Answer to Question #177321 in Molecular Physics | Thermodynamics for swarnajeet kumar

Question

Derive Clausius-Clapeyron equation. Explain why the boiling point of water increases with increasing pressure. 

Answer 1

$\displaystyle \frac{dP}{dT}= \frac{L}{T \Delta v} = \frac{\Delta s}{\Delta v}$

where ${\displaystyle \mathrm {d} P/\mathrm {d} T}$ is the slope of the tangent to the coexistence curve at any point, ${\displaystyle L}$ is the specific latent heat, ${\displaystyle T}$ is the temperature, ${\displaystyle \Delta v}$ is the specific volume change of the phase transition, and ${\displaystyle \Delta s}$ is the specific entropy change of the phase transition.

There is a functional relationship between the phase transition temperature and external pressure, and during the phase transition the derivative ${\displaystyle \left({\partial p \over \partial V} \right) _ {T}}$breaks. Then the isotherms for the substance will have the characteristic form shown in the figure.

The horizontal section of the isotherm corresponding to the phase transition is essential for the derivation. Everything on the left and on the right from this area is in the same phase. We carry out the Carnot cycle with an infinitesimal temperature difference as follows: first, we transfer heat to the body, changing it from state 1 to state 2, then adiabatically cool it to a temperature dT, after which we close the cycle by removing heat and transferring matter in phase 1 followed by adiabatic heating. The work done is equal to the area of ​​the cycle:

$\delta A =dp(V_2 - V_1)$

Transferred heat is equal to

$\delta Q= L m$

According to the Carnot theorem,

$\displaystyle \delta A = \delta Q \frac{T_2 - T_1}{T_1} = \delta Q \frac{\delta T}{T}$

Therefore,

$\displaystyle dp(V_2-V_1) = Lm \frac{\delta T}{T}$

$\displaystyle \frac{dp}{dT} = \frac{Lm}{T\Delta V} = \frac{L}{T\Delta v}$