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Boyle's Law

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There is a substance whose vapourisation and freezing temperature (transition temperature) is same. It is placed at transition temperature with its main constituent being liquid state and a little vapour in a closed container. Now, vapour is pumped out continuosly. The fraction of this substance that will ultimately freeze is $\frac{7}{8}$ . If the latent heat of vaporization is $^{'} x^{'}$ times the latent heat of fusion. Find out value of $^{'} x^{'}$ (Answer upto two digit after decimal point).

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Solution

Let $f \to$ fraction which freezes,
$m =$ mass of substance, $f = \frac{7}{8}$

Mass of substance that freezes $= m f$
Heat lost by freezing substance $= m f L_{F}$

Mass of substance that vaporize $= m (1 - f)$
Heat gained by substance
$= m (1 - f) . L_{V}$

Using principle of calorimetry
$m f L_{F} = m (1 - f) L_{V}$
$\frac{7}{8} L_{F} = \frac{1}{8} L_{V}$
$L_{V} = 7 L_{F}$

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