Assignment 9: First and Second Laws of ThermodynamicsDue: 2:00am on Saturday, November 13, 2010 Note: To understand how points are awarded, read your instructor's Grading Policy. [Switch to Standard Assignment View] A Flexible Balloon A flexible balloon contains 0.320 volume of 7000 of hydrogen sulfide gas . The . Initially the balloon of has a and a temperature of 29.0 first expands isobarically until the volume doubles. Then it expands adiabatically until the temperature returns to its initial value. Assume that the may be treated as an ideal gas with and . Part A What is the total heat Hint A.1 supplied to the gas in the process? Which parts of the process have heat flow? Hint not displayed Hint A.2 Relation of heat capacity and heat Hint not displayed Hint A.3 Find the change in temperature Hint not displayed ANSWER: = 3350 Correct Part B What is the total change in the internal energy Hint B.1 of the gas? Dependence of internal energy on temperature Hint not displayed ANSWER: = 0 Correct Part C What is the total work Hint C.1 done by the gas? How to approach the problem Hint not displayed Hint not displayed ANSWER: = 3350 Correct Part D What is the final volume Hint D.1 ? Relation between volume and temperature is a Recall that, in an adiabatic process with finite changes in volume and temperature, constant. As a result, . ANSWER: = 0.112 Correct Work from an Adiabatic Expansion In this problem you are to consider an adiabatic expansion of an ideal diatomic gas, which means that the gas expands with no addition or subtraction of heat. This applet shows the adiabatic compression and expansion of an ideal monatomic gas with . It will help you to see the qualitative behavior of adiabatic expansions, though your actual calculations will use a slightly different . Assume that the gas is initially at pressure , volume , and temperature . In addition, assume that the temperature of the gas is such that you can neglect vibrational degrees of freedom. Thus, the ratio of heat capacities is . Note that, unless explicitly stated, the variable fact that Part A Find an analytic expression for expansion. Hint A.1 , the pressure as a function of volume, during the adiabatic for an ideal diatomic gas. should not appear in your answers--if needed use the Find the conserved quantity in an adiabatic process Hint not displayed Express the pressure in terms of ANSWER: = and any or all of the given initial values , , and . Correct The fact that is a constant derives from the definition of an adiabatic process as one in in the first law of which no heat flow into or out of the system occurs. Setting which no heat flow into or out of the system occurs. Setting thermodynamics ( your textbook for more details. Part B At the end of the adiabatic expansion, the gas fills a new volume work done by the gas on the container during the expansion. Hint B.1 in the first law of . See ) gives the starting point for this derivation: , where . Find , the How to approach the problem The work done is given by the following general integral: . Using the pressure as a function of volume, done as the gas expands from its initial volume , found in Part A, integrate to find the total work to its final volume Hint B.2 Help with the math Integration gives . Express the work in terms of ANSWER: = , , and . Your answer should not depend on temperature. Correct Part C Find volume Hint C.1 , the change of internal energy of the gas during the adiabatic expansion from volume . to Find the initial internal energy Hint not displayed Hint C.2 Find the final internal energy Hint not displayed Express the change of internal energy in terms of ANSWER: = , , and/or . Correct If you used the hints to solve this part, they lead you on a long, but instructive, path to calculate If you used the hints to solve this part, they lead you on a long, but instructive, path to calculate . A much simpler method you might have used is to apply the first law of thermodynamics, , together with the fact that for an adiabatic process by definition. Basically, the molecules transfer some of their energy and thus also momentum to the walls of the container, causing the expansion. So the temperature decreases, while the volume increases. Of course, some external force will eventually stop the expansion. Internal-Combustion Engine Prototypes Ranking Task Six new prototypes for internal-combustion engines are tested in the laboratory. For each engine, the heat energy input and output per cycle, and the designed number of cycles per second are measured. Part A Rank these engines on the basis of the work they perform per cycle. Hint A.1 How to approach the problem Engine cycles can be thought of as closed thermodynamic cycles, in which the initial and final states of the system are the same. Thus the change in the internal energy during the cycle is zero, allowing you to directly relate the work done to the net heat transfer by applying the first law of thermodynamics to the cycle. Hint A.2 First law of thermodynamics applied to an engine cycle . Therefore, for net heat transfer All closed cycles involve no change in internal energy and net work done becomes . , Based on the net heat transfer in each cycle, you should be able to correctly determine the work done. Hint A.3 Net heat transfer ) is During a simplified engine cycle, the heat transferred to the gas from the hot reservoir ( positive and the heat transferred out of the gas to the cold reservoir ( heat transfer is given by . ) is negative. Thus the net Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: View Correct Part B Rank these engines on the basis of their designed power output. Hint B.1 Calculating power Hint not displayed Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: View Correct Part C Rank these engines on the basis of their thermal efficiency. Hint C.1 Definition of thermal efficiency Hint not displayed Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: View Correct From Hot to Cool: A Change in Entropy In a well-insulated calorimeter, 1.0 Part A of water at 20 is mixed with 1.0 of ice at 0 . What is the net change in entropy of the system from the time of mixing until the moment the . ice completely melts? The heat of fusion of ice is Note that since the amount of ice is relatively small, the temperature of the water remains nearly constant throughout the process. Note also that the ice starts out at the melting point, and you are asked about the change in entropy by the time it just melts. In other words, you can assume that the temperature of the "ice water" remains constant as well. Hint A.1 How to approach the problem Hint not displayed Hint A.2 Description of entropy Hint not displayed Hint A.3 Heat needed to melt the ice Hint not displayed Express your answer numerically in joules per kelvin. Use two significant figures in your answer. ANSWER: = 8.35×10−2 Correct As you would expect, in this spontaneous process the net change in entropy is positive: The entropy increases. This is evident not just from the calculation but also from the fact that a crystal becomes liquid and hence the degree of disorder increases. An Expanding Monatomic Gas We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 120 expands and, in the process, absorbs an amount of heat equal to 1200 equal to 2080 Part A What is the final temperature Hint A.1 of the gas? . . The gas and does an amount of work First law of thermodynamics Hint not displayed Hint A.2 Find the change in internal energy Hint not displayed Hint A.3 Calculate the change in temperature Hint not displayed Use = 8.3145 for the ideal gas constant. ANSWER: ANSWER: = Answer not displayed Various Gas Expansions: pV Plots and Work An ideal monatomic gas is contained in a cylinder with a movable piston so that the gas can do work on the outside world, and heat can be added or removed as necessary. The figure shows various paths that the gas might take in expanding from an initial state whose pressure, volume, and temperature are , , and respectively. The gas expands to a state with final volume . For some answers it will be convenient to generalize your results by using the variable , which is the ratio of final to initial volumes (equal to 4 for the expansions shown in the figure.) The figure shows several possible paths of the system in the pV plane. Although there are an infinite number of paths possible, several of those shown are special because one of their state variables remains constant during the expansion. These have the following names: Adiabiatic : No heat is added or removed during the expansion. Isobaric : The pressure remains constant during the expansion. Isothermal: The temperature remains constant during the expansion. Part A Which of the curves in the figure represents an isobaric process? ANSWER: A B C D Answer not displayed Part B What happens to the temperature of the gas during an isobaric expansion? Hint B.1 Implications of Hint not displayed ANSWER: Temperature increases. Temperature remains constant. Temperature decreases. Answer not displayed Part C Which of the curves in the figure represents an isothermal process? Hint C.1 Ideal gas law again Hint not displayed ANSWER: A B C D Answer not displayed Part D Graphically, the work along any path in the pV plot ____________. ANSWER: is the area to the left of the curve from is the area under the curve from to to Answer not displayed requires knowledge of the temperature Part E Calculate Hint E.1 , the work done by the gas as it expands along path A from to . Expression for Hint not displayed Hint E.2 Find an expression for Hint not displayed Hint E.3 Do the integral Hint not displayed Express ANSWER: in terms of , , and . = Answer not displayed Part F Part not displayed Part G Part G Which of the curves shown represents an adiabatic expansion? Hint G.1 Pressure and volume in an adiabatic expansion Hint not displayed ANSWER: A B C D Answer not displayed A Law for Scuba Divers SCUBA is an acronym for self-contained underwater breathing apparatus. Scuba diving has become an increasingly popular sport, but it requires training and certification owing to its many dangers. In this problem you will explore the biophysics that underlies the two main conditions that may result from diving in an incorrect or unsafe manner. While underwater, a scuba diver must breathe compressed air to compensate for the increased underwater pressure. There are a couple of reasons for this: If the air were not at the same pressure as the water, the pipe carrying the air might close off or collapse under the external water pressure. Compressed air is also more concentrated than the air we normally breathe, so the diver can afford to breathe more shallowly and less often. A mechanical device called a regulator dispenses air at the proper (higher than atmospheric) pressure so that the diver can inhale. Part A Suppose Gabor, a scuba diver, is at a depth of . Assume that: The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose. The temperature of the air is constant (body temperature). The air acts as an ideal gas. Salt water has an average density of around 1.03 , which translates to an increase in pressure of 1.00 for every 10.0 . of depth below the surface. Therefore, for example, at 10.0 , the net pressure is 2.00 What is the ratio of the molar concentration of gases in Gabor's lungs at the depth of 15 meters to that at the surface? The molar concentration refers to calculate . , i.e., the number of moles per unit volume. So you are asked to Hint A.1 Find an equation for calculating concentrations Hint not displayed Hint A.2 Find the pressure underwater Hint not displayed Express your answer numerically to three significant digits. ANSWER: = Answer not displayed Part B If the temperature of air in Gabor's lungs is 37 of air (98.6 ), and the volume is , how many moles must be released by the time he reaches the surface? Let the molar gas constant be given by . = 8.31 Hint B.1 How to approach the problem Hint not displayed Hint B.2 Solve the ideal gas law for moles Hint not displayed Express your answer in moles to three significant figures. ANSWER: = Answer not displayed Part C Part not displayed Part D What type of expansion does the air in the "freediver's" (no gear) lungs undergo as the diver ascends? Hint D.1 How to approach the problem Hint not displayed Hint D.2 Determine the proper prefix Hint not displayed Hint D.3 Possible terms for description of processes Hint not displayed Give the answer as one word. ANSWER: Answer not displayed An Air Conditioner: Refrigerator or Heat Pump? The typical operation cycle of a common refrigerator is shown schematically in the figure . Both the The typical operation cycle of a common refrigerator is shown schematically in the figure . Both the condenser coils to the left and the evaporator coils to the right contain a fluid (the working substance) called refrigerant, which is typically in vapor-liquid phase equilibrium. The compressor takes in low-pressure, low-temperature vapor and compresses it adiabatically to high-pressure, high-temperature vapor, which then reaches the condenser. Here the refrigerant is at a higher temperature than that of the air surrounding the condenser coils and it releases heat by undergoing a phase change. The refrigerant leaves the condenser coils as a high-pressure, high-temperature liquid and expands adiabatically at a controlled rate in the expansion valve. As the fluid expands, it cools down. Thus, when it enters the evaporator coils, the refrigerant is at a lower temperature than its surroundings and it absorbs heat. The air surrounding the evaporator cools down and most of the refrigerant in the evaporator coils vaporizes. It then reaches the compressor as a low-pressure, lowtemperature vapor and a new cycle begins. Part A Air conditioners operate on the same principle as refrigerators. Consider an air conditioner that has 7.00 of refrigerant flowing through its circuit each cycle. The refrigerant enters the evaporator coils in phase equilibrium, with 54.0 of its mass as liquid and the rest as vapor. It flows through the of its mass is vapor. In evaporator at a constant pressure and when it reaches the compressor 95 each cycle, how much heat is absorbed by the refrigerant while it is in the evaporator? The heat of . vaporization of the refrigerant is 1.50×105 Hint A.1 How to approach the problem Hint not displayed Hint A.2 Find the percentage of refrigerant transformed to vapor Hint not displayed Express your answer numerically in joules. ANSWER: = Answer not displayed Part B In each cycle, the change in internal energy of the refrigerant when it leaves the compresser is 1.20×105 . What is the work done by the motor of the compressor? Hint B.1 Adiabatic compression Hint not displayed Express your answer in joules. ANSWER: = Answer not displayed ANSWER: = Answer not displayed Part C Part not displayed Melting Ice with a Carnot Engine A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation of the engine, the heat rejected by the engine melts a mass of ice equal to 2.40×10−2 . Throughout this problem use Part A During this time, how much work Hint A.1 is performed by the engine? for the heat of fusion for water. How to approach the problem Hint not displayed Hint A.2 Temperature conversion Hint not displayed Hint A.3 Calculate the heat rejected Hint not displayed Hint A.4 Calculate the heat absorbed Hint not displayed Hint A.5 Using the first law of thermodynamics Hint not displayed ANSWER: = Answer not displayed Irreversible versus Reversible Processes Part A Which of the following conditions should be met to make a process perfectly reversible? Hint A.1 Reversible processes Hint not displayed Check all that apply. ANSWER: Any mechanical interactions taking place in the process should be frictionless. Any thermal interactions taking place in the process should occur across infinitesimal temperature or pressure gradients. The system should not be close to equilibrium. Answer not displayed Part B Part not displayed Entropy Change of an Expanding Gas Two moles of an ideal gas undergo a reversible isothermal expansion from 2.27×10−2 at a temperature of 29.5 Part A What is the change in entropy Hint A.1 of the gas? . to 4.62×10−2 How to approach the problem Hint not displayed Hint A.2 Calculate the work done by the gas Hint not displayed Hint A.3 Calculating the change in entropy Hint not displayed Express your answer numerically in joules per kelvin. ANSWER: = Answer not displayed Does Entropy Really Always Increase? An aluminum bar of mass 2.00 lake is 15.0 Part A The bar eventually reaches thermal equilibrium with the lake. What is the entropy change the lake? Assume that the lake is so large that its temperature remains virtually constant. Hint A.1 of at 300 is thrown into a lake. The temperature of the water in the . ; the specific heat capacity of aluminum is 900 How to approach the problem Hint not displayed Hint A.2 Find the heat absorbed by the lake Hint not displayed Hint A.3 Entropy change in an isothermal process Hint not displayed Express your answer numerically in joules per kelvin ANSWER: = Answer not displayed Part B Has the entropy of the aluminum bar decreased or increased? Hint B.1 How to approach the question Hint not displayed ANSWER: Since the entropy change of a system is always positive, we can deduce that the entropy of the aluminum bar has increased. Since the final lower temperature of the bar means lower average speed of molecular motion, we can deduce that the entropy of the bar has decreased. We don't have enough information to determine whether the entropy of the aluminum bar has decreased or increased. Answer not displayed Part C Part not displayed Part D Part not displayed Entropy Change in a Free Expansion: A Microscopic View A thin partition divides a thermally insulated vessel into a lower compartment of volume compartment of volume evacuated. Part A When the partition is removed, the gas expands and fills both compartments. How many moles gas were initially contained in the lower compartment if the entropy change of the gas in this free? expansion process is 17.28 of . The lower compartment contains and an upper moles of an ideal gas; the upper part is expansion process is 17.28 Hint A.1 ? How to approach the problem Hint not displayed Hint A.2 Find the number of possible microscopic states of a gas after a free expansion Hint not displayed Hint A.3 Find the microscopic expression for the change in entropy of a gas Hint not displayed Hint A.4 Relating to the gas constant Hint not displayed Express your answer to three significant figures. ANSWER: = Answer not displayed Score Summary: Your score on this assignment is 100%. You received 40 out of a possible total of 40 points.