States of Matter • Topic 3 of 3

Real Gases & Liquid State

The ideal gas equation works beautifully at low pressure and high temperature, but every real gas deviates from it under other conditions. The reason is that two ideal assumptions break down: real molecules do occupy finite volume, and they do exert attractive forces on one another.

The cleanest way to measure deviation is the compressibility factor $Z=\frac{PV}{nRT}$. For an ideal gas $Z=1$ at all conditions. For a real gas:

  • At low pressure, attractive forces dominate, pulling molecules together and reducing volume, so $Z<1$.
  • At high pressure, the finite molecular volume dominates and the gas is harder to compress, so $Z>1$.
  • $\text{H}_2$ and $\text{He}$ (very weak attractions) show $Z>1$ at almost all pressures.

To repair the ideal equation, van der Waals introduced two corrections. The measured pressure is less than ideal because attractions slow molecules approaching the wall, so we add $\frac{an^2}{V^2}$; the free space is less than the container volume because molecules themselves take up room, so we subtract $nb$. This gives the van der Waals equation: $\left(P+\frac{an^2}{V^2}\right)(V-nb)=nRT$, where $a$ measures the strength of attraction and $b$ the effective molecular volume.

When a gas is cooled and compressed enough, attractions win and it liquefies. But there is a limit: above the critical temperature $T_c$, no amount of pressure can liquefy a gas. The pressure needed to liquefy it exactly at $T_c$ is the critical pressure $P_c$, and the volume of one mole then is the critical volume $V_c$. These critical constants relate to the van der Waals constants: $T_c=\frac{8a}{27Rb}$, $P_c=\frac{a}{27b^2}$, $V_c=3b$.

Once liquefied, the substance shows the distinctive properties of liquids:

  • Vapour pressure — the pressure of vapour in equilibrium with its liquid; it rises with temperature, and a liquid boils when its vapour pressure equals the external pressure.
  • Surface tension ($\gamma$) — the inward pull on surface molecules that minimises surface area, causing droplets to be spherical and water to rise in a capillary; it decreases as temperature rises.
  • Viscosity ($\eta$) — a liquid’s resistance to flow, arising from internal friction between layers; it too decreases with rising temperature as molecules gain energy to slide past one another.
Z vs P: real gases dip below Z=1 then rise above it; an ideal gas stays at Z=1Pressure PZ = PV/nRTideal, Z=1Z < 1 (CH4, CO2)Z > 1 (H2, He)
Solved Examples
1
Worked Example
At 300 K and 5 atm, 2.0 mol of a gas occupies 9.0 L. Find its compressibility factor $Z$. ($R=0.0821$)
Solution
  1. Step 1: $Z=\frac{PV}{nRT}$.
  2. Step 2: $Z=\frac{5\times9.0}{2.0\times0.0821\times300}=\frac{45}{49.26}$.
  3. Step 3: $Z\approx0.91$.

Answer: $Z\approx0.91$, so attractive forces dominate ($Z<1$).

2
Worked Example
For a real gas $Z=1.05$ at high pressure. What does this indicate about the dominant effect?
Solution
  1. Step 1: $Z=\frac{PV}{nRT}>1$ means the actual volume exceeds the ideal prediction.
  2. Step 2: This excess volume comes from the molecules’ own finite size (the $b$ term).
  3. Step 3: Hence repulsion / finite molecular volume dominates at high pressure.

Answer: Finite molecular volume (repulsive effect) dominates, making the gas less compressible than ideal.

3
Worked Example
For $\text{CO}_2$, $a=3.6\ \text{L}^2\text{atm mol}^{-2}$ and $b=0.043\ \text{L/mol}$. Estimate the critical temperature. ($R=0.0821$)
Solution
  1. Step 1: $T_c=\frac{8a}{27Rb}$.
  2. Step 2: $T_c=\frac{8\times3.6}{27\times0.0821\times0.043}=\frac{28.8}{0.0953}$.
  3. Step 3: $T_c\approx302\ \text{K}$.

Answer: $T_c\approx302\ \text{K}$ (close to the measured $304\ \text{K}$).

4
Worked Example
The van der Waals constant $b$ for a gas is $0.0428\ \text{L/mol}$. Find its critical volume per mole.
Solution
  1. Step 1: $V_c=3b$.
  2. Step 2: $V_c=3\times0.0428$.
  3. Step 3: $V_c=0.1284\ \text{L/mol}$.

Answer: $V_c\approx0.128\ \text{L/mol}$.

5
Worked Example
A liquid in an open dish boils when its vapour pressure reaches 1 atm at $78^\circ\text{C}$. What happens to its boiling point on a high mountain where external pressure is 0.7 atm?
Solution
  1. Step 1: A liquid boils when vapour pressure equals external pressure.
  2. Step 2: Lower external pressure (0.7 atm) is reached at a lower vapour pressure, hence a lower temperature.
  3. Step 3: Therefore the boiling point falls below $78^\circ\text{C}$.

Answer: The boiling point decreases at lower external pressure.

6
Worked Example
Among $\text{He}$, $\text{N}_2$ and $\text{NH}_3$, which is the easiest to liquefy and why?
Solution
  1. Step 1: Ease of liquefaction tracks the strength of intermolecular attraction (larger van der Waals $a$, higher $T_c$).
  2. Step 2: $\text{NH}_3$ has hydrogen bonding (strongest attraction); $\text{N}_2$ has only weak dispersion; $\text{He}$ has the weakest of all.
  3. Step 3: So $\text{NH}_3$ has the highest critical temperature.

Answer: $\text{NH}_3$ is easiest to liquefy because hydrogen bonding gives it the strongest attractions and highest $T_c$.

Key Points

  • Real gases deviate from ideal behaviour because molecules have finite volume and exert attractive forces; deviation is largest at high pressure and low temperature.
  • Compressibility factor $Z=\frac{PV}{nRT}$: $Z=1$ for an ideal gas; $Z<1$ when attractions dominate (low $P$) and $Z>1$ when molecular volume dominates (high $P$).
  • van der Waals equation: $\left(P+\frac{an^2}{V^2}\right)(V-nb)=nRT$, where $a$ measures attraction and $b$ the molecular volume.
  • A gas cannot be liquefied above its critical temperature $T_c$; critical constants relate to van der Waals constants by $T_c=\frac{8a}{27Rb}$, $P_c=\frac{a}{27b^2}$, $V_c=3b$.
  • Liquid properties: vapour pressure rises with temperature (boiling when it equals external pressure); surface tension and viscosity both decrease as temperature rises.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.For an ideal gas the compressibility factor $Z$ is:
Explanation: By definition $Z=\frac{PV}{nRT}=1$ for an ideal gas under all conditions.
Q2.In the van der Waals equation, the term $\frac{an^2}{V^2}$ corrects for:
Explanation: The pressure-correction term accounts for attractive forces that lower the measured pressure.
Q3.A gas cannot be liquefied by pressure alone if it is above its:
Explanation: Above $T_c$ molecular kinetic energy always exceeds the attractions, so no pressure can liquefy the gas.
Q4.As temperature increases, the surface tension of a liquid:
Explanation: Higher thermal energy weakens the net inward pull on surface molecules, lowering surface tension.
Practice more MCQs → 📝 Assignment · 30 marks