Liquids: viscosity, surface tension, vaporization, vapor pressure and boiling

In gases the intermolecular forces were negligible, and in solids they determine everything. Liquids are in between: the molecules are close together and attract each other, but they can still change places. Every property this week (viscosity, surface tension, vapor pressure, boiling point) is a different face of the same question: how strong are the intermolecular forces, and is the kinetic energy of the molecules enough to overcome them?

Golden rules:

1. Strong IMF = a sticky liquid: high viscosity, high surface tension, low vapor pressure, high boiling point.
2. Vapor pressure depends only on the substance and the temperature. Not on the size of the container or the amount of liquid.
3. The temperature stays constant during a phase change; the heat is computed with $q = n\Delta H$.
4. In Clausius-Clapeyron use kelvin and R = 8.314 J/(mol·K).

Viscosity and surface tension

Viscosity is the resistance of a liquid to flow. When molecules slide past each other they must overcome the intermolecular forces.

• It increases as the IMF get stronger: glycerol (three O–H) is hundreds of times more viscous than water.
• Long molecules that can tangle with each other (oils, polymers) are viscous.
• It decreases as the temperature rises: with more kinetic energy the molecules overcome the attraction more easily. That is why motor oil flows slowly on a cold morning.

Surface tension: the energy needed to increase the surface area of a liquid. A molecule in the interior is pulled in all directions; a molecule at the surface is pulled only inward and sideways. This net inward pull stretches the surface like a membrane: drops become spherical, and a water strider walks on water. Water has a very high surface tension because of H bonds; it decreases as the temperature rises.

Capillary action is a competition between two forces:

• Cohesive forces: among the liquid molecules.
• Adhesive forces: between the liquid and the surface of the container.

Between water and glass adhesion dominates: water rises in a thin tube, and the meniscus is concave. In mercury cohesion dominates: mercury is depressed in the tube, and the meniscus is convex.

Vaporization and the enthalpy of vaporization

In a liquid the kinetic energies of the molecules show a distribution. Sufficiently fast molecules at the surface overcome the attraction and escape into the gas phase: vaporization. The average energy of the remaining molecules drops, which is why vaporization cools (sweating).

Vaporization is endothermic. The heat needed to vaporize 1 mol of liquid at constant temperature and pressure is the enthalpy of vaporization ($\Delta H_{vap}$). The reverse, condensation, releases the same amount of heat:

$\Delta H_{\text{cond}} = -\Delta H_{vap}$

The heat supplied during a phase change does not raise the temperature; it is spent on overcoming the intermolecular forces. That is why the calculation is done with two separate formulas:

• The temperature changes: $q = mc\Delta T$
• The phase changes (T constant): $q = n\,\Delta H_{vap}$

Example 1. Turning 1.00 kg of water into steam at 100 °C:

$1000\ \text{g} \times \frac{1\ \text{mol}}{18.015\ \text{g}} \times 40.7\ \frac{\text{kJ}}{\text{mol}} = 2.26\times10^3\ \text{kJ}$

Heating the same water from 0 to 100 °C needs only $1000 \times 4.184 \times 100 = 418$ kJ. Vaporization consumes more than five times as much energy.

Example 2. At body temperature (37 °C) the $\Delta H_{vap}$ of water is 43.5 kJ/mol. When 10.0 g of sweat evaporates, $10.0/18.015 \times 43.5 = 24.1$ kJ of heat is drawn from the body. $\Delta H_{vap}$ increases slightly as the temperature drops.

Vapor pressure

Put some liquid in a closed container. At first only vaporization happens. As the number of molecules in the vapor increases, some hit back into the liquid and condense. Finally the two rates become equal: dynamic equilibrium. The pressure exerted by the vapor in this state is called the vapor pressure.

The vapor pressure:

• depends on the kind of substance: weak IMF → high vapor pressure → a volatile liquid.
• depends on temperature: it increases exponentially with temperature.
• does not depend on the amount of liquid, the surface area or the volume of the container (as long as vapor and liquid are present together at equilibrium).

If you enlarge the container, the pressure first drops, then more liquid evaporates and the pressure returns to the same value. If the amount is so small that all the liquid evaporates, equilibrium cannot be established; then the pressure is computed with $PV = nRT$ and comes out smaller than the vapor pressure.

Example 3 (does all the liquid evaporate?). 0.100 g of water is placed in an empty, closed 10.0 L container at 25 °C. The vapor pressure of water is 23.8 mmHg.

the most water that can be held in vapor at equilibrium:
  n = PV/RT = (23.8/760) × 10.0 / (0.08206 × 298.15) = 0.0128 mol = 0.231 g
0.100 g < 0.231 g  → ALL the water evaporates, no liquid is left, no equilibrium
actual pressure: n = 0.100/18.015 = 5.55×10^-3 mol
  P = nRT/V = 0.0136 atm = 10.3 mmHg   (< 23.8 mmHg)

If 1.00 g of water were put in the container, 0.231 g of it would evaporate, the rest would remain liquid, and the pressure would be 23.8 mmHg.

The Clausius-Clapeyron equation

The change of vapor pressure with temperature:

$\ln P = -\frac{\Delta H_{vap}}{R}\cdot\frac{1}{T} + B$

The graph of $\ln P$ against $1/T$ is a straight line with slope $-\Delta H_{vap}/R$. For two points:

$\ln\frac{P_2}{P_1} = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Three types of questions come: (a) find $P_2$, (b) find $\Delta H_{vap}$, (c) find $T_2$ (usually the boiling point).

The graphical method: with vapor pressures measured at different temperatures, plot $\ln P$ against $1/T$ and read the slope of the line. For water the slope comes out about $-4895$ K; $\Delta H_{vap} = -\text{slope} \times R = 4895 \times 8.314 = 40.7$ kJ/mol.

Example 4 (find P). The vapor pressure of water at 90.0 °C ($P_1 = 760$ mmHg, $T_1 = 373.15$ K, $\Delta H_{vap} = 40.7$ kJ/mol):

1/T2 − 1/T1 = 1/363.15 − 1/373.15 = 7.379×10^-5 K^-1
ln(P2/760)  = −(40 700 / 8.314) × 7.379×10^-5 = −0.3613
P2 = 760 × e^−0.3613 = 530 mmHg     (measured: 526 mmHg)

Example 5 (find ΔH). The vapor pressure of a liquid is 50.0 mmHg at 20.0 °C and 200 mmHg at 52.0 °C:

$\Delta H_{vap} = \frac{R\ln(P_2/P_1)}{\frac{1}{T_1} - \frac{1}{T_2}} = \frac{8.314 \times \ln 4.00}{\frac{1}{293.15} - \frac{1}{325.15}} = 3.43\times10^4\ \text{J/mol} = 34.3\ \text{kJ/mol}$

PITFALL · units. If you put $\Delta H_{vap}$ in kJ and use R = 8.314 J, the result is off by a factor of a thousand. Either make both J or take R as $8.314\times10^{-3}$ kJ/(mol·K). The pressure unit does not matter, because a ratio is taken; but the two pressures must be in the same unit.

PITFALL · the sign. If $T_2 > T_1$, then $P_2 > P_1$ must come out. If it came out the other way, you mixed up the order in the parentheses.

Boiling

When a liquid is heated, its vapor pressure rises. When the vapor pressure equals the external pressure, vapor bubbles can also form inside the liquid and the liquid boils. The normal boiling point is the boiling temperature when the external pressure is exactly 1 atm.

• On a mountain the external pressure is low → water boils at a lower temperature, and food cooks slowly.
• In a pressure cooker the pressure is high (about 2 atm) → water boils at about 120 °C, and food cooks quickly.
• Even if you keep supplying heat during boiling, the temperature stays constant; the energy goes to vaporization.

Example 6. At what temperature does water boil at an external pressure of 0.700 atm (532 mmHg)? Solve Clausius-Clapeyron for $T_2$:

$\frac{1}{T_2} = \frac{1}{T_1} - \frac{R\ln(P_2/P_1)}{\Delta H_{vap}} = \frac{1}{373.15} - \frac{8.314 \ln(0.700)}{40\,700} \Rightarrow T_2 = 363.3\ \text{K} = 90.1\ ^\circ\text{C}$

Common mistakes