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. $\rho L/T$
• B. $\sqrt{T/\rho L}$
• C. $T/\rho L$
• D. $2\,T/\rho L$

Answer 1

The Correct Option is D

To determine the minimum radius of a drop of liquid evaporating due to its surface energy, we must equate the change in surface energy to the energy required for evaporation. This will ensure that the temperature of the drop remains unchanged.

The surface energy is given by the formula:

dE = T \cdot dA

where dE is the change in energy, T is the surface tension, and dA is the change in surface area.

For a spherical drop of radius r, the surface area A = 4\pi r^2, so the change in surface area when a small volume dV evaporates is:

dA = \frac{d(4\pi r^2)}{dr} \cdot dr = 8\pi r \cdot dr

Now, relate the change in volume to the mass that evaporates:

dV = 4\pi r^2 \cdot dr

Mass dm of the liquid evaporating is:

dm = \rho \cdot dV = \rho \cdot 4\pi r^2 \cdot dr

The heat required for vaporization is:

dQ = L \cdot dm = L \cdot \rho \cdot 4\pi r^2 \cdot dr

For the drop's temperature to remain unchanged, the heat of vaporization must equal the decrease in surface energy:

dE = dQ

Substituting the expressions, we get:

T \cdot 8\pi r \cdot dr = L \cdot \rho \cdot 4\pi r^2 \cdot dr

Solving for r, we get:

8\pi rT = 4\pi r^2 \rho L

Simplifying:

2T = r \rho L

So, the minimum radius r for the drop is given by:

r = \frac{2T}{\rho L}

Thus, the correct answer is 2T/\rho L. This radius ensures that the energy lost through surface tension is sufficient for the vaporization process without changing the temperature of the liquid.