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. ` sqrt((T)/( rho L )) `
B. ` (T)/(rho L ) `
C. `( 2 T ) /( rho L ) `
D. ` (rho L )/( T) `

Explanation

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

### Step 1: Understand Surface Energy
The surface energy (ΔE) of a liquid drop can be expressed in terms of surface tension (T) and the change in area (ΔA):
\[ \Delta E = T \cdot \Delta A \]

### Step 2: Relate Surface Energy to Latent Heat
When the drop evaporates, the change in surface energy is equal to the heat required for evaporation (Q), which can be expressed as:
\[ Q = mL \]
where \(m\) is the mass of the liquid and \(L\) is the latent heat of vaporization.

### Step 3: Express Mass in Terms of Volume
The mass \(m\) can be expressed in terms of the density (\(\rho\)) and volume (V):
\[ m = \rho \cdot V \]
For a spherical drop, the volume \(V\) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus,
\[ m = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 4: Calculate Change in Area
The change in area (ΔA) when the radius changes from \(R\) to \(R - \Delta R\) is:
\[ \Delta A = A_{\text{final}} - A_{\text{initial}} = 4\pi(R - \Delta R)^2 - 4\pi R^2 \]
Expanding this gives:
\[ \Delta A = 4\pi \left( R^2 - 2R\Delta R + \Delta R^2 - R^2 \right) = -8\pi R \Delta R + 4\pi \Delta R^2 \]
For small \(\Delta R\), we can neglect \(\Delta R^2\):
\[ \Delta A \approx -8\pi R \Delta R \]

### Step 5: Set Up the Energy Balance Equation
Equating the change in surface energy to the heat required for evaporation:
\[ T \cdot (-8\pi R \Delta R) = \rho \cdot \frac{4}{3} \pi R^3 \cdot L \]

### Step 6: Simplify the Equation
Cancelling \(\pi\) and rearranging gives:
\[ -8T R \Delta R = \frac{4}{3} \rho R^3 L \]
Dividing both sides by \(\Delta R\) (assuming \(\Delta R \neq 0\)):
\[ -8T R = \frac{4}{3} \rho R^3 L \cdot \frac{1}{\Delta R} \]

### Step 7: Solve for Minimum Radius
Rearranging gives:
\[ R^2 = \frac{6T}{\rho L} \]
Thus, the minimum radius \(R\) is:
\[ R = \sqrt{\frac{6T}{\rho L}} \]

### Final Answer
The minimum radius of the drop for evaporation while maintaining constant temperature is:
\[ R = \sqrt{\frac{6T}{\rho L}} \]