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

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Answer & Solution

Correct Answer

(A) \(\frac{{2T}}{{\rho L}}\)

Detailed explanation

When radius is decrease by \(\Delta R,\)

\(4\pi {R^2}\Delta R\rho L = 4\pi T\left[ {{R^2} - {{\left( {R - \Delta R} \right)}^2}} \right]\)

\( \Rightarrow \rho {R^2}\Delta RL = T\left[ {{R^2} - {R^2} + 2R\Delta R - \Delta {R^2}} \right]\)

\( \Rightarrow \rho {R^2}\Delta RL = T2R\Delta R\,\,\left[ {\Delta R\,is\,very\,small} \right]\)

\( \Rightarrow R = \frac{{2T}}{{\rho L}}\)

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