Gases: pressure and the gas laws

Gases are the most “mathematical” topic in chemistry. All their behavior is described by four variables: pressure $P$, volume $V$, temperature $T$ and the number of moles $n$. A single equation combines them all: $PV = nRT$. Two kinds of questions come in the exam: finding the new value when conditions change (two-state questions) and finding one variable from a single state.

Golden rules:

1. Temperature is always in kelvin. Setting up ratios with °C is the number one cause of wrong answers.
2. The units of R and the units of the data must match. If you use 0.08206, then L, atm, mol, K.
3. In a two-state question, cross out what stays constant. You derive the law you want from the combined law.
4. Check the direction of the result. When the pressure rises the volume must fall, when heated it must expand.

Pressure and its units

Pressure is the force per unit area: $P = F/A$. Its SI unit is the pascal: $1\ \text{Pa} = 1\ \text{N/m}^2$. Atmospheric pressure raises a column of mercury by about 760 mm; the barometer works on this principle.

The pressure of a gas in a closed container is measured with a manometer. In an open-ended manometer, if the mercury level on the gas side is higher, the gas pressure is lower than atmospheric: $P_{\text{gas}} = P_{\text{atm}} - \Delta h$.

Example 1. How many mmHg and kPa is 0.985 atm?

$0.985\ \text{atm} \times \frac{760\ \text{mmHg}}{1\ \text{atm}} = 749\ \text{mmHg} \qquad 0.985 \times 101.325 = 99.8\ \text{kPa}$

The simple gas laws

Charles’s law is a straight line on a graph, and if it is extended to $V = 0$ it crosses the axis at −273.15 °C. This is absolute zero, the zero of the kelvin scale. The relation $V \propto T$ holds only in absolute temperature; that is why you cannot set up ratios with °C.

Example 2 (Boyle). 4.00 L of gas is compressed from 1.00 atm to 2.50 atm (T constant):

$V_2 = \frac{P_1V_1}{P_2} = \frac{1.00 \times 4.00}{2.50} = 1.60\ \text{L}$

Example 3 (Charles). 2.00 L of gas is heated from 20.0 °C to 80.0 °C (P constant):

$V_2 = V_1 \times \frac{T_2}{T_1} = 2.00 \times \frac{353.15\ \text{K}}{293.15\ \text{K}} = 2.41\ \text{L}$

TIP · think in ratios. Instead of memorizing the formula, ask “should the volume grow or shrink?” and write the ratio accordingly. Heating: $V_1 \times \tfrac{\text{larger } T}{\text{smaller } T}$. Compression: $V_1 \times \tfrac{\text{smaller } P}{\text{larger } P}$.

Recognizing by graphs

Sometimes only a graph is given in the exam and you are asked which law it is:

• $V$ against $P$: a hyperbola (inverse proportion), Boyle. If $V$ is plotted against $1/P$, a straight line through the origin.
• $V$ against $T$ (K): a straight line through the origin, Charles. If the axis is °C, the line crosses zero at −273.15 °C.
• $P$ against $T$ (K): a straight line through the origin, Gay-Lussac.
• $V$ against $n$: a straight line through the origin, Avogadro.

The slope of the line depends on the variables held constant. For example, in the Charles graph the slope is $nR/P$; if the pressure is larger, the line is flatter.

The combined gas law

When n is constant, the three laws combine:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

If a variable is constant, it is canceled from both sides and one of the simple laws remains. You can memorize a single formula and derive the rest.

The ideal gas equation

It gathers the four variables in a single equation:

$PV = nRT$

The value of the gas constant $R$ depends on the units:

Standard temperature and pressure (STP): Petrucci uses 0 °C and 1 bar; under this condition 1 mol of ideal gas occupies 22.7 L. Many sources, however, take 0 °C and 1 atm: 22.4 L. Check which one the instructor uses; if in doubt, calculate directly with $PV = nRT$.

$V_m = \frac{RT}{P} = \frac{0.08206 \times 273.15}{1.000} = 22.41\ \text{L/mol} \quad (1\ \text{atm})$

Example 4. How many grams of He are in a 12.0 L balloon at 25.0 °C and 1.50 atm?

$n = \frac{PV}{RT} = \frac{1.50\ \text{atm} \times 12.0\ \text{L}}{0.08206\ \tfrac{\text{L·atm}}{\text{mol·K}} \times 298.15\ \text{K}} = 0.736\ \text{mol}$

$m = 0.7357\ \text{mol} \times 4.003\ \text{g/mol} = 2.94\ \text{g}$

Cancel the units one by one: L and atm cancel, K cancels, mol remains.

Stoichiometry in reactions with gases

When a gas is a product or a reactant, $PV = nRT$ stands at one end of the chain:

$\text{g A} \rightarrow \text{mol A} \xrightarrow{\text{coefficient}} \text{mol gas B} \xrightarrow{V = nRT/P} \text{L gas B}$

Example 5. 10.0 g of $\mathrm{CaCO_3}$ reacts with excess HCl: $\mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2}$. How many L of $\mathrm{CO_2}$ are released at 25 °C and 1.00 atm?

mol CaCO3 = 10.0 / 100.086 = 0.09991 mol  → mol CO2 = 0.09991 (1:1)
V = nRT/P = 0.09991 × 0.08206 × 298.15 / 1.00 = 2.44 L

Only among gases and at the same T and P can the coefficients be used directly as volume ratios (Gay-Lussac’s law of combining volumes). For example, in $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$, 1 L of nitrogen with 3 L of hydrogen gives 2 L of ammonia.

Example 6 (limiting reactant among gases). At the same T and P, 4.0 L of CO and 3.0 L of $\mathrm{O_2}$ are mixed and burned: $\mathrm{2CO + O_2 \rightarrow 2CO_2}$.

CO : 4.0 L ÷ 2 = 2.0
O2 : 3.0 L ÷ 1 = 3.0     → CO is limiting
CO2 formed  = 4.0 L
O2 used     = 2.0 L,  O2 left = 1.0 L
final mixture: 4.0 L CO2 + 1.0 L O2 = 5.0 L  (there was 7.0 L at the start; the number of moles decreased)

More worked examples

Example 7 (manometer). In an open-ended manometer the mercury is 18 mm higher on the gas side than on the atmosphere side. The barometer reads 752 mmHg. The gas pushes the mercury less, so its pressure is lower than atmospheric:

$P_{\text{gas}} = 752 - 18 = 734\ \text{mmHg} = \frac{734}{760}\ \text{atm} = 0.966\ \text{atm}$

Example 8 (number of molecules). There is $\mathrm{O_2}$ at 745 mmHg in a 3.00 L container at 25 °C. How many molecules?

P = 745 / 760 = 0.9803 atm
n = PV/RT = 0.9803 × 3.00 / (0.08206 × 298.15) = 0.1202 mol
N = 0.1202 × 6.022×10^23 = 7.24×10^22 molecules

We first converted mmHg to atm, because R = 0.08206 is in atm.

Which formula do I use?

Common mistakes