Imagine a room full of different types of bouncing balls — red, blue, green — all zipping around and hitting the walls. Each ball hits the wall on its own , not caring what colour the others are. The total push on the wall is just the sum of pushes from each colour separately. That is Dalton's law: in an ideal gas mixture, each gas exerts its own pressure independently, and total pressure is their sum .
Definition Partial pressureThe partial pressure p i p_i p i i i i i i i would exert alone if it occupied the same volume V V V same temperature T T T
Definition Dalton's Law (1801)For a mixture of non-reacting ideal gases in volume V V V T T T P total = p 1 + p 2 + p 3 + ⋯ = ∑ i p i P_{\text{total}} = p_1 + p_2 + p_3 + \dots = \sum_i p_i P total = p 1 + p 2 + p 3 + ⋯ = ∑ i p i
The core assumption of an ideal gas: molecules have negligible volume and negligible interaction forces. So the molecules of gas A do not feel the molecules of gas B — each gas behaves as if the others weren't there.
Start with the ideal gas law applied to each component separately in the shared volume V V V T T T
p 1 = n 1 R T V , p 2 = n 2 R T V , … p_1 = \frac{n_1 RT}{V}, \qquad p_2 = \frac{n_2 RT}{V}, \qquad \dots p 1 = V n 1 R T , p 2 = V n 2 R T , …
Now add them:
p 1 + p 2 + ⋯ = n 1 R T V + n 2 R T V + ⋯ = ( n 1 + n 2 + … ) R T V = n total R T V p_1 + p_2 + \dots = \frac{n_1 RT}{V} + \frac{n_2 RT}{V} + \dots = \frac{(n_1+n_2+\dots)RT}{V} = \frac{n_{\text{total}}\,RT}{V} p 1 + p 2 + ⋯ = V n 1 R T + V n 2 R T + ⋯ = V ( n 1 + n 2 + … ) R T = V n total R T
Define the mole fraction x i = n i n total x_i = \dfrac{n_i}{n_{\text{total}}} x i = n total n i
Divide the single-gas equation by the total equation:
p i P total = n i R T / V n total R T / V = n i n total = x i \frac{p_i}{P_{\text{total}}} = \frac{n_i RT / V}{n_{\text{total}} RT / V} = \frac{n_i}{n_{\text{total}}} = x_i P total p i = n total R T / V n i R T / V = n total n i = x i
When a gas is collected by displacement of water, it comes out saturated with water vapour . The measured pressure is the gas plus water vapour:
P total = p dry gas + p water vapour P_{\text{total}} = p_{\text{dry gas}} + p_{\text{water vapour}} P total = p dry gas + p water vapour
p dry gas = P atm − p water \boxed{p_{\text{dry gas}} = P_{\text{atm}} - p_{\text{water}}} p dry gas = P atm − p water
where p water p_{\text{water}} p water aqueous tension (saturated vapour pressure of water at that T T T
Worked example Example 1 — total pressure from components2 mol 2\text{ mol} 2 mol 2 _2 2 3 mol 3\text{ mol} 3 mol 2 _2 2 10 L 10\text{ L} 10 L 300 K 300\text{ K} 300 K
Step 1: p N 2 = n R T V = 2 × 0.0821 × 300 10 = 4.93 atm p_{\text{N}_2} = \dfrac{n RT}{V} = \dfrac{2 \times 0.0821 \times 300}{10} = 4.93\text{ atm} p N 2 = V n R T = 10 2 × 0.0821 × 300 = 4.93 atm Why? Treat N2 _2 2
Step 2: p O 2 = 3 × 0.0821 × 300 10 = 7.39 atm p_{\text{O}_2} = \dfrac{3 \times 0.0821 \times 300}{10} = 7.39\text{ atm} p O 2 = 10 3 × 0.0821 × 300 = 7.39 atm
Step 3: P total = 4.93 + 7.39 = 12.32 atm P_{\text{total}} = 4.93 + 7.39 = 12.32\text{ atm} P total = 4.93 + 7.39 = 12.32 atm Why? Dalton's law — pressures add.
Worked example Example 2 — using mole fractionA mixture is 20 % 20\% 20% 2 _2 2 80 % 80\% 80% 2 _2 2 5 atm 5\text{ atm} 5 atm p CO 2 p_{\text{CO}_2} p CO 2
Step 1: x CO 2 = 0.20 x_{\text{CO}_2} = 0.20 x CO 2 = 0.20 Why? Percentage by moles is the mole fraction.
Step 2: p CO 2 = x CO 2 P = 0.20 × 5 = 1 atm p_{\text{CO}_2} = x_{\text{CO}_2} P = 0.20 \times 5 = 1\text{ atm} p CO 2 = x CO 2 P = 0.20 × 5 = 1 atm Why? p i = x i P total p_i = x_i P_{\text{total}} p i = x i P total
Worked example Example 3 — gas over waterH 2 H_2 H 2 27 ∘ 27^\circ 2 7 ∘ = 750 mmHg = 750\text{ mmHg} = 750 mmHg = 27 mmHg = 27\text{ mmHg} = 27 mmHg H 2 H_2 H 2 V = 500 mL V = 500\text{ mL} V = 500 mL
Step 1: p H 2 = 750 − 27 = 723 mmHg = 723 760 = 0.951 atm p_{\text{H}_2} = 750 - 27 = 723\text{ mmHg} = \frac{723}{760} = 0.951\text{ atm} p H 2 = 750 − 27 = 723 mmHg = 760 723 = 0.951 atm Why? Subtract water vapour's partial pressure.
Step 2: n = p V R T = 0.951 × 0.5 0.0821 × 300 = 0.0193 mol n = \dfrac{pV}{RT} = \dfrac{0.951 \times 0.5}{0.0821 \times 300} = 0.0193\text{ mol} n = R T p V = 0.0821 × 300 0.951 × 0.5 = 0.0193 mol Why? Ideal gas law for the dry gas only.
Common mistake Steel-manned common mistakesMistake A: "Add pressures using the volume each gas occupied before mixing."
Why it feels right: volumes seem additive too. The fix: partial pressure is defined at the final shared volume V V V , not the pre-mixing volume. If gases came from different containers, first re-compute each p i = n i R T / V p_i = n_i RT / V p i = n i R T / V new V V V
Mistake B: "Use mole fraction of the wrong quantity (mass fraction)."
Why it feels right: fractions all look similar. The fix: Dalton uses mole fraction. Convert mass → moles first (n = m / M n = m/M n = m / M
Mistake C: "Forget to subtract aqueous tension."
Why it feels right: the barometer reads total pressure, seems like the gas pressure. The fix: over water, measured pressure includes water vapour; subtract aqueous tension to get dry gas.
Mistake D: "Apply Dalton's law to reacting gases."
Why it feels right: they're still gases. The fix: if gases react, n i n_i n i non-reacting mixtures at equilibrium composition.
Recall Feynman: explain to a 12-year-oldPut red bouncy balls and blue bouncy balls in the same box. Each colour bangs on the walls making its own "push." The red balls don't know the blue balls exist and vice-versa. So the total push on the wall is just red's push plus blue's push. If red balls make up one-fifth of all the balls, they cause one-fifth of the total push. That's it!
"P total = sum of the parts, each part is x times the whole."
p i = x i P p_i = x_i P p i = x i P "partial = fraction × total." Think PART ial pressure comes from your PART (mole fraction) of the crowd.
#flashcards/chemistry
What is the partial pressure of a gas in a mixture? The pressure that gas would exert alone if it occupied the same volume at the same temperature.
State Dalton's law of partial pressures. P total = ∑ i p i P_{\text{total}} = \sum_i p_i P total = ∑ i p i for non-reacting ideal gases at fixed
V , T V, T V , T .
Formula linking partial pressure and mole fraction. p i = x i P total p_i = x_i P_{\text{total}} p i = x i P total , where
x i = n i / n total x_i = n_i/n_{\text{total}} x i = n i / n total .
Why do partial pressures add? Ideal-gas molecules don't interact, so collisions (hence pressures) from each gas are independent and additive.
For gas collected over water, how do you find dry gas pressure? p dry = P atm − p water p_{\text{dry}} = P_{\text{atm}} - p_{\text{water}} p dry = P atm − p water (subtract aqueous tension).
What is aqueous tension? The saturated vapour pressure of water at the given temperature.
Does Dalton's law hold for reacting gases? No — only for non-reacting mixtures (moles must stay fixed).
Sum of all mole fractions equals? 1, which guarantees
∑ p i = P total \sum p_i = P_{\text{total}} ∑ p i = P total .
Ideal Gas Equation — Dalton's law is derived directly from P V = n R T PV = nRT P V = n R T Mole Fraction and Concentration Terms — supplies x i x_i x i Kinetic Theory of Gases — explains WHY pressures are additive (independent collisions).Real Gases and van der Waals Equation — where Dalton's law breaks due to intermolecular forces.Vapour Pressure — background for aqueous tension.
gas alone in same V and T
Gases behave independently
Dalton's Law Ptotal = sum pi
Ideal gas law per component
Mole fraction xi = ni/ntotal
Intuition Hinglish mein samjhoDekho, Dalton's law ka core idea bahut simple hai. Maan lo ek box mein do alag-alag gases hain — gas A aur gas B. Ideal gas mein molecules ek doosre ko feel nahi karte (na force, na volume). Iska matlab gas A apni wall par apni pressure banata hai jaise gas B wahan hai hi nahi. Isliye total pressure = gas A ki pressure + gas B ki pressure. Bas! P t o t a l = p 1 + p 2 + … P_{total} = p_1 + p_2 + \dots P t o t a l = p 1 + p 2 + …
Derivation bhi seedha hai: har gas ke liye alag se p i = n i R T / V p_i = n_i RT / V p i = n i R T / V V V V T T T n t o t a l n_{total} n t o t a l P t o t a l = n t o t a l R T / V P_{total} = n_{total}RT/V P t o t a l = n t o t a l R T / V p i = x i P t o t a l p_i = x_i P_{total} p i = x i P t o t a l x i x_i x i p C O 2 = 0.2 × 5 = 1 p_{CO_2} = 0.2 \times 5 = 1 p C O 2 = 0.2 × 5 = 1
Exam ka favourite trick: gas ko water ke upar collect karna. Tab measured pressure mein water vapour ki bhi pressure mili hoti hai. Isliye dry gas ki pressure nikalne ke liye aqueous tension (water ka vapour pressure us temperature par) subtract karo: p d r y = P a t m − p w a t e r p_{dry} = P_{atm} - p_{water} p d r y = P a t m − p w a t er
Do cheezein hamesha yaad rakho: (1) partial pressure hamesha final shared volume par calculate karo, pehle wale container ke volume par nahi. (2) Mole fraction use karo, mass fraction nahi — pehle mass ko moles mein convert karo. Yeh law kinetic theory se aata hai aur real gases mein thoda fail hota hai jahan intermolecular forces strong hoti hain.
