Chemistry 2nd Paper · Classes 11-12 · Chapter 1 · 1.3.7

Dalton's Law of Partial Pressures.

Air is not one gas but a mixture of several. Yet a barometer shows a single pressure. How much of it comes from the N₂, and how much from the O₂? Dalton's answer is simple: each gas pushes its own share, and the shares add up to the total.

1. The gas law works for mixtures too

The ideal gas law is not only for pure gases; it holds for gas mixtures as well. Gases mix in any proportion to give a homogeneous mixture, and in a mixture where nothing reacts, each gas behaves as though it had the container to itself.

John Dalton noticed this with air. Add water vapour to dry air and the total pressure goes up by exactly the pressure the vapour would exert on its own: P_total = P_dry air + P_vapour.

2. Dalton's law of partial pressures

The law: at a given temperature, the partial pressure of a gas in a non-reacting mixture is the pressure it would exert if it alone filled the whole volume of the mixture. The total pressure of the mixture is the sum of the partial pressures.

For a mixture of three gases, P_m = P₁ + P₂ + P₃, and in general P_m = P₁ + P₂ + P₃ + ..., one term for each gas.

The law holds under two conditions: the temperature is fixed, and the gases do not react with each other. These are the only two the book names; it does not make high or low pressure a condition. When an exam asks when the law works, the answer is at constant temperature.

3. The proof, and mole fraction

The proof comes straight from the ideal gas equation. In one container (volume V) at one temperature, each gas obeys PV = nRT separately: P₁V = n₁RT, P₂V = n₂RT, and so on. Adding them all gives P₁ + P₂ + ... = (n₁ + n₂ + ...)RT/V = nRT/V, and nRT/V is the pressure of the whole mixture, P_m.

One important idea follows. At fixed temperature and volume the pressure of a gas is proportional to its number of moles, so in a mixture each gas's partial pressure is proportional to its own moles: for N₂ and H₂, P_N₂ = n₁RT/V, P_H₂ = n₂RT/V, and the total moles n = P_mV/RT. Add molecules to the box below and watch.

A closed box, volume and temperature fixed. Add molecules of N₂ or H₂, then take one gas away and see whether the other's pressure changes.

Mole fraction (X) = moles of the component ÷ total moles of the mixture. The mole fractions of all the components always add up to 1, for example X_N₂ + X_H₂ = 1.

Divide one equation by the other and RT/V cancels: X_N₂ = n₁/n = P_N₂/P_m. So P_N₂ = X_N₂ × P_m: partial pressure = mole fraction × total pressure. Nearly every mixture problem runs on this one line.

With the moles given it is one step. Mix 6 mol of O₂ and 14 mol of N₂ to a total of 200 mm Hg and P_O₂ = (6/20) × 200 = 60 mm Hg. Likewise 5 mol of A and 10 mol of B at a total of 15 atm give P_A = (5/15) × 15 = 5 atm. Given masses, turn them into moles first. With equal masses of CH₄ and O₂, n_CH₄ = w/16 and n_O₂ = w/32, so O₂ holds (1/32)/(3/32) = 1/3 of the total pressure. 28 g of N₂ and 142 g of Cl₂ are 1 and 2 mol, so at a total of 12 atm, P_N₂ = 12/3 = 4 atm.

In a mixture of 1.032 g of O₂ and 0.573 g of CO₂, n_O₂ = 1.032/32 = 0.03225 mol and n_CO₂ = 0.573/44 = 0.01302 mol. So X_CO₂ = 0.01302/0.04527 = 0.287.

If you are given the vessel's volume and temperature instead of the total pressure, find the total pressure first. Problem 1.43: 0.4 g of He, 1.4 g of N₂ and 1.6 g of O₂ in a 10 L cylinder at 27°C. n = 0.1 + 0.05 + 0.05 = 0.2 mol, so P = 0.2 × 0.082 × 300/10 ≈ 0.49 atm (the book uses R = 0.0821 and gets 0.4926). X_He = 0.1/0.2 = 0.5, so P_He = 0.246 atm. The calculator below holds all four examples. Numbers you type go nowhere. The notebook is yours.

A diver's cylinder is the same sum. Problem 1.37: at 250 ft under water the pressure is 8.38 atm. To keep the partial pressure of the O₂ being breathed at 0.21 atm, X_O₂ = 0.21/8.38 = 0.025, so the cylinder holds only 2.5% O₂.

Problem 1.46: a 10 L cylinder at 27°C holds 10.08 g of N₂ and 11.2 L of O₂ measured at STP. n_N₂ = 10.08/28 = 0.36 mol, so P_N₂ = 0.36 × 0.082 × 300/10 = 0.886 atm. n_O₂ = 11.2/22.4 = 0.5 mol, so P_O₂ = 1.23 atm. Adding, P_m = 2.116 atm.

4. The partial pressures of air

In air the mole fractions of N₂, O₂, Ar and CO₂ are 0.7808, 0.2095, 0.0093 and 0.00033. Take the total as 1 atm (760 mm Hg) and each partial pressure in atm is the same number as the mole fraction: N₂ 0.7808 atm = 593.4 mm Hg, O₂ 0.2095 atm = 159.2 mm Hg, Ar 0.0093 atm = 7.1 mm Hg, CO₂ 0.00033 atm = 0.3 mm Hg. Together they make about 760.0 mm Hg. Put each value in its gas's box.

5. Two vessels, one mixture

Now the gases start in separate vessels: N₂ at pressure P₁ in volume V₁, O₂ at pressure P₂ in volume V₂. Open the stopcock between them and each gas spreads through the whole V₁ + V₂. At the same temperature, Boyle's law gives P_N₂ = P₁V₁/(V₁ + V₂) and P_O₂ = P₂V₂/(V₁ + V₂). Adding them, P_m = (P₁V₁ + P₂V₂)/(V₁ + V₂). For three or more gases, P_m = (P₁V₁ + P₂V₂ + P₃V₃ + ...)/(V₁ + V₂ + V₃ + ...).

Problem 1.44: vessel A holds 800 mL at 300 mm Hg and B holds 500 mL at 250 mm Hg, both at 27°C. A holds n = 300 × 0.8/(760 × 0.082 × 300) = 0.01284 mol, which is 7.732×10²¹ molecules. With the stopcock open, P = (300 × 800 + 250 × 500)/1300 = 280.77 mm Hg, and warmed to 40°C, P₁ = 280.77 × 313/300 = 292.94 mm Hg. Problem 1.45: at 25°C, N₂ is 400 mL at 1.0 atm and O₂ is 450 mL at 1.5 atm. The N₂ is n = 1.0 × 0.4/(0.082 × 298) = 0.01637 mol, or 9.858×10²¹ molecules. The total is P = (1.0 × 400 + 1.5 × 450)/850 = 1.265 atm. Anyone who measures 0.9 atm has a result that does not fit Dalton's law, so the observation is wrong. Drag the handles and set up the two vessels yourself.

If the new vessel is not V₁ + V₂, the rule is the same with the new volume underneath. Problem 1.38: 600 mL of N₂ at 1.0 atm and 800 mL of O₂ at 750 mm Hg go into a 1.0 L cylinder. P_N₂ = 1.0 × 0.6/1.0 = 0.60 atm; P_O₂ = (750/760) × 0.8/1.0 = 0.79 atm; the total is 1.39 atm. The mm Hg had to become atm first.

Problem 1.42: a 1.5 L vessel gets 0.2 L of O₂ at 0.8 atm and 0.5 L of N₂ at 0.98 atm. P_O₂ = 0.8 × 0.2/1.5 = 0.107 atm, P_N₂ = 0.98 × 0.5/1.5 = 0.327 atm, a total of 0.434 atm. One more: two gases at 0.8 atm in 1 L and 0.7 atm in 1 L, put into a 1.5 L vessel at the same temperature, give P_m = (0.8 × 1 + 0.7 × 1)/1.5 = 1.0 atm.

6. Gas collected over water

A gas that does not dissolve in water (H₂, O₂) is collected in a jar by pushing the water out. The gas in the jar is then not alone: it is saturated with water vapour. With the water level the same inside and outside, the pressure of the moist gas equals the atmospheric pressure. By Dalton's law P_dry + f = P_atm, so P_dry = P_atm − f, where f is the vapour pressure of water at that temperature. The first job is therefore always to take the vapour pressure away.

Problem 1.40: 600 mL of O₂ is collected over water at 17°C and 770 mm Hg; the vapour pressure of water at 17°C is 14.5 mm Hg. The dry O₂ is at P₁ = 770 − 14.5 = 755.5 mm Hg. At STP, V₂ = 755.5 × 600 × 273/(290 × 760) = 561.48 mL, and the mass = 32 × 561.48/22400 = 0.802 g. In the jar below, take the vapour out and see.

The jar holds the gas and water vapour together. Take the vapour out to get the dry-gas pressure, then take it to STP.

Two more on the same steps. Problem 1.39: 0.058 m³ of H₂ over water at 17°C and 99.99 kPa, vapour pressure 3.27 kPa. P_dry = 99.99 − 3.27 = 96.72 kPa, and at STP V₂ = 96.72 × 0.058 × 273/(290 × 101.325) = 0.0521 m³. Problem 1.41: 580 mL of H₂ at 17°C and 0.95 atm, vapour pressure 3.26 kPa. To put both pressures in one unit, 0.95 atm = 96.26 kPa, so P₁ = 96.26 − 3.26 = 92.96 kPa, and the dry H₂ at STP has V₂ = 500.924 mL.

Now put the five steps in order.

7. Check yourself

Three questions in the exam's style. For the second, the vessel holds 16 g of O₂ and 32 g of CH₄ at 25°C, with a total pressure of 200 mm.

The whole note on eight cards.

Source: Chemistry 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Environmental Chemistry”, section 1.3.7.