TASC TCE 2022-style practice: A carbon dioxide cylinder has a capacity of 0.605\ L and when full contains 400.0\ g of CO_2 gas. (a) Show that there are approximately nine moles of CO_2 in the cylinder. (b) Calculate the pressure (in kPa) inside a full cylinder at 25\ ^C. (c) Calculate the number of CO_2 molecules inside the full cylinder. (M(CO_2) = 44.01\ g mol^-1, R = 8.314\ J K^-1mol^-1, N_A = 6.022 × 10^23\ mol^-1.)

(a) n = m/M = 400.0 / 44.01 = 9.09\ mol, approximately nine moles. (1 mark) (b) Use pV = nRT with V = 0.605\ L = 6.05 × 10^-4\ m^3, n = 9.09\ mol, T = 298\ K. $p = (nRT)/(V) = 9.09 × 8.314 × 2986.05 × 10^-4 = 3.72 × 10^7\ Pa = 3.72 × 10^4\ kPa.$ (2 marks) (c) Number of molecules = n × N_A = 9.09 × 6.022 × 10^23 = 5.47 × 10^24 molecules. (1 mark)

TASC TCE 2022-style practice: A 12.2\ L scuba tank has a pressure of 210\ atm at 30.0\ ^C. (a) Calculate the volume the air would occupy if released to 1.0\ atm at the same temperature. (b) In the water the full tank cools to 16.0\ ^C; calculate the new pressure inside the sealed tank.

(a) Constant temperature, so use Boyle's law p_1V_1 = p_2V_2. $V_2 = (p_1V_1)/(p_2) = (210 × 12.2)/(1.0) = 2.6 × 10^3\ L.$ (1 mark) (b) Constant volume, so use p_1T_1 = p_2T_2 with temperatures in kelvin. T_1 = 303\ K, T_2 = 289\ K. $p_2 = p_1 × (T_2)/(T_1) = 210 × (289)/(303) = 200\ atm.$ (1 mark)

TASC TCE 2021-style practice: What volume of carbon dioxide at 25\ ^C and 100\ kPa (V_m = 24.8\ L mol^-1) is produced when 5.0\ g of calcium carbonate decomposes completely? CaCO_3(s) → CaO(s) + CO_2(g). (M(CaCO_3) = 100.1\ g mol^-1.)

Convert mass to moles: n(CaCO_3) = 5.0 / 100.1 = 0.0500\ mol. (1 mark) Mole ratio of CaCO_3 to CO_2 is 1:1, so n(CO_2) = 0.0500\ mol. (1 mark) $V = n × V_m = 0.0500 × 24.8 = 1.24\ L.$ So about 1.24\ L of carbon dioxide forms. (1 mark)

§-Syllabus dot point

TASChemistrySyllabus dot point

How do we relate the pressure, volume, temperature and amount of a gas?

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

The ideal gas equation, the combined gas law, molar gas volume at standard conditions, and using gas data in stoichiometric calculations, with fully worked TASC-style examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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:

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.

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.

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 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.

TCE 20224 marksA 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)

TCE 20222 marksA

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)

TCE 20213 marksWhat 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)