Version 001 HW 15 Thermodynamics C&J sizemore (21301jtsizemore) 1

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1 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 1 This print-out should have 38 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Adiabatic Expansion Conceptual points A gas is initially inside an insulated vessel at avolumev 1,atemperatureT 1,andapressure P 1. The gas then expands adiabatically to a volume V 2. Which statement best describes what occurs? 1. All of these are true. 2. Work is done by the gas. correct 3. None of these is true. 4. The pressure P 2 is greater than P T 2 > T 1. For a gas that expands adiabatically, and T 2 < T 1, P 2 < P 1, sotheonlytruestatementisthatworkisdone by the gas. Heat Engine Internal Energy 002 part 1 of points Determine the change in the internal energy of a system that absorbs 398 cal of thermal energy while doing 410 J of external work. Correct answer: J. Let : Q = 398 cal and W = 410 J. According to the first law of thermodynamics, U = Q W, where Q is the thermal energy transferred into the system and W is the work done by the system, so U = 398 cal J 1 cal = J. 410 J 003 part 2 of points Determine the change in the internal energy of a system that absorbs 355 cal of thermal energywhile305jofexternalworkisdoneon the system. Correct answer: J. Let : Q = 355 cal and W = 305 J. U = Q W = 355 cal J 1 cal = J J 004 part 3 of points Determine the change in the internal energy of a system that is maintained at a constant volume while 773 cal is removed from the system. Correct answer: J. Since the volume is maintained constant, W = P V = 0 U = Q = 773 cal J 1 cal = J. Steam Internal Energy

2 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 2 [ 005 part 1 of points = 1.2 mol18 g/mol How much work is done by the steam when 6 J/kg 1.2 mol of water at 100 C boils and becomes 1.2 mol of steam at 100 C at 1 ] atm pressure? The universal gas constant is J/K mol. = kj. 1 kg 1000 g J 1 kj 1000 J Correct answer: J. Let : n = 1.2 mol, R = J/K mol, T = 100 C = 373 K m = 18 g/mol, P = 1 atm, and ρ = 1 g/cm 3. Using the ideal gas law P V = nrt, so the work done is mp W = P V = nrt n ρ = 1.2 mol J/K mol373 K 18 g/mol1 atm 1.2 mol 1 g/cm m Pa 100 cm 1 atm = J. 006 part 2 of points Consider the steam to be an ideal gas. Determine the change in internal energy of the steam as it vaporizes. Use a heat of vaporization of J/kg. Correct answer: kj. The heat needed to vaporize the water is Q = nml v, so from the first law of thermodynamics U = Q W = nml v W Air Conditioning Power points On a hot day, a Florida home is kept cool by an air conditioner. The outside temperature is 44.3 C and the interior of the home is 21.9 C. If 110 kj/h of heat is removed from the house, what is the minimum power that must be provided to the air-conditioner? Correct answer: kj/h. Let : dq = 110 kj/h, dt T h = 44.3 C = K, and T c = 21.9 C = K. The coefficient of performance of a refrigerator is defined as C r = Q W = T c T h T c where Q is the quantity of thermal energy removed from the interior of the refrigerator and W is the work done by the motor in the process, so W = Q T h T c T c. The rate at which heat is removed by the air-conditioner is dq dt, so P dw = dq T h T c dt dt T c K K = 110 kj/h K = kj/h. Entropy Change points

3 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 3 A solid substance that has a latent heat of fusion22.7kj/kgmeltsatatemperature370 C. Calculate the change in entropy that occurs when kg of this substance is melted. Correct answer: J/K. Let : m = kg, T m = 370 C = 643 K, and L f = 22.7 kj/kg = J/kg. Assume that the melting occurs so slowly that it can be considered a reversible process; in that case the temperature can be regarded as constant. Q = ml f S = f i = 1 = T m f dq r ds = i T dq r = Q = ml f T m kg22700 J/kg 643 K = J/K. T m Entropy in Compression points 2.13 mol of an ideal gas at 4.6 C are compressed isothermally from an initial volume of 91 cm 3 to a final volume of 34.3 cm 3. What is the change in entropy? Correct answer: J/K. Let : n = 2.13 mol, T = 4.6 C, V 1 = 91 cm 3, and V 2 = 34.3 cm 3. According to the first law of thermodynamics, E = W + Q. For isothermal process of ideal gas, E = 0 and Q = W = nrt ln V1 V 2, so the entropy change is S = Q V2 = nr ln T V 1 = 2.13 mol8.31 J/mol K 34.3 cm 3 ln 91 cm 3 = J/K. Entropy in Expansion points 22.6 mol of an ideal gas is allowed to undergo an isothermal free expansion. If the initial volume is 20 cm 3 and the final volumeis100cm 3,find thechange inentropy. Correct answer: J/K. Let : n = 22.6 mol, V 1 = 20 cm 3, and V 2 = 100 cm 3. Since the two states have the same temperature, we connect them with an isothermal process: V2 S = nr ln V 1 = 22.6 mol8.31j/mol K 100 cm 3 ln 20 cm 3 = J/K. Entropy in Gas Engine points A gasoline engine absorbs 8090 J of heat at 487 C and expels 7590 J at a temperature of 287 C.

4 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 4 Find the change in entropy for the system. Correct answer: J/K. Let : Q 1 = 8090 J, T 1 = 487 C = 760 K, Q 2 = 7590 J, and T 2 = 287 C = 560 K. The entropy gained is S 1 = Q 1 T 1 and the entropy lost is S 2 = Q 2 T 2, so the total change of entropy is S = S 1 S 2 = Q 1 Q 2 T 1 T 2 = 8090 J 760 K 7590 J = J/K. 560 K Heating a House 012 part 1 of points A house loses thermal energy through the exterior walls and roof at a rate of 4740 W when the interior temperature is 21.1 C and the outside temperature is 5.8 C. Calculate the electric power required to maintain the interior temperature at T i if the electric power is used in electric resistance heaters which convert all of the electricity supplied to thermal energy. Correct answer: 4740 W. Q Let : = 4740 W. t Since all the electricity supplied is converted to thermal energy, Q t = E t = P = 4740 W. 013 part 2 of points Find the electric power required to maintain the interior temperature at T i if the electric power is used to operate the compressor of a heat pump with a coefficient of performance equal to 0.3 times the Carnot cycle value. Correct answer: W. Let : T i = 21.1 C = K, T o = 5.8 C = K, and ν = 0.3. For a heat pump the COP is C Carnot = T i K = T i T o K K = Thus to bring 4740 W of heat into the house requires only Q t W P h = = C actual 0.3C carnot = 4740 W = W. Mixing Hot and Cold Water 014 part 1 of points One mole of liquid water at 10 C comes into thermal contact with 2 moles of water at 40 C. Assume the specific heat per mole, c sh, is temperature independent. At what temperature does the water reach thermal equlibrium? C C C C C correct Let : n h = 2 mol, n c = 1 mol, T h = 40 C, and T c = 10 C.

5 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 5 For a thermally insulated system, all the heat from the hotter water flows to the cooler water until they reach thermodynamic equilibrium: n h c sh T T h +n c c sh T T c = 0. The water reaches thermal equilibrium at T = n ht h +n c T c n h +n c = 2 mol40 C+1 mol10 C 2 mol+1 mol = 30 C. 015 part 2 of points Which of the following statements is correct about the entropy change between the final system in the thermal equilibrium state and the original system state where the water consisted of two parts at different temperatures? 1. The 1 mole of water originally at 10 C gains more entropy than was lost by the 2 moles of water originally at 40 C. correct 2. The 1 mole of water originally at 10 C gainslessentropythanwaslostbythe2moles of water originally at 40 C 3. The 1 mole of water originally at 10 C loses more entropy than was gained by the 2 moles of water originally at 40 C. 4. The 1 mole of water originally at 10 C gains an amount of entropy equal to that which was lost by the 2 moles of water originally at 40 C. 5. The 1 mole of water originally at 10 C loses less entropy than was gained by the 2 moles of water originally at 40 C. The colder water gains entropy and the hotter water loses entropy. The system of hot and cold water overall gains entropy. So, the 1 mole of water originally at 10 C gains more entropy than was lost by the 2 moles of water originally at 40 C. Pressure in Gas Engine 016 part 1 of points In a cylinder of an automobile engine, just after combustion, the gas is confined to a volumeof32.7cm 3 andhasaninitialpressure of Pa. The piston moves outward to a final volume of 455 cm 3 and the gas expands without heat loss. If γ = 1.4 for the gas, what is the final pressure? Correct answer: kpa. Let : V 1 = 32.7 cm 3 = m 3, V 2 = 455 cm 3, P 1 = Pa, and γ = 1.4. For an adiabatic process, P 1 V γ 1 = P 2V γ 2 γ V1 P 2 = P 1 V cm = Pa 455 cm = Pa = kpa. 017 part 2 of points How much work is done? Correct answer: J. The work V2 V2 W = P dv = P 1 V 1 V 1 [ 1 = P 1 V 1 1 γ 1 1 = V1 V V1 V Pa m 3 γ dv ] γ 1

6 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 6 [ 32.7 cm ] cm part 2 of points Determine the heat Q ab added to the gas during process = J. AB. AP B 1993 FR part 1 of points One mole of an ideal monatomic gas is taken through the cycle ABCA shown schematically inthediagram. StateAhasvolume0.0167m 3 and pressure Pa, and state C has volume m 3. Process CA lies along the 247 ±1 K isotherm. PV diagram 16 Pressure 10 4 Pa A 247 ±1 K Volume 10 3 m 3 Find the temperature T b of state B. The molar heat capacities for the gas are 20.8 J/mol K and 12.5 J/mol K. Correct answer: K. B C Let : P b = Pa, V b = m 3, and R = J/mol K. Using the ideal gas equation T b = P bv b R = Pa m J/mol K = K. Correct answer: J. For state a Let : P a = Pa, V a = m 3, n = 1 mol, and c p = 20.8 J/mol K. T a = P av a R = Pa m J/mol K = K, so Q = nc p T = 1 mol20.8 J/mol K K K = J. 020 part 3 of points Determine the change U ab = U b U a in the internal energy. Correct answer: J. In an isobaric process the change in internal energy is given by U ab = Q ab W = Q ab P V = Q ab P V b V a = J Pa m m 3 = J. 021 part 4 of points

7 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 7 Determine the work W bc done by the gas on its surroundings during process BC. Correct answer: 0. W = P V and V = 0, so W = part 5 of points The net heat added to the gas for the entire cycle is 2130 J. Determine the net work done by the gas on its surroundings for the entire cycle. Correct answer: 2130 J. A certain quantity of an ideal gas initially at temperature T 0, pressure P 0, and volume V 0 is compressed to one-half its initial volume. As shown, the process may be adiabatic process 1, isothermalprocess 2, or isobaric process 3. Pressure Pacal T 1 T 2 Pressure vs Volume 2 1 Let : Q = 2130 J. For a complete cycle the change in internal energy U is zero, so P 0 T3 3 T 4 W = Q = 2130 J. The work is simply the net heat added to the gas. 023 part 6 of points Determine the efficiency of a Carnot engine that operates between the maximum and minimum temperatures in this cycle. Correct answer: The Carnot efficiency is η = 1 T c T h. The maximum temperature from part 1 and the minimum temperature is isotherm. Therefore η = 1 T a T b = K K = AP B 1998 MC part 1 of points V 0 2 Volume liter Which of the following is true of the mechanical work done on the gas? V 0 1. It is greatest for process 1. correct 2. It is the same for process 1 and 2 and less for process It is the same for process 2 and 3 and less for process It is the same for all three processes. 5. It is greatest for process 3. The work done on the gas is the area under thep V curve; theworkisgreatestforprocess part 2 of points Which of the following is true of the final temperature of this gas? 1. It is the same for processes 1 and 2.

8 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 8 2. It is greatest for process 1. correct 3. It is the same for processes 1 and It is greatest for process It is greatest for process 2. The more work done to the gas, the higher the internal energy, so it has a higher final temperature, and the final temperature is the highest for process 1. Diatomic Gas in Closed Cycle 026 part 1 of points 4.2 L of diatomic gas γ = 1.4 confined to a cylinder are put through a closed cycle. The gas is initially at 1.2 atm and at 310 K. First, its pressure is tripled under constant volume. Then it expands adiabatically to its original pressure and finally is compressed isobarically to its original volume. Determine the volume at the end of the adiabatic expansion. Correct answer: L. Let : V 0 = 4.2 L = m 3 and P f = 3P i. Basic Concepts P V = nrt U = Q W gas U = nc v T Q = nc T Solution: For an adiabatic process we have P B V γ = P B C V γ, C so 3P 0 V γ 0 = P 0V γ. C It follows from this that 1 3P0 γ V C = V 0 P 0 = 4.2 L = L. 027 part 2 of points Find the temperature of the gas at the start of the adiabatic expansion. Correct answer: 930 K. Let : T 0 = 310 K. From the equation of state for an ideal gas we obtain P B V B = nrt B = 3P 0 V 0 = 3nRT 0 Therefore T B = 3T 0 = 3310 K = 930 K. 028 part 3 of points Find the temperature at the end of the adiabatic expansion. Correct answer: K. T B V γ 1 B Let : V B = 4.2 L. = T C V γ 1 C γ 1 VB T C = T B V C 4.2 L = 930 K L = K part 4 of points What is the net work done for this cycle? Correct answer: J.

9 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 9 Let : P 0 = 1.2 atm = Pa. Since for a closed cycle U = 0, net work done for this cycle equals net heat transferred to the gas. In AB this heat is In BC Q AB = nc v T = 5 2 nrt B T A = 5 2 nr3t 0 T 0 = 5nRT 0 = 5P 0 V 0 = Pa m 3 = J. Q BC = 0, since the process is adiabatic, and with the help of the equation of the state for an ideal gas, we have Q CA = nc p T = 7 2 nrt A T C = 7 2 nr T 0 T 0 V C = 7 2 P 0V 0 1 V C V 0 V 0 = Pa m L 4.2 L = J. For the whole cycle Q ABCA = Q AB +Q CA = J J = J. Heating One Mole of Gas points Four moles of a diatomic gas has pressure P and volume V. By heating the gas, its pressure is tripled and its volume is doubled. If this heating process includes two steps, one at constant pressure and the other at constant volume, determine the amount of heat transferred to the gas. 1. Q = 5P V 2. Q = 13.5P V correct 3. Q = 12P V 4. Q = 3P V 5. Q = 10P V Let : n = 4. Basic Concepts: The amount of heat transferred to the gas in a process P is Q = nc P T, where n is amount of gas, C P is heat capacity of the gas in the process P and T is change of temperature. Solution: Q = nc P T isobaric +nc V T isovolumemetric In the isobaric process V doubles so T must double, to 2T 0. In the isovolumemetric process P triples so T changes from 2T 0 to 6T 0 : Q = n 7 2 R2T 0 T 0 +n 5 2 R6T 0 2T 0 7 = nr T 0 = 13.5 nr T 0 = 13.5P V, where P V = nr T 0. Ideal Gas Path points Identify the parameter paths for an ideal gas that are isovolumetric/ isobaric/ isothermal.

10 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 10 V i = 0.5 m 3. P From the first law of thermodynamics U = Q W = Q PV f V i C T 1 > T 2 > T 3 = 11.7 kj 1 kpa0.6 m m 3 = 11.6 kj. A B T 1 T 2 T part 2 of points Find the final temperature of the gas. Correct answer: K. V 1. A / C / B correct 2. B / C / A 3. A / B / C 4. C / B / A 5. C / A / B 6. B / A / C A is isovolumetric since volume is constant. B is isothermal since temperature is constant. C is isobaric since pressure is constant. Internal Energy Change part 1 of points An ideal gas initially at 288 K undergoes an isobaric expansion at 1 kpa. If the volume increases from 0.5 m 3 to 0.6m 3 and 11.7 kjof thermal energy is transferredtothegas,findthechangeinitsinternal energy. Correct answer: 11.6 kj. Let : Q = 11.7 kj, P = 1 kpa, V f = 0.6 m 3, and From the ideal gas law V i T i = V f T f Let : T i = 288 K. T f = V f T i V i = 0.6 m3 288 K 0.5 m 3 = K. Isobaric Work points A balloon, originally of volume V i, inflates at constant pressure, P, to twice its original radius. The amount ofworkdone by theballoonis: 1. 7P V i correct 2. 5P V i 3. 2P V i 4. 3P V i 5. P V i The amount of work done in an isobaric process is W = P V f V i. The initial volume of the balloon is V i r 3 i,

11 Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 11 and the final volume of the balloon is so V f = 8V i, and V f r 3 f = 2r i 3 = 8r 3 i, W = P 8V i V i = 7P V i. Gas Expands Isothermally points 11 mol of an ideal gas expands isothermally at 373 K to 6.1 its initial volume. Find the heat flow into the system. The universal gas constant is 8.31 J/K mol. Correct answer: J. Let : n = 11 mol, R = 8.31 J/K mol, T = 373 K, and V f = 6.1V i. This is an isothermal process, so the internal energy of the ideal gas is constant. Thus the heat flowing into the system is equal to the work done to the system: Vf W = nrt ln V i = 11 mol8.31 J/K mol373 K ln6.1 = J. Adiabatic Expansion 036 part 1 of points 2 mol of an idea gas γ = 1.4 expands slowly and adiabatically from a pressure of 3 atm andavolumeof19ltoafinalvolumeof29l. The universal gas constant is J/K mol. What is the final pressure? Correct answer: atm. Let : P 0 = 3 atm = Pa, V 0 = 19 L = m 3, V 1 = 29 L = m 3, n = 2 mol and R = J/K mol. Since an adiabatic process for an ideal gas is described by P V γ = const, we have So, P 0 V γ 0 = P 1V γ 1 γ V0 P 1 = P 0 V 1 19 L = 3 atm 29 L = atm part 2 of points What is the initial temperature? Correct answer: K. From the equation of state for an ideal gas we have T 0 = P 0V 0 nr = Pa0.019 m 3 2 mol J/K mol = K. 038 part 3 of points What is the final temperature? Correct answer: K. From the equation of state for an ideal gas we have T 1 = P 1V 1 nr = Pa0.029 m 3 2 mol J/K mol = K.

Handout 12: Thermodynamics. Zeroth law of thermodynamics

Handout 12: Thermodynamics. Zeroth law of thermodynamics 1 Handout 12: Thermodynamics Zeroth law of thermodynamics When two objects with different temperature are brought into contact, heat flows from the hotter body to a cooler one Heat flows until the temperatures