# Do volatile solutes affect the vapour pressure of solutions?

If a non-volatile solute is added to water it will reduce vapour pressure as some space is occupied by non-volatile solute but what if the solute density is higher than water does vapour pressure still decrease?

• The vapor pressure of the water will still decrease. The sum of the vapor pressures (of solvent and solute) might increase, though.
– Karsten
Commented Apr 21 at 19:42
• The question title talks about volatile solutes, while the question body does not. Commented Apr 22 at 8:17
• Question specifies nonvolatile. The idea of space is incorrect. Changing the surface area is a kinetic effect not a VP change. That requires a change in the chemical potential of the solvent. A volatile [ideal] solute follows Raoult's Law. A volatile, insoluble substance steam distills. Commented Apr 23 at 19:31

The covered surface explanation is a kinetic explanation while the net evaporation is still ongoing. It does not provide the thermodynamic explanation when at equilibrium. While it slows down evaporation, it also slows down condensation.

Thermodynamic explanation of decreasing of vapor pressure of a compound with its molar fraction is related to its chemical potential.

$$\mu_i = \left(\dfrac{\partial G }{ \partial n_i}\right)_{p,T,n_j,j \ne i} \tag{1}$$

This potential decreases for ideal solutions with the molar fraction

$$\mu_{i,\text{(l)}} = \mu_{i,\text{(l)}}^{\circ} + RT \ln {x_i} \tag{2}$$

and so does the vapor pressure for ideal gas:

$$\mu_{i,\text{(g)}} = \mu_{i,\text{(g)}}^{\circ} + RT \ln {\left(\frac{p_i}{p_{i,0}}\right)} \tag{3}$$

For equilibrium, there must be valid

$$\mu_{i,\text{(l)}} = \mu_{i,\text{(g)}} \tag{4}$$ $$\mu^{\circ}_{i,\text{(l)}} = \mu^{\circ}_{i,\text{(g)}} \tag{4a}$$

and therefore

$$RT \ln {x_i} = RT \ln {\left(\frac{p_i}{p_{i,0}}\right)} \tag{5}$$

From that is implied for ideal solutions

$$p = p_0 \cdot x_i \tag{6}$$

• It does not slow down anything condensation can happen at any active site not only at the liquid surface and depends on the actual vapor pressure in the environment. It slows down mass transfer from the surface. Putting the cover on the pot stops the vapor from leaving and filling the room. Commented Apr 23 at 21:38
• Condensation rate is not independent on type of surface. Commented Apr 24 at 6:39