Let us consider, a capillary tube is vertically dipped in a liquid. The surface tension of the liquid is T and the density of the liquid is \( \rho \). Now the whole arrangement is enclosed in an exhausted bell jar. The liquid evaporates, when the space within the bell jar is saturated then the equilibrium is reached.
Case I:
In case of the liquid which wets the capillary tube, as in case of water, alcohol, etc, the liquid rises up inside the capillary tube over the liquid surface outside the capillary tube, as shown in the Fig.1. The liquid meniscus at the top is concave upwards.
Let, over the horizontal surface \( A \), \( P \) be the saturated vapour pressure, \( \sigma \) be the density of the vapour.
The vapour pressure at \( B \) just outside the concave surface at a height \( h \) from the surface of the liquid outside the capillary tube s given by,
\( P_1=P-h\sigma{g}\tag{1} \)
On other hand, the pressure at \( B \) within the liquid just below the concave surface is given by,
\( P_2=P-h\rho{g}\tag{2} \)
Since the density of the liquid is greater than that of the vapour, ( \( \rho>\sigma \)). Therefore \( P_1>P_2 \).
Now the difference of pressures at the points just outside the meniscus and just inside the meniscus is given by,
\( P_1-P_2=P-h\sigma{g}-P+h\rho{g} \)
or, \( P_1-P_2=h(\rho-\sigma)g\tag{3} \).
The meniscus is almost hemispherical is shape, since the tube is narrow. The radius of the hemisphere is \( r \) which is equal to the radius of the capillary tube.
The excess pressure just above the meniscus over that just bellow it is \( \frac{2T}{r} \).
From equation (3) we can write, \( \frac{2T}{r}=h(\rho-\sigma)g\tag{4} \)
Or, \( hg=\frac{2T}{r}\frac{1}{\rho-\sigma} \)
Form equation (1), we get,
\( \displaystyle{P_1=P-\frac{2T}{r}\frac{\sigma}{\rho-\sigma}} \)
Since, the density of the liquid is greater than that of the vapour ( \( \rho>\sigma \)), therefore the pressure above the concave surface of a liquid is less than the vapour pressure at the horizontal surface of the same liquid by an amount \( \displaystyle{\frac{2T}{r}\frac{\sigma}{\rho-\sigma}} \).
Case II:
In case of the liquid which does not wet the capillary tube, as in case of mercury, the liquid is depressed inside the capillary tube over the liquid surface outside the capillary tube, as shown in Fig.2. The liquid meniscus at the top is convex upwards.
Let, over the horizontal surface \( A \), \( P \) be the saturated vapour pressure, \( \sigma \) be the density of the vapour.
The vapour pressure at \( B \) just outside the convex surface at a depth \( h \) below the surface of the liquid outside the capillary tube s given by,
\( P_1=P+h\sigma{g}\tag{1} \)
The press at B within the liquid, just below the convex surface is given by,
\( P_2=P+h\rho{g}\tag{2} \)Since the density of the liquid is greater than that of the vapour, ( \( \rho>\sigma \) ), so \( P_2>P_1 \).
The difference of pressure at points just inside the meniscus and just outside it is given by,
\( P_2-P_1=P+h\rho{g}-P-h\sigma{g} \)
Or, \( \displaystyle{P_2-P_1=h(\rho-\sigma)g} \)
If \( r \) be the radius of the meniscus, which is nearly hemispherical in shape and the radius is equal to the radius of the capillary tube, then \( \frac{2T}{r} \) is the excess pressure just below the meniscus over that just above it.
So \( \frac{2T}{r}=hg{\rho-\sigma} \)
or, \( hg=\frac{2T}{r(\rho-\sigma)}\tag{3} \)
From equation (1), we can write
\( P_1=P+\frac{2T}{r}\frac{\sigma}{\rho-\sigma} \)or, \( \displaystyle{P_1-P=\frac{2T}{r}\frac{\sigma}{\rho-\sigma}} \)
Since density of the liquid ( \( \rho \) ) is greater than that of the vapour ( \( \sigma \) ), so ( \( P_1-P \) ) is a positive quantity.
So the vapour pressure above the convex surface of a liquid is greater than the vapour pressure of the same liquid by an amount \( \frac{2T}{r}\cdot\frac{\sigma}{\rho-\sigma} \).
Let us consider, \( V \) be the volume of 1 mole of vapour, \( M \) be the molecular weight of the vapour and \( \theta \) be the temperature at absolute scale, then assuming the saturated vapour obeys the perfect gas equation, we can write, \( \displaystyle{PV=R\theta} \).
or, \( V=\frac{R\theta}{P} \)
So the density of the vapour \( \sigma=\frac{M}{V}=\frac{MP}{R\ \theta} \)
The excess vapour pressure on the horizontal surface of the liquid over that on its concave surface is given by,
\( \displaystyle{P-P_1=\frac{2T}{r}\frac{\sigma}{\rho-\sigma}} \)
Now the density of the liquid is greater than that of the vapour ( \( \rho>\sigma \) ), then we can write,
\( P-P_1=\frac{2T}{r}\frac{\sigma}{\rho}\\=\frac{2T}{\rho{r}}\cdot\frac{MP}{R\ \theta} \)