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

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.

The free energy curves of the liquid and solid phases for a substance have been shown in the figure. The phase with the lowest gibbs free energy for a given temperature will be the most stable phase.

For example at T < T_{m} solid phase have lowest gibbs free energy so solid will be stable phase and similarly at T > T_{m} liquid phase will be more stable.

For example at T < T

**At T = T _{m} :**

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

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 , \[ \boxed{ ln P = \frac{\Delta H_{vap} }{R} \Bigg[ \frac{-1}{T} \Bigg] } \tag{3} \]

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

Case 1 | Case 2 |
---|---|

\( \Delta V_m = -ve \) | \( \Delta V_m = +ve \) |

Ex. H_{2}O, Cast Iron |
Ex. Fe, Al |

Given Latent Heat of vapourization = \( 195 KJKg^{-1}\)

And, characteristic gas constant = \( 0.055 KJKg^{-1}K^{-1}\)

Answer : 101.54 °C

\[ \ln{ \Bigg( \frac{2}{1} \Bigg) } = - \frac{ 195 }{0.055} \Bigg( \frac{1}{T_2 + 273} - \frac{1}{76 + 273} \Bigg) \]\[\Rightarrow T_2 = 101.54\]

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.

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.

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.

\( b \) : Volume occupied by gas molecules

## Entropy Maximum principleAs 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 PrincipleAs 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 PrincipleAs 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 PrincipleAs 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. |