{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics Solution Manual for 1100 and 2101

2 applies for segment ca the work can be obtained as

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ODYNAMICS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ____________________________________________________________________________________________ 1. (c) This sign convention for Q and W is discussed in Section 15.3 (see, in particular, Equation 15.1). 2. ∆U = −9.3 × 105 J 3. (c) According to the discussion in Section 15.4, the area under a pressure−volume graph is the work W for any kind of process. Since the graph shows the gas being compressed, work is done on the gas. 4. (b) The work done by a gas is the area under the pressure-volume graph. The areas under curves A and B are the same (4 ‘squares’ each), and each is greater than that under curve C (3 ‘squares’). 5. W = +1.2 × 105 J 6. (a) The first law of thermodynamics states that the heat Q is related to the change ∆U in the internal energy and the work W by Q = ∆U + W. Since all three paths start at A and end at B, the change in the internal energy of each gas is the same. Therefore, the path that involves the greatest amount of work is the one that has greatest amount of heat added to the gas. The drawing shows that the work (which is the area under the pressure-volume graph) is greatest for path 1 and smallest for path 3. 7. (e) The first law of thermodynamics, Q = (UB − UA) + W, can be used to find the heat Q, since UB and UA are given and W is the area under the pressure-volume graph. 8. (e) The internal energy of a monatomic ideal gas depends directly on its temperature (see Equation 14.7). In an isothermal process the temperature does not change. Therefore, the internal energy of the gas does not change. 9. W = −1.81 × 104 J 10. (c) The temperature of the gas increases when its internal energy increases. According to the first law of thermodynamics, ∆U = Q − W. Both heat and work can change the internal energy and, hence, the temperature of the gas. If Q = 0 J, ∆U can still increase if W is a negative number, which means that work is done on the gas. 11. (d) The work done on a monatomic gas during an adiabatic compression is given by (See Equation 15.4) W = 3 nR (Ti − Tf ) . 2 Chapter 15 Answers to Focus on Concepts Questions 765 12. (a) The change in temperature is greatest when the change ∆U in the internal energy of the gas is greatest. According to the first law of thermodynamics, ∆U = Q − W, the change in the internal energy is greatest when heat Q is added and no work (W = 0 J) is done by the gas. The gas does no work when its volume remains constant, so its change in temperature is greater than if the volume had changed. 13. (c) A more-efficient engine produces more work from the same amount of input heat, as expressed by W = eQH (Equation 15.11). Part of the heat QH from the hot reservoir is used to perform work W, and the remainder QC is rejected to the cold reservoir. The conservation of energy states that QH = W + QC (Equation 15.12). If QH is constant and W increases, the heat QC must decrease. 14. (a) The efficiency e of a heat engine depends on the ratio QC/QH through the relation e = 1 − QC/QH (Equation 15.13). Doubling QC and QH does not change this ratio. 15. (d) The efficiency of a Carnot engine is given by e = 1 − (TC/ TH), Equation 15.15, where TC and TH are the Kelvin temperatures of the cold and hot reservoirs. The efficiencies of engines C, B, and A, are, respectively, 0.50, 0.20, and 0.11. 16. W2/ W1 = 1.5 17. (c) The refrigerator uses the work W done by the electrical energy to remove heat QC from its interior and deposit heat QH into the room. In accordance with the conservation of energy, QH = QC + W. Therefore, the heat delivered into the room is greater than the electrical energy used to produce the work. 18. (e) According the conservation of energy, the heat QH delivered to the room equals the heat QC removed from the interior of the refrigerator plus the work W; QH = QC + W. The coefficient of performance is the heat QC removed from the refrigerator divided by the work W done by the refrigerator; COP = QC /W (Equation 15.16). 19. (b) The sun loses heat, so its entropy decreases. The earth, on the other hand, gains heat so its entropy increases. The transfer of heat from the sun to the earth, like the flow of heat from a hot reservoir to a cold reservoir, is irreversible. Therefore, the entropy of the sun-earth system increases. 20. (e) The change ∆S in entropy of the gas is given by ∆S = (Q/T)R (Equation 15.18), where Q is the heat added and T is its temperature (which remains constant since the expansion is isothermal). 766 THERMODYNAMICS CHAPTER 15 THERMODYNAMICS PROBLEMS 1. REASONING Since the student does work, W is positive, according to our convention. Since his internal energy decreases, the change ∆U in the internal energy is negative. The first law of thermodynamics will allow us to determine the heat Q. SOLUTION a. The work is W = +1.6 × 104 J . 4 b. The change in internal energy is ∆U = –4.2 × 10 J . c. Applying the first law of thermodynamics from Equation 15.1, we...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online