If energy increases inside a system, there must be an increase in its entropy?

Question

I was reading Frank Lambert's article Entropy is Simple if We Avoid the Briar Patches, which develops the idea of entropy as energy dispersal (and discourages entropy as disorder). On p13 is this paragraph:

Entropy increase without energy increase

Many everyday examples of entropy increase involve a simple energy increase in a particular 'system' (a part of the totality of 'system plus surroundings').. This energy increase is usually evident from a rise in temperature (caused by more rapidly moving molecules) in the system after some occurrence than before, e.g., when a pan or water in the pan is warmed or when a room is warmed, their entropy increases. Additional energy has been dispersed in them from some outside source, the 'surroundings'. The outside source is often combustion, the chemical reaction of petroleum products — natural gas or fuel oil — with oxygen to yield the lesser energetic carbon dioxide and water plus heat. The energy dispersed from the chemical bonds of gas or oil and oxygen is dispersed tothe slower moving molecules of the pan, the water, or the air of the room. If an energy increase occurs inside a system, there must be an entropy increase in it. More energy has been dispersed within the system and this is what entropy measures.

Is that statement in bold true in general? I know that

$$ \left ( \frac{\partial E}{\partial S} \right )_{V, N} = T > 0,$$

so that internal energy and entropy are positively correlated for changes at constant volume and number of molecules. But for an open system that can change volume and can also exchange other kinds of work with the surroundings, it seems like it wouldn't be true generally.

I realize this may be a somewhat informal article, and therefore the statement might not be intended too literally. I can't tell. So, is it actually possible for an energy increase to be accompanied by an entropy decrease in some thermodynamic processes?

Answer 1

For the closed system you are describing, in you reversibly compress and ideal gas isothermally, its internal energy stays constant and its entropy decreases. If you then just add a little heat at constant volume, the internal energy increases, while, provided not too much heat is added, its entropy increases just a little (but not to offset the decrease from the compression). So, the net effect is an increase in the internal energy of the gas and a decrease in its entropy.

Answer 2

Same answer as above but in equation form. Start with:

$$dU = TdS-PdV$$

Slowly compress, reversibly ($dV <0$): $$dS=\frac{P}{T}dV$$

In doing this we reduce the entropy.

Then add heat but don't change volume: $$dS=\frac{1}{T}dU$$ Which is clearly an increase in entropy.

As long as this system isn't brought back to the initial state, the entropy can be reduced while the energy has increased.

If we do bring it back the initial state, then we find that the change in entropy is zero. The integral of entropy along a closed path is zero (for reversible systems).

I would say, no, in general the statement in bold is not true.