Thermodynamics of pressure If the total pressure on a solid in equilibrium with its vapor is increased, will the vapor pressure of the solid increase?
Check this article:
http://twt.mpei.ac.ru/TTHB/2/KiSyShe/eng/Chapter5/5-8-Phase-changes-at-unequal-phase-pressures.html
By definition of the mechanical equilibrium the total pressure of the gas environment is equal to the hydrostatic pressure of the solid at the interface. At constant temperature any increase in the hydrostatic pressure of the solid will increase its specific Gibbs free energy that means its chemical potential in the solid phase. For the thermodynamic equilibrium between solid and its vapor requires that their mutual chemical potentials should be equal. Chemical potential of the vapour is given by Chemical Potential at the standard state plus RT Ln (Partial Pressure of vapor -fucasity in the nonideal case) . That means spontaneously the vaporization of the solid phase takes place in order to increase its partial pressure in the gaseous phase till the thermodynamic equilibrium is established.
Note: For one component system the specific or molal Gibbs free energy is equal to its chemical potential.
A trickier question than one might immediately think. Let me rephrase it. The vapour pressure is defined as that pressure, at which a solid and its vapour are at equilibrium. So at first sight, the answer must be yes: if the pressure is increased, then solid and vapour are at equilibrium at a higher pressure, so the pressure has increased.
The rub is that, if you merely say that you have increased the pressure, we have no idea what the other thermodynamic variables do. And in general, you will leave the coexistence curve, so that the system you consider will become either all-solid or all-vapour, in whihc case the question loses meaning. So be careful to state that you increase the pressure *along the coexistence and then, indeed, the vapour pressure will increase.
How do you want to increase the pressure? If the solid is a pure compound, in equilibrium with its vapour, and no other substance present, the sublimation pressure (at given temperature) is a fixed property: If you try to compress the system, the vapour condenses, and only the single-phase solid is left.
If you increase pressure by putting the solid into a pressurized fluid (compressed nitrogen, argon, …) the gas molecules may interact with those of the solid compound. Usually this leads to an enhanced vapour pressure. This is the principle underlying “supercritical fluid extraction” processes.
When a solid is in equilibrium with its vapour pressure then the chemical potential in solid and gas phase are equal. That is
µ(s,T, p) = µ°(T) + RT ln(p) + RT ln(phi)
where µ is chemical potential, p is pressure, phi is fugasity coefficient,
T is temperature in Kelvins and R = 8.31451 J/Kmol.
The thermodynamic properties of solids are rather insensitive to pressure up to the moderate pressures (< 200 bar) so if the temperature is kept constant the left side of equation remains unchanged.
If the gas phase is assumed ideal (phi= 1) then the right side will also remain constant. Thus, in one component system the gas phase will disappear if pressure is increased for example by a piston. If there are other components present the partial pressure of the solid will be constant so its concentration (mole fraction) in gas phase will decrease.
If the non-ideal behavior of gas phase is included in consideration, the the
product of partial pressure and fugacity coefficient, that is the fugacity in gas
phase, will remain constant. So the pressure can change a little. It can be
estimated by estimating the change in fugacity by a proper state function.
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