# What happens to energy changes in non-reversible processes?

Take internal energy for example: $$dU\leq TdS-PdV$$ Firstly, I would like to make clear that the following question/argument relies on the following statement being correct: the left hand side of the above equation represents the change in internal energy of the system and the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.

The two equations, above, are in fact inequalities where it only becomes an equality when the system is at equilibrium (when changes are reversible). However, when the system is not in equilibrium the inequality implies that the work done on the system by the surroundings is greater than the internal energy gained by the system; how can this be? Where does the energy difference between the energy put in and the energy gained by the system go?

Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?

$\Delta U = q + w$ and $dU = dq + dw$

This is the first law of thermodynamics. It is not an inequality.

$dU\leq TdS-PdV$ is correct when there is only PV work. It is incorrect when there is other work such as electrical or gravitational.

the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.

The PdV term alone represents the reversible work, the TdS term represents reversible heat

Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?

the system must lose internal energy in a spontaneous change

• The equality assumes equilibrium though doesn't it? This is certainly the case for G: $dG\leq VdP-SdT$. What happens in that instance? Jan 12, 2015 at 13:46
• $\Delta U = q + w$ and $dU = dq + dw$ do not assume equilibrium. Jan 12, 2015 at 13:53
• If you write $dU\leq TdS-PdV$ as an equality, that assumes quasistatic (near equilibrium) and only PV work. Jan 12, 2015 at 13:56
• Sorry, I see now I think. However, $dU \leq TdS-PdV$ is a valid inequality isn't it? Assuming equality at equilibrium? Jan 12, 2015 at 13:56
• I have edited my question now - I see my error. Jan 12, 2015 at 13:57