You certainly know the second law of thermodynamics, Do you know its connection to other laws of physics and chemistry?

Abstract

Since the discovery of the second law of thermodynamics, skeptics have never ceased to challenge the absolute validity of the law, leading to a situation where the claim of violation of the second law has almost become routine today. This situation is also a serious threat to the fundamental status of thermodynamics. In this work, we bring to light evidences to prove the absolute validity of the second law as a fundamental law of physics. For this purpose, we propose a short revisit of the history of the discovery of the second law in order to highlight the connection between the second law and the first law of energy conservation. We then demonstrate that the perpetual motion machine of the second kind also violate the first law of thermodynamics, albeit indirectly, contrary to the common belief. This result confirms the second law is an inviolable fundamental law of physics, just like the law of energy conservation. Denying one of these conjoined twin laws is to deny the other. Any presumed violation of the second law, even a probabilistic one, inevitably violates the laws of energy and mass conservation, and undermines all fundamental laws of physics and chemistry. We conclude by summarizing a perspective of the interpretation of the second law taking into account the multiplicity of paths of random motion.

Appendix

This is a summary of the history of the discovery of the second law of thermodynamics over thirty years starting from 1824.

1.1 (I) Carnot theorem from a thought experiment

Figure 1 illustrates a combination of an irreversible real engine and a reversible Carnot engine, both working between a hot reservoir R1 at temperature ${\theta }_{1}$ and a cold reservoir R2 at temperature ${\theta }_{2}$. The irreversible engine produces a work ${W}^{{\prime}}={Q}_{1}-{Q}_{2}$ in absorbing a heat ${Q}_{1}$ from the hot R1 and emitting a heat ${Q}_{2}$ to R2. Its efficiency is given by ${\eta }^{{\prime}}=\frac{{W}^{{\prime}}}{{Q}_{1}}$ [1]. The reversible engine is working in the inverse direction using a work $W$ to absorb the heat ${Q}_{2}$ from R2 and give a heat ${Q}_{1}$ to R1. Its efficiency is given by $\eta =\frac{W}{{Q}_{1}}$. After a cycle, the two reservoirs recover their initial states, respectively, because they did not obtain nor lose heat.

Now Carnot supposed that the irreversible engine has a larger efficiency than the reversible engine, ${\eta }^{{\prime}}>\eta$, or $\frac{W^{{\prime}}}{{Q}_{1}}>\frac{W}{{Q}_{1}}$, leading to $W^{{\prime}}>W$. In this case, the reversible engine can use a part of $W^{{\prime}}$ to produce the work $W$, so that after a cycle, the only thing changed is the extra work $\Delta W={W}^{{\prime}}-W>0$ produced by the ensemble of two engines (red line in Fig. 1). This means that the ensemble of two engines is a perpetual motion machine of the first kind (PMM1) producing the work $\Delta W$ from nothing, which Carnot thought impossible according to a common belief that mechanical work and motion cannot be created from nothing (a rudimental idea of energy conservation). Conclusion: ${\eta }^{{\prime}}$ can never be larger than $\eta$; which is the first part of Carnot theorem ${\eta }^{{\prime}}\le \eta$ [9].

From the same reasoning, we can prove that all reversible engines must have the same efficiency regardless of their different working substances. This is because if a reversible engine was more efficient than another one, we could combine them in the same way as in Fig. 1 with the more efficient Carnot engine replacing the irreversible one; this would make a PMM1 (absurd). Conclusion: all Carnot engines must have the same efficiency depending only on the temperatures of the reservoirs ${\theta }_{1}$ and ${\theta }_{2}$, which is the second part of Carnot theorem [9].

1.2 (II) Kelvin's efficiency

It is quite logical that the efficiency $\eta$ of Carnot engine depends only on the temperature of the reservoirs. But what is the form of the function $\left( {\theta_{1} { },\theta_{2} { }} \right)$? Kelvin is concerned with this question, and made a calculation [40] which can be illustrated by the connection of two Carnot engines as shown in Fig. 2. The first engine works between two reservoirs R1 at $\theta_{1}$ and R2 at $\theta_{2}$ ($\theta_{1} { } >$ $\theta_{2}$), absorbing $Q_{1}$ from R1, emitting $Q_{2}$ to R2 and doing a work $W_{1} = Q_{1} - Q_{2}$; the efficiency is $\eta_{1} = 1 - \frac{{Q_{2} }}{{Q_{1} }}$. The second engine works between the above reservoir R2 and a third reservoir R3 at $\theta_{3}$ ($\theta_{2} >$ $\theta_{3}$), absorbing $Q_{2}$ from R2, emitting $Q_{3}$ to R3, and doing a work $W_{2} = Q_{2} - Q_{3}$; its efficiency is $\eta_{2} = 1 - \frac{{Q_{3} }}{{Q_{2} }}$. On the other hand, these two engines connected in series can be considered as a third engine, working between the hot reservoir R1 at $\theta_{1}$ and the cold reservoir R3 at $\theta_{3}$, absorbing $Q_{1}$ from R1, emitting $Q_{3}$ to R3, and doing a work $W_{3} = W_{1} + W_{2} = Q_{1} - Q_{3}$; its efficiency is $\eta_{3} = \frac{{W_{3} }}{{Q_{1} }} = 1 - \frac{{Q_{3} }}{{Q_{1} }}$.

The mathematics is the following: from the three efficiencies $1 - \eta_{1} = \frac{{Q_{2} }}{{Q_{1} }}$, $1 - \eta_{2} = \frac{{Q_{3} }}{{Q_{2} }}$ and $1 - \eta_{3} = \frac{{Q_{3} }}{{Q_{1} }}$, it is straightforward to write $\left( {1 - \eta_{1} } \right)\left( {1 - \eta_{2} } \right) = 1 - \eta_{3}$. As all Carnot engines have the same efficiency depending only on the temperature of the reservoirs, we can write $1 - \eta_{1} = g\left( {\theta_{1} ,\theta_{2} } \right)$, $1 - \eta_{2} = g\left( {\theta_{2} ,\theta_{3} } \right)$ and $1 - \eta_{3} = g\left( {\theta_{1} ,\theta_{3} } \right)$, leading to $g\left( {\theta_{1} ,\theta_{2} } \right)g\left( {\theta_{2} ,\theta_{3} } \right) = g\left( {\theta_{1} ,\theta_{3} } \right)$, where $g\left( \cdot \right)$ is any continuous function. The only way to reach this last equation is to let $g\left( {\theta_{1} ,\theta_{2} } \right) = \frac{{f\left( {\theta_{2} } \right)}}{{f\left( {\theta_{1} } \right)}}$, $g\left( {\theta_{2} ,\theta_{3} } \right) = \frac{{f\left( {\theta_{3} } \right)}}{{f\left( {\theta_{2} } \right)}}$ and $g\left( {\theta_{1} ,\theta_{3} } \right) = \frac{{f\left( {\theta_{3} } \right)}}{{f\left( {\theta_{1} } \right)}}$ with certain function $f\left( \theta \right)$. Now let us simply take $f\left( \theta \right)$ as a temperature scale, say, $\left( \theta \right) = T$, which turns out to be the absolute temperature. Kelvin finally got $1 - \eta_{1} = \frac{{T_{2} }}{{T_{1} }}$, $1 - \eta_{2} = \frac{{T_{3} }}{{T_{2} }}$ et $1 - \eta_{3} = \frac{{T_{3} }}{{T_{1} }}$, or in general, for a Carnot engine working between $T_{1}$ and $T_{2}$ ($T_{1} > T_{2}$): $\eta = 1 - \frac{{T_{2} }}{{T_{1} }}$ [40].

1.3 (III) Clausius’ discovery

Clausius' work [10] starts with a Carnot engine working between $T_{1}$ and $T_{2}$. He writes $\eta = 1 - \frac{{Q_{2} }}{{Q_{1} }} = 1 - \frac{{T_{2} }}{{T_{1} }}$, or $\frac{{Q_{2} }}{{Q_{1} }} = \frac{{T_{2} }}{{T_{1} }}$, and $0 = \frac{{Q_{1} }}{{T_{1} }} - \frac{{Q_{2} }}{{T_{2} }}$. With a change of notation $\delta Q_{1} = Q_{1} ,$ and $\delta Q_{2} = - Q_{2}$, he obtained $\frac{{\delta Q_{1} }}{{T_{1} }} + \frac{{\delta Q_{2} }}{{T_{2} }} = 0$. Notice that $\delta Q$ is a heat absorbed by the engine from a reservoir (see Fig. 

Fig. 3

Examples of the cycle of Carnot engine in the $T - S$ diagram. The horizontal segments represent the contact of the engine with the reservoirs to absorb heat. The vertical segments represent adiabatic compression (left sides) and expansion (right sides). Left: a cycle starting from and ending at an initial state A after two contacts with $T_{1}$ and $T_{2}$, leading to $\frac{{\delta Q_{1} }}{{T_{1} }} + \frac{{\delta Q_{2} }}{{T_{2} }} = 0$. Right: a cycle starting from and ending at the initial state A after the contacts with N reservoirs at different temperature $T_{i}$, each giving $\delta Q_{i}$ to the engine, leading to $\mathop \sum \nolimits_{i = 1}^{N} \frac{{\delta Q_{i} }}{{T_{i} }} = 0$

3). Obviously, if the engine is in contact with N reservoirs during a cycle, and absorbs $\delta Q_{i}$ from a reservoir at $T_{i}$, we necessarily have $\mathop \sum \nolimits_{i = 1}^{N} \frac{{\delta Q_{i} }}{{T_{i} }} = 0$ (Fig. 3).

Clausius understood at once that $\frac{\delta Q}{T}$ is the variation of a variable of state, he denoted by S ($\delta S = \frac{\delta Q}{T}$) and coined the name entropy, because when the engine comes back to its initial state after a cycle and a series of variations $\frac{{\delta Q_{i} }}{{T_{i} }}$, this variable also comes back to its initial value with zero variation $\mathop \sum \nolimits_{i = 1}^{N} \frac{{\delta Q_{i} }}{{T_{i} }} = \mathop \sum \nolimits_{i = 1}^{N} \delta S_{i} = 0$. If N is vary large, we can write a closed line integral ${\oint }dS = 0$ over a cycle.

Clausius made the above analysis with Carnot engine. What does happen with a real irreversible engine working between the same reservoirs with $\eta ^{\prime} = 1 - \frac{{Q_{2} }}{{Q_{1} }}$? Considering $\eta^{{\prime}} < \eta$ and $\eta { } = 1 - \frac{{T_{2} }}{{T_{1} }}$, we get $1 - \frac{{Q_{2} }}{{Q_{1} }} \le 1 - \frac{{T_{2} }}{{T_{1} }}$ or $\frac{{Q_{1} }}{{T_{1} }} - \frac{{Q_{2} }}{{T_{2} }} < 0$. The same formal change as above yields $\frac{{\delta Q_{1} }}{{T_{1} }} + \frac{{\delta Q_{2} }}{{T_{2} }} < 0$. With N reservoirs during a cycle, we have $\sum_{i = 1}^{N} \frac{{\delta Q_{i} }}{{T_{i} }} < 0$ or ${\oint }\frac{\delta Q}{T} < 0$. Suppose that the variation of entropy of the engine at each equilibrium contact with a reservoir is $\delta S_{i}$, we have, after a cycle, $\sum_{i = 1}^{N} \delta S_{i} = 0$, or ${\oint }dS = 0$ for large N. Compare ${\oint }dS = 0$ to ${\oint }\frac{\delta Q}{T} < 0$, it is straightforward to write $dS > \frac{\delta Q}{T}$ for irreversible engine, or ${\varvec{dS}} \ge \frac{{\user2{\delta Q}}}{{\varvec{T}}}$ for any engine, reversible or irreversible. This inequality is the mathematical expression of the second law of thermodynamics. For an isolated system without exchange of heat with the exterior, $\delta Q = 0$, and $dS \ge 0$. This is why the second law is often referred to as the law of increasing entropy. Notice that the entropy of an open system exchanging heat with environment has $dS \ge \frac{\delta Q}{T}$ and can decrease ($dS < 0$) if the system loses heat $\delta Q < 0$ (like our body); this decrease does not violate the second law. Again, the violation of the second law implies the decreases in entropy in an isolated system.

1.4 (IV) Perpetual motion machine of the second kind

As shown in Fig. 4, a perpetual motion machine of the second kind (PMM2) is capable of extracting a heat $Q_{3}$ from a cold reservoir R2 at $T_{2}$ and transform it entirely into a mechanical work $W_{3} = Q_{3}$. This cycle of PMM2 does not violate the first law of energy conservation.

Fig. 4

Illustration of how a perpetual machine of the second kind (PMM2) allows a flow of heat $Q_{1} = Q_{2} + Q_{3}$ from a colder body to a hotter body without any other effects upon the environment

However, the work $W_{3}$ can be used by a Carnot engine in an inverse cycle, as shown in Fig. 4, to extract a heat $Q_{2}$ from the colder reservoir R2 and to transfer a heat $Q_{1}$ to a hotter reservoir R1 at $T_{1}$, with $Q_{1} = Q_{2} + W_{3} = Q_{2} + Q_{3}$. During this cycle, the reservoir R2 loses an entropy $\Delta S_{2} = - \frac{{Q_{2} + Q_{3} }}{{T_{2} }}$, and the reservoir R1 receives an entropy $\Delta S_{1} = \frac{{Q_{1} }}{{T_{1} }} = \frac{{Q_{2} + Q_{3} }}{{T_{1} }}$, giving a total entropy change $\Delta S = \Delta S_{1} + \Delta S_{2} = Q_{1} \left( {\frac{1}{{T_{1} }} - \frac{1}{{T_{2} }}} \right) < 0$, which violates the Clausius expression of the second law Eq. (4) $\Delta S \ge 0$ for the isolated system composed of the two reservoirs and the engines in Fig. 4.

In order to show how this violation of the second law by PMM2 enables PMM1, we introduce a second Carnot engine in Fig. 4 (see Fig. 5), which absorbs $Q_{1}^{{\prime}}$ during a cycle from the reservoir R1, does a work $W$ to the environment, and emits a heat $Q_{2}^{{\prime}}$ to the reservoir R2. Its efficiency is $\eta { } = 1 - \frac{{T_{2} }}{{T_{1} }} = 1 - \frac{{Q_{2}^{{\prime}} }}{{Q_{1}^{{\prime}} }}$, leading to $\frac{{Q_{1}^{{\prime}} }}{{T_{1} }} = \frac{{Q_{2}^{{\prime}} }}{{T_{2} }}$. During the same cycle, the system composed of the two engines in Fig. 4 transfers a heat $Q_{1}$ from R2 to R1. The ensemble of the three engines in Fig. 5 (surrounded by red line) is equivalent to an global engine which in one cycle absorbs a heat $Q_{1}^{{\prime}} - Q_{1}$ from R1, does a work W, and emits a heat $Q_{2}^{{\prime}} - Q_{1}$ to R2, leading to an efficiency $\eta ^{\prime}{ } = 1 - \frac{{Q_{2}^{{\prime}} - Q_{1} }}{{Q_{1}^{{\prime}} - Q_{1} }}$. It is easy to prove that $\frac{{Q_{2}^{{\prime}} - Q_{1} }}{{Q_{1}^{{\prime}} - Q_{1} }} < \frac{{Q_{2}^{{\prime}} }}{{Q_{1}^{{\prime}} }}$, making $\eta^{{\prime}} > \eta$ for any $Q_{1} > 0$. It turns out that the global engine in Fig. 5 is more efficient than the Carnot engine (violating the first part of Carnot theorem). We can of course use it to construct a PMM1 with the same tricks of Fig. 1.

Fig. 5

Illustration of an engine (red line) having an efficiency $\eta ^{\prime}{ } = 1 - \frac{{Q_{2}^{{\prime}} - Q_{1} }}{{Q_{1}^{{\prime}} - Q_{1} }}$ larger than the efficiency of the Carnot engine $\eta { } = 1 - \frac{{T_{2} }}{{T_{1} }} = 1 - \frac{{Q_{2}^{{\prime}} }}{{Q_{1}^{{\prime}} }}$

In similar way, we can also create a PMM1 using a PPM2 associated with an irreversible engine replacing the second Carnot engine in Fig. 5. In this case, the entropy increase created with the system of two engines of Fig. 4, $\Delta S = Q_{1} \left( {\frac{1}{{T_{1} }} - \frac{1}{{T_{2} }}} \right)$, must be larger than $\frac{{Q_{2}^{{\prime}} }}{{T_{2} }} - \frac{{Q_{1}^{{\prime}} }}{{T_{1} }}$, i.e., $Q_{1} > \frac{{\left( {\frac{{Q_{2}^{{\prime}} }}{{T_{2} }} - \frac{{Q_{1}^{{\prime}} }}{{T_{1} }}} \right)}}{{\left( {\frac{1}{{T_{1} }} - \frac{1}{{T_{2} }}} \right)}}$.