Vapour Pressure, Surface Tension and Viscosity: Quick Revision

A brief note on Vapour Pressure, Surface Tension and Viscosity. Read it before a test. Try each question at the end before you open its answer.

Vapour pressure and boiling

• Vapour pressure: the pressure of the vapour in equilibrium with its liquid. It depends on the liquid and the temperature only, not on the amount of liquid or the vessel.
• Weak forces between molecules mean a high vapour pressure, a low boiling point and a small $\Delta H_{\text{vap}}$.
• A liquid boils when its vapour pressure equals the outside pressure. The normal boiling point is the value at $101.325\ \text{kPa}$.
• $\ln\dfrac{P_2}{P_1} = -\dfrac{\Delta H_{\text{vap}}}{R}\left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)$, with $T$ in K and $\Delta H$ in J/mol.
• Trouton's rule: $\Delta H_{\text{vap}}/T_b$ is about $85$ to $88\ \text{J K}^{-1}\ \text{mol}^{-1}$ for liquids without hydrogen bonds.

Surface tension

• $\gamma$ is a force per length (N/m) and an energy per area (J/m²). Strong forces mean a high $\gamma$, and $\gamma$ falls as $T$ rises.
• Excess pressure: $\dfrac{2\gamma}{r}$ in a drop or an air bubble, and $\dfrac{4\gamma}{r}$ in a soap bubble.
• Capillary rise: $h = \dfrac{2\gamma\cos\theta}{\rho g r}$. For $\theta > 90^\circ$ the liquid falls.
• Drop count: $\dfrac{\gamma_1}{\gamma_2} = \dfrac{\rho_1n_2}{\rho_2n_1}$.

Viscosity

• $F = \eta A\dfrac{v}{z}$. The unit is Pa s. $1\ \text{poise} = 0.1\ \text{Pa s}$, and $1\ \text{cP} = 1\ \text{mPa s}$.
• For a liquid $\eta$ falls as $T$ rises. For a gas it rises.
• Poiseuille: $Q = \dfrac{\pi\Delta Pr^4}{8\eta L}$. Stokes: steady speed $\dfrac{2r^2(\rho_s - \rho)g}{9\eta}$.
• Ostwald viscometer: $\dfrac{\eta_1}{\eta_2} = \dfrac{\rho_1t_1}{\rho_2t_2}$.

A worked example

A liquid with $\gamma = 40\ \text{mN/m}$ and $\rho = 850\ \text{kg/m}^3$ wets glass completely. Find its rise in a tube of radius $0.25\ \text{mm}$, and the radius that gives a rise of $20\ \text{mm}$.

1. The rise. $h = \dfrac{2\gamma}{\rho g r} = \dfrac{2 \times 0.040}{850 \times 9.81 \times 0.25 \times 10^{-3}} = 38.4\ \text{mm}$.
2. The radius. The rise is proportional to $1/r$, so $r = 0.25 \times \dfrac{38.4}{20} = 0.48\ \text{mm}$.

Practice with new numbers

A program to try

The program finds Trouton's ratio of three liquids.

Mistakes to avoid

• Celsius in the Clausius-Clapeyron equation. Use kelvin.
• $\Delta H$ in kJ/mol with $R$ in J. Change to J/mol.
• The diameter in a capillary or Stokes formula. Use the radius.
• $\dfrac{2\gamma}{r}$ for a soap bubble. It has two surfaces, so use $\dfrac{4\gamma}{r}$.
• A flow rate that goes as $r$. It goes as $r^4$.

Check yourself

1. Does the vapour pressure of a liquid depend on the size of the vessel?

2. The radius of a tube is halved. By what factor does the laminar flow rate change at the same pressure difference?

3. Why does mercury fall in a glass tube?

Sources

• OpenStax, Chemistry 2e, Chapter 10, sections 10.2 and 10.3.
• CRC Handbook of Chemistry and Physics, for the vapour pressures, surface tensions and viscosities.