Clausius Clapeyron Equation

Using the appropriate Maxwell relation gives,

\[ \Bigg(\frac{\partial P}{\partial T}\Bigg)_{P} = \Bigg(\frac{\partial S}{\partial V}\Bigg)_{T} \]

This equation is also known as Clausius Clapeyron Equation, and these equation help in understanding a dis-continuous phase-transition between 2 phases of matter of a single constituent. Thus, the Clausius Clapeyron Equation used to estimate vapor pressure as a function of temperature or to find the heat of the phase transition from the vapor pressures at two temperatures. These phases can be vapor and solid for sublimation or solid and liquid for melting.

Derivation of Clausius Clapeyron Equation

At T = T_m :
\[ dG^s = dG^l \] \[ \implies V^s dP - S^s dT = V^l dP - S^l dT \tag{as dG = Vdp -SdT} \] \[ \implies (V^s - V^l) dP = (S^s - S^l) dT \] \[ \implies \boxed{ \frac{dP}{dT} = \frac{(S^s - S^l)}{(V^s - V^l)} } \tag{1} \]

Case 1 : Solid \( \rightarrow \) liquid

At equilibrium, \( \Delta S = \frac{\Delta H}{T} \) , Clausius Clapeyron Equation can also be written as - \[ \boxed{ \frac{dP}{dT} = \frac{\textrm{Latent Heat}}{T \times \Delta v }} \tag{2} \] Note : From equ (2), We can say curve for the solid-liquid transition will be a linear line as LH, T, \( \Delta v \) is constant.

Case 2 : Solid \( \rightarrow \) Gas or Liquid \( \rightarrow \) Gas

Here volume of the gaseous phase will be much higher compare to liquid or solid phase. So for this case \( \Delta v = v_{g} - v_{l/s} = v_{g} \)

So for a molar idea gas, PV = RT \( \Rightarrow V = \frac{RT}{P} \), On subtituting \( \Delta v = v_{2} \) in above equation \[ \frac{dP}{dT} = \frac{\Delta H_{vap} P}{R \times T^2 }\] \[ \Rightarrow \frac{dP}{P} = \frac{\Delta H_{vap} P}{R \times T^2 }dT\] On Integrating both side , \[ \boxed{ ln P = \frac{\Delta H_{vap} }{R} \Bigg[ \frac{-1}{T} \Bigg] } \tag{3} \] Note : From equ (3), We can say curve for the solid-gas or liquid-gas transition will be exponential as \( P = e^{\frac{\textrm {-LH} }{RT} } \).

Equ. (3) can also be written as -

\[ \boxed{ \ln{ \Bigg( \frac{P_{2}}{P_{1}} \Bigg) } = - \frac{ \Delta H_{vap} }{R} \Bigg( \frac{1}{T_2} - \frac{1}{T_1} \Bigg) } \tag{4} \]

Note :
1) \( \Delta H_{S \rightarrow V} = \Delta H_{S \rightarrow L} + \Delta H_{L \rightarrow V} \)

Unary Phase diagram for different cases

Case 1	Case 2
\( \Delta V_m = -ve \)	\( \Delta V_m = +ve \)
Ex. H2O, Cast Iron	Ex. Fe, Al

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Question :Assume \(CCl_4\) is boils at 76 °C at 1 atm Then the boiling point at 2 atm will be __________________ °C ?
Given Latent Heat of vapourization = \( 195 KJKg^{-1}\)
And, characteristic gas constant = \( 0.055 KJKg^{-1}K^{-1}\)

Show Answer

Coefficient of Volume Expensivity \( (\beta) \)

Also known as Isobaric Thermal Expansivity, which tells us that how a substance volume will respond to changes in temperature under constant pressure condition. \[ \boxed{ \beta = \frac{1}{V} \Bigg( \frac{\partial V}{\partial T} \Bigg)_{P} } \] The \( (1/V) \) term makes it an Intensive property.

Isothermal Compressibility \( (k_{T}) \)

It help to quantify, how much a substace is compressible. The isothermal compressibility \( (k_{T}) \) is defined by the fractional differential change in volume due to a change in pressure at constant T. \[ \boxed{ k_{T} = - \frac{1}{V} \Bigg( \frac{\partial V}{\partial P} \Bigg)_{T} } \] The -ve sign is used to make \( (k_{T}) \) a positive value as increase in pressure will lead to a decrease in volume and the \( (1/V) \) term makes it an Intensive property.

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Important relation between \( (\beta) \) and \( (k_{T}) \)

1) \[ \frac{\beta}{k_{T} } = \Bigg( \frac{\partial P}{\partial T} \Bigg)_{V} \] 2) \[ C_p - C_v = \frac{T[v\beta^2]}{k_T} \]

van der waals equation

This equation is basically a modification of the Ideal Gas Law which states that gases consist of point masses that undergo perfectly elastic collision. Here will consider the effect atoms size and their interaction forces unlike Idea Gas Law.

\[ \Big( P + \Big( \frac{a}{v^2} \Big) \Big) . \Big( v-b \Big) = RT \]

\({a}/{v^2} \) : Intermolecular Interaction
\( b \) : Volume occupied by gas molecules

For n-Moles

\[ \boxed{ \Big( P + \Big(\frac{an^2}{v^2}\Big) \Big) . \Big( v-nb \Big) = RT} \]

Conditions for calculating a and b

\[ \Bigg[ \Big( \frac{\partial P}{\partial v} \Big)_{T=T_{cr}} \Bigg] = 0 \hspace{1cm} \textrm{ ; }\hspace{1cm} \Bigg[ \Big( \frac{\partial^2 P}{\partial v^2} \Big)_{T=T_{cr}} \Bigg] = 0\]

Minimum , Maximum energy principle

Entropy Maximum principle As we have, \[ \frac{ \delta Q}{T} \leq dS \] For an isolated system \( \delta Q = 0 \) \[ \Rightarrow dS \leq 0 \] This principle states that for a closed system, having a fixed internal energy and volume, the equilibrium state will have maximum entropy	Minimum Internal Energy Principle As we have, \[dU \leq TdS - PdV\] Taking time differential \[ \frac{dU}{dt} = T\frac{dS}{dt} - P\frac{dV}{dt} \] \[ \Bigg( \frac{dU}{dt}\Bigg)_{S,V} \leq 0 \]This principle states that for a closed system at constant S and V, the equilibrium state will have manimun internal energy.
Minimum Helmholtz Energy Principle As we have, \[dF \leq PdV - SdT\] Taking time differential \[ \frac{dF}{dt} = P\frac{dV}{dt} - S\frac{dT}{dt} \] \[ \Bigg( \frac{dF}{dt}\Bigg)_{V,T} \leq 0 \]This principle states that for a closed system at constant V and T, the equilibrium state will have manimun helmholtz free energy.	Minimum Gibbs Energy Principle As we have, \[dG \leq VdP - SdT\] Taking time differential \[ \frac{dG}{dt} = V\frac{dP}{dt} - S\frac{dT}{dt} \] \[ \Bigg( \frac{dF}{dt}\Bigg)_{P,T} \leq 0 \]This principle states that for a closed system at constant P and T, the equilibrium state will have manimun gibbs free energy.