Part 1 · Chapter 4

States of Matter: Gases & Liquids

The tug-of-war between intermolecular forces and thermal motion — and the laws that govern gases and liquids that result

Fundamentals of Chemistry Prof. Mithun Mondal Reading time ≈ 44 min
i What you'll learn
  • The intermolecular forces that decide whether matter is a gas, liquid or solid.
  • The gas laws and how they combine into the ideal gas equation \(PV=nRT\).
  • Dalton's law of partial pressures and the kinetic molecular theory behind it.
  • The three molecular speeds, the Maxwell distribution, and Graham's law of diffusion.
  • Why real gases deviate from ideality, and the van der Waals equation that corrects for it.
  • Critical temperature, liquefaction, and the properties of the liquid state.
Section 4-1

Intermolecular Forces vs Thermal Energy

The state of matter is settled by a contest. Intermolecular forces pull molecules together; thermal energy drives them apart. When forces win, matter is solid; when thermal energy wins, it is gas; the balance between them is liquid.

The attractive forces — collectively van der Waals forces — come in a few flavours: weak, ever-present London dispersion forces (from instantaneous dipoles, strengthening with molecular size); dipole–dipole forces between polar molecules; dipole–induced dipole forces; and the much stronger hydrogen bond.

Why size matters. London forces grow with the number of electrons, so the noble gases and halogens boil at progressively higher temperatures down their groups — larger, more polarizable atoms hold together more firmly.
Section 4-2

The Gas Laws

Four experimental laws relate the pressure, volume, temperature and amount of a gas, each holding the other quantities fixed.

LawHolds constantRelation
Boyle's\(T,\ n\)\(PV=\text{const}\)
Charles's\(P,\ n\)\(V/T=\text{const}\)
Gay-Lussac's\(V,\ n\)\(P/T=\text{const}\)
Avogadro's\(P,\ T\)\(V/n=\text{const}\)
V → P → PV = const
Boyle's law — \(P\) vs \(V\)
T (K) → V → V ∝ T
Charles's law — \(V\) vs \(T\)
! Always use the Kelvin scale

Charles's and Gay-Lussac's laws are linear only in absolute temperature. Extrapolating Charles's line to zero volume points to \(-273.15\ ^\circ\text{C}=0\ \text{K}\) — absolute zero. Always convert \(^\circ\text{C}\) to \(\text{K}\) before using a gas law.

Section 4-3

The Ideal Gas Equation

Combining the four laws gives a single equation of state for a hypothetical ideal gas — one with point-mass molecules and no intermolecular forces.

⚖️
The ideal gas equation
\(PV=nRT\)

with the gas constant \(R=8.314\ \text{J mol}^{-1}\text{K}^{-1}=0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}\). Rewriting \(n=\tfrac{m}{M}\) gives the density form \(PM=\rho RT\), and for a fixed sample the combined law \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\).

Section 4-4

Dalton's Law of Partial Pressures

In a mixture of non-reacting gases, each gas exerts the pressure it would alone — its partial pressure — and the total is their sum.

Partial pressures
\[ P_{\text{total}}=P_1+P_2+\cdots,\qquad P_i=x_i\,P_{\text{total}} \]
The partial pressure of a component equals its mole fraction \(x_i\) times the total pressure.
Gas collected over water. A gas bubbled through water carries water vapour with it, so the dry-gas pressure is \(P_{\text{gas}}=P_{\text{total}}-P_{\text{water vapour}}\) — subtract the aqueous tension before any calculation.
Section 4-5

The Kinetic Molecular Theory

The gas laws are explained by a simple microscopic model. Its postulates: gas molecules are tiny points in ceaseless random motion; collisions are perfectly elastic; there are no intermolecular forces; and the average kinetic energy is proportional to the absolute temperature.

🎱
Pressure and energy from motion
\(PV=\tfrac{1}{3}\,nM\,u_{\text{rms}}^2,\qquad \bar{E}_k=\tfrac{3}{2}\,k_B T\ \text{per molecule}\)

Pressure is the drumbeat of molecules striking the walls. Since average kinetic energy depends only on temperature, all gases at the same \(T\) share the same mean molecular energy — heavier molecules simply move more slowly.

Section 4-6

Molecular Speeds & the Maxwell Distribution

Molecules do not all move at one speed; they share a distribution. Three speeds summarize it: the most probable speed (the peak), the average speed, and the root-mean-square speed.

molecular speed → fraction low T high T
Raising temperature broadens the curve and shifts the peak to higher speed
The three speeds
\[ u_{\text{rms}}=\sqrt{\frac{3RT}{M}},\quad u_{\text{avg}}=\sqrt{\frac{8RT}{\pi M}},\quad u_{\text{mp}}=\sqrt{\frac{2RT}{M}} \]
They sit in a fixed ratio \(u_{\text{mp}}:u_{\text{avg}}:u_{\text{rms}}=1:1.128:1.224\), so \(u_{\text{mp}}
Section 4-7

Graham's Law of Diffusion

Lighter gases move faster, so they diffuse (spread) and effuse (escape through a pinhole) more quickly. Graham found the rate varies inversely with the square root of molar mass.

💨
Graham's law
\(\dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1}}=\sqrt{\dfrac{d_2}{d_1}}\)

At the same temperature and pressure, the rate of diffusion is inversely proportional to \(\sqrt{M}\). This is how hydrogen escapes a balloon faster than air, and how uranium isotopes were once separated.

Section 4-8

Real Gases & the van der Waals Equation

Real gases obey \(PV=nRT\) only at low pressure and high temperature. At high pressure their molecules occupy real volume; at low temperature attractions matter. The deviation is tracked by the compressibility factor \(Z=\dfrac{PV}{nRT}\) — exactly 1 for an ideal gas.

ideal Z=1 real gas P → Z
\(Z\) dips below 1 (attractions dominate), then rises above 1 (finite volume dominates)
van der Waals equation
\[ \left(P+\frac{an^2}{V^2}\right)(V-nb)=nRT \]
The term \(\tfrac{an^2}{V^2}\) corrects pressure for attractions; \(nb\) corrects volume for the finite size of molecules.
Section 4-9

Liquefaction & the Critical State

Cooling and compressing a gas eventually liquefies it — but only below a certain temperature. The critical temperature \(T_c\) is the highest temperature at which a gas can be liquefied by pressure alone; above it, no amount of pressure works.

Why ammonia liquefies easily and oxygen does not. A gas with strong intermolecular attractions has a high \(T_c\) and liquefies readily; one with weak attractions (high thermal energy needed) has a low \(T_c\). At the critical point the densities of liquid and vapour become identical and the meniscus vanishes.
Section 4-10

The Liquid State

Liquids have a definite volume but no fixed shape. Three properties capture their behaviour. Vapour pressure is 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 is the energy per unit area of surface, pulling a drop into a sphere. Viscosity measures a liquid's resistance to flow.

① Vapour pressure

Rises with temperature and falls with stronger intermolecular forces; boiling occurs when it equals the surrounding pressure.

② Surface tension

Surface molecules are pulled inward, minimizing area — drops bead, and capillary rise follows. It falls as temperature rises.

③ Viscosity

Internal friction between layers; greater for strong forces and large molecules, and it decreases sharply with heating.

Worked Examples

Putting It to Work

1 A combined-gas-law change

Problem. A gas occupies \(2.0\ \text{L}\) at \(300\ \text{K}\) and \(1.0\ \text{atm}\). Find its volume at \(450\ \text{K}\) and \(2.0\ \text{atm}\).

Solution. Use \(\tfrac{P_1V_1}{T_1}=\tfrac{P_2V_2}{T_2}\):

Working
\[ V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{1.0\times2.0\times450}{300\times2.0}=1.5\ \text{L} \]
2 Moles from the ideal gas law

Problem. How many moles of gas occupy \(4.92\ \text{L}\) at \(2.0\ \text{atm}\) and \(300\ \text{K}\)?

Solution. Rearrange \(PV=nRT\) with \(R=0.0821\):

Working
\[ n=\frac{PV}{RT}=\frac{2.0\times4.92}{0.0821\times300}=0.40\ \text{mol} \]
3 A partial-pressure mixture

Problem. A vessel holds \(2\ \text{mol}\ \ce{N2}\) and \(3\ \text{mol}\ \ce{O2}\) at a total pressure of \(5\ \text{atm}\). Find the partial pressure of oxygen.

Solution. Mole fraction of \(\ce{O2}\) is \(\tfrac{3}{5}\):

Working
\[ P_{\ce{O2}}=x_{\ce{O2}}\,P_{\text{total}}=\frac{3}{5}\times5=3\ \text{atm} \]
4 Root-mean-square speed

Problem. Find the RMS speed of oxygen molecules \((M=32\times10^{-3}\ \text{kg mol}^{-1})\) at \(300\ \text{K}\).

Solution. Use \(u_{\text{rms}}=\sqrt{\tfrac{3RT}{M}}\) with \(R=8.314\):

Working
\[ u_{\text{rms}}=\sqrt{\frac{3(8.314)(300)}{32\times10^{-3}}}\approx 483\ \text{m s}^{-1} \]
5 Comparing rates with Graham's law

Problem. How much faster does hydrogen \((M=2)\) effuse than oxygen \((M=32)\)?

Solution. Apply \(\tfrac{r_{\ce{H2}}}{r_{\ce{O2}}}=\sqrt{\tfrac{M_{\ce{O2}}}{M_{\ce{H2}}}}\):

Working
\[ \frac{r_{\ce{H2}}}{r_{\ce{O2}}}=\sqrt{\frac{32}{2}}=\sqrt{16}=4 \]

Hydrogen effuses four times faster.

6 A compressibility check

Problem. At a certain pressure \(1\ \text{mol}\) of a gas occupies \(0.9\ \text{L}\) where an ideal gas would occupy \(1.0\ \text{L}\) at the same \(P,T\). Find \(Z\) and say which effect dominates.

Solution. Since \(Z=\tfrac{V_{\text{real}}}{V_{\text{ideal}}}\) at fixed \(P,T\):

Working
\[ Z=\frac{0.9}{1.0}=0.9<1 \]

\(Z<1\) means the gas is more compressible than ideal — intermolecular attractions dominate here.

Review

Chapter Summary

✓ Forces vs heat

State of matter is the balance of intermolecular attraction against thermal energy.

✓ Gas laws

Boyle, Charles, Gay-Lussac and Avogadro combine into \(PV=nRT\); use Kelvin.

✓ Mixtures

Dalton: \(P_i=x_iP_{\text{total}}\); subtract aqueous tension for gas over water.

✓ Kinetic theory

\(\bar E_k\propto T\); \(u_{\text{mp}}; Graham's law \(r\propto 1/\sqrt M\).

✓ Real gases

\(Z=\tfrac{PV}{nRT}\); van der Waals corrects for volume and attraction.

✓ Liquids

Vapour pressure, surface tension and viscosity; \(T_c\) sets liquefiability.

Practice

Problems

Convert temperatures to Kelvin, pick the gas law that holds the right quantities constant, and keep units consistent with your value of \(R\). Difficulty rises down the list.

  1. A gas at \(2.0\ \text{atm}\) occupies \(3.0\ \text{L}\). What volume will it occupy at \(1.0\ \text{atm}\), temperature constant?
  2. A balloon of \(1.0\ \text{L}\) at \(27\ ^\circ\text{C}\) is heated to \(127\ ^\circ\text{C}\) at constant pressure. Find the new volume.
  3. Calculate the mass of \(\ce{CO2}\) in a \(10\ \text{L}\) vessel at \(2.0\ \text{atm}\) and \(300\ \text{K}\).
  4. Find the density of nitrogen gas at \(1\ \text{atm}\) and \(273\ \text{K}\).
  5. A flask contains \(\ce{H2}\) and \(\ce{He}\) in mole ratio \(1:3\) at total pressure \(4\ \text{atm}\). Find each partial pressure.
  6. Calculate the RMS, average and most probable speeds of \(\ce{N2}\) at \(300\ \text{K}\).
  7. An unknown gas diffuses \(1.5\) times slower than oxygen. Find its molar mass.
  8. A gas is collected over water at \(25\ ^\circ\text{C}\); total pressure is \(760\ \text{mmHg}\) and the aqueous tension is \(24\ \text{mmHg}\). Find the dry-gas pressure.
  9. At what temperature will the RMS speed of \(\ce{O2}\) equal that of \(\ce{H2}\) at \(300\ \text{K}\)?
  10. Using the van der Waals equation, explain qualitatively why \(Z<1\) at moderate pressures but \(Z>1\) at very high pressures.
  11. Two gases \(\ce{NH3}\) and \(\ce{HCl}\) are released from opposite ends of a \(100\ \text{cm}\) tube simultaneously. How far from the \(\ce{HCl}\) end does the white ring form?
  12. Arrange \(\ce{He},\ \ce{CO2},\ \ce{NH3}\) in order of increasing critical temperature, and justify your ordering with intermolecular forces.
Tip: for any gas calculation, first list \(P,V,n,T\) and mark what changes. If only two variables change, a single gas law suffices; if amount or identity is involved, go straight to \(PV=nRT\). Convert \(^\circ\text{C}\to\text{K}\) and match the units of \(R\) before substituting — most errors here are unit errors, not concept errors.