Understanding PV Diagrams: Analyzing Work, Heat, and Temperature Changes

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The discussion focuses on understanding PV diagrams and their implications for work, heat, and temperature changes in thermodynamic processes. It is established that the work done by the system is greatest along path 1 due to the area under the curve being largest. The relationship between internal energy change (ΔU) and heat (Q) is clarified, emphasizing that ΔU is the same for all paths, allowing for a comparison of Q without considering ΔU. The participants agree that as the temperature increases, the internal energy (U) also increases, which is supported by the PV relationship. The discussion concludes with the acknowledgment that isotherms on the graph indicate higher temperatures as they move upward.
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Homework Statement


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Homework Equations

The Attempt at a Solution


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I was wondering if the answer is all statements are true. I want to know if my reasons are correct:

1) the work done by the system is greatest along path 1 because work done is equal to the area under the path. The path under 1 is clearly largest of the two other paths.

2) if Ub > Ua , then Q is greatest along path b. Since the delta U for the other paths are equal to that of path 1, the equation becomes Q=W for all paths as delta U can be ignored in comparison. Q is largest for path 1 because W is largest.

3)Heat is absorbed by the system because Q is positive since deltaU is positive. When Q is positive, this means that heat is added to the system (same as heat aborbed by the system).
 

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It all looks good to me except for the first part of your statement 2 where you said
lc99 said:
2) if Ub > Ua , then Q is greatest along path b.
There is no path b. But the rest of your statement 2 appears to be correct. You are right that ΔU is the same for all paths, so you don't need to worry about ΔU when comparing paths for Q. But, still, I wouldn't write Q = W.
 
TSny said:
It all looks good to me except for the first part of your statement 2 where you said
There is no path b. But the rest of your statement 2 is right. Also, you are right that ΔU is the same for all paths, so you don't need to worry about ΔU when comparing paths for Q. But, still, I wouldn't write Q = W.
oops , you are right. i meant path 1. Also, I guess it was easier to understand it if i wrote Q=W.

Also, regardless of statement 2, it's true that change in U is the same for all of them, but I am unsure if change is U increases? I would think U increases. Is there any information that would say so?
 
lc99 said:
Also, regardless of statement 2, it's true that change in U is the same for all of them, but I am unsure if change is U increases? I would think U increases. Is there any information that would say so?
Consider the product PV for the initial and final states. Which state has the higher value for PV? What does that tell you about the temperatures for the initial and final states? For an ideal gas, how is U related to temperature?
 
TSny said:
Consider the product PV for the initial and final states. Which state has the higher value for PV? What does that tell you about the temperatures for the initial and final states? For an ideal gas, how is U related to temperature?
Ahh. That makes sense. U is higher in the final state because of the temperature increase due to PV relationship.

Also, is it also correct to say that temperature increases because isotherms tend to be up and into the graph which represent higher temperatures?
 
lc99 said:
Also, is it also correct to say that temperature increases because isotherms tend to be up and into the graph which represent higher temperatures?
Yes.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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