Note that I am not calling the principle of increasing entropy the second law of thermodynamics because it was a consequence two previous states that I called the second law of thermodynamics. This can all be thrown aside as semantics that does not contain any invariant information. Setting aside this debate, we can move on to derive the consequences.
Let us first note, that the first law of thermodynamics however powerful it is, is not directly relate-able to experiments, after all it says that \(U = TdS - PdV,\) but how are we going to construct a knob that we can use to control entropy. We therefore need other kinds of state functions that also have units of energy and can easily be transferred to experiments. We shall introduce these functions by taking legendre transforms of U.
First let us introduce one that is important to chemists but not for physicist.
Consider the function \(H= U +PV\). We now take the differential of both sides to get \(dH = dU + VdP + PdV\) but we know what dU is from the first law, so we that
\(dH = TdS -PdV + VdP +PdV = TdS + VdP \) so H(S,P). Now, we recall that the function we are dealing with is a state function and therefore does not depend on the path in otherwords it is an exact differential.
\(F = AdX + BdY \) is exact when \( \frac{\partial A}{\partial Y})_X= \frac{\partial B}{\partial X})_Y\). We apply this to H to get \( \frac{\partial T}{\partial P})_S = \frac{\partial V}{\partial S})_P \). This is our first example of Maxwell Relation, derived painlessly without jumping into a morass of partials. Notice the natural variables of the state function are those that are held constant on each side the relation. This is how one can tell which state function to use in order to derive the relation.
Our second state function is important to physicists and that is got by taking the following legendre transform \(F= U - TS\). We play the same game as before and take the differential of both sidesto get that \(dF = -SdT - PdV\). This is an exact differential so \( \frac{\partial S}{\partial V})_V= \frac{\partial P}{\partial T})_T \)
The third example is got by considering \(G = F+PV\). This gives a differential of
\(dG = -SdT + VdP\). Applying the condition for exact differential gives \( \frac{\partial S}{\partial P})_T = - \frac{\partial V}{\partial T})_P \). The reader can do the same exercise for the internal energy U.
The purpose of these other state functions is to ease analysis of equilibrium states. If one has control over the temperature and volume of a system and considers those as the important variables then one should not insist with working with the internal energy but instead work with F or the Helmholtz free energy or if the thermodynamic coordinates are T and P the natural state function to consider is the the Gibbs free energy, G.
Let us now consider two situations which one is likely to come upon in thermodynamics and we shall use the principle of increasing entropy and the first law of thermodynamics to draw important conclusions.
Let there be a system coupled with a thermal reservoir at temperature \( T_0\). We shall assume that the system can't expand in volume, but there might be some internal degrees of freedom of the system that are not in equilibrium that allow for heat, Q, to be transferred to the reservoir. These situation might represent a chemical reaction or may a cup of ice and water sitting on table. The cup is the system and surrounding is the reservoir at temperature \( T_0\). Let us further assume that the system and the reservoir make up our universe i.e they both are thermally isolated. We know from the principle of increasing entropy that
\begin{equation}
\Delta S + \Delta S_0 \geq 0 \hspace{10mm} \text{eq.1}
\end{equation}.
S, here is the entropy of the system and \( S_0\) is the entropy of the reservoir. We know that \( S_0 = - \frac{Q}{T_0}\) and plugging this into eq.1 we get
\begin{equation}
Q - T_0 \Delta S \leq 0 \hspace{10mm} \text{eq.2}
\end{equation}
We know from the first law that \( \Delta U = Q \), since there is no change work being done by the system and plugging this into eq.2 we get
\begin{equation}
\Delta(U - T_0 S ) = \Delta F \leq 0 \hspace{10mm} \text{eq.3}
\end{equation}
Therefore for a system coupled to a thermal reservoir at fixed volume minimizes the free helmholtz energy if we obey the principle of increasing entropy. So as the system achieves thermodynamic equilibrium F goes to a minimum.
Let us consider another situation where we have a system that can do work that is calculated by \(P_0 \Delta V\) (this is the only work we will consider). This time the reservoir is a thermal reservoir at temperature \( T_0 \) and a pressure reservoir with pressure at \( P_0 \). Again we will there will be some heat exchange and we apply the principle of increasing entropy i.e eq.1 and follow by the same arguments as before to eq.2. This time we know from the first law that \( \Delta U = Q - P_0 \Delta V\). Solving for Q and plugging into eq.2 we arrive at
\begin{equation}
\Delta (U + P_0V - T_0 S) = \Delta G \leq 0 \hspace{10mm} \text{eq.4}
\end{equation}
But this is precisely the Gibbs free energy. So the principle of increasing entropy implies for a system coupled to a thermal and pressure reservoir that Gibbs Free energy is minimized.
Eq.3 and Eq.4 apply for both reversible and irreversible processes since state functions do not care about history or path but simply the endpoints which are equilibrium states.
Let us first note, that the first law of thermodynamics however powerful it is, is not directly relate-able to experiments, after all it says that \(U = TdS - PdV,\) but how are we going to construct a knob that we can use to control entropy. We therefore need other kinds of state functions that also have units of energy and can easily be transferred to experiments. We shall introduce these functions by taking legendre transforms of U.
First let us introduce one that is important to chemists but not for physicist.
Consider the function \(H= U +PV\). We now take the differential of both sides to get \(dH = dU + VdP + PdV\) but we know what dU is from the first law, so we that
\(dH = TdS -PdV + VdP +PdV = TdS + VdP \) so H(S,P). Now, we recall that the function we are dealing with is a state function and therefore does not depend on the path in otherwords it is an exact differential.
\(F = AdX + BdY \) is exact when \( \frac{\partial A}{\partial Y})_X= \frac{\partial B}{\partial X})_Y\). We apply this to H to get \( \frac{\partial T}{\partial P})_S = \frac{\partial V}{\partial S})_P \). This is our first example of Maxwell Relation, derived painlessly without jumping into a morass of partials. Notice the natural variables of the state function are those that are held constant on each side the relation. This is how one can tell which state function to use in order to derive the relation.
Our second state function is important to physicists and that is got by taking the following legendre transform \(F= U - TS\). We play the same game as before and take the differential of both sidesto get that \(dF = -SdT - PdV\). This is an exact differential so \( \frac{\partial S}{\partial V})_V= \frac{\partial P}{\partial T})_T \)
The third example is got by considering \(G = F+PV\). This gives a differential of
\(dG = -SdT + VdP\). Applying the condition for exact differential gives \( \frac{\partial S}{\partial P})_T = - \frac{\partial V}{\partial T})_P \). The reader can do the same exercise for the internal energy U.
The purpose of these other state functions is to ease analysis of equilibrium states. If one has control over the temperature and volume of a system and considers those as the important variables then one should not insist with working with the internal energy but instead work with F or the Helmholtz free energy or if the thermodynamic coordinates are T and P the natural state function to consider is the the Gibbs free energy, G.
Let us now consider two situations which one is likely to come upon in thermodynamics and we shall use the principle of increasing entropy and the first law of thermodynamics to draw important conclusions.
Let there be a system coupled with a thermal reservoir at temperature \( T_0\). We shall assume that the system can't expand in volume, but there might be some internal degrees of freedom of the system that are not in equilibrium that allow for heat, Q, to be transferred to the reservoir. These situation might represent a chemical reaction or may a cup of ice and water sitting on table. The cup is the system and surrounding is the reservoir at temperature \( T_0\). Let us further assume that the system and the reservoir make up our universe i.e they both are thermally isolated. We know from the principle of increasing entropy that
\begin{equation}
\Delta S + \Delta S_0 \geq 0 \hspace{10mm} \text{eq.1}
\end{equation}.
S, here is the entropy of the system and \( S_0\) is the entropy of the reservoir. We know that \( S_0 = - \frac{Q}{T_0}\) and plugging this into eq.1 we get
\begin{equation}
Q - T_0 \Delta S \leq 0 \hspace{10mm} \text{eq.2}
\end{equation}
We know from the first law that \( \Delta U = Q \), since there is no change work being done by the system and plugging this into eq.2 we get
\begin{equation}
\Delta(U - T_0 S ) = \Delta F \leq 0 \hspace{10mm} \text{eq.3}
\end{equation}
Therefore for a system coupled to a thermal reservoir at fixed volume minimizes the free helmholtz energy if we obey the principle of increasing entropy. So as the system achieves thermodynamic equilibrium F goes to a minimum.
Let us consider another situation where we have a system that can do work that is calculated by \(P_0 \Delta V\) (this is the only work we will consider). This time the reservoir is a thermal reservoir at temperature \( T_0 \) and a pressure reservoir with pressure at \( P_0 \). Again we will there will be some heat exchange and we apply the principle of increasing entropy i.e eq.1 and follow by the same arguments as before to eq.2. This time we know from the first law that \( \Delta U = Q - P_0 \Delta V\). Solving for Q and plugging into eq.2 we arrive at
\begin{equation}
\Delta (U + P_0V - T_0 S) = \Delta G \leq 0 \hspace{10mm} \text{eq.4}
\end{equation}
But this is precisely the Gibbs free energy. So the principle of increasing entropy implies for a system coupled to a thermal and pressure reservoir that Gibbs Free energy is minimized.
Eq.3 and Eq.4 apply for both reversible and irreversible processes since state functions do not care about history or path but simply the endpoints which are equilibrium states.
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