Gas laws in general chemistry: worked exercises

First-semester lectures began in mid-September — at the Politecnico di Milano on 14 September 2026 — and in General Chemistry, after stoichiometry and nomenclature, the syllabus reaches the gaseous state almost everywhere in October. It is one of the topics most regularly present in the mid-term test (the Italian prova in itinere): where one is scheduled, it falls at the end of October or the start of November — at the Politecnico from 29 October to 2 November 2026, about five weeks from now.

Compared with stoichiometry and redox, the gas laws have a different problem: the physics behind them is simple (pressure, volume, temperature and amount of substance linked by a single equation), but the number of units involved — atmospheres, torr, pascals, bar, litres, cubic metres, degrees Celsius and Kelvin — is what loses the most marks. It is not a problem of theory, it is a problem of conversions done halfway or with the wrong unit.

What the mid-term really asks

On the gaseous state, the typical questions are of three types: applying $PV=nRT$ directly to find a missing quantity; calculating the density or molar mass of an unknown gas; working with gas mixtures and partial pressures (Dalton’s law), often inside a stoichiometry problem where one of the products is gaseous. The historical laws (Boyle, Charles, Gay-Lussac, Avogadro) appear almost always as a conceptual premise or as a special case of the equation of state with one variable held constant, rarely as an exercise in their own right.

The ideal gas laws, one at a time

Before arriving at the unified equation, it is worth seeing where it comes from, because it is exactly the history that explains why $R$ has the form it has:

• Boyle’s law (constant temperature): at constant $T$, the pressure and volume of a fixed quantity of gas are inversely proportional.

• Charles’s law (constant pressure): at constant $P$, the volume is directly proportional to the absolute temperature.

• Gay-Lussac’s law (constant volume): at constant $V$, the pressure is directly proportional to the absolute temperature.

• Avogadro’s law: at the same $P$ and $T$, equal volumes of different gases contain the same number of moles — it is the law that allows $n$ to be introduced into the final equation.

Putting the four together gives the ideal gas equation of state:

where $P$ is the pressure, $V$ the volume, $n$ the moles, $T$ the absolute temperature (Kelvin, always) and $R$ the universal gas constant. The point where almost everyone goes wrong in the first exercise: $T$ must always be in Kelvin, never in Celsius, because the direct proportionality of Charles and Gay-Lussac holds only on the absolute scale, where zero really is zero volume/pressure. The conversion is

The value of R and why the unit matters

$R$ is not a single number: its value changes depending on the units you are using for $P$ and $V$, and taking the wrong value is by far the commonest mistake on this topic.

The practical rule: if you work in atmospheres and litres use $0.0821$; if the problem is in SI units (pascals and cubic metres, or if you need a result in joules to link up with thermodynamics) use $8.314$. Before substituting into the formula, always check that $P$ and $V$ are in units consistent with the chosen $R$ — otherwise the calculation “works” but is wrong by several orders of magnitude, and it is a mistake that in a timed exam you discover only at the end, when it is too late to redo everything.

The pressure conversions you need most often:

and for volume: $1\mathrm{L}=10^{-3}{\mathrm{m}}^3=1\mathrm{d}{\mathrm{m}}^3$.

Molar volume: 22.4 L or 22.7 L?

The molar volume is the volume occupied by one mole of an ideal gas under standard conditions, and it is one of the few numbers worth knowing by heart — but there is a pitfall to be aware of. With the old STP (0 °C, 1 atm), the molar volume is

With the current IUPAC STP (0 °C, 1 bar, not 1 atm) the correct value is instead $22.7L/mol$, because 1 bar is slightly less than 1 atm. In practice most General Chemistry courses in Italy still use 22.4 L with the 1 atm convention, but it is the classic thing to check in your lecturer’s notes before the exam rather than take for granted: writing “22.4 L, standard conditions at 1 atm” in an exercise where the lecturer uses the IUPAC convention can cost a mark for an avoidable ambiguity.

Density and molar mass of a gas

A frequent application of the equation of state is finding the molar mass of an unknown gas, starting from its measured density. Rewriting $n=m/M$ inside $PV=nRT$:

where $d$ is the density of the gas. From here you obtain the unknown molar mass directly:

It is the typical tool for identifying an unknown gas from an experimental measurement of density, pressure and temperature — a type of question that often appears in the mid-term precisely because it tests both the formula and the units in a single step.

Gas mixtures: Dalton’s law

When several gases share the same container, each behaves as if it were alone, occupying the whole volume at its own partial pressure $P_i$. Dalton’s law says that the total pressure is the sum of the partial pressures:

The partial pressure of each component is linked to the mole fraction $x_i$ (the ratio between the moles of that component and the total moles of the mixture):

The most frequent mistake here is confusing partial pressure and total pressure in a later calculation — for example using $P_{\text{tot}}$ in place of $P_i$ when applying $PV=nRT$ to a single component of the mixture, which has to be treated as if it alone occupied the whole volume at its partial pressure, not at the total pressure.

When the gas is no longer ideal

The equation $PV=nRT$ describes an ideal gas: point-like molecules with no interactions between them. It is an excellent approximation at pressures that are not too high and temperatures not too low, but it departs from reality when the molecules are close together (high pressure) or slow (low temperature), conditions in which the molecules’ own volume and the intermolecular forces stop being negligible. The correct model in those cases is the van der Waals equation:

where $a$ corresponds to the attractive intermolecular forces and $b$ to the molecules’ own volume (the term that makes the “available” volume smaller than the volume of the container). For most exercises in a first General Chemistry course it is enough to know that it exists and when it is needed, without doing the sums with $a$ and $b$ — unless your course has covered it explicitly.

The three worked exercises

Exercise 1 — finding the pressure with $PV=nRT$

A rigid container of volume $5.00\mathrm{L}$ holds $0.200\mathrm{mol}$ of an ideal gas at a temperature of $27°\mathrm{C}$. What is the pressure, in atm?

First compulsory conversion: $T=27+273.15=300.15\mathrm{K}$ (in practice often rounded to $300\mathrm{K}$).

Exercise 2 — molar mass from density

An unknown gas has density $1.96g/L$ at $0°\mathrm{C}$ and $1.00\mathrm{atm}$. Calculate the molar mass and decide whether it is ${\mathrm{N}}_2$ ($M=28.0g/mol$), ${\mathrm{O}}_2$ ($M=32.0g/mol$) or $\mathrm{C}{\mathrm{O}}_2$ ($M=44.0g/mol$).

Calculation: $1.96\times 0.0821=0.1609$; $0.1609\times 273.15\approx 43.96g/mol$.

Quick check: $0.0821\times 273.15\approx 22.4$ is the molar volume, so $M=d\times 22.4\approx 44g/mol$: the gas is $\mathrm{C}{\mathrm{O}}_2$. Beware, though: density and molar mass do not distinguish two gases with the same $M$, such as $\mathrm{C}{\mathrm{O}}_2$ and ${\mathrm{N}}_2\mathrm{O}$ — which is why the question gives you a set of candidates.

Exercise 3 — gas mixture and partial pressure

In a $10.0\mathrm{L}$ container at $25°\mathrm{C}$ there are $0.300\mathrm{mol}$ of ${\mathrm{N}}_2$ and $0.100\mathrm{mol}$ of ${\mathrm{O}}_2$. Calculate the partial pressure of each gas and the total pressure.

Total moles: $n_{\text{tot}}=0.300+0.100=0.400\mathrm{mol}$. Temperature: $T=25+273.15=298.15\mathrm{K}$.

Total pressure, treating the mixture as a single ideal gas:

Calculation: $0.400\times 0.0821=0.03284$; $0.03284\times 298.15\approx 9.792$; divided by $10.0$ gives $0.979\mathrm{atm}$.

Mole fractions: $x_{{\mathrm{N}}_2}=0.300/0.400=0.750$; $x_{{\mathrm{O}}_2}=0.100/0.400=0.250$.

Partial pressures: $P_{{\mathrm{N}}_2}=0.750\times 0.979\approx 0.734\mathrm{atm}$; $P_{{\mathrm{O}}_2}=0.250\times 0.979\approx 0.245\mathrm{atm}$.

Check: $0.734+0.245=0.979\mathrm{atm}$, equal to the $P_{\text{tot}}$ calculated above — the sums work out, as they must for Dalton’s law.

The mistakes that cost marks

• Temperature in degrees Celsius instead of Kelvin: it is mistake number one, and with a small absolute $T$ (temperatures near 0 °C) the relative error on the result is enormous.
• Units of $R$ inconsistent with $P$ and $V$: using $0.0821$ with a pressure in pascals, or $8.314$ with a volume in litres, without converting first.
• Confusing partial pressure and total pressure in a chained calculation on a mixture.
• Taking 22.4 L/mol for granted without checking which convention (1 atm or 1 bar) your course uses.
• Applying van der Waals mechanically even under normal conditions, where the ideal approximation is already excellent and only complicates the sums.

Where it leads next

The gaseous state is also the ground where the stoichiometry seen in the first weeks comes back — reactions that produce or consume gases, where you need to convert between grams of reactant and volume of gas produced always passing through the moles. If that part is not yet solid, the four-step method is in the guide stoichiometry: typical mistakes. After the gaseous state, the syllabus typically continues with redox reactions: the method for balancing them with half-reactions is in the post on redox in General Chemistry. If instead it is the mole itself that is not yet clear, the most elementary starting point is in the mole and stoichiometry: the method in Year 3 of high school.

How I work on this topic in lessons

With my university General Chemistry students the first step is always the same: fix the units of $R$ before touching the calculator, and check every result with a second pass of calculations done in a slightly different way (for example recalculating $RT$ separately before dividing). It is a habit that costs thirty seconds more per exercise and saves you from the silliest mistakes — the ones you see only after handing in.

If you want to practise the format of the test, the General Chemistry practice test follows the structure of the written paper: the correct answer is free, the guided step-by-step solution is unlocked with the first lesson. For the overview of the science subjects I teach there is the sciences page, and for university support in general the one on university tutoring. If the gaseous state stays a weak point after trying this method, write to me through contact: the first introductory call is free and lasts half an hour.