One consequence of the pressure difference across a curved liquid-air interface is the wellknown effect that water rises up in a narrow tube in spite of gravity. The word capilla means hair in Latin; if the tube were hair thin, the rise would be very large. To see this, consider a vertical capillary tube of circular cross section (radius a) inserted into an open vessel of water (Fig. 1). The contact angle between water and glass is acute. Thus the surface of water in the ⊳ capillary is concave. This means that there is a pressure difference between the two sides of the top surface. This is given by
P$_i$ – P$_o$ = $\frac{2S}{r}=\frac{2S}{a \ sec \ \theta}$
P$_i$ – P$_o$ = $\frac{2S}{a} \ cos \ \theta$ (1)
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| Fig.1 Capillary rise, (a) Schematic picture of a narrow tube immersed water. (b) Enlarged picture near interface. |
Thus the pressure of the water inside the tube, just at the meniscus (air-water interface) is less than the atmospheric pressure. Consider the two points A and B in Fig. 1(a). They must be at the same pressure, namely
P$_o$ + $\rho gh=P_i =P_A$ (2)
where ρ is the density of water and h is called the capillary rise [Fig. 1(a)]. Using Eq. (1) and (2) we have
$\rho gh=P_i -P_0=\frac{2S}{a} \ cos \ \theta$
The discussion here, and the Eqs.
W = (P$_i$ – P$_o$) 4πr$^2$∆r
and (P$_i$ – P$_o$) = $\frac{2S_{la}}{r}$
make it clear that the capillary rise is due to surface tension. It is larger, for a smaller a. Typically it is of the order of a few cm for fine capillaries. For example, if a = 0.05 cm, using the value of surface tension for water (Table 1),
we find that
$h=\frac{2S}{\rho ga}$
$h=\frac{2(0.073 N.m^{-1})}{(1000 \ kg/m^3)(9.8 \ m/s^2)(5 \times 10^{-4}) \ m}=9.98 \ cm$
Notice that if the liquid meniscus is convex, as for mercury, i.e., if cos θ is negative then from Eq. (9.28) for example, it is clear that the liquid will be lower in the capillary !
Example 9.10
The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at temperature of the experiments is 7.30 × 10$^{-2}$ Nm$^{-1}$ . 1 atmospheric pressure = 1.01 × 10$^5$ Pa, density of water = 1000 kg/m$^3$ , g = 9.80 m s$^{-2}$ . Also calculate the excess pressure.
The excess pressure in a bubble of gas in a liquid is given by 2S/r, where S is the surface tension of the liquid-gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is 4S/r.)
The radius of the bubble is r. Now the pressure outside the bubble Po equals atmospheric pressure plus the pressure due to 8.00 cm of water column.
That is P$_o$ = (1.01 × 10$^5$ Pa + 0.08 m × 1000 kg m$^{–3}$ × 9.80 m s$^{–2}$) = 1.01784 × 10$^5$ Pa
Therefore, the pressure inside the bubble is
P$_i$ = P$_o$ + 2S/r
= 1.01784 × 10$^5$ Pa+ (2 × 7.3 × 10$^{-2}$ Pa m/10$^{-3} m)
= (1.01784 + 0.00146) × 10$^5$ Pa = 1.02 × 10$^5$ Pa
where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical ! (The answer has been rounded off to three significant figures.) The excess pressure in the bubble is 146 Pa.
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