Chemistry · Ch 10 — States of Matter

Dalton's law of Partial Pressure

10.5.6

Dalton's law of Partial Pressure

John Dalton, alongside his atomic theory, formulated the law of partial pressures in 1801 for mixtures of gases that do not react chemically with one another. The pressure exerted by any one gas within such a mixture -- the pressure it would exert if it alone occupied the same container at the same temperature -- is called its partial pressure. Dalton's law states that the total pressure of a mixture of two or more non-reacting gases equals the sum of the partial pressures of each individual gas in the mixture: $P_{Total} = P_1 + P_2 + P_3 + \dots$ (at constant $V$ and $T$ ). Since each gas obeys the ideal gas equation independently ( $P_i = n_iRT/V$ ), the total pressure works out to $P_{Total} = (n_1+n_2+n_3+\dots)\frac{RT}{V} = n_{Total}\frac{RT}{V}$ . The mole fraction of gas 1 in the mixture, $X_1 = n_1/n_{Total}$ , then lets the partial pressure of any component be found directly from the total pressure: $P_1 = X_1 \times P_{Total}$ (and likewise for $P_2$ , $P_3$ , etc.) -- partial pressure equals mole fraction times total pressure. A related idea is Aqueous Tension: when a gas is collected by displacement over water, it becomes mixed with saturated water vapour, so the measured pressure is really the sum of the dry gas's own pressure and the vapour pressure of water at that temperature (its aqueous tension, $P_{ …

Figure Fig 10.13Fig. 10.13: Schematic illustration of Dalton's law of partial pressures

What this figure shows. Two separate containers are shown, one holding only gas 1 (exerting pressure P1 on its own) and one holding only gas 2 (exerting pressure P2 on its own); a plus sign and an arrow lead to a third container where both gases have been combined together, labelled with the total pressure PT = P1 + P2 -- visually showing that when non-reacting gases are mixed in the same container at the same temperature, the resulting total pressure is simply the arithmetic sum of what each gas would have exerted on i …

Misc Problem 10.5Problem: Partial pressures of a three-gas mixture (Dalton's law)

Worked out. A mixture of 28 g N2, 8 g He and 40 g Ne together exert 20 bar total pressure; find each gas's partial pressure. Step 1, moles: n(N2) = 28/28 = 1 mol; n(He) = 8/4 = 2 mol; n(Ne) = 40/20 = 2 mol (using molar masses N2=28, He=4, Ne=20 g/mol). Step 2, mole fractions (total moles = 1+2+2 = 5): x(N2) = 1/5 = 0.2; x(He) = 2/5 = 0.4; x(Ne) = 2/5 = 0.4. Step 3, partial pressures = mole fraction x total pressure: P(N2) = 0.2 x 20 bar = 4 bar; P(He) = 0.4 x 20 bar = 8 bar; P(Ne) = 0.4 x 20 bar = 8 bar (check: 4+8+8 = 20 bar, matching the …

Table Table 10.5Table 10.5: Aqueous tension of water (vapour pressure) vs temperature

Temp (K) | Pressure (bar) | Temp (K) | Pressure (bar)

273.15 | 0.0060 | 295.15 | 0.0260

283.15 | 0.0121 | 297.15 | 0.0295

288.15 | 0.0168 | 299.15 | 0.0331

291.15 | 0.0204 | 301.15 | 0.0372

293.15 | 0.0230 | 303.15 | 0.0418

The table shows the saturated vapour pressure of water (its aqueous tension) climbing steadily and non-linearly as temperature rises, roughly a seven-fold increase in pressure across this 30 K span, which is the underlying r …

Start your 14-day free trial to unlock the full solution →

← Previous · 10.5.5Relation between density, molar mass and pressure Next · 10.6 →Kinetic Molecular Theory of Gases