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Unit 5
Phase Transitions
Phase Transitions
What is meant by Phase of a system ?
It is defined as any homogeneous and physically distinct part of a thermodynamic system which is separated from other parts of the system by a definite boundary.
or
Transformation of a thermodynamic system from one phase to another at a given constant temperature and pressure is called a Phase transition .
What are different types of Phase transition ?
There are two types of Phase transitions
What is a First Order Phase transitions :
It is the type of Phase transition in which physical state of a matter changes from one to the other at constant temperature and pressure . There is a transfer of heat or latent heat associated with transition.
Examples .
1. transition of ice at $0^{\circ} \mathrm{C}$ into water at $0^{\circ} \mathrm{C}$ (solid siate to liquid state)
2. transition of water at $100^{\circ} \mathrm{C}$ to vapour at $100^{\circ} \mathrm{C}$ (Liquid to gaseous state)
Properties of First oder transition :
- Molar Gibb's function $G$ has the same value for both phases in equillibrium .
- Specific entropy ( $\mathrm{S} /$ Volume ) changes
- Specific volume ( V/mass) changes
- Gibbs free enrgy, Entropy S , Volume V , enthalpy H and Specific heat vary with temperature
- First order differntial of G or the slope of G is discontinuos at the transition temperature
i.e $\left(\frac{\partial G_{1}}{\partial T}\right)_{P} \neq\left(\frac{\partial G_{2}}{\partial T}\right)_{P} \&\left(\frac{\partial G_{1}}{\partial P}\right)_{T} \neq\left(\frac{\partial G_{2}}{\partial P}\right)_{T}$
- Density changes discontinuosly.
How differentials of $G$ are discontinuos :
\begin{align}
G &= H - T S = U + P V - T S \\
dG &= dU + P\,dV + V\,dP - T\,dS - S\,dT \\
dG &= dU + dW + V\,dP - dQ - S\,dT \\
dG &= dU + dW - dQ + V\,dP - S\,dT \\
dG &= 0 + V\,dP - S\,dT \\
G &= G(P, T) \\
\Rightarrow dG &= \left(\frac{\partial G}{\partial P}\right)_{T} dP + \left(\frac{\partial G}{\partial T}\right)_{P} dT \\
\Rightarrow V &= \left(\frac{\partial G}{\partial P}\right)_{T}, \quad S = -\left(\frac{\partial G}{\partial T}\right)_{P}
\end{align}
As V and S chnages first derivative of G is discontinuos.
Solution :
Let us consider an isolated system which contains two thermodynamic phases A and B of a matter in equillibrium. The Phase A is characterised by $\left(E_{1}, V_{1}, N_{1}, S_{1}, T_{1}, \mu_{1}, G_{1}, P_{1}\right)$ and the Phase B is characterised by $\left(E_{2}, V_{2}, N_{2}, S_{2}, T_{2}, \mu_{2}, G_{2}, P_{2}\right)$ Where\\
\textbf{Thermodynamic Definitions:}
\begin{align*}
E &= \text{Internal Energy} \\
V &= \text{Volume} \\
N &= \text{Number of Particles} \\
S &= \text{Entropy} \\
T &= \text{Temperature} \\
\mu &= \text{Chemical Potential} \\
G &= \text{Gibbs Free Energy} \\
P &= \text{Pressure}
\end{align*}
\textbf{Since the system is isolated:}
\begin{align*}
V_1 + V_2 &= V \\
N_1 + N_2 &= N \\
E_1 + E_2 &= E
\end{align*}
As the total system is isolated:
\begin{align*}
dV &= 0, \quad dN = 0, \quad dE = 0 \\
dV_1 &= -dV_2 \\
dN_1 &= -dN_2 \\
dE_1 &= -dE_2
\end{align*}
\textbf{As total entropy of the system is: } \( S = S_1 + S_2 \), we have:
\begin{align*}
0 &= dS_1(E_1, V_1, N_1) + dS_2(E_2, V_2, N_2) \\
0 &= \frac{\partial S_1}{\partial E_1} dE_1 + \frac{\partial S_1}{\partial V_1} dV_1 + \frac{\partial S_1}{\partial N_1} dN_1 \\
&\quad + \frac{\partial S_2}{\partial E_2} dE_2 + \frac{\partial S_2}{\partial V_2} dV_2 + \frac{\partial S_2}{\partial N_2} dN_2
\end{align*}
Substituting thermodynamic identities:
\begin{align*}
0 &= \frac{1}{T_1} dE_1 + \frac{P_1}{T_1} dV_1 - \frac{\mu_1}{T_1} dN_1 \\
&\quad + \frac{1}{T_2} dE_2 + \frac{P_2}{T_2} dV_2 - \frac{\mu_2}{T_2} dN_2
\end{align*}
Now using \( dE_2 = -dE_1 \), \( dV_2 = -dV_1 \), \( dN_2 = -dN_1 \):
\begin{align*}
0 &= \frac{1}{T_1} dE_1 + \frac{P_1}{T_1} dV_1 - \frac{\mu_1}{T_1} dN_1 \\
&\quad - \frac{1}{T_2} dE_1 - \frac{P_2}{T_2} dV_1 + \frac{\mu_2}{T_2} dN_1
\end{align*}
Combining terms:
\begin{align*}
0 &= \left( \frac{1}{T_1} - \frac{1}{T_2} \right) dE_1
+ \left( \frac{P_1}{T_1} - \frac{P_2}{T_2} \right) dV_1
+ \left( \frac{\mu_2}{T_2} - \frac{\mu_1}{T_1} \right) dN_1
\end{align*}
Since this must hold for arbitrary \( dE_1, dV_1, dN_1 \), we conclude:
\begin{align*}
\frac{1}{T_1} - \frac{1}{T_2} &= 0 \\
\frac{P_1}{T_1} - \frac{P_2}{T_2} &= 0 \\
\frac{\mu_2}{T_2} - \frac{\mu_1}{T_1} &= 0
\end{align*}
\textbf{Thus:}
\begin{align*}
T_1 &= T_2, \quad P_1 = P_2, \quad \mu_1 = \mu_2
\end{align*}
Hence when two different phases of a mattere are in equillibrium then their temperature , pressure and chemical potentials must be equal.
Which in turn shows that it must be in Thermal equillibrium ( as $\mathrm{T}_{1}, \mathrm{~T}_{2}$ ), Mechanical Equillibrium ( as $\mathrm{P}_{1}=\mathrm{P}_{2}$ ) and Chemical equillibrium ( as $\mu_{1}=\mu_{2}$ ).
\section*{As Gibbs free energy is defined as Chemical Potentials per particles}
\begin{align*}
G_1(P, T) &= \frac{\mu_1(P, T)}{N_1}, \quad
G_2(P, T) = \frac{\mu_2(P, T)}{N_2}
\end{align*}
\textbf{Then chemical equilibrium condition gives:}
\begin{align*}
\mu_1(P, T) &= \mu_2(P, T) \\
\Rightarrow G_1(P, T) N_1 &= G_2(P, T) N_2 \\
\Rightarrow G_1 &= G_2 \quad \text{if } N_1 = N_2 \\
\Rightarrow G_1 > G_2 &\quad \text{if } N_1 < N_2 \\
\Rightarrow G_1 > G_2 &\quad \text{if } N_1 > N_2
\end{align*}
\textbf{As chemical potentials are equal, the total Gibbs free energy is:}
\begin{align*}
G = N_1 G_1 + N_2 G_2
\end{align*}
\textbf{Also,}
\begin{align*}
N = N_1 + N_2
\end{align*}
\textbf{Taking differential of } \( G \):
\begin{align*}
dG &= dN_1 G_1 + N_1 dG_1 + dN_2 G_2 + N_2 dG_2 \\
\Rightarrow 0 &= dN_1 G_1 + N_1 dG_1 - dN_1 G_2 - N_2 dG_1 \\
\quad \text{using } dN_2& = -dN_1,\, dG_2 = dG_1\text{} \\
\Rightarrow 0 &= (N_1 - N_2) dG_1 + (G_1 - G_2) dN_1
\end{align*}
\textbf{Therefore,}
\begin{align*}
N_1 - N_2 &= 0 \\
G_1 - G_2 &= 0
\end{align*}
Explain How a fast order Phase Transition is discontinuos
The first-order phase transitions are those that involve a latent heat. During such a transition, a system either absorbs or releases a fixed (and typically large) amount of energy. Because energy cannot be instantaneously transferred between the system and its environment, first-order transitions are associated with "mixed-phase regimes" in which some parts of the system have completed the transition and others have not. This phenomenon is familiar to anyone who has boiled a pot of water: the water does not instantly turn into gas, but forms a turbulent mixture of water and water vapor bubbles .Mixed-phase systems are difficult to study, because their dynamics are violent and hard to control. However, many important phase transitions fall in this category, including the solid/liquid/gas transitions.\\[0pt]
The first order phase transition occurs between the triple point and critical point (excluding the critical point) ]
Why the First Order ?
According to Ernsfest classification the order of a phase transition is the order of lowest differential of G which shows discontinuity.\\
Hence first order $\left(\frac{\partial G}{\partial T}\right)_{P} \&\left(\frac{\partial G}{\partial P}\right)_{T}$ are discontinuos since entropy and Volume are discontinuos\\
given by the relations :
$$
\because\left(\frac{\partial G}{\partial T}\right)_{P}=-S \&\left(\frac{\partial G}{\partial P}\right)_{T}=V
$$
\section*{HoW ?}
$$
\begin{aligned}
\because G & =H-T S \\
G= & U+P V-T S \\
\Rightarrow d G & =d U+V d P+P d V-S d T-T d S \\
& =d U+P d V+V d P-S d T-d Q \\
& =d U+d W+V d P-S d T-d Q \\
& =d Q+V d P-S d T-d Q \\
& =V d P-S d T \\
\Rightarrow & \left(\frac{\partial G}{\partial P}\right)_{T}=V \quad \&\left(\frac{\partial G}{\partial T}\right)_{P}=-S
\end{aligned}
$$
In 2nd order Phase Transition , the second order differential of Gibbs free energy are discontinous
Since : \textbf{Second Derivatives of Gibbs Free Energy:}
\begin{align*}
\left(\frac{\partial^2 G}{\partial T^2}\right)_P &= -\frac{C_P}{T} \\
\left(\frac{\partial^2 G}{\partial P^2}\right)_T &= -V k_T \\
\left(\frac{\partial^2 G}{\partial P \partial T}\right) &= V \beta_P
\end{align*}
\textbf{Proof:}
1. Temperature Derivative:
\begin{align*}
\frac{\partial}{\partial T} \left( \frac{\partial G}{\partial T} \right)_P
&= \left( \frac{\partial (-S)}{\partial T} \right)_P \\
&= - \left( \frac{\partial S}{\partial T} \right)_P \\
&= - \left( \frac{\partial}{\partial T} \left( \frac{Q}{T} \right) \right)_P \\
&= - \frac{1}{T} \left( \frac{\partial Q}{\partial T} \right)_P \\
&= - \frac{C_P}{T}
\end{align*}
2. Pressure Derivative:
\begin{align*}
\frac{\partial}{\partial P} \left( \frac{\partial G}{\partial P} \right)_T
&= \left( \frac{\partial V}{\partial P} \right)_T \\
&= V \left( \frac{1}{V} \frac{\partial V}{\partial P} \right)_T \\
&= V k_T
\end{align*}
where
\[
k_T = \frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T \quad \text{(Isothermal compressibility)}
\]
3. Mixed Derivative:
\begin{align*}
\frac{\partial}{\partial P} \left( \frac{\partial G}{\partial T} \right)_P
&= - \frac{\partial S}{\partial P} \\
&= V \beta_P
\end{align*}
where
\[
\beta_P = \left( \frac{1}{V} \frac{\partial V}{\partial T} \right)_P \quad \text{(Thermal expansion coefficient)}
\]
Discontinuity Of Specific Heat In Phase Transitions:
Specific heat at constant pressure is :
$$
\begin{aligned}
C_{p} & =\left(\frac{\partial Q}{\partial T}\right)_{P}=T\left(\frac{\partial S}{\partial T}\right)_{P}=-T\left(\frac{\partial}{\partial T}\left(\frac{\partial G}{\partial T}\right)\right)_{P} \\
& =-T\left(\frac{\partial^{2} G}{\partial T^{2}}\right)_{P}
\end{aligned}
$$
Also in terms of Latent heat $\frac{d L}{d T}=\frac{L}{T}+C_{m f}-C_{m i}$. In first order phase transitions $L \neq 0$, thus specific heat at constant temperature has a sharp jump.
In first order transitions : The discontnuity is as shownn below:\\
\includegraphics[max width=\textwidth, center]{2024_12_28_757fe0ed403a0addbd02g-77}
Change in Specific heat in second order Phase transition :\\
\includegraphics[max width=\textwidth, center]{2024_12_28_757fe0ed403a0addbd02g-77(1)}
Phase transitions of the second order show a finite discontinuity in the specific heat $C_{P}$.
What is a DISCONTINUOS PHASE TRANSITION ?
Phase transition in which change in entropy is discontinuos at a particular temperature .This happens due to existence of latent heat\\
$d S=\frac{L}{T}$.\\
It is seen in First order Phase transition.
What is a CONTINUOS PHASE TRANSITION ?
Phase Transition in which the change in entropy is continuos. There is no involvement of Latent heat\\
Conditions for First Order Transitions :
- There are changes in entropy ( $\mathrm{S}_{1}$ is not equal to $\mathrm{S}_{2}$ )
- There are changes in volume
- The first order derivatives of Gibbs function change discontinuously.
- $\left(\frac{\partial G}{\partial T}\right)_{P} \&\left(\frac{\partial G}{\partial P}\right)_{T}$ are discontinuos since
$$
\because\left(\frac{\partial G}{\partial T}\right)_{P}=-S \&\left(\frac{\partial G}{\partial P}\right)_{T}=V
$$
Consider a system consisting of two phases of a pure substance at equilibrium. Under these conditions, the two phases are at the same pressure and temperature. Consider the change of state associated with a transfer of dn moles from phase 1 to phase 2 ( $\mathrm{p}, \mathrm{T}$ remaining constant). That is,
Define a Second Order Phase Transitions :( Continuos Phase Transition)
The second class of phase transitions are the continuous phase transitions, also called secondorder phase transitions. These have no associated latent heat Clausius -Clapeyron Law does not make sense here.\\
Conditions for second Phase Transitions:
Both specific entropy ( $\mathrm{S} / \mathrm{mass}$ ) and specific volume ( Volume/mass) do not change in second-order phase transitions.\\
In 2nd order Phase Transition, the second order differential of Gibbs free energy are discontinous.\\
Since : $\quad\left(\frac{\partial^{2} G}{\partial T^{2}}\right)_{P}=-\frac{C_{p}}{T},\left(\frac{\partial^{2} G}{\partial P^{2}}\right)_{T}=-V k_{T}$ and $\left(\frac{\partial^{2} G}{\partial P \partial T}\right)=V \beta_{p}$
Examples of second-order phase transitions :
- Ferromagnetic transition,
- Superfluid transition
- Bose-Einstein condensation.
Other Types of Phase Transition :
Several transitions are known as the infinite-order phase transitions. They are continuous but break no symmetries. The most famous example is the Berezinsky-Kosterlitz-Thouless transition in the two-dimensional XY model. Many quantum phase transitions in two-dimensional electron gases belong to this class. Apartfrom this, there is another phase transition known as Lambda phase transition]
Clausius-Clepyeron Equation :
Named after Rudolf Claussius (1822-1888, Germany ) \& Benoit Paul Emile Clapeyron ( 1799-1884, Paris ),
An equation that describes how pressure varies with temperature when sytem is in equillibrium betweeen two phases ( first order).
It is applicable for 1) Solid phase -Liquid phase Transition
- Liquid Phase- Vapour phase transition
- Solid Phase - Vapour Phase transition
Derivation:
Let $G_1$ and $G_2$ be the Gibbs functions for two phases of the system which are in equilibrium. Then,
\begin{align*}
G_1(P_1, T_1) &= G_2(P_2, T_2) \\
\Rightarrow G_1(P + dP, T + dT) &= G_2(P + dP, T + dT) \\
\Rightarrow G_1(P, T) + {G_1}{P} dP + {G_1}{T} dT &= G_2(P, T) + {G_2}{P} dP + {G_2}{T} dT \\
\Rightarrow {G_1}{P} dP + {G_1}{T} dT &= {G_2}{P} dP + {G_2}{T} dT \\
\Rightarrow V_1 dP - S_1 dT &= V_2 dP - S_2 dT \\
\text{Since} \quad S &= -{G}{T}, \quad V ={G}{P}
\end{align*}
Now, rearranging:
\begin{align*}
(V_1 - V_2) dP &= (S_1 - S_2) dT \\
\Rightarrow \frac{dP}{dT} &= \frac{S_2 - S_1}{V_2 - V_1} = \frac{\Delta S}{\Delta V}
\end{align*}
Special Cases :
Solid-Liquid Phase:
\begin{align*}
\frac{dP}{dT} &= \frac{L_f/T}{\Delta V} = \frac{L_f}{T \Delta V}
\end{align*}
where $L_f$ is the latent heat of fusion.
Effect of Pressure on Melting Point of Solids:
\textbf{Case (1):} When the volume of the solid in the liquid phase is more than in the solid phase, i.e., $V_2 > V_1$ or $\Delta V > 0$, then
\[
\frac{dP}{dT} > 0
\]
So, the melting point increases with pressure.
Case (2)
When the solid contracts on melting, i.e., $V_2 < V_1$ or $\Delta V < 0$, then
\[
\frac{dP}{dT} < 0
\]
Thus, the melting point decreases with increase in pressure.
\textbf{Exception:} Substances like Ice, Sb (Antimony), and Bi (Bismuth) contract on melting. So, for them at melting point:
\[
\frac{dP}{dT} < 0
\]
Hence, the P-T graph for solid-liquid transition may have positive or negative slope depending on the sign of $\Delta V$.
2.Liquid-Vapour Phase:
\begin{align*}
\frac{dP}{dT} &= \frac{L_V/T}{\Delta V} = \frac{L_V}{T \Delta V}
\end{align*}
where $L_V$ is the latent heat of vaporization.
In this phase transition, volume always increases, i.e., $\Delta V > 0$ or $V_2 > V_1$ at boiling point. Thus,
\[
\frac{dP}{dT} > 0 \quad \text{always}
\]
and the slope of the P-T graph is always positive.
Effect of Pressure on Boiling Point:
Boiling point increases with an increase in pressure.
How does pressure vary with temperature during Liquid-Vapour transition when volume of vapour is much larger than that of the liquid?
Ans:
Let $V_{\text{gas}} = V_2 \gg V_{\text{liquid}} = V_1$
\begin{align*}
\Rightarrow \frac{dP}{dT} &= \frac{L_V}{T \Delta V} = \frac{L_V}{T(V_g - V_l)} \approx \frac{L_V}{T V_g} \\
\Rightarrow \frac{dP}{dT} &= \frac{L_V}{T V_g} \\
\text{Using ideal gas law, } PV_g &= R T \Rightarrow V_g = \frac{R T}{P} \\
\Rightarrow \frac{dP}{dT} &= \frac{L_V P}{R T^2} \\
\Rightarrow \frac{dP}{P} &= \frac{L_V \, dT}{R T^2}
\end{align*}
Integrating both sides:
\begin{align*}
\int_{P_C}^{P} \frac{dP}{P} &= \int_{T_C}^{T} \frac{L_V \, dT}{R T^2} \\
\Rightarrow \ln \frac{P}{P_C} &= -\left( \frac{L_V}{R} \right) \left( \frac{1}{T} - \frac{1}{T_C} \right) \\
\Rightarrow \frac{P}{P_C} &= e^{\left( \frac{L_V}{R} \right) \left( \frac{1}{T_C} - \frac{1}{T} \right)} \\
\Rightarrow P &= P_C \, e^{\left( \frac{L_V}{R} \right) \left( \frac{1}{T_C} - \frac{1}{T} \right)} \\
&= P_C \, e^{\left( \frac{L_V}{R T_C} \right)} \cdot e^{- \left( \frac{L_V}{R T} \right)} \\
\Rightarrow P &= P_0 \, e^{- \frac{L_V}{R T}}
\end{align*}
This shows that pressure decreases exponentially with temperature.
\[
\Rightarrow P = P_0 \, e^{- \frac{L_V}{R T}}
\]
Variation of Vapour Pressure with Temperature
3. Solid - Liquid Phase:
\begin{align*}
\Rightarrow \frac{dP}{dT} &= \frac{\frac{L_s}{T}}{\Delta V} \\
\Rightarrow \frac{dP}{dT} &= \frac{L_s}{T \Delta V} \quad \text{($L_s$ is the latent heat of sublimation)}
\end{align*}
In Solid-Vapour phase transition, the final volume in the vapour phase is always greater than that in the solid phase, thus
\[
V_2 > V_1 \quad \text{or} \quad \Delta V > 0
\]
Then,
\[
\frac{dP}{dT} >
\]
always, and the slope of the $P$-$T$ graph is always positive.
What are the criteria for the stability of a thermodynamic system?
Ans:
Let us consider a thermodynamic system in contact with a heat reservoir at temperature $T$. Suppose it undergoes an infinitesimal irreversible process by extracting heat $dQ$ from the reservoir. If the change in entropy of the system is $dS$ and that of the reservoir is $dS_0$, then the change in entropy of the universe increases for the irreversible process:
\[
dS + dS_0 > 0
\]
\begin{align*}
\Rightarrow dS - \frac{dQ}{T} &> 0 \\
\Rightarrow T dS - dQ &> 0 \\
\Rightarrow T dS - (dU + dW) &> 0 \\
\Rightarrow T dS - dU - P dV &> 0
\end{align*}
Isolated Adiabatic System:
Since $dQ = 0$, the criterion becomes:
\[
dS > 0
\]
Hence, stability is achieved when entropy is maximum.
Constant Volume and Entropy:
Since $dS = 0$ and $dV = 0$, the inequality becomes:
\[
dU< 0
\]
So the system is stable when internal energy is minimum.
Constant Pressure and Entropy:
Since $dP = 0$, $dS = 0$, then:
\begin{align*}
dU + d(PV) & < 0 \\
\Rightarrow dH & < 0
\end{align*}
Thus, stability requires minimum enthalpy.
Constant Volume and Temperature:
Here, $dV = 0$ and $dT = 0$. Then:
\begin{align*}
dU - d(TS) & < 0 \\
\Rightarrow d(U - TS) & < 0 \\
\Rightarrow dF & < 0
\end{align*}
Hence, the Helmholtz free energy must be minimum for stability.
Constant Temperature and Pressure:
Here, $dT = 0$, $dP = 0$:
\begin{align*}
dU + P dV - T dS &< 0 \\
&= dU + d(PV) - d(TS) \\
&= d(H - TS) \\
&= dG < 0
\end{align*}
Thus, the Gibbs free energy must be minimum for the system to be stable.
How does latent heat vary with pressure?
Ans:
Let $S_i$ and $S_f$ be the molar entropies at the initial and final states. When a first-order phase transition occurs, the molar latent heat $L$ is related to the entropies as:
\[
\Delta S = \frac{L}{T} \quad \Rightarrow \quad L = T (S_f - S_i)
\]
Also,
\[
\frac{L}{T} = -\left( {G_f}{T} \right)_P + \left( {G_i}{T} \right)_P
\]
Differentiating both sides with respect to $T$:
\begin{align*}
\frac{1}{T} {L}{T} - \frac{L}{T^2} &= {S_f}{T} - {S_i}{T} \\
\Rightarrow {L}{T} - \frac{L}{T} &= T {S_f}{T} - T {S_i}{T} \\
\Rightarrow {L}{T} - \frac{L}{T} &= {Q_f}{T} - {Q_i}{T} \\
\Rightarrow {L}{T} - \frac{L}{T} &= C_{mf} - C_{mi} \\
\Rightarrow {L}{T} &= \frac{L}{T} + C_{mf} - C_{mi}
\end{align*}
This is also known as the
second latent heat equation
What are the criteria for the stability of a thermodynamic system?
Let us consider a thermodynamic system which is in contact with a heat reservoir at temperature \( T \). Suppose it undergoes an infinitesimal irreversible process by extracting heat \( dQ \) from the reservoir. If the change in entropy of the system is \( dS \), and that of the reservoir is \( dS_0 \), then the entropy of the universe increases for the irreversible process.
\[
dS + dS_0 > 0
\]
\begin{align}
\Rightarrow dS - \frac{dQ}{T} &> 0 \\
\Rightarrow T\,dS - dQ &> 0 \\
\Rightarrow T\,dS - (dU + dW) &> 0 \\
\Rightarrow T\,dS - dU - P\,dV &> 0
\end{align}
Stability criterion for an isolated adiabatic system:
Since \( dQ = 0 \), we have:
\[
dS > 0
\]
Thus, the system is stable when entropy is maximum.
Stability criterion for a system with constant volume and entropy:
In this case, \( dS = 0 \), \( dV = 0 \), so:
\[
dU < 0
\]
Hence, the system is stable when internal energy is minimum.
Stability criterion for constant pressure and entropy:
Here, \( dP = 0 \), \( dS = 0 \):
\begin{align}
dU + d(PV) &< 0 \\
\Rightarrow dH &< 0
\end{align}
The system is stable when enthalpy \( H \) is minimum.
Stability criterion for constant volume and temperature:
\( dV = 0 \), \( dT = 0 \):
\begin{align}
dU - d(TS) &< 0 \\
\Rightarrow d(U - TS) &< 0 \\
\Rightarrow dF &< 0
\end{align}
The system is stable when Helmholtz free energy \( F \) is minimum.
Stability criterion for constant temperature and pressure:
\( dT = 0 \), \( dP = 0 \):
\begin{align}
dU + P\,dV - T\,dS &< 0 \\
dU + d(PV) - d(TS) &< 0 \\
d(U + PV - TS) &< 0 \\
d(H - TS) &< 0 \\
dG &< 0
\end{align}
Thus, the system is stable when Gibbs free energy \( G \) is minimum.
How does latent heat vary with temperature?
Let \( S_i \) and \( S_f \) be the molar entropies of the initial and final states. For a first-order phase transition, the molar latent heat \( L \) is related to entropy as:
\begin{align}
dS &= \frac{L}{T} \Rightarrow L = T(S_f - S_i) \\
\Rightarrow \frac{L}{T} &= S_f - S_i
\end{align}
Also note:
\[
\frac{L}{T} = -\left( \frac{\partial G_f}{\partial T} \right)_P + \left( \frac{\partial G_i}{\partial T} \right)_P
\]
Differentiating both sides with respect to \( T \):
\begin{align}
\frac{1}{T} \frac{dL}{dT} - \frac{L}{T^2} &= \frac{dS_f}{dT} - \frac{dS_i}{dT} \\
\Rightarrow \frac{dL}{dT} - \frac{L}{T} &= T\left( \frac{dS_f}{dT} - \frac{dS_i}{dT} \right) \\
\Rightarrow \frac{dL}{dT} - \frac{L}{T} &= \frac{dQ_f}{dT} - \frac{dQ_i}{dT} \\
\Rightarrow \frac{dL}{dT} - \frac{L}{T} &= C_{mf} - C_{mi} \\
\Rightarrow \frac{dL}{dT} &= \frac{L}{T} + C_{mf} - C_{mi}
\end{align}
This is also known as the
second latent heat equation
What is a second order Phase Transition ?
Ans: It is a type of Phase transition in which Physical state of the matter does not change.Changes occur in atomic arrangements, spins , etc .
Properties of Second Order Phase Transition:
- The molar Gibbs free energy \( G \) and its first derivatives with respect to \( T \) and \( P \) are continuous, but higher-order derivatives are not:
\[
g_i = g_f
\]
- Specific entropy and specific volume are continuous, but specific heats are discontinuous:
\[
s_i = s_f, \quad v_i = v_f
\]
Thus, the first derivatives of Gibbs free energy are continuous:
\[
\left( \frac{\partial G_1}{\partial T} \right)_P = \left( \frac{\partial G_2}{\partial T} \right)_P
\]
Proof:
From the first-order phase transition:
\begin{align}
\frac{L}{T} &= S_f - S_i
\end{align}
But also:
\begin{align}
\frac{L}{T} &= -\left( \frac{\partial G_f}{\partial T} \right)_P + \left( \frac{\partial G_i}{\partial T} \right)_P
\end{align}
Now, for a second-order phase transition, the latent heat \( L = 0 \), hence:
\begin{align}
-\left( \frac{\partial G_f}{\partial T} \right)_P + \left( \frac{\partial G_i}{\partial T} \right)_P &= 0 \\
\Rightarrow \left( \frac{\partial G_i}{\partial T} \right)_P &= \left( \frac{\partial G_f}{\partial T} \right)_P
\end{align}
Also, continuity of Gibbs free energy with respect to pressure implies:
\[
\left( \frac{\partial G_1}{\partial P} \right)_T = \left( \frac{\partial G_2}{\partial P} \right)_T
\]
It is discontinuous in symmetry arrangement (such as crystal structure), but continuous in the physical state of the body.
Examples :
- Trnasition of Liquid Helium 1 to Liquid Helim2
- Trnasition from Nonferromagnetic state to Ferromagnetic state
- Suprconducting transition in a zero field
- Bose Einstein Condensate
FIRST ORDER PHASE TRANSITION PARAMETERS VS TEMPERATURE
1. Gibbs Free energy vs Temperature
2. Entropy vs temperature
3. Volume vs Temperature
4. Enthalpy vs Temperature
5. specific heat at constant volume vs temperature
What are Ehrenfest Equations? Derive the Ehrenfest Equations:
These are the equations to determine the slope of $P$ vs $T$ graphs, i.e., to find the value of $\frac{dP}{dT}$ in a second-order phase transition.\\
There are two Ehrenfest equations.
Derivation of First Ehrenfest Equation:
Let the phases of matter be denoted by $i$ (initial) and $f$ (final). In a second-order phase transition, entropy is continuous:
\begin{align}
S_i &= S_f \\
S_i + dS_i &= S_f + dS_f \\
dS_i &= dS_f
\end{align}
Since $S = S(T, P)$, we write:
\begin{align}
dS &= \frac{\partial S}{\partial T} dT + \frac{\partial S}{\partial P} dP
\end{align}
Therefore,
\begin{align}
\frac{\partial S_i}{\partial T} dT + \frac{\partial S_i}{\partial P} dP &= \frac{\partial S_f}{\partial T} dT + \frac{\partial S_f}{\partial P} dP \\
T \frac{\partial S_i}{\partial T} dT + T \frac{\partial S_i}{\partial P} dP &= T \frac{\partial S_f}{\partial T} dT + T \frac{\partial S_f}{\partial P} dP \\
C_{P,i} dT + T \frac{\partial S_i}{\partial P} dP &= C_{P,f} dT + T \frac{\partial S_f}{\partial P} dP
\end{align}
Now, using the Maxwell relation:
\[
\left( \frac{\partial S}{\partial P} \right)_T = -\left( \frac{\partial V}{\partial T} \right)_P
\]
We get:
\begin{align}
C_{P,i} dT - T \left( \frac{\partial V_i}{\partial T} \right) dP &= C_{P,f} dT - T \left( \frac{\partial V_f}{\partial T} \right) dP
\end{align}
Multiplying and dividing by $V$:
\begin{align}
C_{P,i} dT - T V_i \beta_i dP &= C_{P,f} dT - T V_f \beta_f dP
\end{align}
Where volume expansivity is defined as:
\[
\beta = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P
\]
Rewriting:
\begin{align}
(C_{P,f} - C_{P,i}) dT &= T V (\beta_f - \beta_i) dP \\
\frac{dP}{dT} &= \frac{C_{P,f} - C_{P,i}}{T V (\beta_f - \beta_i)}
\end{align}
Since $V_i = V_f$ in second-order phase transitions, we take $V$ as common.
Derivation of Second Ehrenfest Equation:
Since volume is continuous in both phases:
\begin{align}
dV_i &= dV_f
\end{align}
But $V = V(T, P)$, so:
\begin{align}
dV &= \left( \frac{\partial V}{\partial T} \right)_P dT + \left( \frac{\partial V}{\partial P} \right)_T dP
\end{align}
Multiplying and dividing by $V$:
\begin{align}
dV &= V \beta dT - V k dP
\end{align}
Where:
\[
\beta = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P, \quad
k = -\frac{1}{V} \left( \frac{\partial V}{\partial P} \right)_T
\]
Now, equating for both phases:
\begin{align}
V \beta_i dT - V k_i dP &= V \beta_f dT - V k_f dP \\
(\beta_f - \beta_i) dT &= (k_f - k_i) dP \\
\frac{dP}{dT} &= \frac{\beta_f - \beta_i}{k_f - k_i}
\end{align}
