Assume that a drop of liquid evaporates by a | vedclass

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(A) Let the radius of the drop be $R$. The volume of the drop is $V = \frac{4}{3}\pi R^3$. When the radius decreases by $\Delta R$,the mass evaporated

Vedclass · Accepted Answer

$\frac{2T}{\rho L}$

Vedclass · Answer

(A) Let the radius of the drop be $R$. The volume of the drop is $V = \frac{4}{3}\pi R^3$. When the radius decreases by $\Delta R$,the mass evaporated is $\Delta m = \rho \Delta V = \rho (4\pi R^2 \Delta R)$.The energy required for evaporation is $\Delta E = \Delta m L = 4\pi R^2 \Delta R \rho L$.The surface area of the drop is $A = 4\pi R^2$. The change in surface energy is $\Delta U = T \Delta A = T [4\pi R^2 - 4\pi (R - \Delta R)^2]$.Expanding the term: $\Delta U = 4\pi T [R^2 - (R^2 - 2R\Delta R + \Delta R^2)] = 4\pi T [2R\Delta R - \Delta R^2]$.Neglecting the higher-order term $\Delta R^2$,we get $\Delta U \approx 8\pi T R \Delta R$.Equating the energy required for evaporation to the decrease in surface energy: $4\pi R^2 \Delta R \rho L = 8\pi T R \Delta R$.Solving for $R$: $R = \frac{8\pi T}{4\pi \rho L} = \frac{2T}{\rho L}$.

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

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