evaporation condensation boiling and saturated vapour

Evaporation is the change of a liquid into vapour from its free surface: the fastest molecules leave the liquid, so during evaporation the internal energy of the liquid decreases, while condensation is the reverse process and releases heat.

In a closed vessel both processes occur at once, and when as many molecules evaporate as return in the same time, dynamic equilibrium is reached and the vapour is called saturated. The saturated vapour pressure $p_{sat}$ depends only on temperature and does not depend on volume as long as liquid remains in the vessel.

Boiling is vaporisation throughout the whole volume of the liquid; it begins when the saturated vapour pressure becomes equal to the external pressure, which is why at $p = 101$ kPa water boils at $100$ °C, while in a pressure cooker it boils at a higher temperature. During boiling the temperature of the liquid stays constant, and all the supplied heat goes into vaporisation. The heat absorbed in vaporisation or released in condensation of a mass $m$ is found from the specific latent heat of vaporisation $L$:

$$Q = Lm.$$

Water has already been heated to $100$ °C and boils at normal pressure. How much heat is needed to turn $m = 0.25$ kg of this water completely into vapour? The specific latent heat of vaporisation of water is $L = 2.26\cdot 10^6$ J/kg.

Water of mass $m = 0.25$ kg is first heated from $20$ °C to $100$ °C and then completely vaporised. Compare the heat of heating $Q_1$ with the heat of vaporisation $Q_2$ and find their ratio. The specific heat capacity of water is $c = 4200$ J/(kg·°C), $L = 2.26\cdot 10^6$ J/kg.

A sealed rigid vessel contains a little liquid water and its saturated vapour at $20$ °C; the vapour pressure is $p_{sat} = 2.3$ kPa. The vessel is heated to $40$ °C, where $p_{sat} = 7.4$ kPa. What happens to the vapour pressure and to the mass of vapour, and will boiling start? To estimate the density of saturated vapour use $\rho = \frac{pM}{RT}$, with $M = 0.018$ kg/mol and $R = 8.31$ J/(mol·K).

Safe home experiment on evaporation (no heating). Place two saucers with equal volumes of water ($20$ ml each) side by side at room temperature: one narrow, one wide. Put a third wide saucer in a weak air stream from a desk fan, or fan it with a sheet of cardboard. Every $10$ minutes for an hour, measure the mass of the saucers and record the results, then plot $m(t)$ and compare the slopes: the larger the surface area and the stronger the air movement, the faster the evaporation. Study boiling and its dependence on pressure only in a simulation (for example, PhET "States of Matter" or "Under Pressure"): never heat sealed vessels. In the simulation raise the external pressure and note the temperature at which boiling starts, then lower the pressure and repeat; record both temperatures.

Show that saturated vapour is the result of dynamic equilibrium and that its pressure grows non-linearly with temperature.

A diagram: a closed vessel with liquid at the bottom and vapour above; two arrows of equal length labelled "evaporation" and "condensation". Next to it a graph of $p_{sat}(T)$: a smoothly rising curve with marked points at $20$ °C ($2.3$ kPa), $40$ °C ($7.4$ kPa) and $100$ °C ($101$ kPa).

A player loses $0{,}10\ \mathrm{kg}$ of sweat in 20 minutes. The sweat is mostly water; at skin temperature about $35^\circ\mathrm{C}$ its specific latent heat of vaporization is about $2{,}4\times10^6\ \mathrm{J/kg}$. Evaporation carries away $Q=mL=0{,}10\ \mathrm{kg}\times2{,}4\times10^6\ \mathrm{J/kg}=2{,}4\times10^5\ \mathrm{J}.$

The scenario exercises evaporation below boiling: faster water molecules leave the liquid surface, so the remaining liquid cools because internal energy is spent on latent heat $L$. The rate is limited by diffusion of vapour away from the skin and by the surrounding air not being saturated.

To cook pasta, a pot contains $1{,}5\ \mathrm{L}$ of water at $20^\circ\mathrm{C}$ at normal pressure $1\ \mathrm{atm}$, so it boils at $100^\circ\mathrm{C}$. An electric stove delivers $1{,}5\ \mathrm{kW}$ useful power. With $c=4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$ and $m=1{,}5\ \mathrm{kg}$, heating to boiling needs $Q_1=mc\Delta T=1{,}5\ \mathrm{kg}\times4180\ \mathrm{J\,kg^{-1}\,K^{-1}}\times80\ \mathrm{K}\approx5{,}0\times10^5\ \mathrm{J},$ giving $t_1=Q_1/P\approx5{,}0\times10^5\ \mathrm{J}/1500\ \mathrm{W}\approx334\ \mathrm{s}$. Boiling away $0{,}20\ \mathrm{kg}$ then needs $Q_2=mL=0{,}20\ \mathrm{kg}\times2{,}26\times10^6\ \mathrm{J/kg}=4{,}52\times10^5\ \mathrm{J}.$

This exercises boiling and saturated vapour: at $100^\circ\mathrm{C}$ the saturated vapour pressure of water equals the external $1\ \mathrm{atm}$, so vapour bubbles can grow inside the liquid instead of being crushed. The plateau in temperature during boiling is explained by heat $Q=mL$ increasing internal energy and doing expansion work $W=p\Delta V$ rather than raising kinetic energy per molecule.

On a cold morning a student breathes onto a window at $5^\circ\mathrm{C}$. Exhaled air is near $37^\circ\mathrm{C}$ and nearly saturated, with saturation vapour pressure about $6{,}3\ \mathrm{kPa}$; at $5^\circ\mathrm{C}$ saturated vapour pressure is only about $0{,}87\ \mathrm{kPa}$. If $0{,}002\ \mathrm{kg}$ of water condenses, latent heat released is $Q=mL=0{,}002\ \mathrm{kg}\times2{,}4\times10^6\ \mathrm{J/kg}=4{,}8\times10^3\ \mathrm{J}.$

This exercises condensation and saturated vapour: cooling below the dew point makes the actual vapour pressure exceed the saturation value, so vapour molecules stick to the glass and release latent heat, warming the surface. The same saturation curve explains why fog forms when air cools at night.