Gas pressure humidity and phase equilibrium

Phase equilibrium between a liquid and its vapor is a state in which, over the same time, as many molecules leave the liquid surface as return to it. Vapor in such equilibrium is called saturated, and the pressure it exerts is the saturated vapor pressure $p_0$ (also $p_{\text{sat}}$). In the particle model, vapor pressure comes from molecules hitting the walls; at equilibrium the concentration of vapor molecules at a given temperature is fixed, so $p_0$ does not depend on volume: on compression the excess vapor simply condenses. As temperature rises, $p_0$ increases rapidly because a larger fraction of molecules has enough energy to escape the liquid. Humidity is described by the partial pressure of water vapor $p$ and the relative humidity $\varphi = \frac{p}{p_0}\cdot 100\%$. When moist air cools at constant $p$, the dew point is reached — the temperature at which $p = p_0(T)$ and condensation begins; boiling occurs when $p_0(T) = p_{\text{external}}$.

At $t = 20\,^\circ\text{C}$ the saturated vapor pressure of water is $p_0 = 2.33\ \text{kPa}$. The partial pressure of water vapor in the air is $p = 1.40\ \text{kPa}$. Find the relative humidity $\varphi$.

A room of volume $V = 60\ \text{m}^3$ has temperature $t = 20\,^\circ\text{C}$ and relative humidity $\varphi = 60\%$. The saturated vapor pressure at this temperature is $p_0 = 2.33\ \text{kPa}$, the molar mass of water is $M = 18\ \text{g/mol}$, and $R = 8.31\ \text{J/(mol}\cdot\text{K)}$. Estimate the mass of water vapor in the room.

In a pressure cooker the pressure above the water is about $p_{\text{external}} \approx 203\ \text{kPa}$, which corresponds to $p_0(T) \approx 203\ \text{kPa}$ at $T \approx 120\,^\circ\text{C}$. At an altitude of 2 km the atmospheric pressure is about $79.5\ \text{kPa}$, and water boils there at about $93.5\,^\circ\text{C}$. (a) Explain these temperatures using the boiling condition and phase equilibrium. (b) Why does food cook faster in a pressure cooker than in an open pot at an altitude of 2 km?

Safe self-study activity (table and simulation). 1. In a spreadsheet, make a table of $p_0(T)$ for water from $0$ to $30\,^\circ\text{C}$ (for example: $0\,^\circ\text{C}$ — 0.61 kPa, $10\,^\circ\text{C}$ — 1.23 kPa, $20\,^\circ\text{C}$ — 2.33 kPa, $30\,^\circ\text{C}$ — 4.24 kPa). 2. Plot $p_0$ versus $T$. 3. Set the partial pressure $p = 1.40\ \text{kPa}$ and for each $T$ compute $\varphi(T) = \frac{p}{p_0(T)}\cdot 100\%$; find the temperature at which $\varphi = 100\%$ — this is the dew point. 4. Check the qualitative behavior in a simulation (for example, PhET “States of Matter” or “Gas Properties”): reduce the volume while liquid is present and verify that the vapor pressure does not grow without limit. Do not heat closed vessels and do not create excess pressure in household containers.

Show that the saturation curve $p_0(T)$ is a boundary: in the presence of liquid the vapor pressure cannot exceed $p_0(T)$, and the intersection of the horizontal line $p = \text{const}$ with this curve on cooling gives the dew point.

A graph with $T$ on the horizontal axis and $p$ on the vertical axis: the rising curve $p_0(T)$; a horizontal line $p = 1.40\ \text{kPa}$; the intersection point is marked as the dew point, and to its left is the condensation region (fog, dew). Inset: a particle-level sketch of the liquid surface where the evaporation and condensation arrows are equal in magnitude at phase equilibrium.

Air at $25^\circ\text{C}$ has relative humidity $60\%$. The saturated vapor pressure of water at $25^\circ\text{C}$ is $3.17\,\text{kPa}$, so the actual water-vapor partial pressure is $p_v = 0.60 \times 3.17\,\text{kPa} \approx 1.90\,\text{kPa}$. If the air cools to $17^\circ\text{C}$, where the saturated vapor pressure is about $1.94\,\text{kPa}$, the vapor becomes saturated ($p_v \approx p_{\text{sat}}$) and further cooling makes water condense as dew or fog. The dew point is therefore about $17^\circ\text{C}$.

This uses the phase equilibrium between liquid water and its vapor: relative humidity compares the actual vapor pressure $p_v$ with the saturated vapor pressure $p_{\text{sat}}(T)$; condensation begins when $p_v$ reaches $p_{\text{sat}}$ at the dew-point temperature.

In an open pan at sea level, water boils when its saturated vapor pressure equals the external pressure $101.3\,\text{kPa}$; this happens at $100^\circ\text{C}$. In a pressure cooker the pressure is raised to $2.0\,\text{atm}$ ($202.6\,\text{kPa}$). Water then boils at about $120^\circ\text{C}$, because its saturated vapor pressure must again match the higher external pressure. Food cooks faster at this higher temperature.

This is phase equilibrium for a liquid–vapor boundary: boiling occurs when $p_{\text{sat}}(T) = p_{\text{external}}$, so increasing gas pressure raises the boiling temperature. The saturated vapor pressure is a property of the liquid and depends strongly on temperature.

A tennis player loses $1.5\,\text{kg}$ of sweat in $1\,\text{h}$. The latent heat of vaporization of water at skin temperature is about $L_v \approx 2.4\times10^6\,\text{J/kg}$. Evaporating all that sweat removes $Q = m L_v = 1.5 \times 2.4\times10^6 = 3.6\times10^6\,\text{J}$, giving an average cooling power $P = Q/t = 3.6\times10^6\,\text{J} / 3600\,\text{s} = 1000\,\text{W}$. On a dry day (relative humidity $30\%$) most sweat evaporates and cools the body; on a humid day (relative humidity $90\%$) the air's water-vapor partial pressure is close to saturation, net evaporation slows, so sweat drips and cooling is weaker.

The skin surface is in dynamic phase equilibrium with water vapor: evaporation rate depends on how far the vapor pressure in air is below the saturated vapor pressure at skin temperature. High humidity reduces that difference, slows evaporation, and lowers the cooling power carried by the latent heat.