The gas laws and PV = nRT: the units that make it work, the one rearrangement, and a gas produced by a reaction

The ideal gas law PV = nRT links a gas's pressure, volume, amount and temperature, and almost every wrong answer with it is a unit: the temperature must be in kelvin, and R must match the pressure and volume units you use. With pressure in kilopascals and volume in litres, R = 8.314 kPa·L/(mol·K); with atmospheres, R = 0.08206 L·atm/(mol·K). Below, the law is solved once for volume, run on a gas produced by a reaction with every line shown, and followed by the wrong line, the two-state gas laws it contains, and a problem to try.

The one law and its units

Four quantities, one constant, one equation. "Ideal" means the gas's molecules are treated as points that do not attract one another, which is a good approximation at everyday pressures and temperatures and fails at high pressure or very low temperature, where real gases condense. For a Grade 12 problem the approximation is the law.

Symbol	Quantity	Unit to use
P	pressure	kPa (or atm)
V	volume	L
n	amount of gas	mol
R	the gas constant	8.314 kPa·L/(mol·K) with kPa; 0.08206 L·atm/(mol·K) with atm
T	temperature	K

Two rules before any number is written. First, kelvin: T in K = T in °C + 273.15, always, before anything else. Second, pick the R that matches your pressure unit. The two values are the same constant in different clothes (1 kPa·L is 1 joule, so 8.314 kPa·L/(mol·K) is the 8.314 J/(mol·K) of the physics course), and using one with the other's pressure unit is a common error on this unit.

The one rearrangement. There are not four formulas to memorise. There is one law and one move: divide both sides by whatever multiplies the unknown. Solving for volume, divide by P: V = nRT ÷ P. Solving for amount, divide by RT: n = PV ÷ (RT). Pressure and temperature are the same move. See the move once and you never need the other three written down.

A size check to carry everywhere. One mole of any gas near room conditions fills about 25 L, roughly a large bucket. If your answer says a tenth of a mole fills 0.2 L, it is wrong before you recalculate anything, and the error is a unit.

In Ontario this is the gas half of SCH4U's Gas laws and stoichiometry (prerequisite review); the SCH4U course map shows where it sits, and every chemistry idea worked, in course order is the section this page belongs to.

Worked example: the gas a reaction makes

The problem: 10.0 g of calcium carbonate is heated until it decomposes completely, CaCO₃ → CaO + CO₂. What volume of carbon dioxide is collected at 25.0 °C and 101.3 kPa? The problem states its molar mass, CaCO₃ 100.09 g/mol, and the constant, R = 8.314 kPa·L/(mol·K). The data carry three significant figures, so the answer will.

1. Check the equation. It is balanced as written, one of each. So n(CO₂) = n(CaCO₃); the mole ratio is 1 : 1.
2. Grams to moles. n = 10.0 g ÷ 100.09 g/mol = 0.09991 mol. Keep the fourth digit; round at the end.
3. Celsius to kelvin. T = 25.0 + 273.15 = 298.15 K.
4. Rearrange for the unknown. V = nRT ÷ P.
5. Substitute with units. V = 0.09991 mol × 8.314 kPa·L/(mol·K) × 298.15 K ÷ 101.3 kPa.
6. Multiply, then divide. 0.09991 × 8.314 = 0.8307; 0.8307 × 298.15 = 247.67; 247.67 ÷ 101.3 = 2.4449 L.
7. Report. V = 2.44 L to three significant figures.

The unit check is a line of its own: kPa·L/(mol·K) × mol × K ÷ kPa leaves L. If the units do not cancel to litres, a number is in the wrong unit, and the arithmetic is irrelevant until that is fixed.

The size check: a tenth of a mole, a bit under two and a half litres, about a tenth of a bucket. Right size.

The wrong line. Writing T = 25.0 in the formula gives V = 0.8307 × 25.0 ÷ 101.3 = 0.205 L, about twelve times too small, because 298 K is about twelve times 25. A second wrong line: R = 0.08206 with the pressure still in kilopascals gives V = 0.09991 × 0.08206 × 298.15 ÷ 101.3 = 0.0241 L, about a hundred times too small, because 101.3 kPa is one atmosphere and the formula was told it was a hundred of them. Either error is caught by the size check: a tenth of a mole of gas cannot fill a fifth of a litre at room temperature, let alone a tablespoon.

Molar volume, and the STP wrinkle

The volume of one mole of ideal gas depends only on the conditions, and the law computes it.

At STP, standard temperature and pressure, taken here as 0 °C and 101.325 kPa (1 atm), the definition the open text this page is checked against uses:

• V = nRT ÷ P = 1 mol × 8.314 kPa·L/(mol·K) × 273.15 K ÷ 101.325 kPa = 2271 ÷ 101.325 = 22.41 L → 22.4 L

At SATP, standard ambient temperature and pressure, 25 °C and 100 kPa:

• V = 1 × 8.314 × 298.15 ÷ 100 = 2479 ÷ 100 = 24.79 L → 24.8 L

Many Canadian courses work at SATP, which is closer to a classroom, and some references define STP at 100 kPa rather than 101.325 kPa, which moves the 22.4 to 22.7. The question always states its conditions; use the stated ones and the stated standard, and do not carry a memorised molar volume from one standard into a problem set at another. The wrong line here is 22.4 L for a mole of gas at room temperature; at 25 °C a mole is 24.8 L, about a tenth larger, and the difference shows up as a 10 % error in every answer built on it.

The two-state gas laws are PV = nRT twice

Boyle's law, Charles's law and the combined gas law are not separate facts. With the amount of gas fixed, nR is a constant, so PV ÷ T is the same number before and after any change:

• Combined gas law: P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂
• Hold T constant and it cancels: Boyle's law, P₁V₁ = P₂V₂
• Hold P constant: Charles's law, V₁ ÷ T₁ = V₂ ÷ T₂
• Hold V constant: Gay-Lussac's law, P₁ ÷ T₁ = P₂ ÷ T₂

One worked line, for Charles's law. A balloon holds 2.00 L at 20.0 °C and is warmed to 60.0 °C at constant pressure. T₁ = 293.15 K, T₂ = 333.15 K.

• V₂ = V₁ × T₂ ÷ T₁ = 2.00 L × 333.15 K ÷ 293.15 K = 2.27 L

The volume grows by about 14 %, because the kelvin temperature grew by about 14 %.

The wrong line. V₂ = 2.00 × 60 ÷ 20 = 6.00 L, the Celsius version. Tripling the Celsius reading is not tripling the temperature, and the balloon does not triple. Every two-state problem has this trap, and the kelvin conversion as the first written line is what avoids it.

Stoichiometry with gases: where the mole goes

The three moves of stoichiometry are unchanged: to moles, across the ratio, back from moles. What changes with a gas is where the moles come from and where they go. A gas's amount comes from PV = nRT instead of from a mass, and a gas's answer goes back through V = nRT ÷ P instead of through a molar mass. The calcium carbonate example above was exactly that: a mass in, a mole ratio across, a volume out. The mole-ratio half of stoichiometry, with the limiting reactant and percent yield, is its own page.

One shortcut, valid only when every gas is at the same temperature and pressure: volumes of gases are in the mole ratio directly, because equal volumes hold equal moles. Two litres of hydrogen react with one litre of oxygen to make two litres of steam, as the coefficients of 2 H₂ + O₂ → 2 H₂O say. The moment a solid or a liquid enters the problem, or the conditions differ, go back to moles.

The ideal gas law on one screen

• The law: PV = nRT. Kelvin, always: K = °C + 273.15.
• Match R to the pressure unit: 8.314 kPa·L/(mol·K) with kPa; 0.08206 L·atm/(mol·K) with atm.
• The one rearrangement: divide both sides by whatever multiplies the unknown. V = nRT ÷ P; n = PV ÷ (RT).
• The size check: a mole of gas near room conditions is about 25 L.
• Molar volume: 22.4 L at STP (0 °C, 101.325 kPa); 24.8 L at SATP (25 °C, 100 kPa). Use the standard the question states.
• Two states, fixed amount: P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂; hold one quantity and the named law falls out.
• A gas in a reaction: moles in from PV = nRT or a mass, across the ratio, out through V = nRT ÷ P.

Why one example is worked in full before any problem is set: Sweller and Cooper (1985) found that students who studied worked examples and then solved matching problems learned the procedure faster than students who only solved problems. The matching problem is next, and chapter 3 of the free playbook is the habit of redoing it cold.

Try one cold: hydrogen from magnesium

Cover everything above. A strip of magnesium, 2.50 g (24.31 g/mol), reacts completely with excess hydrochloric acid: Mg + 2 HCl → MgCl₂ + H₂. What volume of hydrogen is collected at 22.0 °C and 99.5 kPa? Then a second, shorter one: what amount of gas fills 5.00 L at 100 kPa and 25 °C? Write the kelvin line first in both.

Check

• n(Mg) = 2.50 g ÷ 24.31 g/mol = 0.10284 mol; the ratio Mg : H₂ is 1 : 1, so n(H₂) = 0.10284 mol
• T = 22.0 + 273.15 = 295.15 K
• V = nRT ÷ P = 0.10284 × 8.314 × 295.15 ÷ 99.5
• 0.10284 × 8.314 = 0.8550; 0.8550 × 295.15 = 252.4; 252.4 ÷ 99.5 = 2.536 L → 2.54 L
• Size check: a tenth of a mole, two and a half litres, right size
• Second problem: n = PV ÷ (RT) = 100 × 5.00 ÷ (8.314 × 298.15) = 500 ÷ 2479 = 0.202 mol

If you got about 0.19 L for the hydrogen, the 22.0 went in as the temperature. If you got about 0.025 L, R was the atmospheres value with the pressure in kilopascals.

Where this sits in the course

In the SCH4U pack this is the gas half of Gas laws and stoichiometry (prerequisite review), the last of the 24 ideas and the one every later calculation assumes. The idea of a gas's amount returns in Dynamic equilibrium and the equilibrium constant, where a gas-phase equilibrium's concentrations are moles per litre of a gas, and in Le Chatelier's principle, where an increase in pressure shifts a gas-phase equilibrium toward the side with fewer moles of gas; gas-phase equilibria and pressure shifts is that page. It returns once more in Electrolytic cells and Faraday's laws, where a gas evolved at an electrode is measured by its volume and converted back to moles of electrons.

How to practise it

Ten problems with the unknown moved around: solve for V, then for n, then for P, then for T, with the units deliberately mixed so that one problem gives the pressure in atmospheres and the next in kilopascals. The point is to make the unit check and the kelvin line a habit rather than a thing you remember on a good day. Rohrer and Taylor (2007) found that students who practised problems shuffled across types did worse during the practice and better on a test a week later than students who practised in blocks, because the shuffled group practised the decision. Why mixed practice beats twenty of the same has the reasoning. Where practice problems come from and the order that teaches has the sources, and how to study a calculation chapter has the routine: the worked example redone cold, then the steps faded one at a time until you can write the whole thing from the problem statement.

Keep the error log by kind of error, not by problem: "forgot kelvin", "wrong R", "ratio skipped". After a week the log stops growing, and that is the measurement that the law is yours.