Assume that a drop of liquid evaporates by decreases 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.

Options

A. (a) `rhoL//T`
B. (b) `sqrt(T//rhoL)`
C. (c) `T//rhoL`
D. (d) `2T//rhoL`

Explanation

To find the minimum radius of a drop of liquid that can evaporate by decreasing its surface energy while keeping its temperature constant, we can follow these steps:

### Step 1: Understand the Surface Energy Change
The surface energy (U) of a spherical drop of radius R is given by:
\[ U = T \times A \]
where \(T\) is the surface tension and \(A\) is the surface area. For a sphere, the surface area \(A\) is:
\[ A = 4\pi R^2 \]
Thus, the surface energy becomes:
\[ U = T \times 4\pi R^2 \]

### Step 2: Determine the Change in Surface Energy
When a small portion of the drop evaporates, the radius decreases by a small amount \(dR\). The new radius is \(R - dR\). The new surface area is:
\[ A' = 4\pi (R - dR)^2 \]
The change in surface energy (\(dU\)) is:
\[ dU = U' - U = T \times 4\pi (R - dR)^2 - T \times 4\pi R^2 \]
Using the binomial expansion for small \(dR\):
\[ (R - dR)^2 \approx R^2 - 2RdR \]
Thus, we can write:
\[ dU \approx T \times 4\pi (R^2 - 2RdR) - T \times 4\pi R^2 = -8\pi T R dR \]

### Step 3: Relate Energy Change to Mass Evaporated
The mass (\(m\)) of the liquid that evaporates can be expressed as:
\[ m = \rho \times V \]
where \(V\) is the volume of the evaporated portion. The volume change due to the decrease in radius can be approximated as:
\[ V = \frac{4}{3}\pi (R^3 - (R - dR)^3) \approx \frac{4}{3}\pi (3R^2 dR) \]
Thus, the mass of the evaporated portion is:
\[ m = \rho \times \frac{4}{3}\pi (3R^2 dR) = 4\pi \rho R^2 dR \]

### Step 4: Energy Required for Evaporation
The energy required to evaporate this mass is given by:
\[ dQ = mL = 4\pi \rho R^2 dR \times L \]

### Step 5: Set Up the Energy Balance Equation
For the process to occur at constant temperature, the decrease in surface energy must equal the energy required for evaporation:
\[ -8\pi T R dR = 4\pi \rho R^2 dR \times L \]
Cancelling \(4\pi dR\) from both sides (assuming \(dR \neq 0\)):
\[ -2T = \rho R L \]

### Step 6: Solve for Minimum Radius \(R\)
Rearranging the equation gives:
\[ R = \frac{2T}{\rho L} \]

### Final Result
Thus, the minimum radius \(R\) of the drop for evaporation to occur at constant temperature is:
\[ R = \frac{2T}{\rho L} \]