Negative work

[Henry_Phan_4L]

Can negative work be done on a system? Does increasing the volume count as negative work?

[Jonas Talandis]

Negative Work is definitely a thing. All it means is that the system is doing work on the surroundings instead of getting work done on itself. This happens when the piston moves out for example.

[MadelynNguyen1F]

Yes, negative work is when work leaves the system.

[Mukil_Pari_2I]

Yes, if the force being applied is in the opposite direction of the displacement, the work will be negative.

[davidryan3f]

Yes negative work will occur when the system does work on its surroundings rather than the surroundings doing work oil the system.

[Tinisha 1G]

Negative work is represented by a system doing work on its surroundings. When work is done on system, it is positive work.

[Tony Chung 2I]

yes, it is a thing. This will happen when system does work on the surrounding.

[Jasmine Chow 1F]

Yes you can have negative work on a system. An example of this would be the system doing work on its surroundings. Increasing the volume does count as negative work by the system.

[EllaBerry]

Yes, negative work is done when a system does work on its surroundings.

[madisondesilva1c]

If the work is negative, how would the isobaric work equation change? Would the integral now be of a positive quantity instead of a negative one?

[Vincent Li 4L]

If the work is negative, how would the isobaric work equation change? Would the integral now be of a positive quantity instead of a negative one?

For an isobaric expansion/compression, the pressure is constant (isobaric: deltaP = 0), so the we would use the equation w = -P*deltaV. If the change in volume is positive, then the total value will be negative. If we're talking in terms of the integral, you are correct in saying it would then be a positive quantity. Since the equation has ln(Vf/Vi), where Vf is final volume and Vi is initial volume, this means that any time a gas expansion occurs, the final volume will be greater than the initial volume. This makes the natural log of their ratio a positive value, and because there is also a negative sign in the equation, work will then be negative.

[Megan_Ervin_1F]

If the work is negative that means the system DID the work so it lost the energy as work

if the work was positive that means that work was done ON the system so it gained energy as work