# Clausius inequality validity and conclusion

In Wikipedia it is said that the inequality holds only for reversible processes. Why is that?

And also from the statement we arrive at this conclusion:

" In other words, the Clausius statement states that it is impossible to construct a device whose sole effect is the transfer of heat from a cool reservoir to a hot reservoir."

How do reach this conclusion? Is the conclusion an interpretation of the in-equation, or arises from intuition ?

Thanks

• The equality holds for reversible processes. For irreversible processes, $$dS > dQ/T$$ For me the connection is an intuitive one, as a change in internal energy of the system is related to a change in entropy of the system. For a completely reversible process energy and entropy can be exchanged freely. The Carnot cycle demonstrates this net neutral entropy exchange. An irreversible process on the other hand, i.e. any non-theoretical process, will be unable to return to the exact configuration it started in and the entropy will have increased. Commented May 27, 2021 at 18:40
• Is it wrong to simply an say that entropy is an indicator of how erratic is the movement of particles in a system? The more erratic or chaotic the movement of the particles become the more entropy increases. Which means whenever we give heat to the system (while preventing the absorbed heat to get away in form of heat released from the system or work done on the environment) then entropy increases. Which is to say that temperature increase=entropy increase. Is it wrong to think of it that way? And does reversibility or irreversibility plays a role here? Commented May 27, 2021 at 18:51
• Just an FYI you should accept an answer and upvote any answers that were helpful in solving your problems! Commented Oct 25, 2023 at 1:29

In Wikipedia it is said that the inequality holds only for reversible processes. Why is that?

First, the Clausius theorem applies to engine cycles, not individual processes. Second, the equality applies to reversible cycles and the inequality applies to irreversible cycles.

" In other words, the Clausius statement states that it is impossible to construct a device whose sole effect is the transfer of heat from a cool reservoir to a hot reservoir."

This is not the Clausius theorem. This statement is based on the Clausius' statement the second law or

No refrigeration or heat pump can operate without a net work input.

The Clausius theorem and inequality applies to heat engines, not heat pumps and refrigerators. It is given by

$$\int\frac{dQ}{T}\le 0$$

where $$dQ$$ is the heat entering the system at any point in the cycle and $$T$$ is the temperature at the point of entry to the system.

The inequality applies to an irreversible cycle of a real heat engine and says that the entropy given to the environment during the cycle is larger than the entropy transferred to the engine by the environment. Since entropy transfer is always in the form of heat, it means that for an irreversible cycle less heat is available to perform work.

Hope this helps.

• In my judgment, you need to make it clear that the Clausius inequality applies to all individual processes carried out in a closed system. Please clarify the distinction between the Clausius theorem and the Clausius inequality. Commented May 27, 2021 at 20:08
• @ChetMiller But isn't the integral for a complete cycle of processes? Regarding the distinction between the theorem and the inequality, I thought they were related, as shown here hyperphysics.phy-astr.gsu.edu/hbase/thermo/clausius.html#c1 Commented May 27, 2021 at 21:21
• It is satisfied by every step of a cycle, and by the sum of the integrals over the cycle. Commented May 27, 2021 at 21:26
• @ChetMiller Yes, that's why I said in my answer that "\$dQ is the heat entering the system at any point in the cycle". In any case, could you please tell me the distinction between the theorem and the inequality? Commented May 27, 2021 at 21:29
• My understanding is that the Theorem relates to cycles and the inequality is the mathematical relationship for a process in a closed system. Commented May 27, 2021 at 21:39

The Clausius inequality becomes an equality for reversible processes because reversible processes are defined as those during which the system is always at equilibrium with its surroundings (or at most has infinitesimal deviations therefrom so that the process can actually proceed). I find it is helpful to remind oneself where the Clausius inequality comes from in the first place. A succinct relationship between the reversible and irreversible work done on or by a system can be easily argued, $$\begin{gather*} \delta w - \delta w_{rev} \geq 0 \end{gather*}$$ In other words, we will always get the most work out of a reversible process. If we then write out the first law in its differential form, \begin{align*} dE &= \delta q + \delta w \\ &= \delta q_{rev} + \delta w_{rev} \end{align*} where in the latter we have noted that the first law holds for any process, reversible or irreversible. Then, we can say, \begin{align*} \delta w - \delta w_{rev} &= \delta q_{rev} - \delta q \\ \implies \delta q_{rev} - \delta q &\geq 0 \\ \implies \delta q_{rev} &\geq \delta q \\ \implies dS &\geq \frac{\delta q}{T} \end{align*} Since the process that involves the most work is a reversible process, we similarly conclude that the process involving the minimal exchange of heat is the reversible process, which gives exactly the Clausius entropy, $$\begin{gather*} dS = \frac{\delta q_{rev}}{T} \end{gather*}$$ From the Clausius inequality, we thus know that the reversible processes are those that will give the smallest changes in entropy.

To your second question, we can show that heat must exclusively flow from a body with higher temperature to a body with lower temperature using the Clausius inequality. To accomplish this, we simply focus on some system of interest that is at a temperature of $$T_{sys}$$. The system and all of its surroundings constitute the entire universe, which we can reasonably assume is an isolated system. It is relatively straightforward to argue that in general, $$\begin{gather*} \delta q_{sys} = -\delta q_{surr} \end{gather*}$$ But we now turn our attention to the entropy, \begin{align*} &dS_{sys} + dS_{surr} = dS \\ \implies &\frac{\delta q_{sys}}{T_{sys}} + \frac{\delta q_{surr}}{T_{surr}} \geq 0 \\ \implies &\frac{\delta q_{sys}}{T_{sys}} - \frac{\delta q_{sys}}{T_{surr}} \geq 0 \\ \implies &\frac{\delta q_{sys}}{T_{sys}} \geq \frac{\delta q_{sys}}{T_{surr}} \\ \implies &\frac{T_{surr}}{T_{sys}} \delta q_{sys} \geq \delta q_{sys} \end{align*} where we have invoked the second law and Clausius inequality such that $$dS \geq \frac{\delta q}{T} \geq 0$$ for isolated systems. We are almost done. Consider the case where $$\delta q_{sys} > 0$$. What does this imply about the relative temperatures? In order for this inequality to hold, we must have $$T_{surr} > T_{sys}$$. In other words, if heat is flowing into the system, then the surroundings must be at a higher temperature than the system. Consider the other case where $$\delta q_{sys} < 0$$. Now what does this imply about the temperatures? Now, we must flip the inequality which gives $$T_{surr} < T_{sys}$$. In other words, if the system loses heat, then the system must be at a higher temperature than the surroundings. Finally, if the system does not gain or lose heat, then $$T_{sys} = T_{surr}$$, and the system is already at thermal equilibrium. Thus, our assertion about heat flowing from hotter to colder regions/objects is entirely correct. It turns out that the reason processes proceed in the direction they do is often nothing more than the machinations of entropy!