In this paper, the mathematical formulation of \(2^{nd}\) law of thermodynamics has been explained, and the mathematical formulation of \(1^{st}\) law has been revisited from this noble perspective. It is not claimed that the \(2^{nd}\) law of thermodynamics is a redundant of the \(1^{st}\) law, rather I shown here how we can extract the mathematical formulation of the \(2^{nd}\) law from the mathematical formulation of the \(1^{st}\) law of thermodynamics. The Clausius statement of the \(2^{nd}\) law of thermodynamics is, it is impossible to construct a device whose sole effect is the transfer of heat from a cool reservoir to a hot reservoir. An alternative statement of the law is, “All spontaneous processes are irreversible” or, “the entropy of an isolated system always increases”. Having strong experimental evidences, this empirical law is obvious, which tells us the arrow of time and the direction of spontaneous changes.
The \(1^{st}\) law of thermodynamics is often claimed as a version of the law of conservation of energy, adapted to the thermodynamic systems, which can be formulated as:
\[\delta Q_{rev}=dU-\delta W_{rev},\] which states that the amount of heat accumulated by a closed system is spent to change its internal energy, and to do some work by the system on its surroundings. Here \(\delta{\mathrm{W}}_{\mathrm{rev}}\) is the differential work done by the surroundings on the system. So work done by the system on its surroundings is \((-\delta W_{rev})\), hence, \(\delta Q_{rev}=dU+p_{ext}dV\). This law is a very common-sense law, adapted to the thermodynamic system, which is the total energy of an isolated system e.g., the universe is conserved.The aim of this paper is to show the mathematical formulation of the \(2^{nd}\) law of thermodynamics as a consequence of the \(1^{st}\) law.
Here, \(\delta Q_{rev}\) and \(\delta W_{rev}\) are non-exact differentials and \(dU\) is an exact differential. Let \(X(U,V)\) is the integrating factor, so that \(X\delta Q_{rev}\) becomes exact, so, \(X\delta Q_{rev}=XdU+Xp_{ext}dV\) is an exact differential equation.
Definition 1. \(dZ = MdP + NdQ\) will be an exact differential equation provided that \(\left(\frac{\partial M}{\partial Q}\right)_P=\left(\frac{\partial N}{\partial P}\right)_Q\).
Assumption 1. Let us assume that the integrating factor \(X\) is a function of only internal energy \(U\). Joule experimentally shown that the internal energy of an ideal gas is only the function of its temperature, independent of pressure or volume. So, \(X = f(T)\). As, \(X\delta Q_{rev}=XdU+Xp_{ext}dV\) is an exact differential equation, so from Definition 1, we have \[\left(\frac{\partial X}{\partial V}\right)_U=\left(\frac{\partial Xp_{ext}}{\partial U}\right)_V.\] The above partial differential equation in general has more than one solution. The general solution can’t be known as the Internal Energy as a function of temperature and the boundary conditions are not known. However, a particular solution for \(X\) can be obtained assuming \(X\), the integrating factor as a function of only internal energy \(U\).
From Assumption 1, we have \[X = f(T)\] \[\therefore \left(\frac{\partial X}{\partial V}\right)_U=0,\] i.e.,
\[\left(\frac{\partial X}{\partial V}\right)_U=\left(\frac{\partial Xp_{ext}}{\partial U}\right)_V=0,\] i.e., \[Xp_{ext} = \text{function of }\,\,V = g\left(\frac{T}{p_{ext}}\right)\] for ideal gas.The functional equation \(f(T)p_{ext}=g\left(\frac{T}{p_{ext}}\right)\) has a unique solution \(f(y)=g(y)=\frac{1}{y}\), so, the integrating factor is \(X=f(T)=\frac{1}{T}\).
Now, \(\frac{\delta Q_{rev}}{T}\) is an exact differential. For reversible process, temperature of the system is same as the temperature of the surroundings at any particular instant of time, so, \(\frac{\delta Q_{rev}}{T_{surr}}\) is an exact differential and is known as differential change in entropy.
It can be shown that maximum work delivered to surroundings for isothermal gas expansion can be obtained using a reversible path [1], so,
\[(-W)_{irrev}W_{rev}\] and \[\Delta U=Q_{irrev}+W_{irrev}=Q_{rev}+W_{rev}\] so, \[Q_{rev}>Q_{irrev}.\] Being a state function,Theorem 1. The 2nd Law of Thermodynamics is a consequence of the 1st Law of Thermodynamics.
The above argument proves the existence of the equation
\[\oint \frac{\delta Q}{T_{surr}}\leq 0\] for global systems, where the equality holds for reversible cases and for irreversible changes, the equality does not hold.
Let a system (Figure 1) is isolated and spontaneously changes from \(A\) to \(B.\) The system is then brought into contact with a heat source of same temperature as \(B\) at that particular instant of time and reversibly brought back from \(B\) to \(A.\) (It would be necessary and sufficient to show that ”changes from \(A\) to \(B\) is irreversible” to prove the above statement.)
Proof of Theorem 1. Let us first assume that spontaneous change from \(A\) to \(B\) is reversible, so, the total cyclic path is also reversible as all the elements of the path is reversible [2]. Hence
\[\oint \frac{\delta Q}{T_{surr}}=0 \ \Rightarrow \oint^{B}_{A}\frac{\delta Q_{spontaneous}}{T_{surr}}+\oint^{A}_{B}\frac{\delta Q_{rev}}{T_{surr}}=0.\] But \[\oint^{B}_{A}\frac{\delta Q_{spontaneous}}{T_{surr}}=0,\,\, \text{as}\,\,\, Q_{spontaneous}=0\,\, \text{(isolated)}.\] Therefore \[ \oint \frac{\delta Q}{T_{surr}}=\oint^{A}_{B}\frac{\delta Q_{rev}}{T_{surr}}=0,\] which is not true, because, \(Q_{rev}(B\rightarrow A)\neq 0\) (not isolated), so, \[\oint \frac{\delta Q}{T_{surr}} \neq 0.\] That is a contradiction, so, our initial assumption was incorrect, i.e., spontaneous change from \(A\) to \(B\) must be irreversible. Consequently, the cycle of heating and cooling back to the same point \(A\) is an irreversible cycle, as, minimum portion of this path is proved to be irreversible \((A\rightarrow B)\).So,
\[\oint \frac{\delta Q}{T_{surr}}< 0 \ \Rightarrow \oint^{B}_{A}\frac{\delta Q_{spontaneous}}{T_{surr}}+\oint^{A}_{B}\frac{\delta Q_{rev}}{T_{surr}}< 0\ \Rightarrow \oint^{A}_{B}\frac{\delta Q_{rev}}{T_{surr}}0,\] which is the change in entropy for the spontaneous process indicating that the entropy always increases in isolated system for spontaneous processes to occur.From the above argument it is proved that “All spontaneous processes are irreversible”. Also ”the entropy of an isolated system always increases”.
The engine is said to be reversible if in the reversed cycle it can work as a heat pump, i.e., it takes \(\Delta Q_2\) amount of heat from the cold reservoir and releases \(\Delta Q_1\) amount of heat to the hot reservoir, provided that it consumes the useful work converted in the first cycle. That means, after completing two cycles, the engine is back to the old state, so, \(\frac{\Delta Q_1}{T_1}\) + \(\frac{\Delta Q_2}{T_2} = 0\).
But the equality does not hold for actual engines (irreversible), as we have proved that all spontaneous processes are irreversible. That means, after completing the two cycles, the engine is not back to the old state, so, \(\frac{\Delta Q_1}{T_1}\) + \(\frac{\Delta Q_2}{T_2} < 0\), implies \( 1 + \frac{T_1}{\Delta Q_1} \times \frac{\Delta Q_2}{T_2} < 0 \) implies \(\frac{\Delta Q_2}{\Delta Q_1} < -\frac{T_2}{T_1}\).
The efficiency of a heat engine is defined by the ratio of work done per cycle and the heat energy it gains per cycle, i.e., \(\frac{\Delta Q_1+\Delta Q_2}{\Delta Q_1} = 1 + \frac{\Delta Q_2}{\Delta Q_1}< 1-\frac{T_2}{T_1}\), which is clearly less than 100% as we know \(T_1,T_2\nless 0\) from the \(3^{rd}\) law of thermodynamics.
That is the evidence of the alternative formulation of the \(2^{nd}\) law of thermodynamics used in the literature, ”There does not exist any heat engine that does nothing but absorb heat energy from one single reservoir and convert it into work”.