Properties of Matter 4 Question 4

6. 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 $T$, denstiy of liquid is $\rho$ and $L$ is its latent heat of vaporisation

(2013 Main)

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

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

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

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

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Answer:

Correct Answer: 6. (d)

Solution:

  1. Decrease in surface energy = heat required in vaporisation.

$$ \begin{aligned} & & S & =4 \pi r^{2} \\ & \therefore & d S & =2(4 \pi r) d r \\ & \therefore & T(d S) & =L(d m) \\ & \therefore & T(2)(4 \pi r) d r & =L\left(4 \pi r^{2} d r\right) \rho \\ & \therefore & r & =\frac{2 T}{\rho L} \end{aligned} $$



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