Thermodynamics - change in internal energy of the system for the path

AI Thread Summary
The discussion revolves around calculating the change in internal energy of an ideal gas system during various thermodynamic paths. The first part establishes that with 200J of heat added and 80J of work done by the system, the change in internal energy can be determined using the equation ΔQ = ΔU + ΔW. Participants also explore the implications of different paths, noting that work is done by the system in one scenario and questioning the reasoning behind the chosen equations. The importance of providing clear reasoning and understanding the physics behind the calculations is emphasized for better comprehension and grading. Overall, the conversation highlights the complexities of thermodynamic processes and the need for thorough explanations in problem-solving.
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Thermodynamics -- change in internal energy of the system for the path

Homework Statement


A system comprising of an ideal gas is taken from State a to State b, as shown in Figure 9 below, along th epath acb. 200J of heat flows into the system and 80J of work is done by the system along the path acb.

1) Find the change in internal energy of the system for the path acb.

2)Suppose the system were taken from State a to State b along the path adb instead. find the work done if 140J of heat flows into the system along this path. Is work done on or by the system

3)If the work done on the gas is 50J for the curved path ba, what is the heat flowing into or out of the system for this path?

4)If the internal energy of the gas at State a is 40J, what is the internal energy of the gas

5) If the internal energy of the gas is 90J at State d, what is the value of the heat flowing into the system along the path ad?

Homework Equations


ΔQ = ΔU+ΔW

The Attempt at a Solution


In the posted picture

For Part 2...work is done by the system.
 

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Great!
Was there a question in all that?

For those who don't want to download the images:

Path AC is constant volume PA to PC
Path CB is constant pressure from VA=VC to VB > VA
Path BA is a concave-down curve... looks like an arc of a circle.
 
Woa so everything's good then?? :D


I also tried another one and this was my attempt at it:
 

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Woa so everything's good then?? :D
I don't know - is it? It's your post.
Was there something about it that you were unsure of?
Do you have some reason to doubt what you did?

Did you want me to check your work?

What I can do is check you reasoning and see if anything looks like it went awry from the POV of someone maybe marking it.
 
I wasn't sure if I worked it correctly...I won't mind if you checked my reasoning to see if everything's correct though.

Thanks :)
 
... only you don't show any reasoning.
It looks lie you are just plugging numbers into equations.

i.e. I see you have: ##\Delta Q=\Delta U + \Delta W## ... why did you choose that one?
http://en.wikipedia.org/wiki/Internal_energy#Internal_energy_changes
... it helps you get good marks if you write a short comment about the physics you are using.
i.e. "work done by the system lowers internal energy, heat flow into the system increases it."
 
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