By or On the System? [ENDORSED]

[Alejandra_Vasquez_1B]

Hello, my question is how would I be able to identify whether the work was done by a system or on the system? Would I have to identify what type of reaction is occurring in order to determine what did the work?

[Annie_Amundson_1L]

Following up on that question, on page 8 of the course reader the notes say that one way to change the energy of a system is "Do work on the system (compress a piston) or let the system do work on the surroundings (piston moves out)". Can someone please explain to me what this means? How can a system do work, and how is work done on a system?

[Amber_Carlton_3D]

According to the first law of thermodynamics:
-The internal energy of an isolated system remains constant.
-Work that is done on the system is positive and is written out like this: change in U = q + w . This means that W, the system, is positive, as well as the heat added to the system, causing a positive internal energy and an increase in temperature.
-For work that is done by a system, w is subtracted from q, and the heat added to the system is still positive. The equation can now look like this: U = q - w .

For q(heat): If heat is transferred into a system, it is positive. If heat is transferred out of a system, it is negative.
For w(work): If work is done on a system, w is positive. If work is done by a system, w is negative.

Also, work done by a system is not a state function, while work done on a system is (internal energy). This means that work done by a system is dependent on how changes were brought about, while the path of work done on a system is only dependent on the current state of the system, not the path that it took to reach that change.

[Vandan_Patel_3L]

Work is either done ON the system by a force and thus the system would gain energy and if work is done BY the system it would lose energy and thus the internal energy of the system can be increased or decreased based on a net gain or net loss.

[Amy Ko 3C]

"Do work on the system (compress a piston) or let the system do work on the surroundings (piston moves out)"

This means that if we define work as W= -PΔV, we can figure out the sign of W to plug intp ΔU = Q + W
If the piston moves out, then there is an increase in volume. Thus, W = -PΔV and ΔV is positive. W is negative. The energy of the system has decreased by an amount of W. The work is done by the system.
If the piston is compressed, there is a decrease in volume. Thus, W = -PΔV and ΔV is negative. W is positive. The energy of the system has increased by an amount of W. The work is done on onto the system.