576_PartUniversity Physics Solution

576_PartUniversity Physics Solution - The First Law of...

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The First Law of Thermodynamics 19-11 process db d b d b UQ W Δ= 0, db W = as already noted. 30.0 J 8.0 J 22.0 J. b d UU U = −= + Then 22.0 J db d b db QUW + =+ (positive; heat absorbed). EVALUATE: The signs of our calculated ad Q and db Q agree with the problem statement that heat is absorbed in these processes. 19.44. IDENTIFY: . UQW Δ=− SET UP: 0 W = when 0. V EXECUTE: For each process, . QU W =Δ + No work is done in the processes ab and dc , and so 450 J bc abc WW == and 120 J. ad adc The heat flow for each process is: for , 90 J. ab Q = For , bc 440 J 450 J 890 J. Q =+= For 180 J 120 J 300 J. ad Q For 350 J. dc Q = Heat is absorbed in each process. Note that the arrows representing the processes all point in the direction of increasing temperature (increasing U ). EVALUATE: U Δ is path independent so is the same for paths adc and abc . 300 J 350 J 650 J. adc Q 90 J 890 J 980 J. abc Q = Q and W are path dependent and are different for these two paths. 19.45. IDENTIFY: Use p Vn R T = to calculate / . ca TT Calculate U Δ and W and use Δ =− to obtain Q . SET UP: For path ac , the work done is the area under the line representing the process in the pV -diagram. EXECUTE: (a) 53 (1.0 10 J)(0.060 m ) 1.00. (3.0 10 J)(0.020 m ) cc c aa a Tp V V × = × . = (b) Since , = 0 U for process abc . For ab , 0 V Δ = and 0. ab W = For bc , p is constant and 3 (1.0 10 Pa)(0.040 m ) 4.0 10 J. bc Wp V =Δ= × = × Therefore, 3 abc W × Since 0, U 3 QW 3 4.0 10 J × of heat flows into the gas during process abc . (c) 55 3 3 1 2 (3.0 10 Pa 1.0 10 Pa)(0.040 m ) 8.0 10 J. W + × = + × 3 ac ac + × EVALUATE: The work done is path dependent and is greater for process ac than for process abc , even though the initial and final states are the same. 19.46. IDENTIFY: For a cycle, 0 U and . = Calculate W . SET UP: The magnitude of the work done by the gas during the cycle equals the area enclosed by the cycle in the pV -diagram. EXECUTE: (a) The cycle is sketched in Figure 19.46. (b) 44 3 3 (3.50 10 Pa 1.50 10 Pa)(0.0435 m 0.0280 m ) 310 J. W × = + More negative work is done for cd than positive work for ab and the net work is negative. 310 J. W = − (c) 310 J. Since 0, Q < the net heat flow is out of the gas. EVALUATE: During each constant pressure process V = Δ and during the constant volume process 0. W = Figure 19.46 19.47. IDENTIFY: Use the 1st law to relate tot Q to tot W for the cycle. Calculate ab W and bc W and use what we know about tot W to deduce ca W
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19-12 Chapter 19 (a) SET UP: We aren&t told whether the pressure increases or decreases in process bc . The two possibilities for the cycle are sketched in Figure 19.47. Figure 19.47 In cycle I, the total work is negative and in cycle II the total work is positive. For a cycle, 0, U Δ= so tot tot QW = The net heat flow for the cycle is out of the gas, so heat tot 0 Q < and tot 0. W < Sketch I is correct. (b) EXECUTE: tot tot 800 J WQ == tot ab bc ca WWWW =++ 0 bc W = since 0. V ab Wp V since p is constant. But since it is an ideal gas, p Vn RT Δ ( ) 1660 J ab b a Wn R T T =− = tot 800 J 1660 J 2460 J ca ab WWW =−= = EVALUATE: In process ca the volume decreases and the work W is negative.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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576_PartUniversity Physics Solution - The First Law of...

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