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The Gaseous State — Notes

6 sections 5 min readBeta

1. Intermolecular forces and the gaseous state

This chapter is for formative assessment only (not in the annual theory paper). A gas is described by pp, VV, TT and nn. The state of a substance depends on the balance between thermal energy (which keeps molecules apart) and intermolecular forces (which pull them together): in a gas thermal energy dominates. Forces in brief: dispersion (all molecules, weak, grows with size), dipole–dipole (polar molecules), hydrogen bonding (H on N, O, F), ion–dipole. Units: 1 atm = 101.325 kPa = 760 mmHg = 1.01325 bar; T=t+273.15T = t + 273.15.

Table of intermolecular forces by type, particles involved and strength
ForceBetweenStrength
DispersionAll moleculesWeak, grows with size
Dipole–dipolePolar moleculesModerate
Hydrogen bondH on N/O/F and a lone pairStrongest of the three
Ion–dipoleIon and polar moleculeStrong

Key terms

Intermolecular force

An attraction between neighbouring molecules, much weaker than a chemical bond.

Dispersion force

Attraction between temporary dipoles; present in all molecules.

Liquefaction

Conversion of a gas into a liquid by cooling or compression.

2. Boyle’s and Charles’s laws

Boyle: at constant nn and TT, p∝1/Vp \propto 1/V, so p1V1=p2V2p_1V_1 = p_2V_2. pp against VV is a hyperbola; pp against 1/V1/V is a straight line of slope nRTnRT.

Charles: at constant nn and pp, V∝TV \propto T (in kelvin), so V1/T1=V2/T2V_1/T_1 = V_2/T_2. VV against tt in °C is a straight line meeting V=0V = 0 at −273.15 °C.

Combined: p1V1T1=p2V2T2\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}.

One mole of an ideal gas at 300 K and 400 K.
Dashed parts are extrapolated; real gases liquefy before reaching this temperature.

Formulas

  • Boyle
    p1V1=p2V2p_1V_1 = p_2V_2
  • Charles
    V1T1=V2T2\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}
  • Combined gas law
    p1V1T1=p2V2T2\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}

Key terms

Isotherm

A curve showing p against V at one temperature.

Absolute zero

0 K = −273.15 °C, where the extrapolated volume of an ideal gas would be zero.

3. Gay-Lussac’s law, Avogadro’s law and the ideal gas equation

Gay-Lussac (pressure law): at constant nn and VV, p∝Tp \propto T. Avogadro: at the same TT and pp, V∝nV \propto n; equal volumes of gases have equal numbers of molecules. Together: pV=nRTpV = nRT.

R depends on units: 0.08314 L bar K⁻¹ mol⁻¹; 0.0821 L atm K⁻¹ mol⁻¹; 8.314 J K⁻¹ mol⁻¹; 62.36 L mmHg K⁻¹ mol⁻¹. (1 L bar = 100 J.)

Table of values of the gas constant in four units
UnitsR
L bar K⁻¹ mol⁻¹0.08314
L atm K⁻¹ mol⁻¹0.0821
J K⁻¹ mol⁻¹8.314
L mmHg K⁻¹ mol⁻¹62.36

Formulas

  • Ideal gas equation
    pV=nRTpV = nRT

Key terms

Ideal gas

A hypothetical gas that obeys pV = nRT at all pressures and temperatures.

Gas constant R

The constant in pV = nRT, equal to pV/nT; its value depends on the units.

4. Density, molar mass and Dalton’s law

Density and molar mass: d=pMRTd = \dfrac{pM}{RT} and M=dRTpM = \dfrac{dRT}{p}; at fixed pp and TT density is proportional to molar mass.

Dalton’s law: ptotal=p1+p2+…p_\text{total} = p_1 + p_2 + \dots, with pi=xi ptotalp_i = x_i\,p_\text{total} and xi=ni/nx_i = n_i/n.

Gas collected over water: pgas=ptotal−pHX2Op_\text{gas} = p_\text{total} - p_{\ce{H2O}}. Vapour pressure of water (mmHg): 17.5 at 20 °C, 23.8 at 25 °C, 26.7 at 27 °C, 31.8 at 30 °C, 55.3 at 40 °C.

Table of the vapour pressure of water at five temperatures
T / °C2025273040
Vapour pressure of water / mmHg17.523.826.731.855.3

Formulas

  • Density of a gas
    d=pMRTd = \dfrac{pM}{RT}
  • Partial pressure
    pi=xi ptotalp_i = x_i\,p_\text{total}

Key terms

Partial pressure

The pressure a gas in a mixture would exert if it alone filled the container at the same temperature.

Mole fraction

Moles of a component divided by total moles.

Aqueous tension

The vapour pressure of water at the temperature of the experiment.

5. Deviations from ideality and the compressibility factor

Real gases deviate because (i) molecules have a finite volume (important at high pp) and (ii) there are attractions (important at low TT and moderate pp). A real gas is closest to ideal at low pressure and high temperature.

Z=pVnRTZ = \dfrac{pV}{nRT}. Z=1Z = 1 ideal; Z<1Z < 1 attractions dominate; Z>1Z > 1 molecular volume dominates. Z against p: H₂ and He at 273 K start at 1 and rise; N₂ dips slightly then rises; CO₂ dips far below 1 then rises.

Schematic curves, not measured data.

Formulas

  • Compressibility factor
    Z=pVnRTZ = \dfrac{pV}{nRT}

Key terms

Compressibility factor

Z = pV/nRT, a measure of departure from ideal behaviour.

Real gas

A gas that obeys pV = nRT only approximately, best at low pressure and high temperature.

6. van der Waals equation and liquefaction

(p+an2V2)(V−nb)=nRT\left(p + \dfrac{an^2}{V^2}\right)(V - nb) = nRT. a (L² bar mol⁻²) corrects for attractions (larger a: easier to liquefy); b (L mol⁻¹) is the excluded volume per mole, about four times the actual volume of the molecules. Examples: He a = 0.0346, b = 0.0238; CO₂ a = 3.640, b = 0.0427.

Liquefaction: above the critical temperature Tc no pressure liquefies a gas; the critical pressure pc is the pressure needed at Tc. Tc: CO₂ 304.1 K, NH₃ 405.5 K, Cl₂ 417 K, O₂ 154.6 K, N₂ 126.2 K, H₂ 33.2 K. Gases with large a have high Tc.

Table of critical temperature and critical pressure of five gases
GasTc / Kpc / bar
CO₂304.173.8
NH₃405.5113.5
O₂154.650.4
N₂126.233.9
H₂33.213.0

Formulas

  • van der Waals equation
    (p+an2V2)(V−nb)=nRT\left(p + \dfrac{an^2}{V^2}\right)(V - nb) = nRT

Key terms

Excluded volume

The volume around each molecule into which the centres of other molecules cannot enter; b per mole.

Critical temperature

The highest temperature at which a gas can be liquefied by pressure.

Critical pressure

The pressure needed to liquefy a gas at its critical temperature.

Exam tips

Common mistakes

  • Using temperature in °C, or mixing units of R with the units of p and V.
  • Forgetting to subtract the vapour pressure of water for a gas collected over water.
  • Saying Z < 1 means repulsion dominates (it is attraction).
  • Calling b the actual volume of molecules (it is about four times that volume).
  • Thinking a gas can always be liquefied by pressure alone (not above Tc).

How marks are usually awarded

  • Numericals (2–3 marks): formula (1), substitution with correct units (1), answer with unit (1).
  • Deviation questions: two assumptions that fail (1) + conditions for ideal behaviour (1).
  • van der Waals: equation (1) + meaning of a and b (1).