The capillary action lets a liquid rise in a narrow tube to a certain height. In this, the liquid gains some potential energy. According to the Conservation of Energy, the energy must come from somewhere. So where does the energy come from?

One possible solution that I thought was that the molecules of the capillary exert forces on the liquid molecules. These forces are responsible for potential energy being stored in the liquid molecules. The molecules could consume this potential energy and rise in the tube. But I don't know whether this is correct or not. Could anyone tell me if this is correct or some other solution exists?


I had the same question and found the following excellent answer, which I've restated here with a correction:

Capillary action is minimizing the Helmholtz free energy F of the system.

We can actually find a formula for the capillary height that does this once we have accounted for the three components of energy that change under capillary action:

With $\gamma_{L}$ and $\gamma_{G}$ as the surface tensions at the liquid-solid and gas-solid interfaces respectively (defined by $\gamma = \frac{\delta F}{\delta A})$ then:

If the liquid rises by an amount δh, then we have:

  1. $\gamma_{L}$ 2πr δh increase in the surface energy due to the liquid-solid contact
  2. $\gamma_{G}$ 2πr δh decrease in the surface energy due to the gas-solid contact
  3. $\rho g$ $\pi r^2$ h δh increase in the gravitational potential energy of the system where ρ is the difference between the liquid and gas densities, because we're also displacing air and changing the potential energy of that little parcel of air as well.

Overall change in energy $\delta F = \delta h(\pi r^2 \rho g h + 2\pi r(\gamma_L - \gamma_G))$.

At equilibrium $\delta f = 0$, so $r \rho g h + (\gamma_L - \gamma_G) = 0$ and we have

$$h = \frac{\gamma_G - \gamma_L}{\rho g r}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.