Calculate the volume of 1.00 mol of liquid water at 20°C (at which its density is 998 kg/m3), and compare that with the volume occupied by 1.00 mol of water at the critical point, which is 56 X 10^-6m³. Water has a molar mass of 18.0 g/mol.

We can calculate the volume from the equation

$\mathbf{V}\mathbf{=}\frac{\mathbf{m}}{\mathbf{\rho }}$

where m is the mass andis density.

Calculate mass m from the relation between the mass m, the molar mass M and the number of moles n by

$\begin{array}{r}\mathbf{m}\mathbf{=}\mathbf{nM}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ m=1\mathrm{mol}\times 18\times {10}^{-3}\mathrm{kg}/\mathrm{mol}\\ =18\times {10}^{-3}\mathrm{kg}/\mathrm{mol}\end{array}$

Substitute the value of m to calculate volume

$\begin{array}{r}V=\frac{m}{\rho }\\ =\frac{18\times {10}^{-3}\mathrm{kg}/\mathrm{mol}}{998\mathrm{kg}/{\mathrm{m}}^3}\\ =18\times {10}^{-6}{\mathrm{m}}^3\end{array}$

The volume of 1 mol of liquid water at $20°C$is$18\times {10}^{-6}{\mathrm{m}}^3$

When comparing the result with the volume of 1 mol at the critical point we would find the volume at the critical point is larger than the volume at$20°C$

Because the liquid before the critical point tends to make a phase transition from liquid to vapour and temperature still increase. According to Ideal gas law, as the temperature of the vapour increases the volume increases

pV = nRT

So the volume at critical point larger than the volume at T =$20°C$