The ideal gas equation and molar volume (QCE Chemistry Unit 2)

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

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.

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.

Examples in context

Example 1. LPG cylinder pressure on a Carnarvon basin gas well-head. A $9 \, \text{kg}$ propane cylinder for a remote Carnarvon well-head heater contains propane at $5.00 \, \text{bar}$ ($500 \, \text{kPa}$) and $30^{\circ}\text{C}$. Using $pV = nRT$ with $R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1}$, $n = 9{,}000 / 44.10 = 204 \, \text{mol}$, $T = 303 \, \text{K}$: $V = nRT/p = 204 \times 8.314 \times 303 / 500{,}000 = 1.03 \, \text{m}^3$ of gas-phase volume if vented. Cylinder steel is rated to $25 \, \text{bar}$; the design margin lets operators leave cylinders in $45^{\circ}\text{C}$ summer sun without rupture risk, a routine safety calculation for QCAA application contexts.

Example 2. Hydrogen pilot at Gladstone H2-Hub. Stanwell's Gladstone hydrogen pilot stores green $\text{H}_2$ in cryogenic tanks at $-253^{\circ}\text{C}$ ($20 \, \text{K}$). At ambient $25^{\circ}\text{C}$ the same mass occupies a far larger volume: $25^{\circ}\text{C}$ corresponds to $298 \, \text{K}$, so the ratio $V_2 / V_1 = T_2 / T_1 = 298 / 20 = 14.9$ (at constant pressure). Hence $1 \, \text{m}^3$ of liquid hydrogen expands to $\sim 850 \, \text{m}^3$ of gas at SLC (the additional factor comes from the liquid density), explaining why storage choice between liquid and compressed gas is a core engineering decision.

Try this

Q1. State the ideal gas equation, defining each variable and its SI unit. [3 marks]

• Cue. $pV = nRT$; $p$ in Pa, $V$ in m$^3$, $n$ in mol, $R = 8.314 \, \text{J K}^{-1}\, \text{mol}^{-1}$, $T$ in K.

Q2. Calculate the volume occupied by $0.500 \, \text{mol}$ of $\text{N}_2$ at $50.0^{\circ}\text{C}$ and $120 \, \text{kPa}$. [3 marks]

• Cue. $V = nRT/p = 0.500 \times 8.314 \times 323.15 / 120{,}000 = 0.0112 \, \text{m}^3 = 11.2 \, \text{L}$.

Q3. A sealed $2.00 \, \text{L}$ flask holds nitrogen at $200 \, \text{kPa}$ and $25^{\circ}\text{C}$. (a) Calculate moles. (b) Calculate mass. (c) Predict pressure if the flask is heated to $200^{\circ}\text{C}$. [2+2+3 marks]

• Cue. (a) $n = 200{,}000 \times 0.00200 / (8.314 \times 298.15) = 0.161 \, \text{mol}$. (b) $4.52 \, \text{g}$. (c) $p_2 = 200 \times 473/298 = 318 \, \text{kPa}$.