← Unit 2: Molecular interactions and reactions
Topic 1: Intermolecular forces and gases
Apply the ideal gas equation (PV = nRT) and the concept of molar volume at standard conditions to calculate amounts of gases under varying conditions of temperature and pressure
A focused answer to the QCE Chemistry Unit 2 dot point on PV = nRT. Sets out the ideal gas equation with QCAA's preferred units, derives molar volume at standard laboratory conditions (24.79 L/mol at SLC), and works through calculations linking pressure, volume, temperature and amount of gas.
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What this dot point is asking
QCAA wants you to use the ideal gas equation PV = nRT and the molar volume concept to convert between moles, mass and volume of gases at any specified temperature and pressure. This is the calculation foundation for gas stoichiometry and feeds the EA directly.
The answer
The ideal gas equation combines the three two-variable gas laws into one expression that links pressure (P), volume (V), absolute temperature (T) and amount of gas in moles (n). Combined with molar mass, it converts gas measurements into chemical amounts.
R is the universal gas constant, 8.314 J/(mol K), which is the value to use when P is in kPa, V is in L and T is in K.
Units for QCE Chemistry
| Quantity | Symbol | Preferred units |
|---|---|---|
| Pressure | P | kPa |
| Volume | V | L |
| Temperature | T | K (must always be Kelvin) |
| Amount of substance | n | mol |
| Gas constant | R | 8.314 J/(mol K) |
QCAA's chemistry data sheet supplies R in J/(mol K) and defines standard conditions for chemistry. With these units, products PV come out in joules, matching the energy units of nRT.
Conversions you will use often:
- 1 atm = 101.325 kPa.
- 1 mL = 0.001 L (divide mL by 1000).
- T(K) = T(degrees C) + 273.15 (use 273 in most QCE problems).
Molar volume of a gas
The molar volume V_m is the volume occupied by 1 mol of an ideal gas at a specified pressure and temperature.
| Condition set | Symbol | T | P | V_m |
|---|---|---|---|---|
| Standard laboratory conditions | SLC | 25 degrees C (298 K) | 100 kPa | 24.79 L/mol |
| Standard temperature and pressure (older) | STP | 0 degrees C (273 K) | 100 kPa | 22.71 L/mol |
| IUPAC 1982 STP | STP | 0 degrees C (273 K) | 101.325 kPa | 22.41 L/mol |
QCAA uses SLC (25 degrees C, 100 kPa, V_m = 24.79 L/mol) as the default in current syllabus. The older STP (22.4 L/mol at 0 degrees C, 1 atm) is sometimes quoted in textbooks; check which the question specifies.
Molar volume depends only on T and P, not on the identity of the gas. One mole of H_2 at SLC occupies 24.79 L. One mole of CO_2 at SLC occupies 24.79 L. This is Avogadro's law: equal volumes of gases at the same T and P contain equal numbers of molecules.
Rearranging PV = nRT
Solve for any single variable:
- n = PV / (RT)
- V = nRT / P
- P = nRT / V
- T = PV / (nR)
Combine with mass: m = n x M, where M is the molar mass.
Worked example: gas density
What is the density of nitrogen gas, N_2, at SLC?
Approach 1 (molar volume). 1 mol of N_2 has mass 28.02 g and occupies 24.79 L at SLC.
Density = mass / volume = 28.02 / 24.79 = 1.13 g/L.
Approach 2 (PV = nRT). Density = (P x M) / (R x T) for ideal gases. With P in kPa, M in g/mol, R = 8.314 and T in K, the units work out to g/L.
(100 x 28.02) / (8.314 x 298) = 2802 / 2477.6 = 1.131 g/L.
Both methods give the same answer to 3 significant figures.
Worked example: finding molar mass of a gas
A gas sample of mass 0.825 g occupies 600 mL at 75 degrees C and 95.0 kPa. Calculate its molar mass.
n = PV / (RT). T = 348 K, V = 0.600 L, P = 95.0 kPa.
n = (95.0 x 0.600) / (8.314 x 348) = 57.0 / 2893.3 = 0.01970 mol.
M = m / n = 0.825 / 0.01970 = 41.9 g/mol.
A molar mass of about 42 g/mol with no further information might suggest propene (C_3H_6, 42 g/mol) or similar.
When does the ideal gas equation fail?
PV = nRT assumes the kinetic theory of an ideal gas: negligible particle volume, no intermolecular forces, perfectly elastic collisions. Real gases deviate:
- At high pressure: particle volume becomes a significant fraction of total volume, so observed V is larger than nRT/P predicts.
- At low temperature (near boiling point): intermolecular forces cause the gas to "stick" together, so observed PV is smaller than predicted.
At conditions typical of QCE problems (around 100 kPa, around room temperature), ideal-gas behaviour is a very good approximation, error around 1 percent.
Common traps
Forgetting Kelvin. Using degrees C in PV = nRT gives wrong answers and sometimes negative results. Always convert.
Wrong R for the unit set. Use R = 8.314 J/(mol K) with kPa and L. Using R = 0.0821 L atm /(mol K) with kPa will be off by a factor of 100.
Treating SLC and STP as the same. They are not. SLC is 25 degrees C, 100 kPa, V_m = 24.79 L/mol. The classic 22.4 L/mol value uses old STP (0 degrees C, 1 atm). The current syllabus uses SLC.
Confusing V_m with M. V_m is litres per mole (a volume). M is grams per mole (a mass). They are independent and both needed when converting between mass and volume of a gas.
Reporting too many significant figures. Match the input data, typically 3 sig fig in QCAA problems.
In one sentence
The ideal gas equation PV = nRT (with R = 8.314 J/(mol K), P in kPa, V in L, T in K) lets you convert between any of the four gas variables for a specified amount of gas; at standard laboratory conditions (25 degrees C and 100 kPa) one mole of any ideal gas occupies 24.79 L, so molar volume converts directly between volume and amount of gas without needing the identity of the gas.
Past exam questions, worked
Real questions from past QCAA papers on this dot point, with our answer explainer.
2024 QCAA-style4 marksA steel cylinder of volume 8.50 L contains oxygen gas at 25 degrees C and a pressure of 850 kPa. (a) Calculate the amount (in mol) of oxygen in the cylinder. (b) Calculate the mass of oxygen, in grams.Show worked answer →
A 4-mark answer needs the rearranged equation, correct unit handling and the molar mass step.
(a) PV = nRT. Rearrange: n = PV / (RT).
Convert: T = 25 + 273 = 298 K. P = 850 kPa, V = 8.50 L. Use R = 8.314 J/(mol K), with P in kPa and V in L the product PV is in J.
n = (850 x 8.50) / (8.314 x 298) = 7225 / 2477.6 = 2.92 mol.
(b) Mass. M(O_2) = 32.00 g/mol. m = n x M = 2.92 x 32.00 = 93.4 g.
Markers reward the temperature conversion, correct rearrangement, and unit-consistent use of R. Significant figures should match the input data (3 sig fig here).
2023 QCAA-style3 marksCalculate the volume occupied by 0.250 mol of carbon dioxide gas (a) at standard laboratory conditions (SLC: 25 degrees C, 100 kPa), and (b) at 0 degrees C and 100 kPa. Use R = 8.314 J/(mol K).Show worked answer →
A 3-mark answer needs both volumes calculated correctly.
(a) At SLC. V = nRT/P = (0.250 x 8.314 x 298) / 100 = 619.4 / 100 = 6.19 L.
Alternative: V_m at SLC = 24.79 L/mol, so V = 0.250 x 24.79 = 6.20 L. (Either method scores.)
(b) At 0 degrees C. T = 273 K. V = (0.250 x 8.314 x 273) / 100 = 567.4 / 100 = 5.67 L.
Volume falls because temperature is lower at the same pressure (Charles's law). Identity of gas does not affect ideal-gas volume.
Markers reward correct Kelvin conversion, correct application of PV = nRT, and recognition that for ideal gases the molar volume depends only on T and P, not on identity.
Related dot points
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