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//rhoL)`
B. `T//rhoL`
C. `2T//rhoL`
D. `rhoL//T`

Answer

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-by-Step Solution:

1. **Understanding the Problem**:
We have a drop of liquid with radius \( r \). When it evaporates, it loses some of its surface area, which leads to a decrease in surface energy. We need to find the minimum radius \( r \) such that this process can occur without changing the temperature of the liquid.

2. **Surface Area Calculation**:
The surface area \( A \) of a sphere (the drop) is given by:
\[ A = 4\pi r^2 \]

3. **Change in Surface Area**:
When the radius decreases by a small amount \( dr \), the new radius becomes \( r - dr \). The new surface area \( A' \) is:
\[ A' = 4\pi (r - dr)^2 = 4\pi (r^2 - 2r \, dr + (dr)^2) \]
The change in surface area \( dA \) is:
\[ dA = A' - A = 4\pi (r^2 - 2r \, dr + (dr)^2) - 4\pi r^2 = -8\pi r \, dr + 4\pi (dr)^2 \]
For small \( dr \), we can neglect the \( (dr)^2 \) term:
\[ dA \approx -8\pi r \, dr \]

4. **Change in Surface Energy**:
The change in surface energy \( dE \) due to the change in surface area is given by:
\[ dE = T \cdot dA = T \cdot (-8\pi r \, dr) = -8\pi T r \, dr \]

5. **Volume of Liquid Evaporated**:
The volume \( dV \) of liquid that evaporates when the radius decreases from \( r \) to \( r - dr \) is:
\[ dV = 4\pi r^2 \, dr \]
The mass \( dm \) of the evaporated liquid is:
\[ dm = \rho \cdot dV = \rho \cdot (4\pi r^2 \, dr) \]

6. **Energy Required for Evaporation**:
The energy required to evaporate this mass of liquid is given by:
\[ dE = dm \cdot L = \rho \cdot (4\pi r^2 \, dr) \cdot L \]

7. **Equating Energy Changes**:
Since the decrease in surface energy is used for the evaporation, we can set the two expressions for \( dE \) equal:
\[ -8\pi T r \, dr = \rho \cdot (4\pi r^2 \, dr) \cdot L \]
Dividing both sides by \( dr \) (assuming \( dr \neq 0 \)):
\[ -8\pi T r = 4\pi \rho r^2 L \]

8. **Simplifying the Equation**:
Cancel \( 4\pi \) from both sides:
\[ -2T = \rho r L \]
Rearranging gives:
\[ r = \frac{2T}{\rho L} \]

9. **Conclusion**:
The minimum radius \( r \) of the drop for it to evaporate by decreasing its surface energy while keeping the temperature constant is:
\[ r_{\text{min}} = \frac{2T}{\rho L} \]