# How does capillary action work in tubes which are closed at the top?

In all textbooks the examples and descriptions use tubes which are open at both ends. Our teacher remarked that if it were closed at one end like a test tube (albeit a thin one) even then capillary rise would occur. I'm not sure how that is possible. Will the rise be less because it has to compress the air already in it? Will it rise at all?

• It will rise less because it has to compress the air already in it. Otherwise, no fundamental difference. – Chet Miller Apr 16 '16 at 19:46

## 1 Answer

The capillary action results from the competition of multiple effects:

• Wetting of the capillary walls by the liquid (owing to a "favorable" liquid/wall surface tension)

• Tendency to minimize the interfacial area between liquid and air (owing to an "unfavourable" liquid/air surface tension)

• Gravity that is pulling down on the liquid column

• The various pressure constraints the liquid is subject to

In a capillary, the first two effects give rise to the formation of a meniscus as depicted in the figure below The angle $\theta$ is called associated to the wetting angle. For simplicity, we will say that it is $\theta = 0$ meaning that the liquid wets as much as possible the capillary. In that case the meniscus is a half sphere with diameter equal to the width $w$ of the capillary.

Next, we invoke (without proof) the Laplace equation relating the pressures at curved interface i.e. the pressure inside $P_i$ and outside $P_o$ the liquid at the meniscus interface:

\begin{equation} P_o - P_i = \frac{4\gamma_{air/l}}{w} \end{equation}

We can figure out $P_i$ by using the hydrostatic equilibrium which imposes that at the free surface level of the container, the pressure in the liquid ought to be $P_{atm}$ and that $P_{atm} = P_i+\rho_l g h$.

We thus end up with the equation:

\begin{equation} P_o -(P_{atm}-\rho_l g h) = \frac{4\gamma_{air/l}}{w} \end{equation} If we assume that the trapped air in the tube is describable as an ideal gas and that there is thermal equilibrium, we have that $P_o = P_c(h) = P_{atm}(1-\frac{\pi w^2 h}{4 V_c^0})^{-1} \approx P_{atm}(1+\frac{\pi w^2 h}{4 V_c^0})$ where $V_c^0$ is the initial volume of the capillary.

This gives eventually, as a first approximation:

\begin{equation} P_{atm}(1+\frac{\pi w^2 h}{4 V_c^0}) -(P_{atm}-\rho_l g h) = \frac{4\gamma_{air/l}}{w} \end{equation}from which we get the estimate:

\begin{equation} h = \frac{4\gamma_{air/l}/w}{\rho_l g + \frac{P_{atm}\pi w^2}{4V_c^0}} \end{equation}

The term $\frac{P_{atm}\pi w^2}{4V_c^0}$ makes the height at which the liquid climbs smaller than the with the two ends open and this owing to the compression of the air trapped in the tube.