energy can be changed from one form to another, but it cannot be created or destroyed. D. Heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy. 9. What is relation between thermodynamic probability and entropy? A. thermodynamic probability increases with entropy B. thermodynamic probability decreases with entropy C. thermodynamic probability does not change with change of entropy D. none of the above 10. At the most probable state of a system the entropy of the system is A. minimum B. maximum C. constant D. none of the above 29 11. Which of the statements below is/are TRUE about the second law of thermodynamics? I. More energy is wasted in the universe as heat. II. Energy is reversible. III. Disorder in the universe decreases because heat is constant. IV. Energy is irreversible A. I only B. I and II C. I and III D. I and IV 12. The entropy of an isolated system can never ____.. A. be zero B. decrease C. increase D. none of the above 13. The entropy of an isolated system always and becomes at the state of equilibrium. A. decreases, maximum B. decreases, minimum C. increases, maximum D. increases, minimum 14. Which of the following can be considered as an application of the second law of thermodynamics? A. maximum temperature obtainable from two finite bodies B. transfer of heat between two objects C. mixing of two liquids D. all of the above 15. Why do engines cannot reach 100% efficiency? A. because it needs additional pressure B. because friction generates heat C. because some parts are not functional D. all of the above B. TRUE OR FALSE: Write the word true if the statement is correct, false if otherwise. _____ 1. When a gas expands, we can say that positive work is being done on the gas. _____ 2. In an adiabatic process, no heat enters or leaves the system. _____ 3. Being able to decode the information shown in a PV diagram allows us to make statements about the change in internal energy, heat transferred, and work done on a gas. _____ 4. When a gas expands at constant pressure for a small change in volume, the work done is dW=PdV. _____ 5. When the gas expands, energy is transferred to the gas so the energy of the gas increases due to positive work. _____ 6. The law of conservation of energy states that energy can either be created or destroyed. 30 _____ 7. The transformation of chemical energy into kinetic and heat energy by a car engine is a manifestation of the first law of thermodynamics. _____ 8. Machines are 100% efficient. _____ 9. Entropy is a measure of disorder. The process of disorder can be reversed naturally. _____ 10. Energy cannot be transferred but it can be transformed. 31 REFERENCES Alastre-Dizon, Marivelle T., Malabanan, Katherine C. and Bautista, Donald B. Science and Technology for the Future. Makati, Philippines: Diwa Scholastic Press Inc., 2004. ―Definition and Mathematics of Work.‖ The Physics Classroom, https://www.physicsclassroom.com/class/energy/Lesson-1/Definition- and-Mathematics-of-Work Donev et al. ―- Pressure volume diagram‖. Energy Education.4 January 2019. https://energyeducation.ca/encyclopedia/Pressure_volume_diagram. ―First Law of Thermodynamics‖. University Physics Volume 2. OpenStax, https://openstax.org/books/university-physics-volume-2/pages/3-3-first- law-of-thermodynamics ―Module 14 Thermodynamics.‖ SlideShare, 25 Aug. 2013, www.slideshare.net/dionesioable/module-14-thermodynamics. Padua, A., and Crisostomo, R. Practical and Explorational Physics. Quezon City: Vibal Publishing House Inc., 2003. ―Pressure-volume work‖ Khan Academy, https://www.khanacademy.org/science/chemistry/thermodynamics- chemistry/internal-energy-sal/a/pressure-volume-work ―PV Diagrams‖ https://silo.tips/download/lesson-42c-pv-diagrams. Silverio, A. General Physics 1. Exploring Life through Science. Phoenix Publishing House.,2012 Teaching Guide for Senior High School (Physical Science) pp.66-77 ―Thermodynamics.‖ NASA, 05 May 2015, https://www.grc.nasa.gov/www/k- 12/airplane/thermo.html ―Thermodynamics - Definition, Equations, Laws, Meaning, Formulas: Basics of Thermodynamics.‖ BYJUS, BYJU'S, 18 Sept. 2020, byjus.com/physics/thermodynamics/. 32 ―What are PV Diagrams?‖ Khan Academy, https://www.khanacademy.org/science/physics/thermodynamics/laws -of-thermodynamics/a/what-are-pv-diagrams ―What Is the Work Done When a Sample of Gas Expands from 12.5L to 17.2L against a Pressure of 1.29 Atm?‖ Socratic.org, 21 Feb. 2017, https://socratic.org/questions/what-is-the-work-done-when-a-sample- of-gas-expands-from-12-5l-to-17-2l-against-a What is the Second Law of Thermodynamics? May 22, 2015. Jim Lucas - Live Science Contributor, https://www.livescience.com/50941- second-law-thermodynamics.html https://courses.lumenlearning.com/boundless-biology/chapter/components- and- structure/#:~:text=The%20primary%20function%20of%20the,in%20and%2 0out%20of%20cells. http://physics.bu.edu/~duffy/py105/Idealgas.html http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/idegas.html https://byjus.com/physics/reversible-irreversible-processes/ https://www.quipper.com/ph 33 DEPARTMENT OF EDUCATION Division of Negros Oriental SENEN PRISCILLO P. PAULIN, CESO V Schools Division Superintendent FAY C. LUAREZ, TM, Ed.D., Ph.D. OIC - Assistant Schools Division Superintendent Acting CID Chief NILITA L. RAGAY, Ed.D. Assistant Schools Division Superintendent ROSELA R. ABIERA Education Program Supervisor – (LRMS) ARNOLD R. JUNGCO PSDS – Division Science Coordinator MARICEL S. RASID Librarian II (LRMDS) ELMAR L. CABRERA PDO II (LRMDS) ANDRE ARIEL B. CADIVIDA JIMMA C. PORSUELO Writers/Illustrators/Lay-out Artists _________________________________ QUALITY ASSURANCE TEAM ARNOLD D. ACADEMIA ZENAIDA A. ACADEMIA LIEZEL A. AGOR MARY JOYCEN A. ALAM-ALAM EUFRATES G. ANSOK JR. JOAN Y. BUBULI LIELIN A. DE LA ZERNA ADELINE FE D. DIMAANO RANJEL D. ESTIMAR VICENTE B. MONGCOPA FLORENTINA P. PASAJINGUE THOMAS JOGIE U. TOLEDO DISCLAIMER The information, activities and assessments used in this material are designed to provide accessible learning modality to the teachers and learners of the Division of Negros Oriental. The contents of this module are carefully researched, chosen, and evaluated to comply with the set learning competencies. The writers and evaluator were clearly instructed to give credits to information and illustrations used to substantiate this material. All content is subject to copyright and may not be reproduced in any form without expressed written consent from the division. 34 SYNOPSIS ANSWER KEY This self-learning kit discusses thermodynamics in a broader aspect. Learners will explore fundamental applications of thermodynamics in real life. The discussions herein are contextualized and thus meet the standards of the K12 curriculum. In using this learning kit, learners are expected to develop their scientific abilities and enhance their resourcefulness and creativity as they engage in various activities included herein. Hence, this learning kit serves as their way of expanding their knowledge of the things in nature. Come and let us make learning fun. ABOUT THE AUTHORS ANDRE ARIEL B. CADIVIDA finished Bachelor of Science in Biology at Negros Oriental State University Main Campus in 2013. He is currently teaching at Cansal-ing Provincial Community High School as a senior high teacher, library designate and the focal person of the senior high department. He is currently completing Master of Arts in Science Teaching at Negros Oriental State University Graduate School. JIMMA C. PORSUELO, MAED-General Science obtained her Master of Arts in Education major in General Science at Central Currently, she is a senior high school teacher at JBCMHS Main teaching Science and Research subjects. 35