Question

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is $\rho$ and L is its latent heat of vaporization.
A. $\sqrt{\dfrac{T}{\rho L}}$
B. $\dfrac{T}{\rho L}$
C. $\dfrac{2T}{\rho L}$
D. $\dfrac{\rho L}{T}$

Answer 1

We assume a liquid drop which evaporates by decrease in its surface energy so as to keep its temperature constant. We know that when the drop evaporates it loses some of its mass due to the change in energy which is the product of the lost mass and latent heat of vaporization. By finding the change in energy and the lost mass we can solve the question.

Formula used:
$\Delta E=T\times \Delta A$
$m=\rho \times \Delta V$
$\Delta E=m\times L$

Complete step by step answer:
In the question we are given a liquid drop which evaporates by decrease in its surface energy for the temperature to remain constant.
Given that ‘T’ is the surface tension, ‘$\rho$’ is the density and ‘L’ is the latent heat of vaporization.
Let us consider a liquid drop of radius ‘R’. So when we provide a heat energy ‘Q’ to the drop it loses some mass and thus its radium will become $\left( R-\Delta R \right)$, where ‘$\Delta R$’ is the change in radius.
So when the liquid drop shrinks, we know that the system will lose some energy. Let ‘$\Delta E$’ be the change in energy. Then we know that,
$\Delta E=T\times \Delta A$, were ‘$\Delta A$’ is the change in surface area, which can be given as
$\Delta A=4\pi {{R}^{2}}-4\pi {{\left( R-\Delta R \right)}^{2}}$
Therefore we will get the change in energy as,
$\therefore \Delta E=T\times \left( 4\pi {{R}^{2}}-4\pi {{\left( R-\Delta R \right)}^{2}} \right)$
We can take the common terms outside the bracket.
$\Rightarrow \Delta E=T\times 4\pi \left( {{R}^{2}}-{{\left( R-\Delta R \right)}^{2}} \right)$
Since $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$, we can write the above equation as,
$\Rightarrow \Delta E=T\times 4\pi \left( \left( R+R-\Delta R \right)\left( R-R+\Delta R \right) \right)$
By solving this, we will get
$\Rightarrow \Delta E=T\times 4\pi \left( \left( 2R-\Delta R \right)\left( \Delta R \right) \right)$
$\Rightarrow \Delta E=T\times 4\pi \left( 2R\times \Delta R-\Delta {{R}^{2}} \right)$
Since ‘$\Delta R$’ is very small we can neglect $\Delta {{R}^{2}}$. Therefore we will get,
$\therefore \Delta E=T\times 4\pi \left( 2R\times \Delta R \right)$
This is the decrease in energy. We know that this energy will evaporate an amount of mass of the liquid drop. Let ‘m’ be the mass evaporated, then we have
$m=\rho \times \Delta V$, were ‘$\Delta V$’ is the change in volume which is given as,
$\Delta V=\dfrac{4}{3}\pi {{R}^{3}}-\dfrac{4}{3}\pi {{\left( R-\Delta R \right)}^{3}}$
By substituting this in the equation to find the amount of mass evaporated, we will get
$\Rightarrow m=\rho \left( \dfrac{4}{3}\pi {{R}^{3}}-\dfrac{4}{3}\pi {{\left( R-\Delta R \right)}^{3}} \right)$
Since $\left( \dfrac{4}{3}\pi \right)$is common, we can take this term outside the bracket.
$\Rightarrow m=\rho \times \dfrac{4}{3}\pi \left( {{R}^{3}}-{{\left( R-\Delta R \right)}^{3}} \right)$
Since $\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$, the above equation will become,
$\Rightarrow m=\rho \times \dfrac{4}{3}\pi \left( R-R+\Delta R \right)\left( {{R}^{2}}+{{\left( R-\Delta R \right)}^{2}}+R\left( R-\Delta R \right) \right)$
We know that, ${{\left( a-b \right)}^{2}}=\left( {{a}^{2}}-2ab+{{b}^{2}} \right)$. Hence,
$\Rightarrow m=\rho \times \dfrac{4}{3}\pi \left( \Delta R \right)\left( {{R}^{2}}+{{R}^{2}}+\Delta {{R}^{2}}-2R\Delta R+{{R}^{2}}-R\Delta R \right)$
By solving this we will get,
$\Rightarrow m=\rho \times \dfrac{4}{3}\pi \left( \Delta R \right)\left( 3{{R}^{2}}+\Delta {{R}^{2}}-3R\Delta R \right)$
$\Rightarrow m=\rho \times \dfrac{4}{3}\pi \left( 3{{R}^{2}}\Delta R+\Delta {{R}^{3}}-3R\Delta {{R}^{2}} \right)$
Since ‘$\Delta R$’ is small we can cancel ‘$\Delta {{R}^{2}}$’ and ‘$\Delta {{R}^{3}}$’. Thus we will get,
$\Rightarrow m=\rho \times \dfrac{4}{3}\pi \left( 3{{R}^{2}}\Delta R \right)$
$\therefore m=\rho \times 4\pi \times {{R}^{2}}\Delta R$
Therefore this is the mass that gets decreased due to evaporation.
In the question we are asked to calculate the minimum radius for the given process to occur.
We know that the change in energy can be given as,
$\Delta E=m\times L$, where ‘L’ is the latent heat of vaporization.
We have $\Delta E=T\times 4\pi \left( 2R\times \Delta R \right)$and $m=\rho \times 4\pi \times {{R}^{2}}\Delta R$
Therefore we can write,
$\therefore T\times 4\pi \left( 2R\times \Delta R \right)=\left( \rho \times 4\pi \times {{R}^{2}}\Delta R \right)L$
By solving this we will get,
$\Rightarrow T\times 2R=\rho \times {{R}^{2}}\times L$
$\Rightarrow T\times 2=\rho \times R\times L$
$\therefore R=\dfrac{2T}{\rho \times L}$
Therefore we get the minimum radius as, $\dfrac{2T}{\rho L}$.
Hence the correct answer is option C.

Note:
We know that the surface of a fluid will shrink into its minimum surface area possible. This tendency of the liquid surface is known as surface tension. It can also be defined as the tension on the surface film of a liquid due to the attraction of molecules on the surface by bulk of the liquid, which will cause the surface area to minimize.