Physics Solution Manual for 1100 and 2101

U can be evaluated by remembering that the internal

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Unformatted text preview: er of nutritional calories is Calorie (5.03 × 105 J ) 14186 J = 8. 1.20 × 102 nutritional calories REASONING During both parts of this two-step process, the internal energy change ∆U of the gas is the difference between the work W done on or by the gas and the heat Q that flows into or out of it, according to the first law of thermodynamics: ∆U = Q − W (Equation 15.1). The total change in the internal energy of the gas is the sum of the changes that occur during each process: ∆U total = ∆U1 + ∆U 2 (1) The first step is isochoric (constant volume), and thus involves no work: W1 = 0 J. Heat is added to the gas, so Q1 is positive: Q1 = +5470 J. In the second step, the volume decreases (∆V is negative) at a constant pressure of P2 = 3.45×105 Pa, so we will use W2 = P2 ∆V (Equation 15.2) to determine the amount of work W2 done during that step. The second step is also adiabatic, involving no transfer of heat, so that Q2 = 0 J. SOLUTION Applying the first law of thermodynamics ∆U = Q − W (Equation 15.1) to Equation (1), we obtain ∆U total = ∆U1 + ∆U 2 = ( Q1 − W1 ) + ( Q2 − W2 ) (2) As noted in the REASONING, W1 = 0 J and Q2 = 0 J. Substituting these values and W2 = P2 ∆V (Equation 15.2), into Equation (2) yields ∆U total = ( Q1 − 0 J ) + ( 0 J − P2 ∆V ) = Q1 − P2 ∆V (3) Substituting the given values, we obtain the overall internal energy change of the gas: ( )( ) ∆U total = Q1 − P2 ∆V = +5470 J − 3.45 ×105 Pa −6.84 ×10−3 m3 = +7830 J 770 THERMODYNAMICS In obtaining this result, we have used the fact that the volume change ∆V is negative, because the volume of the gas decreases. 9. SSM REASONING According to Equation 15.2, W = P ∆V, the average pressure P of the expanding gas is equal to P = W / ∆V , where the work W done by the gas on the bullet can be found from the work-energy theorem (Equation 6.3). Assuming that the barrel of the gun is cylindrical with radius r, the volume of the barrel is equal to its length L multiplied by the area (π r2) of its cross section. Thus, the change in volume of the expanding gas is ∆V = Lπ r 2 . SOLUTION The work done by the gas on the bullet is given by Equation 6.3 as 2 2 W = 1 m(vfinal − vinitial ) = 1 (2.6 × 10 −3 kg)[(370 m/s) 2 − 0] = 180 J 2 2 The average pressure of the expanding gas is, therefore, P= W 180 J = = 1.2 ×107 Pa ∆V (0.61 m)π (2.8 ×10−3 m)2 10. REASONING Equation 15.2 indicates that work W done at a constant pressure P is given by W = P∆V. In this expression ∆V is the change in volume; ∆V = Vf – Vi, where Vf is the final volume and Vi is the initial volume. Thus, the change in volume is ∆V = W P (1) The pressure is known, and the work can be obtained from the first law of thermodynamics as W = Q – ∆U (see Equation 15.1). SOLUTION Substituting W = Q – ∆U into Equation (1) gives ∆V = W Q − ∆U ( +2780 J ) − ( +3990 J ) = = = −9.60 × 10 −3 m 3 P P 1.26 × 105 Pa Note that Q is positive (+2780 J) since the system gains heat; ∆U is also positive (+3990 J) since the internal energy of the system increases. The change ∆V in volume is negative, reflecting the fact that the final volume is less than the initial volume. Chapter 15 Problems 771 11. REASONING When a gas expands under isobaric conditions, its pressure remains constant. The work W done by the expanding gas is W = P (Vf − Vi), Equation 15.2, where P is the pressure and Vf and Vi 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 (Vf − Vi) for the final volume gives Vf = W 480 J + Vi = + 1.5 × 10−3 m3 = 4.5 × 10−3 m3 P 1.6 × 105 Pa 12. REASONING The work done in the process is equal to the "area" under the curved line between A and B in the drawing. From the graph, we find that there are about 78 "squares" under the curve. Each square has an "area" of ( 2.0 ×104 Pa ) ( 2.0 ×10−3 m3 ) = 4.0 ×101 J SOLUTION a. The work done in the process has a magnitude of ( ) W = ( 78 ) 4.0 ×101 J = 3100 J b. The final volume is smaller than the initial volume, so the gas is compressed. Therefore, work is done on the gas so the work is negative . 13. REASONING For segment AB, there is no work, since the volume is constant. For segment BC the process is isobaric and Equation 15.2 applies. For segment CA, the work can be obtained as the area under the line CA in the graph. 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 15.2 and the data in the drawing, we obtain ch = c.0 × 10 Pa hc.0 × 10 7 5 W = P V f −...
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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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