Answer, Key Homework 6 David McIntyre 1


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1 Answer, Key Homework 6 David McIntyre 1 This printout should have 0 questions, check that it is complete. Multiplechoice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. Chapters 0 and 1 problems. 001 part 1 of 1 0 points A kg ingot of metal is heated to 17 C and then is dropped into a beaker containing 0.46 kg of water initially at C. If the final equilibrium state of the mixed system is 5.4 C, find the specific heat of the metal. The specific heat of water is 4186 J/kg C. Correct answer: J/kg C. Given : m w 0.46 kg, m x kg, c w 4186 J/kg C, T w C, T i 17 C, and T f 5.4 C. Because the thermal energy lost by the ingot equals the thermal energy gained by the water, m x c x T i T f m w c w T f T w, c x m w c w T f T w m x T i T f 0.46 kg 4186 J/kg C kg 5.4 C C 17 C 5.4 C J/kg C. 00 part 1 of 1 0 points 7 g of water at 70 C is poured into a 5 g aluminum cup containing 66 g of water at 9 C. What is the equilibrium temperature of the system? Assume the specific heat of aluminum is 0.15 cal/g C. Correct answer: C. Given : m h 7 g, m c 66 g, m Al 5 g, c w 1 cal/g C, c Al 0.15 cal/g C, T c 9 C, and T h 70 C. The heat lost by hot water equals the heat gained by cold metal plus the water. m h c w T h T f m Al c Al T f T c + m c c w T f T c m h c w + m Al c Al + m c c w T f m h c w T f + m Al c Al T c + m c c w T c T f m h c w T h + m Al c Al T c + m c c w T c m h c w + m Al c Al + m c c w Since m h c w T h + m Al c Al T c + m c c w T c 7 g 1 cal/g C 70 C + 5 g 0.15 cal/g C 9 C + 66 g 1 cal/g C 9 C cal and m h c w + m Al c Al + m c c w 7 g 1 cal/g C + 5 g 0.15 cal/g C + 66 g 1 cal/g C cal/ C, then the equilibrium temperature will be cal T f cal/ C C, 00 part 1 of 1 10 points A calorimeter contains 410 ml of water at 48 C and 4 g of ice at 0 C. Find the final temperature of the system. Correct answer: 5.89 C.
2 Answer, Key Homework 6 David McIntyre Given : V 410 ml, m w 410 g, m ice 0.04 kg, c w 1 cal/g C, L f 000 J/kg, T w 48 C, T i 0 C, and T 48 C. The 410 ml m v of water cools to 0 C, giving off Q w m w c w T 410 g 1 cal/g C 48 C cal heat. If all of the ice melts, it absorbs Q i m ice L f 0.04 kg 000 J/kg 0.89 cal/j cal heat. Since Q w > Q i, all of the ice melts and also warms up to a temperature above 0 C: Thus Q lostwater Q meltice + Q gained m w c w T w T f Q i + m ice c w T f T ice m w c w T w m w c w T f Q i + m ice c w T f 0 m w c w T w Q i m ice c w + m w c w T f T f m w c w T w Q i m ice c w + m w c w 410 g1 cal/g C48 C cal 0.04 kg1 cal/g C g1 cal/g C cal cal 4 cal/ C cal/ C 5.89 C. Calculate the work done by the gas during this process. Correct answer: J. Given : P Pa, V 1 4 cm, V 60.5 cm, m kg, and L v J/kg. W P V P V V Pa60.5 cm 4 cm 1 m 100 cm J. 005 part of 0 points Find the amount of heat added to the water to accomplish this process. Correct answer: 9040 J. Q m L v kg J/kg 9040 J. 006 part of 0 points Find the change in internal energy. Correct answer: J. U Q W 9040 J J J. 004 part 1 of 0 points 4 cm of water is boiled at atmospheric pressure to become 60.5 cm of steam, also at atmospheric pressure. 007 part 1 of 4 points Two moles of helium gas initially at 5 K and 0. atm are compressed isothermally to 1.57 atm.
3 Answer, Key Homework 6 David McIntyre Find the final volume of the gas. Assume the helium to behave as an ideal gas. Correct answer: m. Given : n mol, R J/K mol, T f 5 K, and P f 1.57 atm. From the ideal gas law, P f V f n R T V f n R T P f mol J/K mol5 K 1.57 atm10100 Pa/atm m. 008 part of points Find the work done by the gas. Correct answer: kj. Given : P i 0. atm and T i 5 K. The initial volume was V i n R T P i mol J/K mol5 K 0. atm10100 Pa/atm m and the work done was W P dv n R T ln Vf V i mol J/K mol5 K m 1 kj ln m 1000 J kj. 009 part of points Find the thermal energy transferred. Correct answer: kj. From the first law of thermodynamics, since U 0, Q W kj. 010 part 1 of 5 points A gas is taken through the cyclic process described by the figure. Let a 9, b 9. P kpa A B V m Find the net energy transferred to the system by heat during one complete cycle. Correct answer: 40.5 kj. Given : C P 9000 Pa and V 9 m. The change in internal energy is U cycle Q cycle + W cycle 0 Q cycle W ABCA area enclosed in P V diagram 1 9 m 9000 Pa J 40.5 kj. Alternate Solution:
4 Answer, Key Homework 6 David McIntyre 4 For each step the work is the negative of the area under the curve on the PV diagram: W AB [.6 kpa 10.5 m 6.5 m kpa.6 kpa 10.5 m 6.5 m ] 6.4 kj W BC 0 kj W CA.6 kpa 6.5 m 10.5 m 14.4 kj Q cycle W ABC 6.4 kj + 0 kj kj 1 kj. 011 part of 5 points If the cycle is reversed that is, the process follows the path ACBA, what is the net energy transferred by heat per cycle? Correct answer: 40.5 kj. If the cycle is reversed, then Q cycle W ACBA W ABCA 40.5 kj. 01 part 1 of 0 points An ideal gas initially at K undergoes an isobaric expansion at kpa. If the volume increases from 1.6 m to 4 m and 15.9 kj of thermal energy is transferred to the gas, find the change in its internal energy. Correct answer: 11.1 kj. Given : P kpa, V f 4 m, V i 1.6 m, T i K. and From the first law of thermodynamics U Q W Q P V f V i 15.9 kj kpa4 m 1.6 m 11.1 kj. 01 part of 0 points Find the final temperature of the gas. Correct answer: 8.5 K. From the ideal gas law V i T i V f T f T f V f T i V i 4 m K 1.6 m 8.5 K. 014 part 1 of 0 points One mole of an ideal gas does 519 J of work on the surroundings as it expands isothermally to a final pressure of 0.8 atm and volume of L. Determine the initial volume. R J/K mol. Correct answer: L. Given : n 1 mol, W 519 J, R J/K mol, P f 0.8 atm, and V f L. By the ideal gas law, P V n R T. The work done in an isothermal process is Vf W n R T ln V i Vf P V ln V i Vf ln W V i P V [ ] V f W exp V i P V [ ] W V i V f exp P V Since
5 Answer, Key Homework 6 David McIntyre 5 W P V 519 J 0.8 atm10100 L1e V i L exp [ ] L. 015 part of 0 points Determine the temperature of the gas. Correct answer: K. According to the equation of state for an ideal gas, the temperature is T f P f V f n R 0.8 atm10100 L1e 1 mol J/K mol K. 016 part 1 of 0 points Air in the cylinder of a diesel engine at 1.8 C is compressed from an initial pressure of 0.81 atm and of volume of 79 cm to a volume of 86 cm. Assuming that air behaves as an ideal gas γ 1.40 and that the compression is adiabatic and reversible, find the final pressure. Correct answer: atm. Given : P i 0.81 atm, V i 79 cm, V f 86 cm. Using the relation we find that P i V γ i γ P f V γ f, Vi P f P i V f 79 cm 0.81 atm 86 cm atm. 017 part of 0 points and 1.40 Find the final temperature under the same assumptions as above. Correct answer: C. Since Given : T i 1.8 C 94.8 K. P V n R T is always valid during the process and since no gas escapes from the cylinder, Therefore, P i V i T i T f P f V f P i V i T i P f V f T f n R atm 86 cm 0.81 atm 79 cm 94.8 K K C. 018 part 1 of 4 points 4. L of diatomic gas γ 1.4 confined to a cylinder are put through a closed cycle. The gas is initially at 1 atm and at 0 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. Given : V 0 4. L m and P f P i. Basic Concepts P V n R T U Q W gas U n C v T
6 Answer, Key Homework 6 David McIntyre 6 Q n C T Solution: For an adiabatic process we have so P B V γ B P C V γ C, P 0 V γ 0 P 0 V γ C. It follows from this that V C 1 P0 γ V 0 P 0 4. L L. 019 part of 4 points Find the temperature of the gas at the start of the adiabatic expansion. Correct answer: 990 K. Given : T 0 0 K. From the equation of state for an ideal gas we obtain P B V B n R T B P 0 V 0 n R T 0 Therefore T B T 0 0 K 990 K. 00 part of 4 points Find the temperature at the end of the adiabatic expansion. Correct answer: 7.94 K. T B V γ 1 B T C T C V γ 1 C γ 1 VB T B V C 4. L 990 K L 7.94 K part 4 of 4 points What is the net work done for this cycle? Correct answer: J. Given : P 0 1 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 n C v T 5 n R T B T A 5 n R T 0 T 0 5 n R T 0 5 P 0 V Pa m 17. 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 n C p T For the whole cycle 7 n R T A T C 7 n R T 0 T 0 V C 7 P 0 V 0 1 V C V 0 V Pa m L 4. L J. Q ABCA Q AB + Q CA 17. J J J.
7 Answer, Key Homework 6 David McIntyre 7 0 part 1 of 0 points A cylinder contains.46 mol of helium gas at a temperature of 85 K. How much heat must be transferred to the gas to increase its temperature to 60 K if it is heated at constant volume? The molar specific heat at constant volume, c v, of helium is 1.5 J/mol/K. Correct answer: J. Given : n.46 mol, T i 85 K, and T f 60 K. For the constantvolume process, the work done is zero. Therefore for the heat Q 1 transferred to the gas during the process we have Q 1 U n R T n c v T f T i, Q 1.46 mol 1.5 J/mol/K 60 K 85 K J. slowly and adiabatically from a pressure of 4 atm and a volume of 1 L to a final volume of L. What is the final pressure? Correct answer: atm. Given : P 0 4 atm Pa, V 0 1 L 0.01 m, and V 1 L 0.0 m. Basic Concepts P V γ const P V n R T Solution: Since an adiabatic process for an ideal gas is described by P V γ const, we have P 0 V γ 0 P 1 V γ 1, so γ V0 P 1 P 0 V 1 1 L 4 atm L atm part of 0 points How much thermal energy must be transferred to the gas at constant pressure to raise the temperature to 60 K? The molar specific heat at constant presure, c p, of helium is 0.8 J/mol/K. Correct answer: 160. J. For the thermal energy Q transferred to the gas at constant pressure, Q n c p T.46 mol 0.8 J/mol/K 17 K 160. J. 04 part 1 of 0 points Two moles of an idea gas γ 1.4 expands 05 part of 0 points What is the initial temperature? Correct answer: K. From the equation of state for an ideal gas we have T 0 P 0 V 0 n R Pa 0.01 m mol J/K mol K. 06 part of 0 points What is the final temperature? Correct answer: 18. K.
8 Answer, Key Homework 6 David McIntyre 8 From the equation of state for an ideal gas we have T 1 P 1 V 1 n R Pa 0.0 m mol J/K mol 18. K. 07 part 1 of 4 0 points Assume that a molecule has f degrees of freedom. Consider a gas consisting of such molecules. Determine its total thermal energy. 1. U f n R T. U f n R T. U f n R T 4. U f n R T correct 5. U f n R T C v 1 n du dt According to the equipartition theorem, the amount of energy per each degree of freedom is k B T, so that U N f k B T f n R T 08 part of 4 0 points Find its molar specific heat at constant volume. 1. C v f R. C v 1 f R correct. C v f R 4. C v 1 4 f R. 5. C v 1 f R The specific heat is given by C v 1 du 1 n dt f R. 09 part of 4 0 points Find its molar specific heat at constant pressure. 1. C p 1 f + R. C p f + R. C p 1 f + 1 R 4. C p 1 f + R 5. C p 1 f + R correct Since C p C v + R, we have C p C v + R 1 f + R. 00 part 4 of 4 0 points Determine the ratio γ C p C v 1. γ f + f. γ f + f. γ f + correct f 4. γ f + f 5. γ f + 1 f Taking the magnitudes of C p and C v from the previous sections, we obtain γ C p f + C v f.
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