JEE Chemistry States of Matter Notes Set 04

Critical State and Liquefaction

Compress a gas at constant temperature and, below a certain temperature, its volume falls until liquid appears; the isotherm then has a flat stretch where vapour and liquid coexist, followed by a steep rise when only liquid remains. As the temperature is raised the flat stretch shrinks and finally shrinks to a point, the critical point, where the liquid and vapour become identical in density and every other property.
1. Critical temperature \( T_c \): the temperature above which a gas cannot be liquefied by pressure alone.
2. Critical pressure \( P_c \): the least pressure needed to liquefy the gas at \( T_c \).
3. Critical volume \( V_c \): the volume of one mole at \( T_c \) and \( P_c \).

Deviation from Ideal Behaviour

Real gases depart from ideality because their molecules have a finite size and attract one another. The departure is measured by the compressibility factor \( Z=\frac{PV_m}{RT}=\frac{V_{m,real}}{V_{m,ideal}} \).
1. Z = 1 for an ideal gas.
2. Z less than 1 means attraction dominates and the gas is more compressible than an ideal gas (most gases at low and moderate pressure).
3. Z greater than 1 means molecular size dominates and the gas is harder to compress (all gases at high pressure; H\(_2\) and He at all pressures).
A gas behaves most ideally at low pressure and high temperature.

The van der Waals Equation

Two corrections are made to the ideal gas equation. Molecules occupy space, so the volume free for motion is \( V-nb \), where b is the excluded volume per mole, equal to four times the actual volume of the molecules in a mole, \( b=4N_A\cdot\frac43\pi r^3 \). Attraction weakens the impacts on the wall, so the ideal pressure is the observed pressure plus a term proportional to the square of the density, \( \frac{an^2}{V^2} \). Together:
\[ \left(P+\frac{an^2}{V^2}\right)(V-nb)=nRT \]
The constant a measures the strength of attraction and is larger for polar molecules and for larger nonpolar molecules (water > carbon dioxide > argon; krypton > argon > neon > helium). A large a means easy liquefaction and a high boiling point. The constant b tracks molecular size. Units: a in atm L\(^2\) mol\(^{-2}\), b in L mol\(^{-1}\).

Z from the van der Waals Equation

For one mole:
1. Low pressure (large \( V_m \)), b negligible: \( Z=1-\frac{a}{V_mRT} \), less than 1.
2. High pressure, a negligible: \( Z=1+\frac{Pb}{RT} \), greater than 1.
3. Very low pressure and high temperature: both corrections vanish and Z approaches 1.
4. H\(_2\) and He (a almost zero): \( Z=1+\frac{Pb}{RT} \) at all pressures.

Virial Equation and Boyle Temperature

The virial equation expands Z in powers of the molar density: \( Z=1+\frac{B}{V_m}+\frac{C}{V_m^2}+\frac{D}{V_m^3}+\dots \). B, C, D are the second, third and fourth virial coefficients; they depend on temperature but not on density. B describes interactions between pairs of molecules, C between triplets. Expanding the van der Waals result gives \( B=b-\frac{a}{RT},\quad C=b^2,\quad D=b^3 \).
B is zero at the Boyle temperature \( T_B=\frac{a}{Rb} \), where attraction and size cancel and the gas obeys Boyle's law over a wide pressure range. Below \( T_B \), B is negative; above it, positive.

Critical Constants

Written as a cubic in \( V_m \), the van der Waals equation has three roots at a given T and P. At the critical point all three roots coincide, so the cubic must equal \( (V_m-V_c)^3 \). Comparing coefficients (or setting \( \partial P/\partial V \) and \( \partial^2P/\partial V^2 \) to zero) gives
\[ V_c=3b,\qquad P_c=\frac{a}{27b^2},\qquad T_c=\frac{8a}{27Rb} \]
From these, \( b=\frac{RT_c}{8P_c},\ a=3P_cV_c^2=\frac{27R^2T_c^2}{64P_c} \), the compressibility factor at the critical point is \( Z_c=\frac{P_cV_c}{RT_c}=\frac38 \), and \( \frac{T_B}{T_c}=\frac{27}{8} \). When V_c, T_c and P_c are all measured, use T_c and P_c to find a and b, as V_c is the least accurate.

Worked Example 1. Find the critical constants of HCl from a = 371.843 kPa dm\(^6\) mol\(^{-2}\) and b = 40.8 cm\(^3\) mol\(^{-1}\).
Answer:
\( P_c=\frac{a}{27b^2}=8.27 \) MPa.
\( T_c=\frac{8a}{27Rb}=324.8 \) K (R = 8.314 kPa dm\(^3\) K\(^{-1}\) mol\(^{-1}\)).
\( V_c=3b=122.4 \) cm\(^3\) mol\(^{-1}\).

Reduced Equation of State

Reduced variables are \( T_r=\frac{T}{T_c},\ P_r=\frac{P}{P_c},\ V_r=\frac{V_m}{V_c} \). Substituting the critical constants into the van der Waals equation removes a, b and R:
\[ \left(P_r+\frac{3}{V_r^2}\right)(3V_r-1)=8T_r \]
So all van der Waals gases follow one curve in reduced variables (law of corresponding states).

Worked Example 2. For gases with a = 4.0, 8.0, 6.0 and 12.0 atm L\(^2\) mol\(^{-2}\) and b = 0.027, 0.030, 0.032, 0.027 L mol\(^{-1}\), which has the highest critical temperature?
Answer:
\( T_c\propto a/b \): 148, 267, 188 and 444 respectively.
The fourth gas has the highest \( T_c \).