Question

Calculate the percentage of free volume available in 1 mole gaseous water at 1 atm and ${100}^{o}C$. Density of liquid $H_2$ at ${100}^{o}C$ is $0.958g/ml$.

Answer 1

For gaseous water $PV=nRT$

Thus, volume occupied by 1 mole gaseous water can be derived as

$T=273+100=373K$

$V=nRT/P=\frac{1\times 0.0821\times 373}{1}$

$=30.62$ litre

Also Volume of 1 mole of liquid water = mass/density

$=18/0.958$

$=18.79$ mL

$=18.79\times {10}^{-3}$ litre

Thus Volume Percentage occupied by water molecules in gaseous state

$=\frac{18.79\times {10}^{-3}}{30.62}\times 100$

$=0.0614$

% of free volume $=100-0.0614$

$=99.9386$