Physics Solution Manual for 1100 and 2101

Thus the change is u 3 nr t 2 3 2 100 mol 831

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Unformatted text preview: Vi 5 −3 hc m 3 − 2 .0 × 10 −3 m 3 h= +2 .1 × 10 3 J 772 THERMODYNAMICS 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 (1.0 × 105 Pa)(1.0 × 10–3 m3) = 1.0 × 102 J The work, then, is 2 W = – 15 (1.0 × 10 J) = −1.5 × 10 3 J ______________________________________________________________________________ 14. REASONING The pressure P of the gas remains constant while its volume increases by an amount ∆V. Therefore, the work W done by the expanding gas is given by W = P∆V (Equation 15.2). ∆V is known, so if we can obtain a value for W, we can use this expression to calculate the pressure. To determine W, we turn to the first law of thermodynamics ∆U = Q − W (Equation 15.1), where Q is the heat and ∆U is the change in the internal energy. ∆U is given, so to use the first law to determine W we need information about Q. According to Equation 12.4, the heat needed to raise the temperature of a mass m of material by an amount ∆T is Q = c m ∆T where c is the material’s specific heat capacity. SOLUTION According to Equation 15.2, the pressure P of the expanding gas can be determined from the work W and the change ∆V in volume of the gas according to P= W ∆V Using the first law of thermodynamics, we can write the work as W = Q − ∆U (Equation 15.1). With this substitution, the expression for the pressure becomes P= W Q − ∆U = ∆V ∆V (1) Using Equation 12.4, we can write the heat as Q = c m ∆T , which can then be substituted into Equation (1). Thus, P= Q − ∆U c m ∆T − ∆U = ∆V ∆V ( ) 1080 J/ ( kg ⋅ C° ) 24.0 × 10−3 kg ( 53.0 C° ) − 939 J = = 3.1×105 Pa −3 3 1.40 × 10 m Chapter 15 Problems 773 15. SSM REASONING The work done in an isobaric process is given by Equation 15.2, W = P∆V ; therefore, the pressure is equal to P = W / ∆V . In order to use this expression, we must first determine a numerical value for the work done; this can be calculated using the first law of thermodynamics (Equation 15.1), ∆U = Q − W . SOLUTION Solving Equation 15.1 for the work W, we find W = Q − ∆U = 1500 J–(+4500 J) = − 3.0 ×103 J Therefore, the pressure is P= W −3.0 ×103 J = = 3.0 × 105 Pa ∆V −0.010 m3 The change in volume ∆V, which is the final volume minus the initial volume, is negative because the final volume is 0.010 m3 less than the initial volume. 16. REASONING AND SOLUTION The rod's volume increases by an amount ∆V = βV0∆T, according to Equation 12.3. The work done by the expanding aluminum is, from Equation 15.2, 5 –6 W = P∆V = PβV0∆T = (1.01 × 10 Pa)(69 × 10 /C°)(1.4 × 10 –3 3 2 m )(3.0 × 10 C°) = 2.9 J 17. SSM WWW REASONING AND SOLUTION The first law of thermodynamics states that ∆U = Q − W . The work W involved in an isobaric process is, according to Equation 15.2, W = P∆V . Combining these two expressions leads to ∆U = Q − P∆V . Solving for Q gives (1) Q = ∆U + P∆V Since this is an expansion, ∆V > 0 , so P∆V > 0 . From the ideal gas law, PV = nRT , we have P ∆V = nR ∆T . Since P∆V > 0 , it follows that nR ∆T > 0 . The internal energy of an ideal gas is directly proportional to its Kelvin temperature T. Therefore, since nR ∆T > 0 , it follows that ∆U > 0 . Since both terms on the right hand side of Equation (1) are positive, the left hand side of Equation (1) must also be positive. Thus, Q is positive. By the convention described in the text, this means that heat can only flow into an ideal gas during an isobaric expansion 774 THERMODYNAMICS 18. REASONING AND SOLUTION According to the first law of thermodynamics, the change in internal energy is ∆U = Q – W. The work can be obtained from the area under the graph. There are sixty squares of area under the graph, so the positive work of expansion is c h c h W = 60 1.0 × 10 4 Pa 2.0 × 10 –3 m 3 = 1200 J Since Q = 2700 J, the change in internal energy is ∆U = Q – W = 2700 J – 1200 J = 1500 J 19. REASONING AND SOLUTION Since the pan is open, the process takes place at constant (atmospheric) pressure P0. The work involved in an isobaric process is given by Equation 15.2: W = P0∆V. The change in volume of the liquid as it is heated is given according to Equation 12.3 as ∆V = βV0∆T, where β is the coefficient of volume expansion. Table 12.1 bg gives β = 207 × 10 –6 C ° –1 for water. The heat absorbed by the water is given by Equation 12.4 as Q = cm∆T, where c = 4186 J/ ( kg ⋅ C° ) is the specific heat capacity of liquid water according to Table 12.2. Therefore, P0 β Pβ W P0 ∆V P0 β V0 ∆T = = = =0 Q cm∆T cm∆T c(m / V0 ) c ρ where ρ = 1.00 × 10 3 kg / m 3 is the density of the water (see Table 11.1). Thus, we find ( )( ) 1.01 × 105 Pa...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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