📂Thermal Physics

Clausius Inequality

Theorem

In a cyclic process the following inequality holds. $$ \oint {{\delta Q} \over {T}} \le 0 $$

In particular, for a reversible process the following holds.

$$ \oint {{\delta Q_{\text{rev}}} \over {T}} = \oint dS = 0 $$

Here $\delta$ is the inexact differential, $T$ is temperature, and $S$ is entropy.

Explanation

A cyclic process is one in which the state of the system is the same at the beginning and at the end of the process. If the whole cycle is reversible, its closed integral is always $0$, and one writes expressions like $Q := Q_{\text{rev}}$.

Proof ¹

Reversible system

Heat and temperature in a Carnot engine: $$ \frac{Q_{h}}{Q_{l}} = \frac{T_{h}}{T_{l}} $$

In a Carnot engine, during one cycle heat $Q_{h}$ is absorbed and heat $Q_{l}$ is rejected; thus heat by itself is not a conserved quantity in a cyclic process. However, considered together with temperature, $$ \frac{Q_{h}}{Q_{l}} = \frac{T_{h}}{T_{l}} \implies \frac{Q_{h}}{T_{h}} = \frac{Q_{l}}{T_{l}} $$ so for a reversible system $$ \sum_{\text{Cycle}} {\frac{ \Delta Q_{\text{rev}} }{ T }} = {\frac{ Q_{h} }{ T_{h} }} + {\frac{ - Q_{l} }{ T_{l} }} = 0 $$ holds. Converting this into integral form yields: $$ \oint {{\delta Q_{\text{rev}}} \over {T}} = 0 $$

Irreversible system

First, consider dividing the cycle into a finite number of points. Suppose certain parts of the cyclic process are connected to reservoirs at temperature $T_{k}$ and that heat $\delta Q_{k}$ is supplied.

First law of thermodynamics: $$ d U = \delta Q + \delta W $$

Since there is no change in internal energy, $d U = 0$, and the total work obtainable according to the first law of thermodynamics, $\Delta W$, can be written as follows. $$ \Delta W = \sum_{k} \delta Q_{k} \qquad \cdots 🤔 $$

Now imagine that at each point the heats $Q_{k}$ pass through some Carnot engine once before being transferred. The initial heat source at temperature $T$ is connected to every $k$-th Carnot engine, and the $k$-th Carnot engine produces work $\delta W_{k}$ at temperature $T$ and transfers heat $\delta Q_{k}$ to the specific part of the cyclic process. The reason for this somewhat contrived setup is to concentrate the transfers of work and heat into $T \to T_{k}$ so as to avoid cumbersome calculations. As a result, because after passing through the Carnot engines the ratio of heat to temperature from the source at temperature $T$ must equal the ratio of heat to temperature that reaches the reservoir at $T_{k}$, $$ {\frac{ \delta Q_{k} + \delta W_{k} }{ T }} = \frac{\delta Q_{h}}{T_{h}} = \frac{\delta Q_{l}}{T_{l}} = {\frac{ \delta Q_{k} }{ T_{k} }} $$ and rearranging gives: $$ \delta W_{k} = \delta Q_{k} \left( {\frac{ T }{ T_{k} }} - 1 \right) $$ Summing both sides yields the total work per cycle $\left( \Delta W + \sum \delta W_{k} \right)$ as follows. $$ \begin{align*} \sum_{\text{Cycle}} \delta W_{k} =& \sum_{\text{Cycle}} \delta Q_{k} \left( {\frac{ T }{ T_{k} }} - 1 \right) \\ =& T \sum_{\text{Cycle}} {\frac{ \delta Q_{k} }{ T_{k} }} - \sum_{\text{Cycle}} \delta Q_{k} \\ =& T \sum_{\text{Cycle}} {\frac{ \delta Q_{k} }{ T_{k} }} - \Delta W & \qquad \because 🤔 \\ \implies \Delta W + \sum \delta W_{k} =& T \sum_{\text{Cycle}} {\frac{ \delta Q_{k} }{ T_{k} }} \end{align*} $$

Second law of thermodynamics:

• Clausius: There is no process whose sole result is the transfer of heat from a colder to a hotter body.
• Kelvin: There is no process whose sole result is the complete conversion of heat into work.

According to Kelvin’s statement of the second law, there is no process that converts heat completely into work, hence $\Delta W + \sum \delta W_{k} \le 0$, and $$ T \sum_{\text{Cycle}} {\frac{ \delta Q_{k} }{ T_{k} }} \le 0 $$ holds. $T > 0$ is an absolute temperature, so it can be canceled without reversing the inequality, and $$ \sum_{\text{Cycle}} {\frac{ \delta Q_{k} }{ T_{k} }} \le 0 $$ Expressing this in integral form yields the Clausius inequality. $$ \oint {{\delta Q} \over {T}} \le 0 $$

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