Gas density, gas mixtures, kinetic theory, effusion and real gases

Last week we knew gases only through $PV = nRT$. This week we draw three practical results from this equation (density, molar mass and mixtures), then look at kinetic theory, which explains why the equation works. We close the week with when the ideal gas breaks down and the basic concepts of thermochemistry.

Golden rules:

1. Density and molar mass are two faces of the same equation: $d = \frac{PM}{RT}$.
2. In a mixture, each gas behaves as if it were alone. Partial pressures add up.
3. In the speed formula, R = 8.314 and M is in kg/mol. This single trap is half the question.
4. A heavy gas is slow: the speed is proportional to $1/\sqrt{M}$.

Gas density and molar mass

Put $n = m/M$ into the ideal gas equation: $PV = \frac{m}{M}RT$. Rearranged:

$d = \frac{m}{V} = \frac{PM}{RT} \qquad\Longleftrightarrow\qquad M = \frac{dRT}{P} = \frac{mRT}{PV}$

The consequences:

• At the same T and P, the density of gases is proportional to their molar masses. $\mathrm{CO_2}$ (44) is heavier than air (about 29) and sinks to the ground; He (4) rises.
• When a gas is heated its density falls (T is in the denominator). That is why a hot air balloon flies.

Example 1. The density of $\mathrm{N_2}$ at 0 °C and 1.00 atm:

$d = \frac{1.00 \times 28.014}{0.08206 \times 273.15} = 1.25\ \text{g/L}$

Example 2 (Dumas method). A volatile liquid is vaporized at 100.0 °C and fills a 0.500 L flask at 0.987 atm. After cooling, 1.012 g of substance is found in the flask.

$M = \frac{mRT}{PV} = \frac{1.012 \times 0.08206 \times 373.15}{0.987 \times 0.500} = 62.8\ \text{g/mol}$

Combined with the empirical formula it gives the molecular formula: if the empirical formula is $\mathrm{CH_3O}$ (31.03), then $k = 2$, and the formula is $\mathrm{C_2H_6O_2}$.

Gas mixtures: Dalton’s law

Each gas exerts its own pressure “without seeing” the other gases in the container. The pressure that a component would exert alone is called its partial pressure. Dalton’s law:

$P_{\text{tot}} = P_A + P_B + P_C + \dots \qquad P_A = \frac{n_A RT}{V}$

Mole fraction: $x_A = \frac{n_A}{n_{\text{tot}}}$. The mole fractions add up to 1, and

$P_A = x_A \cdot P_{\text{tot}}$

Example 3. In a 10.0 L container at 27 °C there are 1.00 g $\mathrm{H_2}$ and 5.00 g $\mathrm{N_2}$.

n(H2) = 1.00 / 2.016  = 0.4960 mol
n(N2) = 5.00 / 28.014 = 0.1785 mol      n_tot = 0.6745 mol
P(H2) = 0.4960 × 0.08206 × 300.15 / 10.0 = 1.22 atm
P(N2) = 0.1785 × 0.08206 × 300.15 / 10.0 = 0.440 atm
P_tot = 1.66 atm        x(H2) = 0.4960 / 0.6745 = 0.735
check: 0.735 × 1.661 = 1.22 atm

Collecting gas over water

A gas produced in the laboratory is usually collected in an inverted container filled with water. The collected gas is saturated with water vapor. When the levels inside and outside the container are equalized, the total pressure equals the atmospheric pressure:

$P_{\text{gas}} = P_{\text{atm}} - P_{\mathrm{H_2O}}(T)$

The vapor pressure of water depends only on temperature (23.8 mmHg at 25 °C) and is given from a table.

Example 4. $\mathrm{2KClO_3 \rightarrow 2KCl + 3O_2}$. At 25 °C, 0.512 L of gas is collected over water, with a total pressure of 758 mmHg. How many grams of $\mathrm{KClO_3}$ decomposed?

P(O2) = 758 − 23.8 = 734.2 mmHg = 0.9661 atm
n(O2) = 0.9661 × 0.512 / (0.08206 × 298.15) = 0.02022 mol
n(KClO3) = 0.02022 × 2/3 = 0.01348 mol  × 122.55 g/mol = 1.65 g

PITFALL · not subtracting the water vapor. If you take the total pressure as the pressure of $\mathrm{O_2}$, you find about 3% too many moles. If the vapor pressure is given in the exam, it is there to be used.

The kinetic molecular theory of gases

The kinetic theory explains the gas laws with these assumptions:

1. A gas consists of a very large number of molecules moving constantly and randomly.
2. The volume of the molecules themselves is negligible next to the volume of the container.
3. There is no attraction or repulsion between molecules; collisions are elastic (no energy is lost).
4. The average kinetic energy of the molecules is proportional only to the absolute temperature.

For one mole of gas the average translational kinetic energy is:

$\overline{E_k} = \frac{3}{2}RT$

At 25 °C this is $1.5 \times 8.314 \times 298.15 = 3718$ J/mol, whichever gas it is.

The source of pressure is molecules hitting the wall. When heated, the molecules speed up and hit more often and harder: at constant volume the pressure rises (Gay-Lussac). When the volume is reduced, the same molecules hit more often (Boyle).

Molecular speeds

Since kinetic energy is $\tfrac12 mu^2$, at the same temperature light molecules are faster. The root-mean-square speed:

$u_{rms} = \sqrt{\frac{3RT}{M}} \qquad R = 8.314\ \tfrac{\text{J}}{\text{mol·K}},\ M\ \text{in kg/mol}$

Unit check: since $\text{J} = \text{kg·m}^2/\text{s}^2$, the inside of the root comes out as m²/s². If you put M in g/mol, the units do not work.

The speeds show a distribution (the Maxwell-Boltzmann distribution). When the temperature rises, the curve shifts right and spreads out; the curve of heavy gases is narrower and to the left.

Diffusion, effusion and Graham’s law

• Diffusion: the spreading of a gas into another gas (a smell spreads through a room).
• Effusion: the escape of a gas into a vacuum through a very small hole.

Graham’s law: under the same conditions, the ratio of the effusion rates of two gases is inversely proportional to the square root of the ratio of their molar masses:

$\frac{\text{rate}_A}{\text{rate}_B} = \sqrt{\frac{M_B}{M_A}} \qquad \frac{t_A}{t_B} = \sqrt{\frac{M_A}{M_B}}$

Example 5. How many times faster does $\mathrm{H_2}$ effuse than $\mathrm{O_2}$?

$\sqrt{\frac{31.998}{2.016}} = 3.98$

In uranium enrichment the ratio between $^{235}\mathrm{UF_6}$ and $^{238}\mathrm{UF_6}$ is only 1.004; that is why the process needs thousands of stages.

Real gases (nonideal gases)

The ideal gas assumptions break down in two situations:

1. High pressure: the molecules come close to each other, and their own volume can no longer be neglected.
2. Low temperature: the molecules slow down and the attraction between them becomes important (approaching liquefaction).

A gas behaves most ideally at high temperature and low pressure. A way to measure this is the compressibility factor $Z = \frac{PV}{nRT}$: for an ideal gas $Z = 1$.

The van der Waals equation adds two corrections:

$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT \qquad\Rightarrow\qquad P = \frac{nRT}{V - nb} - \frac{an^2}{V^2}$

• $b$: the volume of the molecules themselves (L/mol). The free volume drops to $V - nb$, and the pressure rises.
• $a$: the attraction between molecules. A molecule hitting the wall is pulled back by its neighbors inside, and the pressure drops.

A gas with a large $a$ value (polar or large molecules) deviates more. For $\mathrm{N_2}$ ($a = 1.370$, $b = 0.0387$) at 1.00 mol, 1.00 L, 300 K: ideal 24.6 atm, van der Waals 24.2 atm. The difference is small because nitrogen is small and nonpolar.

Introduction to thermochemistry

Next week’s topic is heat and energy. First the language:

• System: the part we study (the reaction mixture). Surroundings: everything else.
• Open system: exchanges matter and energy (an uncovered cup). Closed system: only energy (a closed bottle). Isolated system: neither (an ideal thermos).
• Energy: the capacity to do work. It exists as kinetic (motion) and potential (position, bonds). The SI unit is the joule: $1\ \text{J} = 1\ \text{kg·m}^2/\text{s}^2$.
• The old unit is the calorie: $1\ \text{cal} = 4.184\ \text{J}$ (exact). The food “Calorie” is really a kcal: 250 kcal = 1046 kJ.
• Heat ($q$): energy transferred because of a temperature difference. Work ($w$): energy transferred by a force over a distance.

Common mistakes