Class 11 States of Matter — Gas Law Numericals Solved
States of Matter is where Class 11 students first meet a set of formulas that all look alike. Boyle, Charles, Gay-Lussac and the combined gas law are really one relationship seen from four angles, and once you see that, the numericals stop being memory work. This guide gives every formula, says which quantity is held constant, and works the standard question types with the arithmetic written out in full.
The four simple gas laws
Each law describes what happens when you change two quantities and hold the rest fixed. The amount of gas (n) is constant throughout unless stated otherwise.
Charles's law (P constant): V1/T1 = V2/T2 — volume is directly proportional to absolute temperature
Gay-Lussac's law (V constant): P1/T1 = P2/T2
Avogadro's law (P, T constant): V1/n1 = V2/n2
Every T in every one of those equations is in kelvin: T(K) = t(°C) + 273.15 (273 is accepted in most school working). This is not optional. Charles's law fails completely in Celsius, and it is the reason the chapter defines an absolute temperature scale at all.
Worked example 1 — Boyle's law. A gas occupies 2.0 L at 1.0 bar. It is compressed at constant temperature to 0.50 L. Find the new pressure.
P1V1 = P2V2
(1.0 bar)(2.0 L) = P2(0.50 L)
P2 = 2.0 ÷ 0.50 = 4.0 bar
Check the direction: volume was cut to one quarter, so pressure should rise four times. It did.
Worked example 2 — Charles's law. 300 mL of a gas at 27 °C is heated to 127 °C at constant pressure. Find the new volume.
T1 = 27 + 273 = 300 K, T2 = 127 + 273 = 400 K
V2 = V1 × (T2/T1) = 300 × (400 ÷ 300) = 400 mL
Had you used 27 and 127 directly, you would have got 300 × 127/27 ≈ 1411 mL — completely wrong, and a very common exam slip.
The combined gas law
When pressure, volume and temperature all change for a fixed amount of gas, the three laws merge into one:
This one equation contains all three simple laws. Cancel T and you have Boyle; cancel P and you have Charles; cancel V and you have Gay-Lussac. If you remember only one formula from this section, remember this one.
Worked example 3 — combined gas law. A sample of gas occupies 500 mL at 27 °C and 1.0 bar. What volume will it occupy at 0 °C and 2.0 bar?
T1 = 300 K, T2 = 273 K
V2 = (P1V1T2) ÷ (T1P2)
= (1.0 × 500 × 273) ÷ (300 × 2.0)
Numerator = 500 × 273 = 136 500
Denominator = 300 × 2.0 = 600
V2 = 136 500 ÷ 600 = 227.5 mL
Both changes squeeze the gas, so a volume well below 500 mL is expected.
The ideal gas equation
Density form: d = P M / R T
The value of R you use must match your pressure and volume units:
| Value of R | Use when |
|---|---|
| 0.0821 L atm K−1 mol−1 | P in atm, V in litres |
| 0.0831 L bar K−1 mol−1 | P in bar, V in litres |
| 8.314 J K−1 mol−1 | SI units — P in pascal, V in m³ |
A note on standard conditions, because textbooks differ. Current NCERT and IUPAC take STP as 273.15 K and 1 bar, giving a molar volume of 22.7 L mol−1. Older books use 273.15 K and 1 atm, giving the familiar 22.4 L mol−1. Both are correct for their own definition — check which pressure your question specifies before quoting a molar volume.
Worked example 4 — PV = nRT. What mass of oxygen gas occupies 5.0 L at 2.0 atm and 300 K? (M of O2 = 32.0 g mol−1)
n = PV ÷ RT = (2.0 × 5.0) ÷ (0.0821 × 300)
Numerator = 10.0
Denominator = 0.0821 × 300 = 24.63
n = 10.0 ÷ 24.63 = 0.406 mol
mass = n × M = 0.406 × 32.0 = 13.0 g
Dalton's law of partial pressures
Gas collected over water: pdry gas = Ptotal − aqueous tension
Water vapour contributes its own pressure (the aqueous tension at that temperature), so it must be subtracted before the dry gas pressure is used in any other formula.
Worked example 5 — Dalton's law. A gas is collected over water at 300 K. The total pressure is 1.000 bar and the aqueous tension at 300 K is 0.035 bar. Find the pressure of the dry gas.
pdry = 1.000 − 0.035 = 0.965 bar
Graham's law of diffusion
Note the swap: the lighter gas diffuses faster, so the molar masses appear the other way round from the rates. Writing the formula down carefully is half the battle.
Worked example 6 — Graham's law. How many times faster does hydrogen diffuse than oxygen at the same temperature and pressure? (M of H2 = 2.016, M of O2 = 32.0 g mol−1)
rH₂ / rO₂ = √(32.0 ÷ 2.016) = √15.873 = 3.98
Hydrogen diffuses about four times faster than oxygen — the classic answer.
Mistakes that cost marks
- Temperature in Celsius. This is the number one error in the whole chapter. Convert to kelvin before you write anything else.
- Mixing mL and L, or bar and atm. In Boyle, Charles and the combined law the units cancel, so mL on both sides is fine — but in PV = nRT they must match your chosen R.
- Inverting Graham's law. Rates ratio equals the square root of the inverse molar mass ratio. If your heavier gas came out faster, you inverted it.
- Forgetting to subtract aqueous tension for a gas collected over water.
- Applying a law when its "constant" quantity is not constant. Read the question for the words "at constant temperature" or "at constant pressure"; if two things change, use the combined gas law.
- Quoting 22.4 L when the question says 1 bar. At 1 bar and 273.15 K the molar volume is 22.7 L mol−1.
Which law, at a glance
| Law | Held constant | Relationship |
|---|---|---|
| Boyle's law | T, n | P1V1 = P2V2 |
| Charles's law | P, n | V1/T1 = V2/T2 |
| Gay-Lussac's law | V, n | P1/T1 = P2/T2 |
| Avogadro's law | P, T | V ∝ n |
| Combined gas law | n only | P1V1/T1 = P2V2/T2 |
| Ideal gas equation | — | PV = nRT |
| Dalton's law | V, T | Ptotal = Σ pi |
| Graham's law | P, T | r ∝ 1/√M |
Learn the middle column first. Once you can name what is held constant, choosing the right formula takes two seconds and the rest is arithmetic.
Check your P–V–T answers instantly. The free Combined Gas Law calculator takes any five of P1, V1, T1, P2, V2, T2 and solves for the sixth — so it also handles Boyle, Charles and Gay-Lussac problems as special cases.
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