The Gaseous
State

H2 Chemistry (9476), JC1

By the end of this topic, you can answer

1

Why must temperature be in kelvin, never in °C, when you use the gas laws?

2

Why does a sealed packet of crisps puff up on a plane, and why do tyres look flatter on a cold morning?

3

How can you find the molar mass of a gas just by weighing a flask of it?

4

Divers breathe compressed air. How much of its pressure comes from the oxygen alone?

5

The particles in a gas do not all move at the same speed. What does the spread of speeds look like, and how does it change on heating?

6

Why does ammonia turn into a liquid easily under pressure, while helium hardly does?

Recall: what you know from O levels

1Describe the arrangement and motion of particles in a gas.

Far apart, randomly arranged, moving quickly and randomly in all directions.[1]

2Why can a gas be compressed but a liquid hardly at all?

Gas particles have large empty spaces between them, so they can be pushed closer together.[1]

3What volume does 1 mol of gas occupy at room temperature and pressure?

24.0 dm³ (see The Mole Concept).[1]
1

Gases and their Units

1.1Properties of gases

  • Much lower density than solids and liquids
  • Take the volume and shape of the container
  • Highly compressible, exert pressure equally in all directions

All because gas particles are far apart with much weaker intermolecular forces than in liquids.

Pressure: the force exerted by gas particles colliding with a surface, per unit area. p ↑ with n ↑, T ↑ and V ↓.
1

Gases and their Units

1.2S.I. units and conversions

QuantityS.I. unitConversions
pressure, pPa (N m⁻²)1 atm = 101 325 Pa = 760 mm Hg, 1 bar = 10⁵ Pa
volume, Vm³1 dm³ = 1 L = 10⁻³ m³, 1 cm³ = 1 mL = 10⁻⁶ m³
temperature, TKT / K = T / °C + 273
amount, nmoln = mass / Mr

Your turnConvert into S.I. units: (a) 25 °C, (b) 250 cm³, (c) 2.00 atm.

(a)
25 + 273 = 298 K[1]
(b)
250 × 10⁻⁶ m³ = 2.50 × 10⁻⁴ m³[1]
(c)
2.00 × 101 325 Pa = 2.03 × 10⁵ Pa[1]
2

The Gas Laws

Boyle's law (fixed n and T)

PredictThe gas is squeezed into half the volume at constant temperature. What happens to its pressure?

At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.

pV = k    p₁V₁ = p₂V₂

2

The Gas Laws

Charles's law (fixed n and p)

PredictThe gas is heated from 300 K to 600 K at constant pressure (the piston is free to move). What happens to the volume?

The volume of a fixed mass of gas at constant pressure is directly proportional to its temperature in kelvin.

V / T = k    V₁/T₁ = V₂/T₂

NotePlotted against T in °C, the line meets V = 0 at −273 °C (0 K). Always use kelvin in calculations.
2

The Gas Laws

Avogadro's law (fixed p and T)

Under the same temperature and pressure, equal volumes of all gases contain equal numbers of molecules.

V / n = k    V₁/n₁ = V₂/n₂

The size of the molecules does not matter: the volume of a gas is almost all empty space.

2

The Gas Laws

2.1Explaining the gas laws with kinetic theory

  • Boyle: V ↓ → particles travel a shorter distance between collisions with the walls → less time between collisions → more frequent collisions with the walls → p ↑
  • Charles: T ↑ → particles have higher KE and move faster → hit the walls more often and with greater force → walls move out (V ↑) until p is restored
  • Avogadro / Dalton: p depends on the no. of collisions per second, which is proportional to the no. of gas molecules

AnalogyHail on a tin roof. The noise gets louder if the hailstones land more often (a smaller roof, Boyle), land harder (faster stones, Charles) or if there are more of them (Avogadro).

2

The Gas Laws

Worked example: BoyleA balloon holds 600 cm³ of gas. Its pressure falls to one-third at constant temperature. Find its new volume.

p₁V₁ = p₂V₂  →  p × 600 = p3 × V₂

V₂ = 3 × 600 = 1800 cm³

Worked example: Charles32.0 cm³ of CO₂ is collected at 0 °C and 1 atm. Find its volume at r.t.p. (20 °C, 1 atm).

T₁ = 273 K, T₂ = 293 K  (use kelvin, never °C)

V₂ = 32.0 × 293273 = 34.3 cm³

2

The Gas Laws

Your turn2.00 dm³ of a gas at 27 °C is heated to 127 °C at constant pressure. Find its new volume.

T₁ = 27 + 273 = 300 K, T₂ = 127 + 273 = 400 K[1]
Apply V₁T₁ = V₂T₂
V₂ = 2.00 × 400300 = 2.67 dm³[1]

ChallengeWhy is the answer not 2.00 × 127/27 = 9.41 dm³?

V is proportional to T only when T is in kelvin. 127 °C is not 127 ÷ 27 times as hot as 27 °C.[1]
2

The Gas Laws

Quick checkFor a fixed mass of gas at constant temperature, which graph is a straight line through the origin?

Ap against V
Bp against 1/V
CpV against p
DV against T / °C
NoteB: p = k × (1/V). p against V is a curve, pV against p is a horizontal line, and V against T / °C does not pass through the origin (it meets V = 0 at −273 °C).
3

The Ideal Gas Equation

pV = nRT
R = 8.31 J K⁻¹ mol⁻¹, all quantities in S.I. units

An ideal gas obeys pV = nRT exactly at all pressures and temperatures.

DescribeHold two of n, p, V and T constant. What does pV = nRT become?

n, T constantp ∝ 1/VBoyle
n, p constantV ∝ TCharles
p, T constantV ∝ nAvogadro
n, V constantp ∝ T
3

The Ideal Gas Equation

Finding Mr

Substitute n = mMr into pV = nRT:

pV = mMrRT  →  Mr = mRTpV

Finding density

Then use density, ρ = mV:

Mr = ρRTp  →  ρ = pMrRT

NoteBoth are pV = nRT in another form, derived by substitution: learn the method, not two more equations.
3

The Ideal Gas Equation

Worked example: Mr of a gas0.10 g of a gas occupies 83.1 cm³ at 1.0 × 10⁵ Pa and 27 °C. Find its Mr.

T = 27 + 273 = 300 K,  V = 83.1 × 10⁻⁶ m³

Mr = mRTpV = 0.10 × 8.31 × 3001.0 × 10⁵ × 83.1 × 10⁻⁶

Mr = 30.0

Your turn0.320 g of a gas occupies 246 cm³ at 1.00 × 10⁵ Pa and 23 °C. Find its Mr and suggest what the gas is.

T = 23 + 273 = 296 K, V = 246 × 10⁻⁶ m³
Mr = 0.320 × 8.31 × 2961.00 × 10⁵ × 246 × 10⁻⁶ = 32.0[1]
The gas is O₂.[1]
3

The Ideal Gas Equation

Worked example: Molar volume at s.t.p.Find the volume of 1 mol of an ideal gas at s.t.p. (0 °C, 1 bar).

V = nRTp = 1 × 8.31 × 27310⁵

V = 2.27 × 10⁻² m³ = 22.7 dm³

Worked example: DensityFind the density of H₂S (Mr 34.1) at 20 °C and 2.00 atm.

ρ = pMrRT = 2.00 × 101 325 × 34.18.31 × 293

ρ = 2840 g m⁻³ = 2.84 g dm⁻³

Exam tipWith Mr in g mol⁻¹, the density comes out in g m⁻³. Divide by 1000 for g dm⁻³.
3

The Ideal Gas Equation

Worked example: Change of conditionsA gas at 25 °C is heated until both its pressure and volume double (n constant). Find the final temperature.

n constant, so p₁V₁T₁ = p₂V₂T₂

T₂ = T₁ × p₂p₁ × V₂V₁ = 298 × 2 × 2 = 1192 K

Your turn2.00 dm³ of gas at 27 °C and 100 kPa is compressed to 0.500 dm³ and heated to 127 °C. Find its final pressure.

Apply p₁V₁T₁ = p₂V₂T₂, with T₁ = 300 K and T₂ = 400 K[1]
p₂ = 100 × 2.000.500 × 400300 = 533 kPa[1]
3

The Ideal Gas Equation

Quick checkWhich set of units must be used in pV = nRT with R = 8.31 J K⁻¹ mol⁻¹?

AkPa, dm³, °C
BPa, m³, K
Catm, dm³, K
DPa, cm³, K
NoteB: R = 8.31 J K⁻¹ mol⁻¹ only works with S.I. units: Pa, m³, K and mol.
4

Dalton's Law of Partial Pressures

In a mixture of gases that do not react, each gas behaves as if it were the only gas present. The total pressure is the sum of the partial pressures.

ptotal = pA + pB + …

Partial pressure: the pressure the gas would exert alone.

xA = nA / ntotal   mole fraction of A

pA = xA × ptotal

4

Dalton's Law of Partial Pressures

Worked example: Partial pressuresA cylinder holds 0.5 mol He, 2.0 mol Ne and 2.5 mol Ar at a total pressure of 100 kPa. Find each partial pressure.

ntotal = 0.5 + 2.0 + 2.5 = 5.0 mol

xHe = 0.1,  xNe = 0.4,  xAr = 0.5

pHe = 10 kPa,  pNe = 40 kPa,  pAr = 50 kPa

Your turnDry air is 78% N₂, 21% O₂ and 1% Ar by moles. Find the partial pressure of O₂ when the total pressure is 101 kPa.

pO₂ = mole fraction of O₂ × total pressure = 0.21 × 101 = 21.2 kPa[1]
4

Dalton's Law of Partial Pressures

Worked example: Connecting flasksA 1 dm³ flask of He at 2 kPa is connected to a 2 dm³ flask of Ne at 1 kPa at constant temperature. Find the final pressure.

Open the tap: each gas spreads through both flasks, as if the other gas were not there.

So each gas expands from its own flask to 3 dm³, and p1V1 = p2V2 for each:

He: 2 × 1 = pHe × 3, so pHe = 0.667 kPa

Ne: 1 × 2 = pNe × 3, so pNe = 0.667 kPa

ptotal = pHe + pNe = 1.33 kPa

5

Kinetic Theory of Gases

5.1Assumptions for an ideal gas

  1. Gas particles have negligible volume compared to the volume of the container (point masses)
  2. Gas particles exert negligible intermolecular forces on one another
  3. Collisions between particles are perfectly elastic: no loss of kinetic energy
  4. Particles are in constant, random motion

Average kinetic energy of the particles ∝ T (in kelvin). The first two are the 2 main features of an ideal gas.

5

Kinetic Theory of Gases

5.2Maxwell–Boltzmann distribution

DescribeThe curve shows how many particles have each kinetic energy. Where are most particles?

When temperature increases

  1. The peak shifts to the right: the most probable speed or energy increases
  2. The curve flattens: a larger spread of speeds, more particles at high speeds
  3. The area under the curve stays constant: the total no. of particles is unchanged
5

Kinetic Theory of Gases

Quick checkThe temperature of a gas is raised. Which statement about its Maxwell–Boltzmann curve is not correct?

AThe peak moves to higher energy.
BThe peak becomes lower.
CThe area under the curve increases.
DMore particles have high energies.
NoteC: the area is the total number of particles, which does not change when the gas is heated.
6

Real Gases

6.1Deviation from ideality

DescribeHow does the real gas differ from the ideal line?

For an ideal gas pV / nRT = 1 at all p. Real gases deviate most at high p and low T.

Deviation is due to (1) IMF between particles and (2) the volume of the particles. Volume of gas = volume of particles + empty space, so a real gas occupies more volume than an ideal gas.

6

Real Gases

Why pV dips: IMF significant

  1. A particle about to hit the wall is attracted by the particles around it
  2. → hits the wall with less force → p lower than ideal → pV ↓

Why pV rises at high p: particle volume significant

  1. Volume of the gas particles is no longer negligible compared to the volume of the container
  2. → volume of real gas > ideal gas → pV ↑
6

Real Gases

Worked exampleRank CH₄, HCl and HF in order of decreasing ideality at the same conditions.

  1. CH₄ is non-polar: only weak id–id
  2. HCl is polar: pd–pd as well as id–id
  3. HF: strong hydrogen bonding
  4. Stronger IMF → greater deviation ∴ ideality CH₄ > HCl > HF

Your turnWhich deviates more from ideal behaviour, NH₃ or CH₄? Explain.

NH₃ deviates more. NH₃ molecules form hydrogen bonds with each other, which are stronger than the instantaneous dipole–induced dipole interactions between CH₄ molecules.[1]
The intermolecular forces of attraction between NH₃ molecules are less negligible, so NH₃ deviates more from ideal behaviour.[1]
6

Real Gases

6.2Conditions for ideal behaviour

  1. Low pressure: the gas occupies a large volume, so the volume of the particles is negligible compared to the container. Particles are far apart, so IMF are negligible
  2. High temperature: particles have high kinetic energy, enough to overcome the IMF
Answer structureState both conditions and explain each one: low p (particle volume and IMF negligible) and high T (KE enough to overcome IMF).
6

Real Gases

6.3Effect of IMF

  • Stronger IMF → greater deviation. H₂O (H bonds) > CO₂ > H₂ (weak id–id). H₂'s id–id are so weak that its pV never dips below the ideal value
  • Polar molecules: pd–pd or H bonds are stronger than id–id of non-polar molecules of similar Mr
  • Large electron cloud: stronger, more extensive IMF, and larger particle volume
6

Real Gases

6.4Effect of temperature

PredictThe same gas, CO₂, is heated from 500 K to 1100 K. Will it deviate more or less?

T ↑ → particles have higher KE and move faster → IMF become insignificant → pV approaches the ideal value → less deviation.

At very low T the particles move slowly, so IMF dominate and the gas condenses to a liquid, far from ideal.

6

Real Gases

Quick checkUnder which conditions does a gas behave most like an ideal gas?

Ahigh pressure, low temperature
Blow pressure, high temperature
Chigh pressure, high temperature
Dlow pressure, low temperature
NoteB: at low p the particles' volume and IMF are negligible, and at high T the particles have enough KE to overcome the IMF.

Exit ticket

1State Boyle's law.

At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.[1]

2Find the volume of 0.500 mol of gas at 300 K and 1.00 × 10⁵ Pa.

V = 0.500 × 8.31 × 300 / 10⁵ = 1.25 × 10⁻² m³ = 12.5 dm³[1]

3State the 2 main features of an ideal gas.

Particles have negligible volume compared to the container,[1]
and exert negligible intermolecular forces on one another.[1]

4Why does pV / nRT rise above 1 at very high pressure?

The volume of the particles is no longer negligible compared to the container, so the real volume is larger than ideal.[1]