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Clausius inequality states that $\oint {\delta Q\over T}$ equals zero for a system undergoing a reversible cycle, whereas it can’t be greater than zero for an irreversible cycle.

But everywhere, I see people and resources mentioning that “it follows” that for irreversible, it is strictly negative. HOW??

Mostly, they mention that since entropy changes for irreversible processes are positive, the above follows, except that this follows from the above once the above is justified! Else, give me an independent justification of this argument!

Otherwise, it is mentioned that irreversible processes involve dissipation, so entropy generation. But wait! An irreversible process doesn’t have to be dissipative. Say, irreversibility is due to non-quasi-staticity. Now what?

Similar is the case with the efficiencies of irreversible engines working between two temperatures.

Carnot’s theorem merely says that an engine working between two temperatures can’t be more efficient than a reversible engine working between the same temperatures.

I know that this implies that all reversible engines working between these temperatures have the same efficiency. But HOW does this imply that an irreversible engine between the same two temperatures will have a strictly less efficiency? Still people and resources blatantly mention this without any justification.

Someone please resolve this!

Qmechanic
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2 Answers2

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But everywhere, I see people and resources mentioning that “it follows” that for irreversible, it is strictly negative. HOW??

When a process in a cycle is irreversible, it generates entropy. In order to complete a cycle all thermodynamic properties, including entropy, must be returned to their original state. That requires the system to get rid of the entropy generated by transferring it to the surroundings. The only way to transfer entropy to the surroundings is by heat. Heat out of the system is negative. Ergo $\oint {\delta Q\over T}<0$ for an irreversible cycle.

Hope this helps.

Bob D
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I show the Clausius equality for cyclic reversible process

It is not obvious from first law of thermodynamics that $dQ,dU,dW$ are differentials for the integration.

It doesnt has to be an ideal gas, all needed is to assume the integrals exist in Riemann sense and certain functions are absolutely continuous. let $\epsilon , a >0 $,

$T_{n}>T_{n-1}>a,|T_{n}-T_{n-1}| < \epsilon$

Using the first law of thermodynamics, \begin{align} \mathrm Q_2-Q_1 & = \mathrm U_2 - U_1 +\mathrm W_2 - W_1\\ \mathrm \int_{Q_{n-1}}^{Q_n} dQ &= \int_{T_{n-1}}^{T_n} m C_V \mathrm dT + \int_{V_{n-1}}^{V_n} P \mathrm dV\\ \sum_{n=1}^{n=N} \frac{1}{T_{n-1}}\mathrm \int_{Q_{n-1}}^{Q_n} dQ &=\sum_{n=1}^{n=N}\frac{m}{T_{n-1}} \int_{T_{n-1}}^{T_n} C_V \mathrm dT + \sum_{n=1}^{n=N}\frac{1}{T_{n-1}}\int_{V_{n-1}}^{V_n} P \mathrm dV\\ \end{align}

Because of the following inequalities:

$$|\frac{1}{T_{n-1}}\int_{Q_{n-1}}^{Q_n} dQ-\int_{Q_{n-1}}^{Q_n} \frac{1}{T} dQ| \le \int_{Q_{n-1}}^{Q_n} |\frac{1}{T}-\frac{1}{T_{n-1}}| dQ \le \int_{Q_{n-1}}^{Q_n} \frac{\epsilon}{a^2} dQ $$

$$|\frac{1}{T_{n-1}}\int_{T_{n-1}}^{T_n} C_VdT-\int_{T_{n-1}}^{T_n} \frac{1}{T} C_VdT| \le \int_{T_{n-1}}^{T_n} |\frac{1}{T}-\frac{1}{T_{n-1}}| C_VdT \le \int_{T_{n-1}}^{T_n} \frac{\epsilon}{a^2} C_VdT $$

$$|\frac{1}{V_{n-1}}\int_{V_{n-1}}^{V_n} PdV-\int_{V_{n-1}}^{V_n} \frac{1}{T} PdV| \le \int_{V_{n-1}}^{V_n} |\frac{1}{T}-\frac{1}{V_{n-1}}| PdV \le \int_{V_{n-1}}^{V_n} \frac{\epsilon}{a^2} PdV $$

letting $\epsilon \to 0$ we have $$\int \frac{1}{T} dQ=m\int \frac{1}{T} C_VdT+\int \frac{1}{T} PdV$$

$$\frac{df}{dQ}=\frac{1}{T}$$

$$\frac{dg}{dT}=\frac{C_V}{T}$$

$$\frac{dh}{dV}=\frac{P}{T}$$

if $f,g,h$ are absolutely continuous then $$\oint \frac{1}{T} dQ=m\oint \frac{1}{T} C_VdT+\oint \frac{1}{T} PdV=0$$

ibnAbu
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