c. Applying the first law of thermodynamics from Equation 15.1, we find that c h c h.


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1 Week 11 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution is the same, but you will need to repeat part of the calculation to find out what your answer should have been. WebAssign Problem 1: In moving out of a dormitory at the end of the semester, a student does of work. In the process, his internal energy decreases by. Determine each of the following quantities (including the algebraic sign): (a) W, (b) U, and (c) Q. 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 = J. b. The change in internal energy is U = J. c. Applying the first law of thermodynamics from Equation 15.1, we find that c h c h. Q = U + W = J J = J WebAssign Problem : In exercising, a weight lifter loses kg of water through evaporation, the heat required to evaporate the water coming from the weight lifter s body. The work done in lifting weights is. (a) Assuming that the latent heat of vaporization of perspiration is, find the change in the internal energy of the weight lifter. (b) Determine the minimum number of nutritional Calories of food ( ) that must be consumed to replace the loss of internal energy. REASONING AND SOLUTION a. For the weight lifter U = Q W = ml v W = (0.150 kg)( J/kg) J = J
2 b. Since 1 nutritional calorie = 4186 J, the number of nutritional calories is 5 1Calorie J = nutritional calories 4186 J WebAssign Problem : A gas, while expanding under isobaric conditions, does 480 J of work. The pressure of the gas is, and its initial volume is. What is the final volume of the gas? REASONING When a gas expands under isobaric conditions, its pressure remains constant. The work W done by the expanding gas is W = P (V f V i ), Equation 15., where P is the pressure and V f and V i are the final and initial volumes. Since all the variables in this relation are known, we can solve for the final volume. SOLUTION Solving W = P (V f V i ) for the final volume gives V W 480 J = + V = m = m P Pa f i 5 WebAssign Problem 4: The pressure and volume of a gas are changed along the path ABCA. Using the data shown in the graph, determine the work done (including the algebraic sign) in each segment of the path: (a) A to B, (b) B to C, and (c) C to A. REASONING For segment AB, there is no work, since the volume is constant. For segment BC the process is isobaric and Equation 15. applies. For segment CA, the work can be obtained as the area under the line CA in the graph.
3 SOLUTION a. For segment AB, the process is isochoric, that is, the volume is constant. For a process in which the volume is constant, no work is done, so W = 0 J. b. For segment BC, the process is isobaric, that is, the pressure is constant. Here, the volume is increasing, so the gas is expanding against the outside environment. As a result, the gas does work, which is positive according to our convention. Using Equation 15. and the data in the drawing, we obtain c f i h c h c h c h W = P V V 5 = Pa m m = J c. For segment CA, the volume of the gas is decreasing, so the gas is being compressed and work is being done on it. Therefore, the work is negative, according to our convention. The magnitude of the work is the area under the segment CA. We estimate that this area is 15 of the squares in the graphical grid. The area of each square is ( Pa)( m ) = J The work, then, is W = 15 ( J) = J WebAssign Problem 5: Five moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 70 to 90 K. Determine (a) the work done (including the algebraic sign) by the gas, and (b) the change in its internal energy. REASONING Since the gas is expanding adiabatically, the work done is given by W = nr T T. Once the work is known, we can use the first Equation 15.4 as i f law of thermodynamics to find the change in the internal energy of the gas. SOLUTION a. The work done by the expanding gas is i f W = nr T T = 5.0 mol 8.1 J/ mol K 70 K 90 K = J b. Since the process is adiabatic, Q = 0, and the change in the internal energy is U = Q W = J = J WebAssign Problem 6: A monatomic ideal gas expands from point A to point B along the path shown in the drawing. (a) Determine the work done by the gas. (b) The
4 temperature of the gas at point A is 185 K. What is its temperature at point B? (c) How much heat has been added to or removed from the gas during the process? REASONING a. The work done by the gas is equal to the area under the pressureversusvolume curve. We will measure this area by using the graph given with the problem. b. Since the gas is an ideal gas, it obeys the ideal gas law, PV = nrt (Equation 14.1). This implies that PA VA / TA = PB VB / T B. All the variables except for T B in this relation are known. Therefore, we can use this expression to find the temperature at point B. c. The heat Q that has been added to or removed from the gas can be obtained from the first law of thermodynamics, Q = U + W (Equation 15.1), where U is the change in the internal energy of the gas and W is the work done by the gas. The work W is known from part (a) of the problem. The change U in the internal energy of U = U U = nr T T, where the gas can be obtained from Equation 14.7, B A B A n is the number of moles, R is the universal gas constant, and T B and T A are the final and initial Kelvin temperatures. We do not know n, but we can use the ideal gas law (PV = nrt ) to replace nrt B by P B V B and to replace nrt A by P A V A. SOLUTION a. From the drawing we see that the area under the curve is 5.00 squares, where 5 5 each square has an area of work W done by the gas is Pa.00 m = J. Therefore, the 5 6 W = 5.00 squares J/square = J b. In the Reasoning section, we have seen that PA VA / TA = PB VB / T B. Solving this relation for the temperature T B at point B, using the fact that P A = P B (see the graph), and taking the values for V B and V A from the graph, we have that
5 P BV B V B B A A P AV A V A.00 m 10.0 m T = T = T = ( 185 K ) = 95 K c. From the Reasoning section we know that the heat Q that has been added to or removed from the gas is given by Q = U + W. The change U in the internal U = U U = nr T T. Thus, the heat can be energy of the gas is expressed as B A B A Q = U + W = nr T T + W We now use the ideal gas law (PV = nrt ) to replace nrt B by P B V B and nrt A by P A V A. The result is B Q = P V P V + W B B A A Taking the values for P B, V B, P A, and V A from the graph and using the result from part a that W = J, we find that the heat is Q = P V P V + W B B A A = Pa 10.0 m Pa.00 m J = J WebAssign Problem 7: Three moles of a monatomic ideal gas are heated at a constant volume of 1.50 m. The amount of heat added is. (a) What is the change in the temperature of the gas? (b) Find the change in its internal energy. (c) Determine the change in pressure. REASONING AND SOLUTION a. The amount of heat needed to raise the temperature of the gas at constant volume is given by Equations 15.6 and 15.8, Q = n C V T. Solving for T yields Q J T = = = K nc (.00 mol ) ( R ) v b. The change in the internal energy of the gas is given by the first law of thermodynamics with W = 0, since the gas is heated at constant volume: A U = Q W = J 0 = J c. The change in pressure can be obtained from the ideal gas law, (.00 mol) R ( K ) nr T P = = =. 10 Pa V 1.50 m
6 WebAssign Problem 8: MultipleConcept Example 6 provides a review of the concepts that play roles here. An engine has an efficiency of 64% and produces 5500 J of work. Determine (a) the input heat and (b) the rejected heat. REASONING AND SOLUTION a. The efficiency is e = W / QH, where W is the magnitude of the work and Q H is the magnitude of the input heat. It follows that Q H W 5500 J = = = e J b. The magnitude of the rejected heat is QC = QH W = 8600 J 5500 J = 100 J
7 Practice conceptual problems:. A gas is compressed isothermally, and its internal energy increases. Is the gas an ideal gas? Justify your answer. REASONING AND SOLUTION The internal energy of an ideal gas is proportional to its Kelvin temperature (see Equation 14.7). In an isothermal process the temperature remains constant; therefore, the internal energy of an ideal gas remains constant throughout an isothermal process. Thus, if a gas is compressed isothermally and its internal energy increases, the gas is not an ideal gas. 5. (a) Is it possible for the temperature of a substance to rise without heat flowing into it? (b) Does the temperature of a substance necessarily have to change because heat flows into or out of it? In each case, give your reasoning and use the example of an ideal gas. REASONING AND SOLUTION a. It is possible for the temperature of a substance to rise without heat flowing into the substance. Consider, for example, the adiabatic compression of an ideal gas. Since the process is an adiabatic process, Q = 0. The work done by the external agent increases the internal energy of the gas. Since the internal energy of an ideal gas is proportional to the Kelvin temperature, the temperature of the gas must increase. b. The temperature of a substance does not necessarily have to change because heat flows into or out of it. Consider, for example, the isothermal expansion of an ideal gas. Since the internal energy of an ideal gas is proportional to the Kelvin temperature, the internal energy, U, remains constant during an isothermal process. The first law of thermodynamics gives U = Q W = 0, or Q = W. The heat that is added to the gas during the isothermal expansion is used by the gas to perform the work involved in the expansion. The temperature of the gas remains unchanged. Similarly, in an isothermal compression, the work done on the gas as the gas is compressed causes heat to flow out of the gas while the temperature of the gas remains constant. 10. Suppose you want to heat a gas so that its temperature will be as high as possible. Would you heat it under conditions of constant pressure or constant volume? Why? REASONING AND SOLUTION According to Equation 14.7, the Kelvin temperature T of the gas is related to its internal energy U by U = ( / ) n R T. The change in the internal energy is given by the first law of thermodynamics (Equation 15.1), U = Q W. It is desired to heat a gas so that its temperature will be as high as possible. If the process occurs at constant pressure, so that the volume of the gas increases, work is done by the gas. The available heat is used to do work and to increase the internal
8 energy of the gas. On the other hand, if the process is carried out at constant volume, the work done is zero, and all of the heat increases the internal energy of the gas. From Equation 14.7, the internal energy is directly proportional to the Kelvin temperature of the gas. Since the internal energy increases by a greater amount when the process occurs at constant volume, the temperature increase is greatest under conditions of constant volume. Therefore, if it is desired to heat a gas so that its temperature will be as high as possible, you should heat it under conditions of constant volume. 18. Air conditioners and refrigerators both remove heat from a cold reservoir and deposit it in a hot reservoir. Why, then, does an air conditioner cool the inside of a house while a refrigerator warms the house? REASONING AND SOLUTION In a refrigerator, the interior of the unit is the cold reservoir, while the warmer exterior of the room is the hot reservoir. An air conditioner is like a refrigerator, except that the room being cooled is the cold reservoir, and the outdoor environment is the hot reservoir. Therefore, an air conditioner cools the inside of the house, while a refrigerator warms the interior of the house.. When water freezes from a less ordered liquid to a more ordered solid, its entropy decreases. Why doesn t this decrease in entropy violate the second law of thermodynamics? REASONING AND SOLUTION When water freezes from a lessordered liquid to a moreordered solid, its entropy decreases. This decrease in entropy does not violate the second law of thermodynamics, because it is a decrease for only one part of the universe. In terms of entropy, the second law indicates that the total change in entropy for the entire universe must be either zero (reversible process) or greater than zero (irreversible process). In the case of freezing water, heat must be removed from the water and deposited in the environment. The entropy of the environment increases as a result. If the freezing occurs reversibly, the increase in entropy of the environment will exactly match the decrease in entropy of the water, with the result that S u n i v e r s e = S w a t e r + S e n v i r o n m e n t = 0. If the freezing occurs irreversibly, then the increase in entropy of the environment will exceed the decrease in entropy of the water, with the result that S u n i v e r s e = S w a t e r + S e n v i r o n m e n t > 0.
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