Gas laws and molar volume: TCE Chemistry (Tasmania)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

The ideal gas equation $pV = nRT$ links pressure, volume, moles and temperature; rearrange it for whichever quantity is unknown. At a stated standard temperature and pressure, one mole of any gas occupies a fixed molar volume $V_m$ , so $n = V/V_m$ converts a gas volume directly to moles for stoichiometry. Temperature must always be in kelvin.

What this dot point is asking

TASC expects you to find the moles of a gas from its measured $p$ , $V$ and $T$ , then feed that into a mole-ratio calculation, and to use the simple gas laws when one variable is held constant.

The ideal gas equation

Ideal gas equation

pV = nRT

p

is pressure,

V

is volume,

n

is moles,

T

is temperature in kelvin, and

R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}

. To use this value of

R

, work in SI units:

p

in pascals (

\text{Pa}

V

in cubic metres (

\text{m}^3

). Convert

\text{kPa} \to \text{Pa}

\times 1000

and

\text{L} \to \text{m}^3

\times 10^{-3}

Temperature must be absolute: $T(\text{K}) = \theta(^\circ\text{C}) + 273$ . Watch your units of volume and pressure carefully because a single missed conversion (litres for cubic metres, or kPa for Pa) shifts the answer by a factor of $1000$ .

The combined gas law

When a fixed amount of gas changes conditions, the moles cancel and you get the combined gas law:

Combined gas law

\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}

Special cases: Boyle's law (

p_1V_1 = p_2V_2

at constant

T

); Charles' law (

V_1/T_1 = V_2/T_2

at constant

p

); and Gay-Lussac's law (

p_1/T_1 = p_2/T_2

at constant

V

Molar gas volume and Avogadro's law

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. TASC commonly uses $V_m \approx 24.8\ \text{L mol}^{-1}$ at $25\ ^\circ\text{C}$ and $100\ \text{kPa}$ , or $22.7\ \text{L mol}^{-1}$ at $0\ ^\circ\text{C}$ and $100\ \text{kPa}$ . Always use the molar volume that matches the conditions on the data sheet.

Equal volumes contain equal moles

At the same temperature and pressure, equal volumes of any gases contain equal numbers of molecules (Avogadro's law). So for reactions between gases the volume ratio equals the mole ratio from the balanced equation, and you can work directly with volumes without converting to moles.

Using gas data in reactions

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

Mass of reactant from a gas volume

Find the mass of zinc that reacts with excess hydrochloric acid to produce $250\ \text{mL}$ of hydrogen gas at $25\ ^\circ\text{C}$ and $100\ \text{kPa}$ , given the equation $\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$ , $V_m = 24.8\ \text{L mol}^{-1}$ and $M(\text{Zn}) = 65.38\ \text{g mol}^{-1}$ .

Step 1: Convert the gas volume to litres

Volumes from gas-collection experiments are often given in millilitres, but the molar volume $V_m$ is in litres per mole. Dividing by 1000 converts the units before any further calculation:

V(\text{H}_2) = 250\ \text{mL} = 0.250\ \text{L}.

Step 2: Find the moles of hydrogen gas

At $25\ ^\circ\text{C}$ and $100\ \text{kPa}$ the molar volume is $24.8\ \text{L mol}^{-1}$ , so dividing the volume by this value gives the moles of hydrogen produced:

n(\text{H}_2) = \frac{V}{V_m} = \frac{0.250}{24.8} = 1.008 \times 10^{-2}\ \text{mol}.

Step 3: Apply the stoichiometric mole ratio

From the balanced equation the ratio of zinc to hydrogen is $1:1$ , so exactly as many moles of zinc reacted as moles of hydrogen were produced. No scaling is needed here:

n(\text{Zn}) = 1.008 \times 10^{-2}\ \text{mol}.

Step 4: Convert moles of zinc to mass

Multiply the moles of zinc by its molar mass to find the mass that dissolved:

m(\text{Zn}) = nM = 1.008 \times 10^{-2} \times 65.38 = 0.659\ \text{g}.

Final answer: 0.659 g of zinc reacts to produce 250 mL of hydrogen under these conditions.

Real gases and the limits of the model

The ideal gas model assumes the particles have negligible volume of their own and no attractive forces between them. These assumptions hold well at low pressure and high temperature, where particles are far apart and moving fast. They break down at high pressure and low temperature, where particles are forced close together: their own volume becomes significant and intermolecular attractions pull them slightly together, so a real gas occupies a smaller volume than the ideal equation predicts. For TASC calculations you treat gases as ideal unless told otherwise, but you should be able to state when the model is least reliable.

Mixtures of gases

In a mixture, each gas exerts its own partial pressure as if it alone occupied the container, and the total pressure is the sum of the partial pressures (Dalton's law). When a gas is collected over water, the measured pressure includes the saturated water vapour pressure, so subtract the vapour pressure to find the partial pressure of the dry gas before using $pV = nRT$ . This correction is a common feature of gas-collection experiments.

Forgetting to convert to kelvin (or mixing units)

The ideal gas equation requires absolute temperature. Using degrees Celsius directly, especially small or negative values, gives a badly wrong answer; always add $273$ . The second classic slip is leaving $V$ in litres or $p$ in $\text{kPa}$ while using $R = 8.314$ . Either convert to $\text{m}^3$ and $\text{Pa}$ , or use the molar-volume route $n = V/V_m$ to avoid the unit trap entirely. A third slip is forgetting to subtract the water vapour pressure when a gas is collected over water.

In the exam, check which standard conditions and molar volume the question uses, convert temperatures to kelvin, keep units consistent with $R$ , subtract any water vapour pressure for gas collected over water, and link gas volumes to moles through the mole ratio.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

A carbon dioxide cylinder has a capacity of

0.605\ \text{L}

and when full contains

400.0\ \text{g}

\text{CO}_2

gas. (a) Show that there are approximately nine moles of

\text{CO}_2

in the cylinder. (b) Calculate the pressure (in

\text{kPa}

) inside a full cylinder at

25\ ^\circ\text{C}

. (c) Calculate the number of

\text{CO}_2

molecules inside the full cylinder. (

M(\text{CO}_2) = 44.01\ \text{g mol}^{-1}

R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}

N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}

Show worked answer →

(a) $n = m/M = 400.0 / 44.01 = 9.09\ \text{mol}$ , approximately nine moles. (1 mark)

(b) Use $pV = nRT$ with $V = 0.605\ \text{L} = 6.05 \times 10^{-4}\ \text{m}^3$ , $n = 9.09\ \text{mol}$ , $T = 298\ \text{K}$ .

p = \frac{nRT}{V} = \frac{9.09 \times 8.314 \times 298}{6.05 \times 10^{-4}} = 3.72 \times 10^{7}\ \text{Pa} = 3.72 \times 10^{4}\ \text{kPa}.

(2 marks)

Practice2 marks

12.2\ \text{L}

scuba tank has a pressure of

210\ \text{atm}

30.0\ ^\circ\text{C}

. (a) Calculate the volume the air would occupy if released to

1.0\ \text{atm}

at the same temperature. (b) In the water the full tank cools to

16.0\ ^\circ\text{C}

; calculate the new pressure inside the sealed tank.

Show worked answer →

(a) Constant temperature, so use Boyle's law $p_1V_1 = p_2V_2$ .

V_2 = \frac{p_1V_1}{p_2} = \frac{210 \times 12.2}{1.0} = 2.6 \times 10^{3}\ \text{L}.

(1 mark)

(b) Constant volume, so use $\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$ with temperatures in kelvin. $T_1 = 303\ \text{K}$ , $T_2 = 289\ \text{K}$ .

p_2 = p_1 \times \frac{T_2}{T_1} = 210 \times \frac{289}{303} = 200\ \text{atm}.

(1 mark)

Practice3 marks

What volume of carbon dioxide at

25\ ^\circ\text{C}

and

100\ \text{kPa}

(

V_m = 24.8\ \text{L mol}^{-1}

) is produced when

5.0\ \text{g}

of calcium carbonate decomposes completely?

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

. (

M(\text{CaCO}_3) = 100.1\ \text{g mol}^{-1}

Show worked answer →

Convert mass to moles: $n(\text{CaCO}_3) = 5.0 / 100.1 = 0.0500\ \text{mol}$ . (1 mark)

Mole ratio of $\text{CaCO}_3$ to $\text{CO}_2$ is $1:1$ , so $n(\text{CO}_2) = 0.0500\ \text{mol}$ . (1 mark)

V = n \times V_m = 0.0500 \times 24.8 = 1.24\ \text{L}.

So about

1.24\ \text{L}

of carbon dioxide forms. (1 mark)

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC