Question 9.1: At 18°C, water rises at a height of 5 cm in a glass capillar...
At 18°C, water rises at a height of 5 cm in a glass capillary of 0.3 mm radius. The density of water at this temperature is 998.6 kg m^{-3}. Based on the given data, calculate the equilibrium vapor pressure of water above a water drop with a radius of 10 nm. The vapor pressure over a plane water surface is 2.1 kPa. Suppose that the surface of water in the capillary is spherical with a zero contact angle at the glass wall.
First we shall determine the surface tension of water from (9.29):
h=\frac{2\gamma }{\rho gr} (9.29)
\gamma =\frac{\rho grh}{2}=73.55 N m^{-1}.The vapor pressure above the water droplet can be determined using the Kelvin equation (9.43)
P_{r} = P_{0} e^{\frac{2 \gamma M}{RTr\rho} } =2.343 kPa.Thus, the vapor pressure of a 10 nm radius water droplet is 11.6 % higher than the equilibrium pressure over a flat water surface.