0.25 mol of an ideal gas is held in a 4.1 dm3 container at 27 °C. What is its pressure? (R = 0.0821 dm3 atm K−1 mol−1)
- T = 27 + 273 = 300 K.
- P = nRT ÷ V = (0.25 × 0.0821 × 300) ÷ 4.1.
- 0.25 × 0.0821 × 300 = 6.16, and 6.16 ÷ 4.1 = 1.5.
Answer: 1.5 atm
Gas laws
MDCAT Chemistry · Chemistry · Gases. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
Four quantities describe a sample of gas: pressure P, volume V, absolute temperature T and number of moles n. Each simple gas law keeps two of them fixed and shows how the other two change together.
The ideal gas equation, PV = nRT, joins all the laws in one line. Temperature must always be in kelvin, and the units of P and V must match the value of R you use.
In a mixture, each gas behaves as if it were alone. Its share of the total pressure is its partial pressure, and it depends on its share of the moles, not of the mass. Lighter gases diffuse and effuse faster than heavier ones.
Rules to remember
- Boyle's law: at constant T, P1V1 = P2V2
- Charles's law: at constant P, V1/T1 = V2/T2 (T in K); the V against T line reaches zero volume at −273 °C
- Avogadro's law: at the same T and P, equal volumes of gases contain equal numbers of molecules (V ∝ n)
- PV = nRT; R = 0.0821 dm3 atm K−1 mol−1 = 8.314 J K−1 mol−1; density d = PM/RT
- Dalton's law: total pressure = sum of the partial pressures; partial pressure = mole fraction × total pressure
- Graham's law: rate1/rate2 = √(M2/M1)
- K = °C + 273; °F = (9/5 × °C) + 32; 1 atm = 760 torr = 101 325 Pa; 1 cal = 4.18 J
Common mistake
Students use the temperature in °C in a gas law. Add 273 first: a rise from 27 °C to 54 °C is 300 K to 327 K, which does not double the volume.
Worked example
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