Question

a system that does work to push back a piston but transfers no heat to the surroundings has
○ w < 0, q = 0, δe < 0
○ w > 0, q < 0, δe = 0
○ w < 0, q > 0, δe = 0
○ w > 0, q = 0, δe > 0

Explanation:

Step1: Recall First Law of Thermodynamics

The first law of thermodynamics is \(\Delta E = q + w\), where \(\Delta E\) is the change in internal energy, \(q\) is heat transferred, and \(w\) is work done.

Step2: Analyze Heat Transfer (\(q\))

The system transfers no heat to surroundings, so \(q = 0\) (since heat transfer out would be \(q < 0\), but here \(q = 0\) for no transfer).

Step3: Analyze Work Done (\(w\))

The system does work to push back a piston: work done by the system means \(w < 0\) (convention: \(w\) is positive when work is done on the system, negative when done by the system).

Step4: Calculate \(\Delta E\)

Substitute \(q = 0\) and \(w < 0\) into \(\Delta E = q + w\): \(\Delta E = 0 + w\), so \(\Delta E < 0\) (since \(w < 0\)).

Answer:

The correct option is: \(w < 0, q = 0, \Delta E < 0\) (the first option in the list, assuming the first option is labeled with this description).