2003年10月7日
UW EPost for CHEM 452
Date: 10/06/2003, 23:32
Author: Dah-Jin Ma
Subject: A quick Question regarding "work"
According to today's lecture, Prof. Keller says that "work" is not a "state function", which means the process should be taken into consideration.
In pressure-volume work lecture, Prof. Keller also demonstrated how we get "zero work is done" by doing a process in a cycle.
I'm wondering is there any relationship between these 2 concepts? I have a hard time understanding the connection here, cuz apparently we don't take the "process" part into consideration when calculating the 2nd example. (the cycle one)
Or, is it because we set the condition in the 2nd example as a "reversible reaction"!?
Thank you for the help!!
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Date: 10/07/2003, 10:32
Author: Sarah L. Keller
Subject: RE: A quick Question regarding "work"
Hi Dah-Jin,
Good question.
Let's think about the cycle first (p 13 of the notes). If there is a reversible compression followed by a reversible expansion, then work total = 69.3J - 69.3J = 0. However, this total does not end up being zero if the cycle is irreversible. Using the numbers from p11-12, if there is an irreversible compression followed by an irreversible expansion, then work total = 100J - 50J = 50J. So, the work expended definitely depended on the process. Work is not a state function.
In contrast, for a state function, we would always get no change in the value if we returned to the beginning of a cycle. We will even see this later in class. The specific example will be that there is no change in E (internal energy, which is a state function) over a particular cycle called a Carnot cycle.
posted by Biochemie on 7:03 下午
2 comments
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