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.

Heat and Thermodynamics PYQ — JEE Main 2013 · Shift 0 · Q69 with Solution

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Correct answer: D

Idea of the problem

A liquid drop evaporates. If its temperature remains unchanged, then the energy needed for vaporization must come entirely from the decrease in surface energy.

So,

\text{loss in surface energy} = \text{latent heat required for evaporated mass}

We find the condition for a drop of radius $r$ .

Surface energy of the drop

For a spherical drop of radius $r$ :

A = 4\pi r^2

Hence surface energy,

U = T \cdot 4\pi r^2

where $T$ is the surface tension.

If the radius decreases by a small amount $dr$ , then

dU = T\,d(4\pi r^2) = 8\pi T r\,dr

Since the radius decreases, $dr<0$ , so the decrease in surface energy is

- dU = -8\pi T r\,dr

Mass evaporated when radius decreases by $dr$

Volume of sphere:

V = \frac{4}{3}\pi r^3

So,

dV = 4\pi r^2 dr

Since $dr<0$ , the decrease in volume is

-dV = -4\pi r^2 dr

Hence mass evaporated is

dm = \rho(-dV)= -4\pi \rho r^2 dr

Latent heat required for this evaporation:

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

Condition for evaporation using only surface energy

For temperature to remain unchanged, the decrease in surface energy must at least supply the latent heat:

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

Since $-4\pi r\,dr >0$ , divide both sides by it:

2T \ge \rho L r

Thus,

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

For this to be just possible, the limiting (critical) radius is

r_{\min/\text{critical}} = \frac{2T}{\rho L}

In the language of the options, the required radius is

\boxed{\frac{2T}{\rho L}}

Checking options

A: $\dfrac{\rho L}{T}$ — wrong dimensions.
B: $\sqrt{\dfrac{T}{\rho L}}$ — wrong dimensions.
C: $\dfrac{T}{\rho L}$ — misses factor $2$ .
D: $\dfrac{2T}{\rho L}$ — correct.

Comparison with stored answer

Stored correct answer: D

Derived answer: D

So they agree.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: D

Idea of the problem

A liquid drop evaporates. If its temperature remains unchanged, then the energy needed for vaporization must come entirely from the decrease in surface energy.

So,

\text{loss in surface energy} = \text{latent heat required for evaporated mass}

We find the condition for a drop of radius $r$ .

Surface energy of the drop

For a spherical drop of radius $r$ :

A = 4\pi r^2

Hence surface energy,

U = T \cdot 4\pi r^2

where $T$ is the surface tension.

If the radius decreases by a small amount $dr$ , then

dU = T\,d(4\pi r^2) = 8\pi T r\,dr

Since the radius decreases, $dr<0$ , so the decrease in surface energy is

- dU = -8\pi T r\,dr

Mass evaporated when radius decreases by $dr$

Volume of sphere:

V = \frac{4}{3}\pi r^3

So,

dV = 4\pi r^2 dr

Since $dr<0$ , the decrease in volume is

-dV = -4\pi r^2 dr

Hence mass evaporated is

dm = \rho(-dV)= -4\pi \rho r^2 dr

Latent heat required for this evaporation:

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

Condition for evaporation using only surface energy

For temperature to remain unchanged, the decrease in surface energy must at least supply the latent heat:

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

Since $-4\pi r\,dr >0$ , divide both sides by it:

2T \ge \rho L r

Thus,

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

For this to be just possible, the limiting (critical) radius is

r_{\min/\text{critical}} = \frac{2T}{\rho L}

In the language of the options, the required radius is

\boxed{\frac{2T}{\rho L}}

Checking options

A: $\dfrac{\rho L}{T}$ — wrong dimensions.
B: $\sqrt{\dfrac{T}{\rho L}}$ — wrong dimensions.
C: $\dfrac{T}{\rho L}$ — misses factor $2$ .
D: $\dfrac{2T}{\rho L}$ — correct.

Comparison with stored answer

Stored correct answer: D

Derived answer: D

So they agree.

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