Calculate the percentage of free volume available in one mole gaseous water at $1.0 a t m$ and $100^{o} C$ . Density of liquid $H_{2} O$ at $100^{o} C$ is $0.958 g / m L$ (as nearest integer)

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Solution

For gaseous water $P V = n R T$
Thus, volume occupied by 1 mole gaseous water can be derived as $(T = 273 + 100 = 373 K)$
$V = \frac{n R T}{P} = \frac{1 \times 0.0821 \times 373}{1} = 30.62 l i t r e$
Also, volume of 1 mole of liquid water $=$ mass/ density
$= \frac{18}{0.958} = 18 - 79 m L = 18.79 \times 10^{- 3} l i t r e$
Thus, volume percentage occupied by water molecules in gaseous state
$= \frac{18.79 \times 10^{- 3}}{30.62} \times 100 = 0.0614$
Percentage of free volume $= 100 - 0.0614 = 99.9386$
so nearest integer value is 100.

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