Question 368: Properties of Matter

Question

A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

Accepted Answer

As surface area decreases so energy is released.
Energy released = $$4\pi {R^2}T\left[ {{n^{1/3}} - 1} \right]$$
where R = n
1/3
r
$$ = 4\pi {R^3}T\left[ {{1 \over r} - {1 \over R}} \right] = 3VT\left[ {{1 \over r} - {1 \over R}} \right]$$

Answer

energy = 4VT $$\left( {{1 \over r} - {1 \over R}} \right)$$ is released.

Answer

energy = 3VT$$\left( {{1 \over r} + {1 \over R}} \right)$$ is absorbed.

Answer

energy = 3VT$$\left( {{1 \over r} - {1 \over R}} \right)$$ is released

Answer

energy is neither released nor absorbed.

Properties of Matter

A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

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