Ch 14.2 Help

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Ch 14.2 Help

Post  periwinkle on Mon Dec 01, 2008 2:04 am

1) A gas does 359.4 J of work at constant temperature ("isothermal"). How much heat is exchanged with the environment (see sheet 8,12') ? Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.
Answer: 359.4
How to solve:

2) An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 1.575 atm and volume VB is 366.3 cm3. The pressure at C is 1 atmosphere. How many calories are exchanged with the environment? (Hint: What is the relationship between the work done and the area of the triangle shown?) Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.
Answer:
How to solve:

3) A heat engine withdraws heat at the rate of 22.72 W from a heat reservoir. What is the power output of the engine if 11.044 W of heat are released into the heat sink (see sheet 16,21) ? Indicate with a positive (negative) sign whether the input heat is always smaller (larger) than the output heat.
Answer: -11.675999999999998
How to solve:

4) A heat engine converts 58.81 % of the input heat into mechanical work. How much heat is released into the environment when the engine does 19.37 kJ of work (see sheet 16,17) ? Indicate with a negative (positive) sign whether the the efficiency of the engine is always (not always) less than 1.
Answer: -13566.575412344839
How to solve:


Last edited by periwinkle on Mon Dec 08, 2008 2:47 am; edited 1 time in total

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Question 1

Post  Kathleen on Wed Dec 03, 2008 2:16 am

Question 1
The answer is obvious from the post above but I'll explain why,
According to sheet 12 of the notes, an ideal gas expands at a constant temperature, and Delta U:T = constant..so if Delta T = 0, then DeltaU = 0...
So, the amount of work done is equal to the amount of heat exchanged

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Question 3 and 4

Post  sedwards on Wed Dec 03, 2008 11:56 am

Anyone know how to do question 3 and 4?

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question 2

Post  hwilson on Thu Dec 04, 2008 12:10 pm

Any ideas on 2?

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Question 3

Post  Kathleen on Thu Dec 04, 2008 2:01 pm

A heat engine withdraws heat at the rate of 22.72 W from a heat reservoir. What is the power output of the engine if 11.044 W of heat are released into the heat sink (see sheet 16,21) ? Indicate with a positive (negative) sign whether the input heat is always smaller (larger) than the output heat.
Answer: -11.675999999999998

So I don't know if this is right but I used the formula from sheet 16 which is W = Qh-QL
With Qh being the input heat and QL being the output heat released into the heat sink
And I basically just did 22.72-11.044 = 11.676
And the answer is negative

Edit: I just did that with my question on maple TA and it was marked correct so that should work

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questions 2 & 4

Post  angelbab on Thu Dec 04, 2008 2:14 pm

I really need help with 2 and 4 can someone pls help?

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Re: Ch 14.2 Help

Post  Guest01 on Fri Dec 05, 2008 1:10 am

I can't really figure out Question 2 (or Question 4)

For Question 2 I'm not sure if what I'm doing is anywhere near correct, here's what I've done so far:

(1.504*101300)*(372.3/1000000) = 56.72184096
146.2 - 56.72184096 = 89.47815904
89.47815904/4.187 = 21.37047027

Any ideas? I haven't really worked on Question 4 yet.

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Question 4

Post  Kathleen on Fri Dec 05, 2008 1:12 am

4) A heat engine converts 58.81 % of the input heat into mechanical work. How much heat is released into the environment when the engine does 19.37 kJ of work (see sheet 16,17) ? Indicate with a negative (positive) sign whether the the efficiency of the engine is always (not always) less than 1.
Answer: -13566.575412344839

From Sheet 17:
e = W/Qh
e = efficiency = 58.81/100 = .5881
W = work = 19.37kj*1000 = 19370 J
Qh = input heat

.5881 = 19370/Qh
Solve and you find Qh = 32936.57541

Then do W = Qh - QL
Where QL is the output heat
19370 J = 32926.57541 - Qh
Qh = 32926.57541 - 19370
Qh = 13566.57541

Answer is negative because the efficiency which is usually a percent is less than 1 when put back into decimal form

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Question 2

Post  Kathleen on Fri Dec 05, 2008 1:38 am

I can't seem to get question 2 either yet but this is the closest ive gotten

So, I figured you would first have to determine the amount of work done from A->B and then B->C and then add it together.

Using the info, and after converting to SI units you could do

For A->B
W=P*Change in Volume
W=152355.2*(2.722e-4) <- i found that by doing the volume of B - volume of C (100cm^3)
W= 41.4756/4.187 = 9.905

And then it says the work for B->C = 146.2/4.187 = 34.917

and then...if you subtracted the two, which I don't know why you would but if you did you got something around 25. So maybe the problem deals with calculating the work done from each point to point and adding it together?

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Re: Ch 14.2 Help

Post  Guest01 on Fri Dec 05, 2008 2:21 am

How do you know the volume for C is 100 cm3?

Well the energy given is a combined energy, so I'm assuming whatever is missing we need to compensate for, I'm guessing that's why there's an addition to that? But wasn't that compensated for previously? Shouldn't we be subtracting then?

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Re: Ch 14.2 Help

Post  Kathleen on Fri Dec 05, 2008 2:30 am

if you look at the actual picture of it on maple Ta, on the graph it says volume is on x axis and it says 100 where the C is so i assumed that meant it was 100cm^3, so the only change from C->A would be in pressure

It says in the hint to account for the changes from A->b, B->C, and C->A, and you can find the change from A->B, its given for B->c, and I don't know what it would be for C->a because the volume change is 0, so multiplying any pressure by that is 0, which would make work zero (in the sense that W=p*change in volume and if W=P*0 then W=0) but that doesn't seem to quite work. From my other method my value ended up being about 25.01 which was sort of like 25.009 or whatever the value was rounded up but when i tried that method on maple ta for my own numbers it didnt quite work out

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Question 2

Post  sedwards on Fri Dec 05, 2008 11:30 am

I am having the same problem with question number 2 I can't figure it out. If someone knows how to do it please help I am sure all of us would really appreciate it. Thank you.

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Question 2

Post  Jones on Sat Dec 06, 2008 11:39 am

I also need help with number 2 if anyone could help I would really appreciate it I just cant figure it out.

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question 2 answer

Post  ryan240 on Sun Dec 07, 2008 12:35 am

An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 2.492 atm and volume VB is 268.5 cm3. How many calories are exchanged with the environment if 140.7 J of work is done on the gas during the trip B-C (see sheet 7,Cool? (Hint: The heat exchange Q asked for is for the complete "trip" from A to B to C to A. What does the gas equation tell you about the temperature change, and hence about the change in internal energy, for the complete trip, when the gas returns to the same PV-pair in point A (see sheet 3,5))? When calculating the work you must split the complete trip into 3 pieces, WAB+ WBC+WCA.)


Wtotal=WAB+WBC+WCA

WAB=P*deltaV
P=2.492atm*101300=252439.6Pa
At A, V=100cm^3=.000100m^3
At B, V=268.5cm^3=.0002685m^3
so deltaV=.000100-.0002685=-1.685E-4
WAB=(252439.6)(-1.685E-4)=-42.5360726J <--leave this as is, don't convert to calories just yet

WBC is given as 140.7J

WCA=0J because from point C to point A there is no change in volume, it's constant at 100cm^3
and W=P*deltaV so W=P*0=0

Wtotal=-42.5360726J + 140.7J +0J=98.1639274J
convert to calories:98.1639274J/4.187=23.44493131calories

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question 2 answer

Post  ryan240 on Sun Dec 07, 2008 12:40 am

^it should be Qtotal=WAB+WBC+WCA.. NOT Wtotal
my bad.

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due date?

Post  guesss on Sun Dec 07, 2008 1:29 am

when is 14.2 due? monday? or was it due already on friday?

thanksss santa

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Re: Ch 14.2 Help

Post  Kathleen on Sun Dec 07, 2008 1:45 am

It is due on this coming Wednesday actually

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Question 2

Post  Kathleen on Sun Dec 07, 2008 2:04 am

Just a note, The answer for Question 2 is POSITIVE, because on Maple TA it doesnt ask to indicate with a negative/positive sign or anything. Just letting you know..

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question 2

Post  helpp on Sun Dec 07, 2008 4:06 pm

My question 2 is different from the one posted, this is how it reads....

An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 2.786 atm and volume VB is 233.5 cm3. The pressure at C is 1 atmosphere. How many calories are exchanged with the environment? (Hint: What is the relationship between the work done and the area of the triangle shown?) Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.

how do you solve for this question?

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Different Question 2

Post  hwilson on Sun Dec 07, 2008 6:05 pm

I have the same question 2 that you do and I do not know how to figure it out does anyone have any ideas?

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Re: Ch 14.2 Help

Post  imaG on Sun Dec 07, 2008 7:47 pm

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Re: Ch 14.2 Help

Post  Kathleen on Sun Dec 07, 2008 9:17 pm

That's weird that you guys have a different question 2.

I think what you have to do is the same concept as posted above, because since W=p*change in V, even though they now give you a pressure for C, which changes from C to A, the volume still does not change (so I believe, I'm not sure if your picture looks different also). This means that the W will still equal 0. Try using the above formula and ignoring the additional information about C. If this doesnt work then I'll try thinking of something else.

An ideal gas changes its pressure and volume as shown in the PV-diagram starting at point A and returning to it. The pressure PAB is 2.786 atm and volume VB is 233.5 cm3. The pressure at C is 1 atmosphere. How many calories are exchanged with the environment? (Hint: What is the relationship between the work done and the area of the triangle shown?) Indicate with a positive (negative) sign whether heat is put into (taken out of) the gas.

I think you would still do 2.786*101300 = __
233.5 / 1000000 = 1e-4(volume at point A)-2.335e-4 = -1.335e-4
W(AB)= (2.786*101300)*(-1.335e-4) = -#
And then find the Work from B->C
Volume change would be (2.335e-4-1e-4) = 1.335e-4
And then I don't know if you would do a pressure change (as in PB-PC) or just do the 1atm*101300
but it would be Work(BC) = Pressure * 1.335e-4 = ___
and then add those two answers together and divide by 4.187 to put into calories

I don't have a calculator on me so I can't work out the problem right now but I think that should work

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Question 2

Post  sedwards on Sun Dec 07, 2008 10:08 pm

I tried that method for question 2 its not working.

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Re: Ch 14.2 Help

Post  Lali on Sun Dec 07, 2008 11:05 pm

Yeah, question 2 on maple TA reads different for me as well

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Re: Ch 14.2 Help

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Re: Ch 14.2 Help

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