$$\delta S\ge\frac{\delta q}T$$

I don't understand how a quantity (the change in entropy in this case) can be greater than the very thing that defines it ($\mathrm{\frac{\delta q}{T}}$).

Also what is the significance of this inequality?


Your definition of entropy is incorrect. The significance of the Clausius inequality is that it shows that the definition of entropy, i.e. $\mathrm{\delta S=\cfrac{\delta q_{rev}}{T}}$ (note that entropy change is defined for a reversible process) is consistent with observed reality: the entropy of an isolated system does not decrease spontaneously.

We know that $$\mathrm{\delta w \geq \delta w_{rev} \Rightarrow \delta w-\delta w_{rev}\geq 0}$$

and according to the First Law of Thermodynamics

$$\mathrm{\delta U = \delta q + \delta w=\delta q_{rev}+\delta w_{rev}}$$

Therefore $$\mathrm{\delta q_{rev}-\delta q=\delta w-\delta w_{rev} \geq 0}$$

Now using the definition of entropy yields

$$\mathrm{\delta S=\cfrac{\delta q_{rev}}{T} \geq \cfrac{\delta q}{T}}$$

Taking $\mathrm{\delta q=0}$ (since we have an isolated system no heat can flow into it), produces

$$\mathrm{\cfrac{\delta q_{rev}} {T} \geq 0 \Rightarrow \delta S \geq 0}$$

  • 3
    $\begingroup$ this just derives it, not answers the OP's question $\endgroup$ – RE60K Nov 20 '14 at 11:50
  • 1
    $\begingroup$ why is dw(rev) less than dw? $\endgroup$ – RobChem Nov 22 '14 at 17:21
  • $\begingroup$ Also how does this tell you the direction of spontaneous change? $\endgroup$ – RobChem Nov 22 '14 at 17:29

I don't understand how a quantity (the change in entropy in this case) can be greater than the very thing that defines it.

Actually it is not greater, but greater than or equal to:

$$\delta S\ge\frac{\delta q}T$$

If heat is provide irreversibly, the term on LHS is greater otherwise in case of reversible process it is equal.

Also what is the significance of this inequality?

This inequality when rearranged becomes: $$TdS-dq\ge0$$ For constant volume or pressure it becomes(recall $q_v=U,q_p=H$): $$TdS-U\ge0\\TdS-H\ge0$$ both of which say the common equations,(for a spontaneous change): $$dA=U-TdS\le0\\dG=H-TdS\le0$$


The reason that there is so much confusion over this is that the usual form of the Clausius inequality appearing in most references is not completely and properly specified. In an irreversible process, the temperature within the system is not going to be spatially uniform, so there is considerable ambiguity as to what value of the temperature T to use in calculating the integral of dq/T. The correct value to use is the temperature at the interface between the system and surroundings (at which the heat transfer is occurring), $T_I$. So the correct form of the Clausius inequality should more properly read: $$\Delta S \ge \int {\frac{dq}{T_I}}$$ So, for example, if the surroundings consists of a constant temperature bath, the T in the Clausius inequality is the temperature of the bath (i.e., $T_I=T_{bath}$).

If the path between the initial and final equilibrium states of the system is reversible, then the system temperature is uniform throughout and equal to the interface temperature, such that $T_I=T$, and the equality in the Clausius relationship applies. The integral then gives the change in entropy. For any irreversible path between the same two thermodynamic equilibrium states, the value of the integral is less than the integral for a reversible path, $\Delta S$.

  • $\begingroup$ Are all these inequality defined for irreversible process... How to cslculate if temperature is also changing $\endgroup$ – Scáthach Sep 15 '18 at 6:28
  • $\begingroup$ @harambe Are you asking how to calculate the entropy change for an irreversible process, or are you asking how to evaluate the integral on the right hand side of the equation for an irreversible process? $\endgroup$ – Chet Miller Sep 15 '18 at 11:25
  • $\begingroup$ The latter... How to evaluate for irreversible process where there is no thermodynamic equilibrium.. And the system is not in equilibrium with itself so it's temperature is changing $\endgroup$ – Scáthach Sep 15 '18 at 11:28
  • $\begingroup$ @harambe In that case, for most situations, you would have to solve the transient mass-, momentum-, and energy balance partial differential equations for the system, which include heat conduction and viscous momentum transport, to establish the variation in local heat flux and temperature at the boundary; and you would have to integrate this over the area of the interface and over time. $\endgroup$ – Chet Miller Sep 15 '18 at 11:35

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