# C) The vapour in (b) has a density1/1600 of that of boiling water. Estimate the ratio of the mean separation of water molecules at 100°C in the vapour to that in the liquid.

llltkl | College Teacher | (Level 3) Valedictorian

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Density is mass per unit volume. Mathematically, `D = M/V`

Given `D_v/D_w=1/1600`

Where, `D_w` is the density of boiling water, and `D_v` , that of the vapour.

Assuming no loss of vapour to the surroundings (i.e. constant mass),

`rArr (M/V_v)/(M/V_w) = 1/1600`

`rArr V_w/V_v=1/1600`

`rArr V_v/V_w=1600`

Volume has the dimension of` L^3` . Assuming the mean distance of separation between water molecules in any given phase (liquid or vapour) to be l, volume (at that phase) will be given by `l^3` .

In other words, mean distance of separation between water molecules in any given phase will be obtained from cube root of its volume at that phase. Hence we can write,

`(l_v)^3/(l_w)^3=1600`

`rArr ((l_v)/(l_w))^3=1600`

`rArr (l_v)/(l_w)=root(3)(1600) = 11.7.`

Therefore, the ratio of the mean separation of water molecules at 100°C in the vapour to that in the liquid = 11.7.