Apply the ideal gas equation and molar gas volume to calculate gas quantities in reactions.

The ideal gas equation, molar gas volume at standard conditions, and using gas data in stoichiometric calculations.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-30

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What this dot point is asking

You must use the gas laws to find moles of a gas and feed that into stoichiometric calculations.

The ideal gas equation

Temperature must be in kelvin: add 273 to the Celsius value. Keep units consistent with the value of $R$ you use. If $p$ is in kPa and $V$ in litres, then $R = 8.31$ works because the units combine correctly.

Molar gas volume

At a defined standard condition, one mole of any ideal gas occupies the same volume regardless of identity, because the molecules are so far apart that their own size and identity barely matter. A common value is about $24.8\ \text{L mol}^{-1}$ at 25 degrees Celsius and 100 kPa. Always use the molar volume that matches the conditions given on your data sheet.

Using gas data in reactions

Convert a gas volume to moles with $n = V / V_m$ (or with $pV = nRT$ ), use the mole ratio, then convert the product to the quantity asked for.

example

Volume of gas from a reaction

What volume of carbon dioxide at 25 degrees Celsius and 100 kPa forms when $5.0\ \text{g}$ of calcium carbonate decomposes? $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ . ( $M(\text{CaCO}_3) = 100.1$ ; $V_m = 24.8\ \text{L mol}^{-1}$ .)

Step 1: $n(\text{CaCO}_3) = 5.0 / 100.1 = 0.04995\ \text{mol}$ .

Step 2: Mole ratio is 1 to 1, so $n(\text{CO}_2) = 0.04995\ \text{mol}$ .

Step 3: $V = n \times V_m = 0.04995 \times 24.8 = 1.24\ \text{L}$ .

So about $1.24\ \text{L}$ of carbon dioxide is produced.

In the exam, check which standard conditions and molar volume the question uses, convert temperatures to kelvin, keep units consistent with $R$ , and link gas volumes to moles through the mole ratio.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 TASC2 marksA 12.2 L tank of air, which divers use to breathe underwater, has a pressure of 210 atm at 30.0 degrees C. (a) Calculate the volume that this air would occupy if it were released from the tank at 1.0 atm. (b) In the water, the tank and air cool to 16.0 degrees C; calculate the pressure inside the full tank at 16.0 degrees C.

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(a) Constant temperature, so use Boyle's law P1V1 = P2V2. V2 = P1V1 / P2 = (210 x 12.2) / 1.0 = 2562 L, about 2.6 x 10^3 L. (1 mark)

(b) Constant volume, so use Gay-Lussac's law P1/T1 = P2/T2 with temperatures in kelvin. T1 = 30.0 + 273 = 303 K; T2 = 16.0 + 273 = 289 K. P2 = P1 x T2/T1 = 210 x 289/303 = 200 atm. (1 mark)

2022 TASC4 marksA carbon dioxide cylinder has a capacity of 0.605 L and when full contains 400.0 g of CO2 gas. (a) Show that there are approximately nine moles of CO2 in the cylinder. (b) Calculate the pressure (in kPa) inside a full cylinder at 25 degrees C. (c) Calculate the number of CO2 molecules inside the full cylinder.

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(a) n = m / M = 400.0 / 44.01 = 9.09 mol, approximately 9 moles. (1 mark)

(b) Use PV = nRT with V = 0.605 L = 6.05 x 10-4 m3, n = 9.09 mol, T = 298 K, R = 8.314. P = nRT / V = (9.09 x 8.314 x 298) / 6.05 x 10-4 = 22 522 / 6.05 x 10-4 = 3.72 x 10^7 Pa = 3.72 x 10^4 kPa. (2 marks)

2021 TASC2 marksA tyre has a volume of 8.5 L. How many of these car tyres can be inflated to a pressure of 220 kPa from one 50 L gas cylinder of nitrogen, initially at a pressure of 18 000 kPa? Assume the temperature remains constant.

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At constant temperature the total amount of gas is conserved (PV is proportional to moles). Total gas available, expressed as PV at the tyre pressure: P1V1 = 18 000 x 50 = 900 000 kPa L in the cylinder.

The cylinder itself will retain gas at 220 kPa once it can no longer fill a tyre, so gas available to deliver = (18 000 - 220) x 50 = 889 000 kPa L. Each tyre needs 220 x 8.5 = 1870 kPa L.

Number of tyres = 889 000 / 1870 = 475 tyres (rounding down to a whole number, 475). (2 marks: total gas available and division by gas per tyre; accept about 480 if the residual cylinder pressure is neglected.)

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