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2 Thermodynamic questions 
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#1
Mar2508, 03:08 AM

PF Gold
P: 824

1. The problem statement, all variables and given/known data
1. a) Determine the work done on a fluid that expands from i to f as indicated in picture below. b) How much work is performed on the fluid if it is compressed from f to i along the same path? 2. An ideal gas initially at Pi, Vi and Ti is taken through a cycle as shown. a) find the net work done on the gas per cycle b) What is the net energy added by heat to the system per cyle c) Obtain a numerical value for the net work done per cycle for 1.00 mol of gas initially at 0C 2. Relevant equations [tex]\Delta E_{int}= Q+ W [/tex] 3. The attempt at a solution 1. a)Determine the work done on a fluid I thought that work was the area under the curve...but acording to my book for constant pressure.. I think it is [tex]P_i(V_fV_i) [/tex] Now I'm thinking that If I find the are under the curve then it would be=> (41m^3)(2x10^6Pa)+ (21m^3)(6x10^62x10^6)+ 1/2(32m^3)(6x10^62x10^6)= (6x10^6)+(4x10^6)+ (2x10^6)= 1.2x10^7 J [B]b) How much work is performed on the fluid if it is compressed from f to i along the same path? here I'm not sure what they want, when they say that the fluid is compressed along the same path. I assume that would mean that the pressure changes? But how would that look on the graph? Am I supposed to draw a new graph? _______________________________________________________________________ __ 2. a) net work done on the gas per cycle I would say work done is = area of the cyclic cycle thus: (3PiPi)(3ViVi)= 4J b) What is the net energy added by heat to the system per cyle if it is cyclic I think that the net energy added by heat would be = Q= W because the [tex]E_{int}= 0[/tex] thus would it would be 4J ? c) Obtain a numerical value for the net work done per cycle for 1.00 mol of gas initially at 0C Not sure how I would do this if they say that there is 1.00mol of gas at 0C. I know that [tex]Q=nc\Delta T[/tex] but if you don't have the final temp then how do you do the question, and also if you don't have c? I need help on kowing whether I'm correct or not and need help on parts c for #2 and also on part b) of #1. Please help me out. Thank you. 


#2
Mar2508, 08:14 AM

PF Gold
P: 824

um..does anyone know how to do thermodynamic Q's?
not so sure b/c I haven't seen alot of people ask them... Well If anyone is capable and willing to help me out I'd greatly appreciate it 


#3
Mar2508, 11:50 AM

Sci Advisor
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P: 6,677

_______________________________________________________________________ __ When does heat flow into the system? (ie. does heat flow into or out of the system in each of AB, BC, CD, DA?). How much heat flows into/out of in each of these segments (use the first law: dQ = dU + W = dU + PdV ). Add them up (using correct signs) and that gives you the net heat flow. AM 


#4
Mar2508, 12:48 PM

PF Gold
P: 824

2 Thermodynamic questions
_______________________________________________________________________ __ WAB and WBC is = 0 then WBC = 3Pi(3ViVi) and the same is for the WAD just negative then wouldn't it = 0?? AM[/QUOTE] Thank You 


#5
Mar2508, 03:28 PM

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In order to determine the heat flow from B to C, you have to determine the change in temperature using the Ideal gas law and determine the change in internal energy. The heat flow is the change in internal energy + work done (or you can use dQ = nCpdT since pressure is constant) Repeat this for CD and DA (is the heat flow positive or negative?) and add them all together (careful with the signs) to get the total net heat flow. AM 


#6
Mar2508, 06:00 PM

PF Gold
P: 824

For AB There is no work done for when there is no change in volume so W= 0 and [tex]dU= dQ  dW[/tex] W= 0 so [tex]dU= dQ[/tex] and I know that [tex]\Delta U = nC_v\Delta T[/tex] and [tex]PV= nRT [/tex] and so [tex]\DeltaT= \Delta(PV)/nR [/tex] and since V doesn't change but P does then: [tex]\DeltaT= \Delta P V_i /nR[/tex] BC=> pressure constant [tex]\Delta U=QW[/tex] [tex]Q= nC_p \Delta T[/tex] [tex]PV= nRT [/tex] and so [tex]\DeltaT= \Delta(PV)/nR [/tex] CD is the same as AB except [tex]dU= dQ[/tex] [tex]\Delta U = nC_v\Delta T[/tex] and [tex]PV= nRT [/tex] and so [tex]\DeltaT= \Delta(PV)/nR [/tex] [tex]\Delta T= \Delta P 3V_i /nR[/tex] DA is same as BC except that the volume decreases... [tex]Q= nC_p \Delta T[/tex] [tex]PV= nRT [/tex] and so [tex]\DeltaT= \Delta(PV)/nR [/tex] I assume that I substitute the pressure into the equations and then add them..but I'm not sure about the signs...how would I know if it si +/? 


#7
Mar2608, 07:00 AM

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AM 


#8
Mar2608, 07:22 AM

PF Gold
P: 824

and I thought I had to multiply to get the net work like I did in part A just this=> [tex]nC_v \Delta(P V_i)/nR + nC_p \Delta(PV)/nR  nC_v \Delta (PV_i) nCp \Delta(PV)/nR [/tex][ 


#9
Mar2608, 08:34 AM

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The answer can be reduced: [tex]C_v 2P_iV_i/R + C_p6P_iV_i/R  C_v6P_iV_i  C_p2P_iV_i/R = 4P_iV_i(C_p  C_v) /R [/tex] What is CpCv for any gas? AM 


#10
Mar2608, 10:00 AM

PF Gold
P: 824

but that would be their definition of a numercal value? You don't have to use the given temperature of 0 C ? and 1mol ? so it becomes [tex]4P_iV_i[/tex] 


#11
Mar2608, 11:23 AM

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AM 


#12
Mar2608, 11:51 AM

PF Gold
P: 824

PV= nRT PV= 1(8.31J/K*mol)*273K= 2432.43 4(PV)= 9729.72 J Thank You 


#13
Mar2608, 12:22 PM

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Part b), the net heat energy added to the system is a not clear (the word "net" is confusing). If it refers to Qh  the heat flow into the system, it is asking for the heat flow in parts AB and BC only (heat flows out to the cold reservoir in CD and DA). In order to provide that as a numerical value you will need to know Cv. AM 


#14
Mar2808, 06:01 PM

PF Gold
P: 824

Thanks 


#15
Mar3008, 12:13 AM

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For AB the heat in is Cv(2Pi)Vi/R. For BC, Qh is Cp(2Vi)(3Pi)/R. So the total is: [tex]Q_{hAC} = P_iV_i(2C_v + 6C_p)/R = P_iV_i(2C_v + 6(C_v +R))/R = P_iV_i(8C_v + 6)/R[/tex] AM 


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