Topic 3 · The Gaseous State · spread mockup

The Gaseous State

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

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

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
2

The Gas Laws

Boyle's law (fixed n, T)

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

pV = k   p₁V₁ = p₂V₂
p against V
p against 1/V
pV against p

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

p(600) = (p/3)V₂ → V₂ = 1800 cm³

Charles's law (fixed n, p)

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
V against T / K
V against T / °C

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).

V₂ = 32.0 × 293273 = 34.3 cm³  (use kelvin, never °C)

Avogadro's law (fixed p, T)

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

V / n = k   V₁/n₁ = V₂/n₂
V against n
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2

The Gas Laws (continued)

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
3

The Ideal Gas Equation

pV = nRTR = 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.

n, T constantp ∝ 1/VBoyle
n, p constantV ∝ TCharles
p, T constantV ∝ nAvogadro
n, V constantp ∝ T
NoteTwo special forms of pV = nRT, derived by substitution
Mr = mRTpVsubstitute n = mMr into pV = nRT ρ = pMrRTthen use density, ρ = mV

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.

Mr = mRTpV = 0.10 × 8.31 × 3001.0 × 10⁵ × 83.1 × 10⁻⁶ = 30.0

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 = 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 = 2840 g m⁻³ = 2.84 g dm⁻³

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

p₁V₁T₁ = p₂V₂T₂ → T₂ = 4 × 298 = 1192 K
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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 + …Total pressure = sum of the partial pressures. Partial pressure pA: the pressure gas A would exert if it occupied the container alone.
xA = nA / ntotalMole fraction xA: the fraction of all the gas particles that are A.
pA = xA × ptotalA's share of the pressure = A's share of the particles.

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.

  1. ntotal = 0.5 + 2.0 + 2.5 = 5.0 mol
  2. xHe = 0.5 ÷ 5.0 = 0.1
    xNe = 2.0 ÷ 5.0 = 0.4
    xAr = 2.5 ÷ 5.0 = 0.5
  3. pHe = 0.1 × 100 = 10 kPa
    pNe = 0.4 × 100 = 40 kPa
    pAr = 0.5 × 100 = 50 kPa

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.

  1. Each gas spreads through all 3 dm³ as if alone, so p1V1 = p2V2 for each
  2. He: 2 × 1 = pHe × 3, so pHe = 0.667 kPa
    Ne: 1 × 2 = pNe × 3, so pNe = 0.667 kPa
  3. ptotal = pHe + pNe = 0.667 + 0.667 = 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.

5.2Maxwell–Boltzmann distribution

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
kinetic energyno. of particlesT₁T₂ > T₁
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6

Real Gases

6.1Deviation from ideality

For an ideal gas pV / nRT = 1 at all p. Real gases deviate most at high p and low T, because of (1) IMF between particles and (2) the volume of the particles.

ppV/nRT1ideal gasIMF ↓ pVvolume ↑ pV
↓ Dip: 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 ↓
↑ Rise 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 ↑

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

  1. CH₄ has only weak id–id, HCl has pd–pd, HF has strong hydrogen bonding
  2. Stronger IMF → greater deviation → ideality: CH₄ > HCl > HF

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

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
ppV/nRTH₂CO₂H₂Oideal gas

6.4Effect of temperature

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.

ppV/nRT500 K700 K900 K1100 Kideal gas
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