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Thermodynamics Part 1

Section: Physics

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Question : 51 of 100

Marks: +1, -0

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 vaporization.

$\sqrt{\frac{T}{\rho L}}$
$\frac{T}{\rho L}$
$\frac{2T}{\rho L}$
$\frac{\rho L}{T}$

Solution:

surface Tension

= T \;\;

Radius

= R

density

= \rho

Latent heat

= L

Let for

R,

small decrease on

R

is

\triangle R

\therefore

change in surface energy

\Delta E = \;\; \left[4 \pi R^{2} - 4 \pi (R - \Delta R)^{2}\right] \times T

= 4 \pi (R - R + \Delta R)(R + R - 4R) \times T

= 4 \pi \cdot \Delta R \cdot 2 R \cdot T \;\; \therefore \;\; \Delta R \approx 0

= 8 \pi R \cdot \Delta R \cdot T

change in mass

\Delta m \;\; = \frac{9}{3} \pi (R^{3} - (R - \Delta R)^{3}) \times \rho

= \frac{4}{3} \pi (R - R + \Delta R)(R^{2} + (R - \Delta R)^{2} + R(R - \Delta R)) \rho

= \frac{4}{3} \pi \cdot \Delta R (R^{2} + R^{2} + R^{2} - 3 R \Delta R + (\Delta R)^{2}) \rho

= 4 \cdot \pi \cdot \Delta R \cdot R^{2} \cdot \rho

Energy for vaporization

\Delta V = 4 \pi \Delta R \cdot R^{2} \cdot L \rho

8 \pi \cdot R \cdot \Delta R \cdot T = 4 \cdot \pi \cdot \Delta R \cdot R^{2} \cdot L \cdot \rho

or

2 T = R L \rho

or

R = \frac{2 T}{\rho L}

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