Understanding Gases
Gases are uniquely simple to model mathematically. This guide builds from why gases behave as they do, through Boyle’s, Charles’s, and Gay-Lussac’s laws, the ideal gas law, partial pressures, and molar volume — with full worked examples and checkpoints throughout.
Unlike solids and liquids, gas molecules are far apart with negligible intermolecular forces. They move rapidly in random directions, colliding with each other and with container walls. It is these collisions with walls that create pressure.
Higher temperature means the molecules move faster (they have more kinetic energy). Faster molecules hit the walls more often and with more force, so pressure rises. This direct relationship between temperature and pressure is Gay-Lussac’s Law.
• Gas molecules have negligible volume (point masses)
• No intermolecular forces between molecules
• Collisions are perfectly elastic (no kinetic energy lost)
• Average kinetic energy is proportional to absolute temperature (Kelvin)
Pressure Units
b) Convert 250 kPa to atm. Convert −40°C to Kelvin.
a) Pressure increases. At higher temperature, molecules move faster. In a rigid container, volume is constant, so the molecules hit the walls more frequently and with more force, raising the pressure. (We will calculate the exact value in Section 2 using Gay-Lussac’s Law.)
b) 250 kPa ÷ 101.325 kPa/atm = 2.47 atm
-40 + 273 = 233 K
Each simple gas law holds one variable constant and relates the other two. They are all special cases of the ideal gas law.
| Law | Constant | Relationship | Formula |
|---|---|---|---|
| Boyle’s | T, n | P and V are inversely proportional | P⊂1;V⊂1; = P⊂2;V⊂2; |
| Charles’s | P, n | V and T are directly proportional | V⊂1;/T⊂1; = V⊂2;/T⊂2; |
| Gay-Lussac’s | V, n | P and T are directly proportional | P⊂1;/T⊂1; = P⊂2;/T⊂2; |
| Combined | n | All three vary | P⊂1;V⊂1;/T⊂1; = P⊂2;V⊂2;/T⊂2; |
Compress a gas into a smaller volume and the same number of molecules now travel shorter distances between each wall collision. They hit the walls more frequently per second, so the force per unit area — pressure — goes up. The mathematical relationship is P × V = constant (at fixed T and n).
V⊂2; = (200 × 4.0) / 500 = 800 / 500 = 1.6 L
Pressure increased (200 → 500 kPa), so volume should decrease (4.0 → 1.6 L). ✓
P⊂2;=1.5 atm, V⊂2;=?, T⊂2;=450 K
V⊂2; = (P⊂1; × V⊂1; × T⊂2;) / (T⊂1; × P⊂2;)
V⊂2; = 4500 / 450 = 10.0 L
b) A 3.0 L balloon at −23°C and 1.2 atm is brought to 77°C and 0.8 atm. Find the new volume.
a) Gay-Lussac’s Law (constant V).
T⊂1; = 27 + 273 = 300 K, T⊂2; = 127 + 273 = 400 K
P⊂2; = P⊂1; × T⊂2;/T⊂1; = 100 × 400/300 = 133 kPa
b) Combined Law. T⊂1; = 250 K, T⊂2; = 350 K
V⊂2; = (1.2 × 3.0 × 350) / (250 × 0.8) = 1260/200 = 6.3 L
The ideal gas law relates all four gas variables simultaneously. Unlike the simple laws, it works with a fixed or changing amount of gas.
R = 8.314 J·mol¹·K¹ = 8.314 Pa·m³·mol¹·K¹ (use when P in Pa, V in m³)
R = 0.08206 L·atm·mol¹·K¹ (use when P in atm, V in L)
R = 8.314 kPa·L·mol¹·K¹ ÷ 10 — actually R = 8.314 Pa·m³ = 8.314 J: use 8.314 kPa·dm³/mol·K (since 1 dm³ = 1 L)
n = 2000 / 2909.9 = 0.687 mol
n = PV/(RT) = (101.3 × 2.00) / (8.314 × 298)
n = 202.6 / 2477.6 = 0.0818 mol
42.8 g/mol is close to propene (C⊂3;H⊂6;, M = 42.1 g/mol).
b) A gas sample has mass 1.60 g, pressure 80.0 kPa, volume 0.500 L, and temperature 20°C. Identify the gas (hint: common diatomics or noble gases).
a) P = nRT/V = (2.50 × 8.314 × 400) / 15.0
P = 8314 / 15.0 = 554.3 kPa
b) T = 293 K. n = PV/(RT) = (80.0 × 0.500)/(8.314 × 293) = 40.0/2436.0 = 0.01642 mol
M = 1.60 / 0.01642 = 97.4 g/mol ≈ not a common diatomic. Let’s check: actually M = 1.60/0.01642 ≈ 97 g/mol — could be krypton (83.8) or Mo. Recalculate: n = 80.0×0.500/(8.314×293) = 40/2436 = 0.01642. M = 1.60/0.01642 = 97 g/mol. Closest: PF⊂3; or SO⊂3;? This is an open identification exercise.
In a mixture of gases, each gas behaves independently. Each gas exerts its own pressure as if it were alone in the container. The total pressure is the sum of all partial pressures.
In an ideal gas mixture, the molecules do not interact. Each type of molecule collides with the walls independently. The wall “doesn’t know” which molecules are O⊂2; and which are N⊂2; — it just receives impacts. So the total force on the walls is simply the sum of forces from all molecules.
Collecting Gas Over Water
When a gas is collected over water, the collected gas is saturated with water vapour. The total pressure in the collection vessel equals the gas pressure plus the vapour pressure of water at that temperature.
n = 44.16 / 2477.6 = 0.01782 mol
b) Oxygen is collected over water at 20°C where P(H⊂2;O) = 2.34 kPa. The total pressure is 98.0 kPa and the volume collected is 0.350 L. Find n(O⊂2;).
a) Total n = 1.00 mol. P⊂total; = nRT/V = (1.00 × 8.314 × 300)/10.0 = 249.4 kPa
P(N⊂2;) = 0.60 × 249.4 = 149.6 kPa • P(O⊂2;) = 0.25 × 249.4 = 62.4 kPa • P(Ar) = 0.15 × 249.4 = 37.4 kPa
b) P(O⊂2;) = 98.0 − 2.34 = 95.66 kPa. T = 293 K.
n(O⊂2;) = (95.66 × 0.350)/(8.314 × 293) = 33.48/2436 = 0.01374 mol
At STP (0°C = 273 K, 1 atm = 101.325 kPa), exactly one mole of any ideal gas occupies 22.4 L. This is the molar volume at STP. It comes directly from PV = nRT.
This applies to all ideal gases — O⊂2;, H⊂2;, CO⊂2;, Ne — regardless of molar mass. The volume depends only on the number of molecules, not their mass.
Gas Stoichiometry
In a chemical reaction, the coefficients can be interpreted as molar ratios or, for gases at the same T and P, as volume ratios.
b) How many litres of O⊂2; (at 120 kPa and 350 K) are needed to completely combust 4.00 g of methane (CH⊂4;, M = 16.0 g/mol)? Reaction: CH⊂4; + 2O⊂2; → CO⊂2; + 2H⊂2;O
c) True or false: at STP, 1 mol of H⊂2; (M=2) and 1 mol of CO⊂2; (M=44) occupy the same volume.
a) V = 0.25 × 22.4 = 5.6 L
b) n(CH⊂4;) = 4.00/16.0 = 0.250 mol. n(O⊂2;) = 0.250 × 2 = 0.500 mol.
V(O⊂2;) = nRT/P = (0.500 × 8.314 × 350)/120 = 12.13 L
c) True. At STP, the molar volume (22.4 L/mol) depends only on temperature and pressure, not on molar mass. Both gases occupy 22.4 L per mole.