15. Ideal gases

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15.1 The mole

Role	Correct term
SI base quantity	amount of substance
quantity symbol	n
SI base unit	mole
unit symbol	mol

Amount of substance provides a macroscopic measure of how many specified microscopic entities are present. Always name the entity: atom, molecule, ion, electron or formula unit.

One mole of H₂O means one mole of H₂O molecules. It contains two moles of H atoms and one mole of O atoms only after the molecular composition is used.

The mole is a unit, not a particle or a mass unit. Different substances can have the same amount in mol but different masses because their molar masses differ.

One mole of any substance is the amount containing a number of specified entities equal to the Avogadro constant N_A. Thus N_A is the number of entities per mole.

$N_A = 6.02 × 10²³ mol⁻¹$

$n = m/M_m, N = nN_A, n = N/N_A, mass of one entity = M_m/N_A$

Use compatible mass units: if m is in kg, use molar mass M_m in kg mol⁻¹; if m is in g, use g mol⁻¹. N is a dimensionless count, while n is in mol.

0.50 mol of electrons contains N=(0.50)(6.02×10²³)=3.01×10²³ electrons. A gas sample containing 3.4×10²² molecules has n=(3.4×10²²)/(6.02×10²³)=5.65×10⁻² mol.

N_A is not a mass and does not depend on substance. Specify whether N counts atoms, molecules, ions, electrons or formula units; one mole of NaCl means N_A formula units, not N_A grams.

15.2 Equation of state

An ideal gas is a gas that obeys pV ∝ T, where p is pressure, V is volume and T is thermodynamic temperature, across the states considered.

$For a fixed amount of ideal gas: pV/T = constant and p₁V₁/T₁ = p₂V₂/T₂$

Additional fixed quantity	Resulting proportionality
V and amount fixed	p ∝ T
p and amount fixed	V ∝ T
T and amount fixed	pV = constant

A sealed rigid sample heated from 300 K to 450 K has p₂/p₁=T₂/T₁=1.50. This p∝T result follows only because both amount and volume are fixed.

Do not define every ideal gas as merely p∝T: the defining relation contains pV. Use kelvin, not Celsius, in ratios. Real gases approximate ideal behavior best under appropriate conditions; the ideal equation is the model definition.

Given gas quantity	Equation	Constant
amount n in mol	pV = nRT	R = 8.31 J mol⁻¹ K⁻¹
number N of molecules	pV = NkT	k = 1.38 × 10⁻²³ J K⁻¹

Symbol	SI meaning/unit
p	pressure in Pa
V	volume in m³
T	thermodynamic temperature in K
n	amount in mol
N	number of molecules, a count

For n=0.0160 mol, T=282 K and V=1.87×10⁻⁴ m³, p=nRT/V=(0.0160)(8.31)(282)/(1.87×10⁻⁴)=2.01×10⁵ Pa.

If pV=270 J and kT=8.0×10⁻²¹ J per molecule, N=pV/(kT)=270/(8.0×10⁻²¹)=3.4×10²² molecules.

$N=nN_A and k=R/N_A, so NkT=(nN_A)(R/N_A)T=nRT$

n and N are not interchangeable: n is measured in mol and N is a molecule count. Match n with R or N with k, convert cm³/dm³ to m³, and always use kelvin.

Boltzmann constant k=R/N_A, so kT is the thermal energy scale per particle while RT is the corresponding molar scale.

Use k when counting individual particles and R when working with moles; the two forms describe the same thermal physics at different scales.

The average translational energy scale of one molecule uses kT, whereas one mole uses RT.

k and R are not interchangeable numbers; their units and particle-count basis differ.

15.3 Kinetic theory of gases

Basic assumption	Meaning in the ideal model
Molecules are in continuous random motion	Every direction is equally likely in a large sample
Molecular volume is negligible compared with gas volume	Molecules are treated as point particles for the model
No intermolecular forces act except during collisions	Molecular potential energy is taken as zero and motion is uniform between collisions
Collisions are perfectly elastic	Total kinetic energy is conserved in each collision
Collision duration is negligible	Collisions are treated as instantaneous

These are modelling assumptions, not literal properties of every real gas. Ideal behaviour is a better approximation when molecules are far apart, so their own volume and intermolecular forces are negligible.

At very high pressure molecules are close together. Their volume or intermolecular forces may no longer be negligible, so the gas can depart from ideal behaviour.

“No intermolecular forces” means no forces between collisions. During a collision, forces act briefly, change molecular momentum and transfer momentum.

A molecule colliding elastically with a wall reverses its perpendicular momentum. The wall exerts a force on the molecule, so the molecule exerts an equal and opposite force on the wall. Many impacts across the wall area produce pressure.

Take a cube of side L and volume V=L³. One molecule has mass m and x-component of velocity cₓ toward a wall of area A=L².

$momentum change magnitude = 2mcₓ time between successive hits on the same wall = 2L/cₓ average force contribution = (2mcₓ)/(2L/cₓ) = mcₓ²/L$

$F = (m/L)Σcₓ² p = F/A = mΣcₓ²/L³ pV = mΣcₓ² = Nm<cₓ²>$

$Random motion is isotropic: <cₓ²>=<cᵧ²>=<c_z²> and <c²>=<cₓ²>+<cᵧ²>+<c_z²> Therefore <cₓ²>=(1/3)<c²>, so pV=(1/3)Nm<c²>.$

Symbol	Meaning
N	number of molecules
m	mass of one molecule
<c²>	mean of the squared molecular speeds

The factor 1/3 comes from three equivalent squared velocity components. It does not come from three walls. Also <c²> is the mean square speed, not <c>².

$c_rms = sqrt(<c²>) so c_rms²=<c²>$

Step	Operation on all molecular speeds
1	square each speed c
2	find the mean <c²>
3	take the square root

For speeds 2, 3 and 6 m s⁻¹, <c²>=(4+9+36)/3=16.3 m² s⁻², so c_rms=sqrt(16.3)=4.04 m s⁻¹. The arithmetic mean speed is 3.67 m s⁻¹, so the two means are not equal.

$For one ideal gas species, c_rms=sqrt(3kT/m), so c_rms is proportional to sqrt(T). A graph of c_rms against thermodynamic T starts at the origin and rises with decreasing gradient.$

c_rms is a statistical speed scale, not the speed of every molecule. Use thermodynamic temperature in kelvin; do not replace <c²> by <c>².

$pV=(1/3)Nm<c²> and pV=NkT (1/3)Nm<c²>=NkT (1/3)m<c²>=kT (1/2)m<c²>=(3/2)kT Therefore <E_k>=(3/2)kT.$

The average translational kinetic energy per molecule depends only on thermodynamic temperature. At the same T, molecules of different ideal gases have the same average translational kinetic energy.

At T=400 K, <E_k>=(3/2)(1.38×10⁻²³)(400)=8.28×10⁻²¹ J per molecule.

$Because (1/2)m c_rms²=(3/2)kT, c_rms=sqrt(3kT/m). At equal T, the lighter molecule has the greater rms speed even though average translational kinetic energies are equal.$

Do not omit the factor 1/2 from kinetic energy or use Celsius. This is an average per molecule: individual molecules have a distribution of speeds and energies.

15. Ideal gases

Exam analysis

Most tested topics

15.1 The mole

Amount of substance is an SI base quantity measured in moles

Use Avogadro's constant to connect mass, moles and particles

15.2 Equation of state

An ideal gas obeys pV proportional to thermodynamic temperature

Choose the molar or molecular ideal-gas equation

Boltzmann’s constant links particle-scale temperature to molar gas constant

15.3 Kinetic theory of gases

The kinetic model treats an ideal gas as many randomly moving particles

Derive gas pressure from molecular collisions with a wall

Root-mean-square speed is the square root of mean-square speed

Average translational kinetic energy is three-halves kT