Ch 14.2 Help
2 posters
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Re: Ch 14.2 Help
Well can someone with the alternative question 2 post their question again with the answer given so that we can work on it?
Kathleen- Guest
question 2
the question that i posted is from maple ta not the cd so I dont have what the answer should be....
sorry...
sorry...
helpp- Guest
Re: Ch 14.2 Help
I'm sorry, I got the questions and answers from the website that engelmann posted up on blackboard in the beginning of the semester to allow us to do 'practice' questions. So far, the questions have been the same so I didn't realize that #2 was a different on this quiz this time. Although, I still don't know how to do question 2, I'll repost the correct question.
periwinkle- Posts : 22
Join date : 2008-09-17
Question #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 2.503 atm and volume VB is 299.1 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.
krolik- Guest
Question 2
If someone knows how to do this question please post an answer because many of us have this question and don't know how to do it. Thanks.
Jones- Guest
Re: Ch 14.2 Help
Are you able to post the correct answer? What formulas have you used so far that have proven to be incorrect/ineffective?
Guest01- Posts : 133
Join date : 2008-09-19
other question 2 answer
So I didn't have this question so I don't know if my logic is 100% accurate, but this is what I would do if I had this problem:
The question gives a hint: (Hint: What is the relationship between the work done and the area of the triangle shown?)
Since it is a closed cycle, the total change of the internal energy is zero. So by the first law of thermodynamics, Q=W.
The work done by the gas is equal to the area enclosed by the cycle from A-->B-->C-->A.
So Q=W=Area of triangle.
Area of a triangle=(1/2)*(base)*(height)
the base can be represented by the change in pressure from C-->A (press. of C=1atm, press. of A=2.503atm)
base=2.503atm-1atm=1.503atm*101300= 152253.9Pa
the height can be represented by the change in volume from B-->A (vol. of A=.000100m^3, vol. of B=.0002991m^3)
height=.0002991-.000100= 1.991E-4 m^3
W=A=(1/2)(152253.9)(1.991E-4)= 15.15687575J
convert to calories: 15.15687575J/4.187= 3.619984655calories
and it should be negative
Again, I'm not sure if this is the correct way to solve it but it's an idea.
The question gives a hint: (Hint: What is the relationship between the work done and the area of the triangle shown?)
Since it is a closed cycle, the total change of the internal energy is zero. So by the first law of thermodynamics, Q=W.
The work done by the gas is equal to the area enclosed by the cycle from A-->B-->C-->A.
So Q=W=Area of triangle.
Area of a triangle=(1/2)*(base)*(height)
the base can be represented by the change in pressure from C-->A (press. of C=1atm, press. of A=2.503atm)
base=2.503atm-1atm=1.503atm*101300= 152253.9Pa
the height can be represented by the change in volume from B-->A (vol. of A=.000100m^3, vol. of B=.0002991m^3)
height=.0002991-.000100= 1.991E-4 m^3
W=A=(1/2)(152253.9)(1.991E-4)= 15.15687575J
convert to calories: 15.15687575J/4.187= 3.619984655calories
and it should be negative
Again, I'm not sure if this is the correct way to solve it but it's an idea.
ryan240- Guest
question 2 is right!
ryan240 posted the EXACT right answer but it's POSTIVE.
it worked for my on mapla ta. thank you!!
thanks to everyone else too ^^
it worked for my on mapla ta. thank you!!
thanks to everyone else too ^^
luvely- Guest
# 4
Q_H = 19.37 Kj * 1000= 19370 J
e= 58.81%
Q_L= 19370 J / .5881 = 32936.57541
W= Q_H - Q_L
W= 19370 - 32936.57541 = - 113566.57541
e= 58.81%
Q_L= 19370 J / .5881 = 32936.57541
W= Q_H - Q_L
W= 19370 - 32936.57541 = - 113566.57541
lshenry- Guest
# 4
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: solution above
Answer: -13566.575412344839
How to solve: solution above
lshenry- Guest
Question 2
Question 2 now finally worked, so the posted above solution is correct. I wonder why it didn't work the other ten times I tried though. God I HATE Maple TA.
dl- Guest
Re: Ch 14.2 Help
Okay, so I figured out this site doesn't allow links. So this is the new exccel sheet for 14-2 for people that just want answers. DON'T CONVERT ANYTHING!!!!!!!!!!!!!!!!!!!!!!! The questions are on different sheets on the bottom.
http://www. sendspace. com/ file/ i7vsk4
There are 4 spaces (1 after each period and 1 after each backslash) in the link provided so cope and paste and delete the spaces and go download. Good luck on the final
BTW, 14-3 will be coming soon
http://www. sendspace. com/ file/ i7vsk4
There are 4 spaces (1 after each period and 1 after each backslash) in the link provided so cope and paste and delete the spaces and go download. Good luck on the final
BTW, 14-3 will be coming soon
The Show- Guest
different question2
RYAN u were right!!!! but it is positive and it took me four times to finally get it right. THANK U FOR ALL UR HELP!!!!
boo- Guest
How do you plug in answers
First off, thank you for the excel file. i have a huge final this week, and could really use the break from maple ta. that said, how do you plug in the variables?
S- Guest
Re: Ch 14.2 Help
I'm glad the excel is helpinng people First off, all you do is plug in the numbers that it asks for in the "Number" section. Stuff like "first number" means just type in the first number that was given in your problem. It could sometimes say "kilojoules" which means to plug in the number that has kilojoules as its units (I'll make sure there's only one number with that unit in the problem).
There are NO UNIT CONVERSIONS. Just plug in the numbers that the problem gives and the answer will be given in the "answer" section.
Good luck on your final This should alleviate some load off you guys.
There are NO UNIT CONVERSIONS. Just plug in the numbers that the problem gives and the answer will be given in the "answer" section.
Good luck on your final This should alleviate some load off you guys.
The Show- Guest
question 2
my question 2 does not give the Wbc. it just says pressure = 1am so shouldnt that be:
1*101300 = Pressure
Vb-Vc = value-(1x10^-1) = deltaV
and then DeltaV*101300?
For that i would get a positive value then add it to the negative value found for Wab.
Wca = 0
then convert to J by diving by 4.187
I don't know why i keep getting this wrong.
Answer's + right?
1*101300 = Pressure
Vb-Vc = value-(1x10^-1) = deltaV
and then DeltaV*101300?
For that i would get a positive value then add it to the negative value found for Wab.
Wca = 0
then convert to J by diving by 4.187
I don't know why i keep getting this wrong.
Answer's + right?
student- Guest
Re: Ch 14.2 Help
It worked. Thank you, again for all the help-- sometime the Physics teachers don't understand that we have other classes to juggle-- other, hard, science classes.
Thanks again.
Thanks again.
S- Guest
THANK YOU!
To: "The Show"
THANK YOU!
From: An appreciative student with a little less work to do
THANK YOU!
From: An appreciative student with a little less work to do
The Show- Guest
Clarification
Apparently "The Show's #1 Fan" doesn't fit as a user name...so instead of a thank you post from "The Show's #1 Fan", maple ta has a way of being a pain sometimes and it looks like The Show is a loser and is writing himself or herself thank you notes haha Well, just to clarify, that is not the case The Show is amazing and I just wanted to thank that person!!
thankful- Guest
question 2 reply
ryan 240 is 100% correct if you dont trust him u'll be screwed. Thank you and 2 is positive.
nightnig- Guest
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