Evaporation from a capillary tube Consider a capillary tube (say from a liquid / capillary thermometer), that means a tube of small internal diameter which holds liquid by capillary action . The tube is fulfilled with water and closed at one end. How long it would take for water to completely evaporate out through the open end of the tube?  
To be specific, let the diameter of the tube be $d=1$mm, the length of the tube $L=1$m, the environmental temperature $T=300$ K and pressure $P=10^5$ Pa, the relative humidity $f=60$ %, the contact angle a) $\theta=0^{\circ}$ and b) $\theta=180^{\circ}$.  
Does it have any effect on the time when the tube is positioned horizontally or vertically(open end above)?

 A: As a rough answer: you can calculate the rate of evaporation from a water surface for the given temperature, humidity and a parabolic surface area. 
The problem starts a few seconds later: The relative humidity above the water surface will rise close to 100% and now the rate of evaporation is limited by the diffusion of water molecules through the air out of the capillary. If you are not interested in the time dependence one can consider these two cases. After some time the rate of evaporation will be equal to the diffusion rate at the end of the capillary. Additionally the length of the tube compared to the length of the water column can influence the result because it determines the length of the capillary through which the water vapour has to diffuse. To your last question: In theory yes, as you will have less diffusion against the additional gravity, in practice it will only matter if you have a much longer tube. (The opposite is true if the experiment is surrounded by normal air. The mass of air molecules like O$_2$, N$_2$, ... is higher than H$_2$O as Georg pointed out) 
A: OK, i tried to do a rough estimation, supposing that the contact angle is $\theta=\frac{\pi}{2}$
A start point is Fick's law of diffusion 
$$j=-D\frac{dn}{dx}$$ where $D$ $(\frac{m^2}{s})$  is diffusivity of water vapor in air , $n$ $(\frac{1}{m^3})$ is concentration of water vapor in air and $x$- axis lies in the center of the capillary. Using ideal gas law $p=nkT$  
$$j=-\frac{D}{kT}\frac{dp}{dx}$$  If we multiply the last eq. with $m_0$(mass of water molecule) then  
$$\frac{dm}{Sdt}=-\frac{D\mu}{RT}\frac{dp}{dx}$$  where $\mu$ is the molar mass, $R$ is the universal gas constant, $S$ is cross-sectional area of the capillary.
On the other hand, $m=\rho_w Sx$ where $x$ is the length of the water column in the capillary, $\rho_w$ is density of water. So  
$$\frac{dx}{dt}=-\frac{D\mu}{\rho_w RT}\frac{dp}{dx}=-\frac{D}{\rho_w}\frac{d\rho}{dx}$$   
Because the diffusion process is quasistatic, $\frac{d\rho}{dx}=\frac{\rho_s-\rho}{L-x}$ holds.  
Here $\rho_s$ is saturation vapor density over flat water, $\rho=f\rho_s$ is water vapor density in the environment. Finally:  
$$\frac{dx}{dt}=-\frac{D\rho_s(1-f)}{\rho_w(L-x)}$$   
After integrating from $x=0$ to $x=L$ one gets evaporation time:  
$$t=\frac{\rho_w L^2}{2D\rho_s(1-f)}$$  
Now, a numerical estimation:  
$L=1m$
$D=3*10^{-5}\frac{m^2}{s}$
$\rho_w=10^{3}\frac{kg}{m^3}$
$\rho_s=1.8*10^{-2}\frac{kg}{m^3}$
$f=0.6$ 
So $t=7.3$ years
