Work done by system

[arlo]

I am confused with this concept and would like a clarification. I'm using Kaplan's book and this is how I understand this:

Change in internal energy = Q - W and W = work done by system

Kaplan says:
+ W = work done by system (expansion)
- W = work done on system (compression)

Normally, I would say work done BY system would be -W from the system's point. However, since the equation for change in internal energy is Q - W, the negative sign in front of W already indicates work done by system is negative so to plug in work done by system into that equation, we have to say work done by system value will be +W while work done on system will be -W?

This is really confusing so if someone could clarify this for me, I would really appreciate it.

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[aldol16]

Yeah, it's a really counterintuitive way to teach it, which is why I would recommend just adding U = q + w, where work done by system is negative. That simplifies all the signs. Whatever convention you want to use, just keep it consistent.

 

[NextStepTutor_2]

I am confused with this concept and would like a clarification. I'm using Kaplan's book and this is how I understand this:

Change in internal energy = Q - W and W = work done by system

Kaplan says:
+ W = work done by system (expansion)
- W = work done on system (compression)

Normally, I would say work done BY system would be -W from the system's point. However, since the equation for change in internal energy is Q - W, the negative sign in front of W already indicates work done by system is negative so to plug in work done by system into that equation, we have to say work done by system value will be +W while work done on system will be -W?

This is really confusing so if someone could clarify this for me, I would really appreciate it.

The equation U = Q-W is applicable ONLY if you are considering W to be work done by the system from the systems perspective. If there is + Work done by the system, that means the system is losing energy to its surroundings (hence the - sign in the equation) while if the surroundings do work on the system, you end up with Q - (-W) and thus U = Q + W since +Won = -Wby and this would mean energy is entering the system.

its all about the framce of reference YOU choose for your calculations. They system vs the surroundings.

Example 1: Work done ON the system > 0.

Here the first law is written as

U= Q+W


If your frame of reference is the "system", then the work done on the system (W) is positive and the heat that is added to the system is also positive, which means the change in internal energy is also positive by first law of thermodynamics, which means that there is an increase in temperature. This appeals to common sense. Here positive change in internal energy corresponds to increase in temperature

Example 2: Work done BY the system > 0

Here the first law is written as

U=Q−W


If work is applied to the system, the W term becomes negative making two negatives positive, which is identical to equation (1) and heat added to the system is still positive here. Rest of the arguments follow as above

By convention, generally equation (1) is used, but (2) is not wrong.

Hope this helps, good luck!