Ideal Gases

— Chemistry · AS Level (7405) · States of Matter —

★ Exam Tip: Most questions on this chapter come in Paper 1 (MCQs) — the hard ones are there. Paper 2 rarely has stand-alone questions (sometimes a simple pV = nRT). Marks are lost on unit conversions, not on the chemistry!

Scope: “States of matter” is the name of the whole chapter. Covalent / ionic / metallic bonding, giant lattices, simple molecular structures, graphene–graphite–diamond and SiO₂ were already done in Chemical Bonding. Here we do only ideal gases (gaseous state: ideal and real gases).

1. Kinetic Theory of Gases

The kinetic theory tells us what an ideal gas should look like — it is a model (a set of assumptions) of how gas particles behave.

random motion · elastic collisions · hits on walls = p

Particles are in continuous random motion — always moving from place to place.
Particles behave as rigid spheres (think tiny marbles).
Pressure of the gas is due to collisions between the particles and the walls of the container.
All collisions are perfectly elastic ⇒ no loss of kinetic energy (they rebound with the same speed; in real life some energy is always lost, but not in this model).
★ The intermolecular forces between particles are negligible.
★ The volume occupied by the particles themselves is negligible compared to the volume of the container.

Remember: ★5 and ★6 are the VERY important ones — they decide whether a gas is ideal. Exam: “State two assumptions…” (see Q2(a) at the end).

What does ★6 mean?

Imagine collecting all the little marbles (gas particles) and packing them in one corner: that total volume is tiny compared to the container. Also remember: gases are small particles far apart, so the container is mostly empty space.

total volume of the particles ≪ volume of container (assumption ★6)

In reality: no gas follows completely ideal behaviour. Even the smallest atom (H, He, any noble gas) still has temporary dipoles, so IMF are never exactly zero — a gas can get close to ideal, but never 100 %. The assumptions work well enough that our calculations have only a small, acceptable error. Gases with stronger IMF (e.g. steam — H-bonding) are very far from ideal.

Which T and p give the most ideal behaviour?

Use ★5 and ★6: what do they need? Weak IMF and negligible particle volume. So ask: which T and p give me those?

Condition	What happens to the particles	Assumption broken	Ideal?
Low T	slow, come close together ⇒ IMF ↑; gas may condense to a liquid — liquid volume is comparable to the container	★5 and ★6	✗
High p	gas is squashed into a small volume (may even liquefy) ⇒ particle volume comparable to container	★6 (and ★5)	✗
High T	particles move faster, spread out far apart ⇒ IMF very weak	—	✓
Low p	gas free to expand as far as it likes ⇒ particle volume negligible	—	✓

For IDEALITY choose HIGH temperature and LOW pressure

IMF weak, particle volume negligible ⇒ IDEAL

IMF stronger, volume comparable to container ⇒ NON-IDEAL

Learn this sentence (written on the board): At low temperatures or high pressures, the particles get close together, which causes the intermolecular forces to increase and the volume occupied by the gas to become comparable to the volume of the particles → the gas deviates from ideal behaviour.

2. The Ideal Gas Equation

① Ideal gas equationpV = nRT

Derivation is not in the course — just memorise it as it is, and use it whenever needed.

Units — must be exactly these:
p = pressure in Pa | V = volume in m³ | n = number of moles
R = molar gas constant = 8.31 J K⁻¹ mol⁻¹ (given in the paper’s data page — no need to memorise)
T = absolute temperature in K

Common mistake: forgetting to convert kPa → Pa, cm³/dm³ → m³ or °C → K. Lots of students lose marks here (especially the volume conversion!). In this chapter “the problem is always the conversions” — the questions themselves are easy.

Stoichiometry link

In stoichiometry: 1 mol of gas = 24 dm³ at room temperature and pressure (RTP). If the conditions are not RTP, assume the gas is ideal and use pV = nRT to find the volume.

Conversions (remember these!)

Convert	Rule	Logic (so you don’t need to cram)
1 atm	1.013 × 10⁵ Pa	—
kPa → Pa	× 10³	e.g. 102 kPa = 102 × 10³ Pa
1 cm³	10⁻⁶ m³	1 cm = 10⁻² m (÷100) → cube it → 10⁻⁶
1 dm³	10⁻³ m³	1 dm = 10⁻¹ m (deci = 10) → cube it → 10⁻³
m³ → cm³	× 10⁶	just flip the sign of the power
°C → K	K = °C + 273	K → °C: subtract 273

Trick: for cm³ or dm³ → m³ just cube the length conversion. Then you can never forget it.

Mass in the moles formula (n = m ÷ M_r) always goes in grams — not kg.

unit check BEFORE substituting into pV = nRT

Two other forms (③ and ②) — understand, don’t memorise

Just substitute n = m ÷ M_r (or A_r) into pV = nRT, or find n first and then use it. Whatever route you take is fine.

② To find M_r (or A_r) of a gas pV = nRTpV = mM_r/A_r × R × TM_r / A_r = m × R × Tp × V

③ To find the concentration pV = nRTp = nV × R × Tp = c × R × Tc = n/V = concentration in mol m⁻³

Convert c afterwards if needed (mol m⁻³ → mol dm⁻³ etc.).

3. Special Cases: Constant p, V or T

Board note: Learning these is not necessary — you can always derive them from pV = nRT in the exam. But it can speed up Paper 1 MCQs (the long method takes longer). Physics students will find the derivation very easy (same idea as intensity ∝ 1/A²).

Why does it work? n (same gas in a closed container) and R are always constant. If one more of p, V or T is held constant, then the whole side that has all constants is constant. We rearrange so that all constant terms sit together on one side, then replace them by “const”.

Constant pressure pV = nRTVT = nRp = constV ∝ TV₁T₁ = V₂T₂

Constant volume pV = nRTpT = nRV = constp ∝ Tp₁T₁ = p₂T₂

Constant temperature pV = nRTnRT = constpV = constp₁V₁ = p₂V₂

V ∝ T (straight line through origin)

p ∝ T (straight line through origin)

p ∝ 1/V (curve)

Warning: p₁V₁ = p₂V₂ only works if the moles stay constant. If gas is added, removed or combined with another gas (like Q1 below) you can’t use it — go back to pV = nRT. T must be in K in V₁/T₁ and p₁/T₁ too.

4. Past-Paper Questions: pV = nRT

Q13MCQ · conditions for ideality

At room T and p chlorine is not ideal. At which T and p would Cl₂ become more ideal?

Ideal ⇒ pressure ↓, temperature ↑ (weak IMF, tiny particle volume). Pick the lowest p and highest T.

A ✗ 50 kPa, 200 K — T too lowB ✓ 50 kPa, 400 K — low p + high TC ✗ 200 kPa, 200 K — high p, low TD ✗ 200 kPa, 400 K — p too high

Answer: B

Q20MCQ · find T

Iodine vapour: mass 6.35 g, volume 1.247 dm³, p = 1.00 × 10⁵ Pa. If it acts as an ideal gas, T = ?

Convert V to m³ (× 10⁻³). No n given → n = m ÷ M_r. Iodine exists as I₂ ⇒ M_r = 2 × 126.9, not 126.9!

pV = nRTpV = mM_r × R × T(1 × 10⁵)(1.247 × 10⁻³) = 6.352(126.9) × 8.31 × TT = (1×10⁵)(1.247×10⁻³)(2 × 126.9)6.35 × 8.31T ≈ 600 K (board: 599 K)

A ✗ 300 K — what you get if you forget I₂ (use 126.9)B ✓ 600 KC, D ✗ 300 000 K / 600 000 K — V not converted from dm³ to m³ (× 1000 too big)

Answer: B

same moles of He, now spread through 4V

Q21MCQ · two vessels, two unknowns

Glass vessels M and N joined by a closed valve. M has He at 20 °C, 1 × 10⁵ Pa. N is evacuated and has 3 × the volume of M. Valve opened and whole apparatus heated to 100 °C. Final pressure?

Let V_M = V ⇒ total volume after = V + 3V = 4V. Both n and V are unknown ⇒ two unknowns ⇒ solve simultaneously: use the “before” case to get n, then substitute into the “after” case. N was empty, so all the moles are the same moles — they just spread into 4V. V and R cancel.

Before (M only) pV = nRT(1×10⁵)(V) = n × 8.31 × (20 + 273.15)n = 1×10⁵ V8.31 (293.15)

After (4V, 100 °C) pV = nRTp(4V) = n × 8.31 × (100 + 273.15)p(4V) = 1×10⁵ V8.31 (293.15) × 8.31 × (373.15)p = 1×10⁵ × 373.154 × 293.15p = 31 822 Pa

= 3.18 × 10⁴ Pa ⇒ Answer: A (board used 273.15; with 273 you get ≈ 31 827 Pa — same option. Part of the board working is rubbed out.)

Q14MCQ · find volume (in cm³)

0.56 g ethene at 102 kPa and 30 °C. Volume of the vessel?

Ethene = C₂H₄ ⇒ M_r = 2(12) + 4 = 28. kPa → Pa; answer comes out in m³ ⇒ × 10⁶ for cm³.

pV = nRTpV = mM_r × R × T(102 × 10³) V = 0.5628 × 8.31 × (30 + 273.15)V = 4.94 × 10⁻⁴ m³4.94 × 10⁻⁴ × 10⁶ = 494 cm³

A (49), C (48 900), D (494 000) are wrong powers of ten from bad conversions. Answer: B

Q4MCQ · which mass fills 3 dm³

Which mass of gas occupies 3 dm³ at 25 °C and 1 atm? [1 mol of gas = 24 dm³ at 25 °C, 1 atm]

The question gives you the stoichiometry rule, so no pV = nRT needed: moles needed = 3 ÷ 24 = 0.125 mol. Work out moles of each option.

A ✗ 3.2 g O₂: 3.2 ÷ 32 = 0.1 mol (→ 2.4 dm³)B ✗ 5.6 g N₂: 5.6 ÷ 28 = 0.2 mol (→ 4.8 dm³)C ✓ 8.0 g SO₂: 8 ÷ (32 + 32) = 0.125 mol (× 24 = 3.0 dm³)D ✗ 11.0 g CO₂: 11 ÷ (12 + 32) = 0.25 mol (→ 6.0 dm³)

Check with pV = nRT (the board’s start) — also valid:(1.013×10⁵)(3×10⁻³) = n × 8.31 × (25 + 273.15)n = m/M_r ≈ 0.1227 ≈ 0.125

Answer: C

two gases mix ⇒ moles ADD, volumes ADD

Q1MCQ · connecting two flasks

Flask X: 1 dm³ He at 2 kPa. Flask Y: 2 dm³ Ne at 1 kPa. Connected at constant T — final pressure?

Find moles in each flask (T unknown but the same ⇒ just call it T), add the moles and add the volumes, then find p.

Flask XpV = nRT(2×10³)(1×10⁻³) = n × 8.31 × Tn = 28.31 T

Flask YpV = nRT(1×10³)(2×10⁻³) = n × 8.31 × Tn = 28.31 T

Combined: n = 48.31 T, V = 3 dm³ = 3×10⁻³ m³pV = nRTp(3×10⁻³) = 48.31 T × 8.31 × T8.31 and T cancelp = 43 × 10⁻³ = 1333 Pa= 1⅓ kPa

Note: you cannot use p₁V₁ = p₂V₂ here — the moles are changing (two gases combine). Must use pV = nRT. Answer: A

Q22MCQ · identify the gas from M_r

Evacuated glass tube, 200 cm³, filled with gas at 300 K and 101 kPa; mass of tube increases by 1.06 g. Identity of the gas? (Ar, Kr, Ne, Xe)

Mass increase = mass of gas. Find M_r (or A_r) and match it. T already in K, but convert kPa and cm³!

pV = nRTpV = mM_r × R × T(101×10³)(200×10⁻⁶) = 1.06M_r × 8.31 × 300M_r = 1.06 × 8.31 × 300101×10³ × 200×10⁻⁶M_r = 130.8 ≈ Xe (131)

Answer: D (xenon)

5. Comparing the Ideality of Gases

We judge ideality from graphs. First, what do the graphs of an ideal gas look like?

pV ∝ T (n, R const) — rarely used

const T ⇒ nRT const ⇒ horizontal line

pV vs T: n and R are constants ⇒ pV ∝ T ⇒ straight line through the origin (not used much — little information). pV vs p (the one we use): keep T constant ⇒ nRT is constant ⇒ pV = const ⇒ an ideal gas is always a straight horizontal line.

pV vs p at constant T: bigger bulge = bigger deviation (NH₃ most, He least)

Real gases deviate: their line bends away from the horizontal ideal line into a bulge. More bulge ⇒ more non-ideal. Gases nearly straight (He, H₂) are the most ideal; NH₃ has the biggest bulge ⇒ most non-ideal. (Teacher fixed the labels on the printed graph: the CH₄ and CO curves had been swapped. Ignore the “Z > 1 / Z < 1” labels — not in our course. Just look at the bulge!)

Why? — order of ideality He > H₂ > CH₄ > CO > NH₃

Gas	Why it is more non-ideal than the one before
He (most ideal)	Only temporary dipoles and just 2 electrons ⇒ weakest IMF; single tiny atoms (smallest particle volume).
H₂	More non-ideal than He as it exists as a diatomic molecule (board: “stronger temporary dipoles”). Both have 2 e⁻ per particle, so the main difference is that a molecule is bigger than an atom ⇒ its volume is more comparable to the container (★6).
CH₄	More non-ideal than H₂: stronger temporary dipoles because it has more electrons.
CO	More non-ideal than CH₄: it has permanent dipoles (O much more electronegative than C; molecule not symmetrical ⇒ δ+ / δ−) — stronger than temporary dipoles.
NH₃ (least ideal)	Has hydrogen bonding — the strongest IMF of these.

Method (sir’s advice): always look at the intermolecular forces first (temporary < permanent dipole < H-bond; more electrons ⇒ stronger temporary dipoles). Only if IMF can’t explain the difference (like He vs H₂, same electrons) go to particle volume / size — bigger molecules (diatomic, triatomic, P₄…) ⇒ less ideal.

Effect of temperature on ideality

lower T ⇒ bigger deviation (T₃ least ideal, T₁ most)

Same gas, three temperatures: T₃ < T₂ < T₁. T₃ shows the biggest bulge ⇒ least ideal; T₁ is nearest the horizontal ideal line ⇒ most ideal.

When temperature is lowered, the particles come closer together:

volume of the particles is no longer negligible (★6 fails)
intermolecular forces get stronger (★5 fails)

At high T IMF are weaker and the particle volume is more negligible compared with the container ⇒ more ideal.

6. Past-Paper MCQs on Ideality

Strategy for “which gas deviates most / is most ideal”: write the strongest IMF next to each option — the one with the strongest IMF is the least ideal; the one with the weakest is the most ideal.

Q11MCQ · pV vs p graph

pV vs p for an ideal and a non-ideal gas. Which gas shows the greatest deviation from ideality?

A ✓ ammonia — permanent dipoles & hydrogen bonding (strongest)B ✗ ethene — temporary dipolesC ✗ methane — temporary dipolesD ✗ nitrogen — temporary dipoles

Answer: A

not symmetrical ⇒ δ− don't cancel ⇒ permanent dipole

Q8MCQ · least resembles an ideal gas

A ✓ ammonia — hydrogen bondingB ✗ helium — temporary dipolesC ✗ hydrogen — temporary dipolesD ✗ trichloromethane CHCl₃ — permanent dipoles

CHCl₃: three Cl and one H around C ⇒ not symmetrical ⇒ the δ− on the Cl atoms don’t cancel ⇒ overall dipole ⇒ permanent dipole. But a permanent dipole is much weaker than H-bonding, so ammonia wins. (Ammonia always gets blamed!)

Answer: A

Q9MCQ · closest to ideal at RTP

Ammonia (H-bonding ✗) · carbon dioxide ✗ · helium ✓ · oxygen ✗ — helium has only temporary dipoles and just 2 e⁻ (single small atoms), so it is the closest to ideal. Answer: C

Q10MCQ · deviates most

A ✓ HCl — perm dipolesB ✗ He — temp dipolesC ✗ CH₄ — temp dipolesD ✗ N₂ — temp dipoles

Only HCl has a stronger (permanent dipole) IMF ⇒ most deviation. Answer: A

bulge size ⇒ strength of IMF ⇒ identity

Q15MCQ · identify X, Y, Z from pV–p graph

Real gases ammonia, hydrogen, nitrogen plotted as X, Y, Z. Identities?

Y — nearly straight = the most ideal ⇒ fewest temporary dipoles ⇒ H₂
X — shows more bulge ⇒ N₂: also temporary dipoles but much stronger than H₂ because it has more e⁻
Z — biggest bulge ⇒ NH₃ (hydrogen bonding)

Answer: D (X = nitrogen, Y = hydrogen, Z = ammonia)

Q25 — pV against n for an ideal gas at constant T

Q25MCQ · which graph?

pV = nRTRT = constant (T const)pV = n × (const)pV ∝ n

So a straight line through the origin (pV = 0 when n = 0). A has an intercept ✗; B is a falling curve ✗; C horizontal (pV not changing with n) ✗.

Answer: D

7. Structured Question: Kinetic Theory (Q2)

Note: this full exam question is on the board with answers written in; the recording did not cover it, so below are the board answers (plus a few extra mark-scheme-style hints in purple).

Q2 (a)[2] two assumptions of kinetic theory (ideal gas)

Negligible intermolecular forces between particles.
Volume occupied by the particles themselves is negligible compared to the volume of the container.

Q2 (b)[2] conditions where a real gas approaches ideal

(i) low pressure (ii) high temperature

Q2 (c)[3] order of ideal behaviour: ammonia, neon, nitrogen

most ideal → least ideal: neon, nitrogen, ammonia

Nitrogen has more electrons than neon ⇒ stronger temporary dipoles. Ammonia has hydrogen bonding, which is the strongest IMF.

↑T ⇒ KE ↑ ⇒ particles overcome IMF ⇒ spread out

Q2 (d)[2] kinetic-molecular model: why a liquid eventually becomes a gas as T increases

As the temperature is increased the particles move more vigorously. The intermolecular forces between them get weaker and they move further apart until the liquid becomes a gas.

Extra hint: say “kinetic energy increases” and “enough energy to overcome the IMF” in your answer.

Q2 (e)[1] total electrons in one CH₃F molecule

C: 6, 3 × H: 3, F: 96 + 3 + 9 = 18 electrons

(CH₃CH₃ also has 2 × 6 + 6 = 18, so they are iso-electronic.)

Q2 (f)[2] boiling points: CH₃CH₃ 184.5 K, CH₃F 194.7 K

(i) Close similarity: the same number of electrons (18), so the temporary dipole forces are equally strong.

(ii) CH₃F slightly higher: CH₃F is a polar molecule so it has permanent dipole forces in addition.

Q2 (g)calculation: M_r of A

0.148 g of A vaporised at 60 °C; vapour volume 67.7 cm³ at 101 kPa. Use pV = nRT to find M_r.

pV = nRTpV = mM_r × R × T(101×10³)(67.7×10⁻⁶) = 0.148M_r × 8.31 × (60 + 273)M_r = 0.148 × 8.31 × 333101×10³ × 67.7×10⁻⁶M_r = 59.9

✓ Quick Recap — Ideal Gases (Full Chapter Cheat-Sheet)

Ideal gas (kinetic theory): continuous random motion · rigid spheres · pressure = collisions with walls · elastic collisions · ★ negligible IMF · ★ negligible particle volume.
More ideal: HIGH T, LOW p (weak IMF, tiny particle volume). Less ideal: low T / high p ⇒ particles close ⇒ IMF ↑ and volume comparable ⇒ deviation.
pV = nRT p in Pa · V in m³ · n in mol · R = 8.31 J K⁻¹ mol⁻¹ · T in K | n = m/M_r (g).
M_r = mRT / pV | p = cRT (c in mol m⁻³).
Conversions: 1 atm = 1.013 × 10⁵ Pa · kPa × 10³ · 1 cm³ = 10⁻⁶ m³ · 1 dm³ = 10⁻³ m³ · K = °C + 273 · 1 mol = 24 dm³ at RTP.
Const p: V₁/T₁ = V₂/T₂ Const V: p₁/T₁ = p₂/T₂ Const T: p₁V₁ = p₂V₂ (only if n constant).
Graphs (pV vs p): ideal = horizontal; more bulge = more non-ideal. Order: He > H₂ > CH₄ > CO > NH₃. Lower T ⇒ more bulge. pV vs n at const T = straight line through origin.
Non-ideality reasons: compare IMF first (temporary < permanent dipole < H-bond; more e⁻ ⇒ stronger temporary dipoles); then particle size (diatomic etc.).
Traps: I₂ not I · convert V, p, T · mass in grams · two unknowns ⇒ simultaneous · don’t use p₁V₁ = p₂V₂ when moles change.