Dalton's Law of Partial Pressures: Formula, Mole Fraction and Gas Over Water
Every real gas sample you meet in an exam is a mixture: air, a flue gas, a gas collected over water, the reaction mixture in the Haber process. Dalton's law is the tool that lets you treat each gas on its own inside the mixture. It is a Class 11 States of Matter idea, and it comes back in physical chemistry at IIT-JAM, GATE and CSIR-NET level whenever equilibrium constants are written in partial pressures (Kp).
The law in one line
For an ideal gas mixture, the total pressure is the sum of the pressures each gas would exert if it alone filled the container at the same temperature. Each of those pressures is called the partial pressure.
pi = xi × Ptotal, where xi = ni / ntotal
pi = niRT / V
What each symbol means
| Symbol | Meaning | Unit |
|---|---|---|
| Ptotal | Total pressure of the mixture | atm, kPa or mmHg |
| pi | Partial pressure of gas i | same unit as Ptotal |
| xi | Mole fraction of gas i, ni/ntotal; the x values of all gases add up to 1 | no unit |
| R | 0.08206 L atm mol−1 K−1 or 8.314 J mol−1 K−1 | pick the one that matches P and V |
Why does it work? In an ideal gas the molecules do not attract each other, so each gas fills the whole volume and hits the walls as if the others were absent. Only the number of moles of a gas decides its share of the pressure, not its identity or mass. Real gases follow the law closely at low pressure and high temperature.
Worked example 1: partial pressures from moles
A mixture contains 3.00 mol N2, 1.00 mol O2 and 1.00 mol Ar at a total pressure of 2.50 atm. Find the partial pressure of each gas.
Step 1. ntotal = 3.00 + 1.00 + 1.00 = 5.00 mol.
Step 2. x(N2) = 3.00 / 5.00 = 0.600; x(O2) = 1.00 / 5.00 = 0.200; x(Ar) = 0.200. Check: 0.600 + 0.200 + 0.200 = 1.000.
Step 3. p(N2) = 0.600 × 2.50 = 1.50 atm; p(O2) = 0.200 × 2.50 = 0.50 atm; p(Ar) = 0.50 atm.
Check. 1.50 + 0.50 + 0.50 = 2.50 atm, which is the total pressure given.
Worked example 2: a gas collected over water
When hydrogen is collected by bubbling it through water, the gas in the jar is a mixture of hydrogen and water vapour. The barometer reading is the total pressure, so the vapour pressure of water must be subtracted before you use PV = nRT for the hydrogen.
250 mL of H2 is collected over water at 25 °C. The barometric pressure is 755 mmHg. The vapour pressure of water at 25 °C is 23.8 mmHg. Find the moles and mass of H2.
Step 1. p(H2) = 755 − 23.8 = 731.2 mmHg.
Step 2. Convert to atm: 731.2 / 760 = 0.9621 atm. Volume V = 0.250 L. T = 25 + 273.15 = 298.15 K.
Step 3. n = pV / RT = (0.9621 × 0.250) / (0.08206 × 298.15) = 0.2405 / 24.466 = 9.83 × 10−3 mol.
Step 4. Mass = 9.83 × 10−3 mol × 2.016 g mol−1 = 0.0198 g.
Check. At STP-like conditions 1 mol would occupy about 24.5 L here, so 0.250 L should hold roughly 0.250 / 24.5 = 0.0102 mol; the answer is a little lower because 0.9621 atm is below 1 atm. It is consistent.
Worked example 3: two gases in one container
A 10.0 L vessel at 300 K holds 4.00 g of He (4.00 g mol−1) and 28.0 g of N2 (28.0 g mol−1). Find the partial pressure of each gas and the total pressure.
Step 1. n(He) = 4.00 / 4.00 = 1.00 mol; n(N2) = 28.0 / 28.0 = 1.00 mol.
Step 2. p(He) = nRT / V = 1.00 × 0.08206 × 300 / 10.0 = 2.46 atm. p(N2) is the same, 2.46 atm, because the moles are equal.
Step 3. Ptotal = 2.46 + 2.46 = 4.92 atm.
Second route. ntotal = 2.00 mol, so Ptotal = 2.00 × 0.08206 × 300 / 10.0 = 4.92 atm. Both routes agree.
Notice that He is seven times lighter than N2 per molecule, yet it contributes the same pressure. Mass does not matter; the number of moles does.
Common mistakes
- Forgetting the water vapour. A gas collected over water is never at the full barometric pressure. Subtract the vapour pressure first.
- Mixing mass fractions with mole fractions. The partial pressure uses mole fraction. 50% by mass of H2 and O2 is nowhere near 50% by moles.
- Mixing units. If you use R = 0.08206 L atm mol−1 K−1, pressure must be in atm and volume in litres. Convert mmHg to atm by dividing by 760.
- Using °C. Temperature in PV = nRT must be in kelvin.
- Applying it to reacting gases. The law is for gases that do not react with each other. If they react, find the moles after the reaction first.
Where this shows up in exams
| Situation | What to use |
|---|---|
| Composition of a gas mixture given in moles or volume % | xi = ni/ntotal; for ideal gases, volume fraction equals mole fraction |
| Gas collected over water | p(gas) = Patm − p(water vapour) |
| Equilibrium constant Kp | Write each partial pressure as xiPtotal and substitute into the Kp expression |
| Diffusion and effusion of mixtures | Combine with Graham's law for each gas |
The marks, weightage and pattern differ between papers and years, so check the current official syllabus and notification of your exam rather than relying on any fixed number.
Try it yourself: check each partial-pressure answer above with the PV = nRT tool.
Open the Ideal Gas (PV = nRT) Calculator →Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG physical chemistry? ABC Chemistry runs batches through live online across India.