Clausius inequality and irreversible heat transfer

[Hobold]

I don't seem to understand Clausius inequality at all. Really. It was deduced to me that the Clausius inequality is given by

$$dS = \frac{\delta Q_i}{T} > 0$$

where Q_i is the irreversible heat transferred to a system. Though I cannot find a way to prove an assertion my teacher said: through Clausius inequality, the irreversible heat to be transferred is lower or equal than through a reversible process.

It really doesn't make any sense to me, can anyone explain?

 

[Gordianus]

The quantity dS is defined as the ratio of the reversibly exchanged heat to the temperature. Just the opposite way you defined it.

 

[Hobold]

Yeah, that's right, I'm sorry. I meant by Clausius inequality

$$\frac{Q_i}{T} > 0$$

 

[Gordianus]

Clausius's theorem applies to a closed evolution. Just writing Qi/T>0 is simply wrong. Consider an irreversible engine and focus on the released heat. You easily obtain Qi/T<0. However, if you take into account the complete cycle you get sum (Q/T)<0 (I'm sorry, I know nothing about latex)

 

[Andrew Mason]

I don't seem to understand Clausius inequality at all. Really. It was deduced to me that the Clausius inequality is given by

$$dS = \frac{\delta Q_i}{T} > 0$$

This is incorrect. This is not the definition of change in entropy. The change in entropy is uses the reversible heat flow:

$$dS = \frac{\delta Q_{rev}}{T}$$

The change in entropy referred to in the Clausius inequality is the total change in entropy of the system and surroundings during a process. You must use the reversible heat flow for the system and surroundings.

Where the process is irreversible, the total change in entropy will be greater than 0. In order to do the calculation, you must determine the integral of dS for the system on the reversible path between the initial and final states of the system. Then you must do the same for the surroundings.

AM