Question

The surface tension of toluene at 298 K is 0.0284 N/m and its density is 0.866 g/cc. What is the largest radius of the capillary that will permit the liquid to rise $2\times {{10}^{-4}}$m? Assume $\theta $ = 0.

Answer 1

Hint: This question is based on the phenomenon of capillary action. When the liquid in a capillary rises without any external forces or assistance is known as capillary action. It is also known as the capillary effect or capillary motion.

Complete answer:
The rise of liquid in a capillary is due to the intermolecular forces between the surrounding solid surfaces and the liquid. It is further propelled by the surface tension caused by the attractive forces between the molecules of the liquid which keeps them together and the adhesive forces between the wall of the container and the liquid.
Now, the rise of liquid in a capillary is given by the formula
\[h=\dfrac{2\gamma \cos \theta }{dgr}\]
Where h is the height of the liquid column,
$\gamma $ is the surface tension between the liquid and the air given by force/unit,
d is the density of the liquid in the capillary tube given by mass/volume,
r is the radius of the capillary tube,
g is the acceleration due to gravity is given by length/square of time, and
$\theta $ is the angle of contact.
Now, it is given to us that h = $2\times {{10}^{-4}}m$, $\theta $= 0, $\gamma $ = 0.0284 N/m, d = 0.866 g/cc and g = $9.8m/{{s}^{2}}$.
Now, cos 0 =1 and 0.866 g/cc = 866 kg/${{m}^{3}}$.
So by substituting these values in the capillary rise formula we get
\[\begin{align}
  & h=\dfrac{2\gamma \cos \theta }{dgr} \\
 & 2\times {{10}^{-4}}=\dfrac{2\times 0.0284\times 1}{866\times 9.8\times r} \\
 & r=\dfrac{2\times 0.0284}{866\times 9.8\times 2\times 1{{0}^{-4}}} \\
 & r=3.346\times {{10}^{-2}}m \\
\end{align}\]
So, the largest radius of the capillary that will permit the liquid to rise $2\times {{10}^{-4}}$m is $3.346\times {{10}^{-2}}m$.

Note:
It is important to note that the units of all the factors are of the same multitude. In the given question, since the unit of surface tension was in N/m and acceleration due to gravity was in $m/{{s}^{2}}$, we had to convert the unit of density from g/cc to kg/${{m}^{3}}$ by multiplying it by 1000.