Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature r

Q: Why does the question ask for the 'minimum' radius when the inequality gives a maximum limit $r \le \frac{2T}{\rho L}$?

This is a well-known phrasing discrepancy in the original 2013 JEE Advanced question paper. Physically, the inequality $r \le \frac{2T}{\rho L}$ defines the upper bound (maximum radius) for spontaneous evaporation. However, the threshold value itself is $\frac{2T}{\rho L}$, which corresponds to option (d).

Q: Where does the energy for vaporisation come from if the temperature remains constant?

The energy comes entirely from the reduction in surface area. When the surface area of a liquid decreases, surface energy is released. In this microscopic regime, this released energy is sufficient to supply the latent heat of vaporisation.

Q: Why is the change in surface area $8\pi r \, dr$ instead of just $4\pi r^2$?

We are looking at a differential change. The surface area is $A(r) = 4\pi r^2$. When the radius changes by a tiny amount $dr$, the change in area is the derivative $dA = \frac{dA}{dr} dr = 8\pi r \, dr$.

Q: Does this spontaneous evaporation happen for macroscopic water drops?

No, because for macroscopic drops, the radius $r$ is much larger than $\frac{2T}{\rho L}$. This effect is only significant for extremely tiny, microscopic or nanoscopic droplets where the surface-area-to-volume ratio is incredibly high.

Q: What happens if the radius of the drop is larger than $\frac{2T}{\rho L}$?

If $r > \frac{2T}{\rho L}$, the surface energy released by shrinking is not enough to cover the latent heat of vaporisation. In this case, the drop cannot evaporate at a constant temperature without absorbing heat from its surroundings.

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 vaporisation

Options:

(A) $\frac{\rho L}{T}$

(B) $\sqrt{\frac{T}{\rho L}}$

(C) $\frac{T}{\rho L}$

(D) $\frac{2T}{\rho L}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a tiny, pristine spherical liquid drop floating in space.

This drop has a surface tension $T$, density $\rho$, and latent heat of vaporisation $L$.

Normally, when a liquid evaporates, it requires heat energy from its surroundings to break the intermolecular bonds and transition into the vapor phase.

If no heat is supplied from the outside, the liquid's temperature would drop.

However, our problem introduces a fascinating constraint: the temperature remains completely unchanged, and no external heat is supplied!

How is this possible?

The answer lies in the drop's **surface energy**.

As the drop evaporates, its radius decreases, which means its surface area shrinks.

This reduction in surface area releases surface energy, which can then be directly utilized to supply the latent heat required for vaporisation.

Let's mathematically model this beautiful thermodynamic balance.

### The Master Equations

Let the initial radius of the drop be $r$.

The surface area of the spherical drop is:

$$A = 4\pi r^2$$

Since surface tension $T$ is the energy per unit surface area, the total surface energy $E$ of the drop is:

$$E = T \cdot A = 4\pi r^2 T$$

Now, let's assume a tiny layer of thickness $dr$ evaporates.

The new radius becomes $r - dr$.

The decrease in surface area $dA$ is found by differentiating $A$ with respect to $r$:

$$dA = 8\pi r \, dr$$

Consequently, the decrease in surface energy $dU$ is:

$$dU = T \cdot dA = 8\pi r T \, dr$$

This is the energy "freed up" by the shrinking surface.

Next, let's find how much mass $dm$ is contained in this evaporated shell of thickness $dr$.

The volume of this thin shell is:

$$dV = 4\pi r^2 \, dr$$

Multiplying this volume by the density $\rho$ of the liquid gives the mass of the evaporated liquid:

$$dm = \rho \, dV = 4\pi r^2 \rho \, dr$$

To vaporise this mass $dm$, the heat energy $dQ$ required is:

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

### Spontaneous Evaporation Condition

For this evaporation process to occur spontaneously without any external heat source and without lowering the temperature, the energy released by the shrinking surface must be greater than or equal to the heat required for vaporisation:

$$dU \ge dQ$$

Substituting our expressions for $dU$ and $dQ$ into this inequality:

$$8\pi r T \, dr \ge 4\pi r^2 \rho L \, dr$$

We can simplify this by cancelling the common terms $4\pi r \, dr$ from both sides:

$$2T \ge r \rho L$$

Solving for the radius $r$:

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

This elegant inequality tells us that spontaneous evaporation powered solely by surface energy is only possible if the radius of the drop is extremely small—specifically, less than or equal to $\frac{2T}{\rho L}$.

Therefore, the limiting threshold radius is:

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

This matches perfectly with **Option (d)**.

Select Answer:

Visualized Solution

Visualizing the Evaporating Drop

Formulating Surface Energy

Calculating the Decrease in Surface Energy

Mass of the Evaporated Shell

Heat Required for Vaporisation

Energy Conservation Condition

Solving for Radius $r$

Concluding the Threshold Radius

Exploring Further Variations

The Sigma Insight: Surface Tension and Capillary Action

Analyzing the Setup

The Master Equations

Spontaneous Evaporation Condition

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