Equivalent formulations of the second law

We have introduced in Sec.  the conditions that govern equilibrium between two systems that are otherwise isolated from the rest of the Universe. In particular, we have pointed out that the removal of any internal constraint causes the total entropy of the system to increase, simply because the number of states available to the system increases. Now we want to ask the question in a slightly different way.

Let us consider a system in contact with a heat bath and suppose that it is subject to some internal constraints. These constraints may or may not cause the system to have a different pressure or temperature from those of the heath bath. We know that upon the removal of the constraints the total entropy of the system increases:

$$dS_1 + dS_2 > 0,$$ (3.84)

where we use the usual notation of 1 for the system and 2 for the heat bath. The heat bath can be assumed to be in equilibrium whatever the interactions with the system, and so we have:

$$dS_2 = -\frac{{\mathchar'26\mkern-12mu d}Q}{T_2},$$ (3.85)

where $T_2$ is the temperature of the bath and ${\mathchar'26\mkern-12mu d}Q$ is the heat flowing from the bath to the system as the constraints are removed. If the only work on the system is done by the heat bath we have:

$${\mathchar'26\mkern-12mu d}W = -P_2 dV_1,$$ (3.86)

where $dV_1$ is the change of volume of the system. Using the first law for the system, $dE_1 = {\mathchar'26\mkern-12mu d}Q + {\mathchar'26\mkern-12mu d}W$, we have:

$$dE_1 = -T_2 dS_2 - P_2 dV_1 < T_2 dS_1 - P_2 dV_1,$$ (3.87)

where the inequality follows from . Eq.  can be re-arranged as:

$$dE_1 - T_2 dS_1 + P_2 dV_1 < 0.$$ (3.88)

Defining the availability $A = E_1 - T_2 S_1 + P_2 V_1$ we can rewrite :

$$dA < 0,$$ (3.89)

because $T_2$ and $P_2$ are constant. This inequality is particularly useful when one considers constant temperature ($T_1 = T_2$) and constant volume ($dV_1 = 0$) variations, or constant temperature and constant pressure ($P_1 = P_2$) variations. In the former case we have:

$$dE_1 - T_1 dS_1 = d(E_1 - T_1 S_1) = dF_1 < 0$$ (3.90)

and in the latter:

$$dE_1 - T_1 dS_1 + P_1 dV_1 = d(E_1 - T_1 S_1 + P_1 V_1) = dG_1 < 0,$$ (3.91)

where $F_1$ and $G_1$ are the Helmholtz and Gibbs free energy of the system. Eqs.  and  give the conditions for the direction of natural processes for the canonical and the isothermal-isobaric ensembles. A final condition may be expressed by considering constant entropy ($dS_1 = 0$) and constant pressure variations:

$$dE_1 + P_1 dV_1 = d(E_1 + P_1 V_1) = dH_1 < 0,$$ (3.92)

which defines the Enthalpy, $H_1 = E_1 + P_1 V_1$, and gives the condition for equilibrium. The fundamental thermodynamic relation for $H$ is:

$$dH = d(E + PV ) = dE + PdV + VdP = VdP + TdS + \mu dN$$ (3.93)

which gives

$$V = \left( \frac{\partial H}{\partial P} \right )_{S,N}; \quad T ... ...right )_{P,N}; \quad \mu = \left( \frac{\partial H}{\partial N} \right )_{P,S}.$$ (3.94)

The conditions for the directions of natural processes, i.e. processes that follow the removal of internal constraints, can be thus summarised:

$$dS > 0 \quad \quad {\rm Constant} \quad N, V, E$$

$$dE < 0 \quad \quad {\rm Constant} \quad N, V, S$$

$$dF < 0 \quad \quad {\rm Constant} \quad N, V, T$$

$$dG < 0 \quad \quad {\rm Constant} \quad N, P, T$$

$$dH < 0 \quad \quad {\rm Constant} \quad N, P, S$$

The fundamental thermodynamic relation  for the Helmholtz free energy gives $dF = -PdV - SdT + \mu dN$, which is equal to zero for constant temperature, constant volume and constant number of particles variations, so why does  show a negative value instead? The reason is that Eq.  refers to a system with fixed (or absence of) internal constaints, while  states what would happen to a system when physical internal constraints are removed. Similar arguments apply for $G$ and $H$. Note how Eqs. , , ,  allow to express the differential of the thermodynamic potentials $E,F,G,H$ as functions of the differential of two specific thermodynamic variables amongst $S,V,T,P$, with the coefficients being the other two variables (for simplicity, we assume constant number of particles, but the discussion can be generalised to variable number of particles with additional algebra). For example, we have $dE = TdS -P dV$ and so we are considering variations of the energy in terms of explicit variations of the entropy and the volume, $E = E(S,V)$. Of course, there is nothing special with this choice and we could instead decide to consider explicitly the dependence of the energy on volume and temperature, for example, writing $E= E(V,T)$. With such a choice we would have:

$$dE = \left( \frac{\partial E}{\partial V} \right )_T dV + \left( ... ...left [T\left(\frac{\partial P}{\partial T} \right )_V - P \right ] dV + C_V dT.$$ (3.95)

We see that the coefficients multiplying the differential $dV$ involves a combination of thermodynamic variables, and so $V$ and $T$ are not natural variables for the energy, but $V$ and $S$ are. Similar arguments apply for the other thermodynamic potentials, with $T$ and $V$ being the natural variables for $F$; $T$ and $P$ the natural variables for $G$; and $P$ and $S$ the natural variables for $H$. To obtain the equivalence between the first terms on the right hand sides of Eq.  we use the following procedure. Consider the fundamental thermodynamic relation for the entropy:

$$dS = \frac{1}{T} dE + \frac{1}{T}P dV = \frac{1}{T}\left [ \left(... ...\left( \frac{\partial E}{\partial V} \right )_T dV \right ] + \frac{1}{T}P dV =$$

$$\frac{1}{T} \left( \frac{\partial E}{\partial T} \right )_V dT +\... ...{1}{T} \left [ \left( \frac{\partial E}{\partial V} \right )_T + P \right ] dV,$$ (3.96)

from which we get

$$\left(\frac{\partial S}{\partial T} \right )_V = \frac{1}{T} \lef... ...rac{1}{T} \left [ \left( \frac{\partial E}{\partial V} \right )_T + P \right ].$$ (3.97)

If we now differentiate the first term in  w.r.t to $V$ we obtain $\left(\frac{\partial^2 S}{\partial V\partial T} \right )$. Differentiating the second term w.r.t $T$ we obtain $\left(\frac{\partial^2 S}{\partial T\partial V} \right )$, and since $\left(\frac{\partial^2 S}{\partial V\partial T} \right ) = \left(\frac{\partial^2 S}{\partial T\partial V} \right )$ we have:

$$\frac{\partial}{\partial V} \left [ \frac{1}{T} \left( \frac{\par... ...\left [ \left( \frac{\partial E}{\partial V} \right )_T + P \right ] \right\}_V$$

$$\frac{1}{T} \left( \frac{\partial^2 E}{\partial T\partial V} \rig... ...partial T} \right ) + \left (\frac{\partial P}{\partial T} \right )_V \right ],$$ (3.98)

and again with $\left(\frac{\partial^2 E}{\partial T\partial V} \right ) = \left(\frac{\partial^2 E}{\partial V\partial T} \right )$ we obtain:

$$\left( \frac{\partial E}{\partial V} \right )_T = T \left(\frac{\partial P}{\partial T}\right)_V - P .$$ (3.99)