isothermal expansion

[chiddler]

answer is A.

I don't understand why. For any two processes that lead to the same point on such a graph, both their final energies are the same regardless of how they get there, right? It's a state function. So one process can have a lot of work done with one path or very little work done in another leading. Both can lead to the same endpoint.

Since internal energy is the same for any endpoints, we can have

U = Q + (a lot of work done) = a small Q
(such as with path 1)
or

U = Q + (a little bit of work done) = a bigger Q
(such as with path 3)
With this, I say that it depends on the path if work is > or < heat. Why am I wrong?

---

Members don't see this ad.

 

[Baller MD]

If work done by the system or energy output = work done on the system or energy input, volume would increase while pressure is increasing proportionally.

But this is clearly not displayed by the graphs. All three graphs suggests that the system is losing energy to the external environment without any (or very little) energy input.

That's why more work is done by the system than heat flowing in.

That's how I think of it, anyway...

 

[chiddler]

thanks, but i don't understand.

can i have another explanation, please?

 

[MedPR]

thanks, but i don't understand.

can i have another explanation, please?

I started to explain it earlier, but then I realized I couldn't 😀

Where did you find this question? Maybe if I read the explanation I can help. I don't have my physics 2 book with me right now though.

 

[chiddler]

http://www.wikipremed.com/course_videos.php?syl=06&video_code=010303_02

somewhere towards the end. thanks for offering help.