Question 8.13

Volume:
Temperature:

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Amy Zheng 2l
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

Question 8.13

Postby Amy Zheng 2l » Thu Feb 01, 2018 3:02 pm

whats the significance of saying " compressed suddenly (and irreversibly)". Why must they include irreversibly in the question?

Lucian1F
Posts: 87
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

Re: Question 8.13

Postby Lucian1F » Thu Feb 01, 2018 3:05 pm

If it was not irreversibly, some of the equations we use would not work because irreversible reactions occur at constant temperature but if it was not irreversible there would be a change in temperature

Kelly Kiremidjian 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: Question 8.13

Postby Kelly Kiremidjian 1C » Thu Feb 01, 2018 3:11 pm

In addition, when a system is irreversible, we can see it as expanding into a vacuum. This means that the ∆Ssurroundings will be equal to zero because there is no change of entropy in a pressure-less reservoir. However, this is very different from a reversible system, for which the ∆Stotal of the universe if actually equal to zero. All of this information is crucial to know so that you can do the proper calculations and make the proper assumptions in your reasoning.

Matthew Lin 2C
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

Re: Question 8.13

Postby Matthew Lin 2C » Fri Feb 02, 2018 2:01 pm

It is important that the question clarifies that it is an irreversible process because it tells us which equation to use. Because it is irreversible, we can use the equation w=-p(delta V), rather than the equation w=-nRTln(V2/V1) for reversible processes.


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