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Take a sample of phase *I* and slowly transform it into phase
*II*. Some heat *Q* will be absorbed.

- Since this is reversible,
(1) - Since in equilibrium
*P*=_{1}*P*, the process occurs at constant pressure_{2} - Since
*T*=_{1}*T*, it occurs at constant temperature_{2}

On the *P*-*T* diagram phase transition occur along a *line*
*P*(*T*). Let's calculate the tangent to this line *dP*/*dT*.

Chemical potentials are equal:

Differentiating:(2) |

From (2):

or Since we obtained(3) |

- For liquid-gas phase transitions
*q*>0. Since*v*>_{2}*v*,_{1}*dT*/*dP*= (*dP*/*dT*)^{-1}>0. Boiling point temperature*increases*with pressure - For solid-liquid phase transitions
*q*>0 (exception: ). Usually*v*>_{2}*v*, and melting point increases with pressure. For_{1}*water**v*<_{2}*v*, and it_{1}*decreases*--that is why skating & skiing is possible!

Assumptions:

- 1.
- Phase transition heat
*q*does not depend on*T*and*P*: - 2.
- Gas is ideal (
*PV*=*NkT*), and its volume is much grater than of liquid:

If we move from phase *I* to phase *II*, thermodynamic potential *G*
is continuous *(why?)*, but its *first derivative S*
jumps. Another possibility:

**Definition:**- if the
*n*-th derivative of free energy is discontinuous, and all lower derivatives are not, the transition is of*n*-th order. **Phase equilibrium:**- first order phase transition!
**Second order phase transitions:**- Curie point, some liquid crystal phase transitions...
**Practical guide:**- if
*q*=0, the transition is of the second order (or higher!). **First-order-close-to-second-order transitions:***q*small (liquid crystals)

© 1997

Tue Oct 7 22:16:36 EDT 1997