Question 152 of 192

Assume that a drop of a liquid evaporates by a decrease in its surface energy so that its temperature remains unchanged. The minimum radius of the drop for this to be possible is. (The surface tension is $T$, the density of the liquid is$\rho$ and $L$ is its latent heat of vaporisation.)

A$\dfrac{T}{\rho L}$B$\dfrac{2T}{\rho L}$C$\dfrac{\rho L}{T}$D$\sqrt{\dfrac{T}{\rho L}}$

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Answer: (B) $\dfrac{2T}{\rho L}$

Let the radius of the drop at time $t=t$ be $r$ and at an instant $t=t+dt$ be $r-dr$

As surface area is given by, $A=4\pi {r}^{2}$,
$∴$ decrease in surface area is,
$dA=4\pi \left(2rdr\right)$
$∴$ Decrease in surface energy during time $dt$ is,
$dU=TdA=T\cdot 8\pi rdr...\left(1\right)$
Decrease in volume during time $dt$ is,
$dV=4\pi {r}^{2}dr$
$∴$ Decrease in mass during time $dt$ is,
$dm=\rho dV=4\pi \rho {r}^{2}dr$
$∴$ Heat required in vaporisation is,
$dQ=Ldm=4\pi \rho {r}^{2}drL...\left(2\right)$
Apply conservation of energy,
decrease in surface energy$=$heat required in vaporisation.
$\Rightarrow dU=dQ$
$\Rightarrow 8T\pi rdr=4\pi \rho {r}^{2}drL$
$\Rightarrow 2T=\rho rL$
$\Rightarrow r=\dfrac{2T}{\rho L}$
Hence, the minimum radius is $\dfrac{2T}{\rho L}$