10Statistical Thermodynamics: Partition Functions

The tomb of Ludwig Boltzmann in Vienna carries a single line, $S = k\log W$: the entropy of a system counts the ways its molecules can share out its energy. Thermodynamics, as the Year 1 and Year 2 volumes built it, never needs molecules: it relates measured heats, entropies and equilibrium constants to one another. Statistical thermodynamics computes them from the molecules themselves. Its central object is a single sum over the energy levels of one molecule, the partition function; the levels come from spectroscopy, and the standard entropy of nitrogen computed from its spectrum agrees with the tabulated value to a hundredth of a joule per kelvin and per mole. This chapter builds the partition function, splits it into its translational, rotational, vibrational and electronic parts, and turns it into internal energy, entropy and Gibbs energy; Chapter 11 applies it to heat capacities and equilibrium constants.

You already know

The Year 2 volume defined the standard molar entropy, the Gibbs energy and the chemical potential, and tabulated entropies obtained from heat capacities measured down to low temperature. Chapter 1 gave the energy levels of a particle in a box, of the harmonic oscillator and of the rigid rotor, $E_J = hcBJ(J+1)$ with degeneracy $2J + 1$; Chapter 6 measured $B$ and $\tilde\nu$ for diatomic molecules and found that in a homonuclear molecule such as $\ce{N2}$ the nuclear spins make alternate rotational levels unequal.

10.1 Microstates and the Boltzmann distribution

Consider $N$ identical, independent molecules that can occupy levels of energy $\varepsilon_0 = 0 < \varepsilon_1 < \varepsilon_2 < \dots$, with total energy $E$. Saying how many molecules are on each level, $\{n_0, n_1, \dots\}$, describes a distribution; saying which molecule is on which level describes much more.

Definition 10.1 (Microstate, statistical weight)

A microstate of a system of molecules is a complete specification of the state of every molecule. The statistical weight $W$ of a distribution $\{n_i\}$ of $N$ molecules over non-degenerate levels is the number of microstates that realise it:

$$W = \frac{N!}{n_0!\,n_1!\,n_2!\cdots}.$$

Example 10.2 (Three quanta among three oscillators)

Three oscillators share three quanta of energy. The distribution “one oscillator holds all three” can be realised in $3!/(1!\,2!) = 3$ ways, “one holds two, another one” in $3! = 6$ ways and “each holds one” in one way: ten microstates in all, and the most even distribution that still spreads the energy over several levels is the most probable. With $10^{23}$ molecules the weights become so sharply peaked that one distribution, and the ones that differ from it by a negligible amount, carry practically all the microstates.

Remark 10.3

The postulate of statistical thermodynamics is that all the microstates of an isolated system with a given energy are equally probable. The system is then found, overwhelmingly, in the distribution of largest weight: finding it is the task of this section.

For large numbers, Stirling’s approximation $\ln x! \approx x\ln x - x$ gives

Definition 10.4 (Boltzmann distribution, population)

The population $n_i/N$ of a level is the fraction of the molecules that occupy it. The Boltzmann distribution is the distribution of largest weight of independent molecules at thermal equilibrium at temperature $T$, given by the theorem below.

Theorem 10.5 (The Boltzmann distribution)

For independent molecules at equilibrium at temperature $T$, the population of a level of energy $\varepsilon_i$ and degeneracy $g_i$ is

$$\frac{n_i}{N} = \frac{g_i\eu^{-\varepsilon_i/kT}}{q}, \qquad q = \sum_i g_i\eu^{-\varepsilon_i/kT},$$

where $k$ is the Boltzmann constant. Two levels have populations in the ratio $\frac{n_j}{n_i} = \frac{g_j}{g_i}\eu^{-(\varepsilon_j - \varepsilon_i)/kT}$.

Definition 10.6 (Molecular partition function)

The molecular partition function of a molecule at temperature $T$ is the sum

$$q = \sum_{\text{levels}}g_i\eu^{-\varepsilon_i/kT} = \sum_{\text{states}}\eu^{-\varepsilon_s/kT},$$

with energies measured from the lowest level, so that $q \ge 1$.

The partition function counts the states that are thermally accessible: at very low temperature only the lowest level contributes and $q \to g_0$; at very high temperature every state contributes about 1 and $q$ grows without limit. A rule of thumb follows: levels much more than $kT$ above the lowest one are empty, levels within $kT$ of it are populated. At $298.15\,\mathrm{K}$, $kT$ corresponds to $207.2\,\mathrm{cm}^{-1}$, or $RT = 2.479\,\mathrm{kJ}/\mathrm{mol}$.

Method 10.7 (Populations of levels from spectroscopic constants)

1. List the levels with their energies (from the constants, e.g. $hcBJ(J+1)$) and degeneracies ($2J + 1$).
2. Express each energy in units of $kT$; at $298.15\,\mathrm{K}$, divide a wavenumber by $207.2\,\mathrm{cm}^{-1}$.
3. Form the terms $g_i\eu^{-\varepsilon_i/kT}$, add them up to get $q$, and divide. Stop the sum when the terms are negligible.
4. Check: the populations add up to 1; the ratio of two of them is $\frac{g_j}{g_i}\eu^{-(\varepsilon_j-\varepsilon_i)/kT}$.

Example 10.8 (The rotational levels of hydrogen chloride)

For $\ce{H^{35}Cl}$, $B_0 = B_e - \alpha_e/2 = 10.44\,\mathrm{cm}^{-1}$. At $300\,\mathrm{K}$ the level $J$ lies at $10.44\,J(J+1)$ $\mathrm{cm}^{-1}$, i.e. $0.0501\,J(J+1)$ in units of $kT$. The terms $(2J+1)\eu^{-0.0501J(J+1)}$ are 1, 2.71, 3.70, 3.84, 3.31, 2.45, …; their sum is $q = 20.3$ and the populations of $J = 0$ to 4 are 0.049, 0.134, 0.182, 0.189 and 0.163: the most populated level is $J = 3$, not $J = 0$, because the degeneracy $2J + 1$ grows while the Boltzmann factor falls. This is the envelope of the rotation–vibration band of Chapter 6.

10.2 The molecular partition function

Proposition 10.9 (Factorisation)

If the energy of a molecule is a sum of independent contributions, $\varepsilon = \varepsilon^{\mathrm T} + \varepsilon^{\mathrm R} + \varepsilon^{\mathrm V} + \varepsilon^{\mathrm E}$, each state being any combination of a translational, a rotational, a vibrational and an electronic state, then

$$q = q^{\mathrm T}q^{\mathrm R}q^{\mathrm V}q^{\mathrm E}.$$

The separation is an approximation: rotation and vibration interact (the constant $\alpha_e$), and the electronic state sets the rotational and vibrational constants. For a molecule in its ground electronic state at ordinary temperatures the errors are a few hundredths of a joule per kelvin in an entropy. The populations are then computed separately for each kind of motion: the fraction of molecules in a vibrational level $v$ does not depend on their rotation.

10.3 The four contributions

Translation

Definition 10.10 (Thermal wavelength)

The thermal wavelength of a molecule of mass $m$ at temperature $T$ is $\Lambda = \dfrac{h}{\sqrt{2\pi mkT}}$.

Proposition 10.11 (Translational partition function)

For a molecule of mass $m$ free to move in a volume $V$,

$$q^{\mathrm T} = \frac{V}{\Lambda^3}.$$

Example 10.12 (Argon in a flask)

For argon ($M = 39.95\,\mathrm{g}/\mathrm{mol}$) at $298.15\,\mathrm{K}$, $\Lambda = 16.0\,\mathrm{pm}$, a tenth of an atomic diameter. In $1\,\mathrm{dm}^{3}$, $q^{\mathrm T} = 10^{-3}/(1.600\times10^{-11})^3 = 2.44 \times 10^{29}$ translational states are thermally accessible, far more than the $2.4\times10^{22}$ atoms the flask holds at $1\,\mathrm{bar}$: each state is almost always empty, the condition under which the molecules can be counted as below.

Rotation

Definition 10.13 (Characteristic temperatures)

The characteristic rotational temperature of a linear molecule of rotational constant $B$ is $\theta_{\mathrm R} = hcB/k$; the characteristic vibrational temperature of a vibration of wavenumber $\tilde\nu$ is $\theta_{\mathrm V} = hc\tilde\nu/k$.

Definition 10.14 (Symmetry number)

The symmetry number $\sigma$ of a molecule is the number of distinct orientations of the rigid molecule that are reached by proper rotations and are indistinguishable from the starting one (the order of the rotational subgroup of its point group): 1 for $\ce{HCl}$ and $\ce{CO}$, 2 for $\ce{N2}$, $\ce{CO2}$ and $\ce{H2O}$, 3 for $\ce{NH3}$, 12 for $\ce{CH4}$ and $\ce{C6H6}$.

Proposition 10.15 (Rotational partition function)

For a linear molecule, $q^{\mathrm R} = \sum_J(2J+1)\eu^{-\theta_{\mathrm R}J(J+1)/T}$. When $T \gg \theta_{\mathrm R}$,

$$q^{\mathrm R} \approx \frac{T}{\sigma\theta_{\mathrm R}} = \frac{kT}{\sigma hcB},$$

with a relative error of about $\theta_{\mathrm R}/3T$ for $\sigma = 1$ (the exact sum is close to $T/\theta_{\mathrm R} + \frac13$).

For $\ce{N2}$, $B_0 = 1.990\,\mathrm{cm}^{-1}$ and $\theta_{\mathrm R} = 2.86\,\mathrm{K}$: at $298.15\,\mathrm{K}$, $q^{\mathrm R} = 298.15/(2\times2.863) = 52.1$. For $\ce{HCl}$, $\theta_{\mathrm R} = 15.0\,\mathrm{K}$ and $q^{\mathrm R} = 19.8$ (exact sum 20.2). Only $\ce{H2}$ ($\theta_{\mathrm R} = 85\,\mathrm{K}$) and its isotopologues need the exact sum at room temperature. A non-linear molecule with rotational constants $A$, $B$, $C$ has $q^{\mathrm R} = \frac{1}{\sigma}\bigl(\frac{kT}{hc}\bigr)^{3/2}\sqrt{\frac{\pi}{ABC}}$, admitted.

Vibration

Proposition 10.16 (Vibrational partition function)

For a harmonic vibration of wavenumber $\tilde\nu$, with energies measured from the zero-point level,

$$q^{\mathrm V} = \sum_{v\ge0}\eu^{-v\theta_{\mathrm V}/T} = \frac{1}{1 - \eu^{-\theta_{\mathrm V}/T}} .$$

It tends to 1 when $T \ll \theta_{\mathrm V}$ and to $T/\theta_{\mathrm V}$ when $T \gg \theta_{\mathrm V}$. A molecule with several normal modes has the product of one such factor per mode.

For the vibrational quantum of a diatomic molecule this book takes the observed fundamental $G(1) - G(0) = \omega_e - 2\omega_ex_e$ (Chapter 6); at room temperature only the first levels matter, and their spacing is what counts. Stiff, light molecules are frozen: $\ce{N2}$ has $\theta_{\mathrm V} = 3352\,\mathrm{K}$, $q^{\mathrm V} = 1.000013$ at $298.15\,\mathrm{K}$, and only 13 molecules in a million are in $v = 1$. Heavy, soft ones are not: $\ce{I2}$ has $\theta_{\mathrm V} = 307\,\mathrm{K}$, $q^{\mathrm V} = 1.556$, and more than a third of its molecules vibrate.

Electronic states

Proposition 10.17 (Electronic partition function)

With the electronic levels $\varepsilon_j$ (degeneracy $g_j$) measured from the ground level,

$$q^{\mathrm E} = g_0 + g_1\eu^{-\varepsilon_1/kT} + \cdots$$

For most molecules the first excited electronic level lies tens of thousands of wavenumbers up and $q^{\mathrm E} = g_0$: 1 for a closed-shell molecule, 3 for $\ce{O2}$ (ground term ${}^3\Sigma_g^-$), $2J + 1$ for an atom in a level $J$.

This is the definition applied to the electronic levels; the cases where low levels matter are atoms and radicals with a fine structure.

Example 10.18 (Nitric oxide and the chlorine atom)

$\ce{NO}$ has a ${}^2\Pi$ ground term split by spin–orbit coupling into ${}^2\Pi_{1/2}$ (lower) and ${}^2\Pi_{3/2}$, $119.73\,\mathrm{cm}^{-1}$ higher, each doubly degenerate. At $298.15\,\mathrm{K}$, $q^{\mathrm E} = 2 + 2\eu^{-119.73/207.22} = 2 + 2\times0.561 = 3.12$: 36 % of the molecules are in the upper component. The chlorine atom has its ${}^2P_{1/2}$ level $882.35\,\mathrm{cm}^{-1}$ above the ${}^2P_{3/2}$ ground level: $q^{\mathrm E} = 4 + 2\eu^{-4.258} = 4.03$.

10.4 From partition functions to thermodynamics

Theorem 10.19 (Internal energy)

For $N$ independent molecules, the internal energy measured from its value at $T = 0$ is

$$U - U(0) = NkT^2\Bigl(\frac{\partial\ln q}{\partial T}\Bigr)_V = -N\Bigl(\frac{\partial\ln q}{\partial\beta}\Bigr)_V .$$

The contributions of the four kinds of motion add up.

For translation, $q^{\mathrm T} \propto T^{3/2}$ gives $\frac32NkT$; for the rotation of a linear molecule at high temperature, $q^{\mathrm R} \propto T$ gives $NkT$; a vibration gives $Nk\theta_{\mathrm V}/(\eu^{\theta_{\mathrm V}/T} - 1)$, which is $NkT$ only when $T \gg \theta_{\mathrm V}$. The classical equipartition of energy, $\frac12kT$ per quadratic term, is the high-temperature limit of these results; Chapter 11 follows the heat capacities as each motion freezes out.

Definition 10.20 (Canonical partition function)

The canonical partition function of a system at temperature $T$ and volume $V$ is the sum $Q = \sum_s\eu^{-E_s/kT}$ over all the states $s$ of the whole system, of energies $E_s$.

Proposition 10.21 (Canonical partition function of an ideal gas)

For $N$ independent molecules, $Q = q^N$ if they are distinguishable (localised in a crystal), and $Q = q^N/N!$ for an ideal gas of identical molecules.

Definition 10.22 (Statistical entropy)

The statistical entropy of an isolated system is $S = k\ln W$, where $W$ is the number of its microstates; for a system at temperature $T$, $W$ is the weight of the dominant distribution.

Theorem 10.23 (Entropy and the other functions)

For a system at temperature $T$ with canonical partition function $Q$,

$$S = \frac{U - U(0)}{T} + k\ln Q, \qquad A - A(0) = -kT\ln Q, \qquad p = kT\Bigl(\frac{\partial\ln Q}{\partial V}\Bigr)_T .$$

For an ideal gas this gives $pV = NkT$ and $G - G(0) = -NkT\ln(q/N)$.

Theorem 10.24 (Sackur–Tetrode equation)

The molar entropy of a monatomic ideal gas of molar mass $M$ at temperature $T$ and pressure $p$ is

$$S_{\mathrm m} = R\Bigl[\ln\Bigl(\frac{kT}{p\Lambda^3}\Bigr) + \frac52\Bigr], \qquad \Lambda = \frac{h}{\sqrt{2\pi mkT}},\ m = \frac{M}{N_A}.$$

For argon at $298.15\,\mathrm{K}$ and $1\,\mathrm{bar}$, $kT/p = 4.116 \times 10^{-26}\,\mathrm{m}^{3}$ and $\Lambda = 1.600 \times 10^{-11}\,\mathrm{m}$: $S_{\mathrm m} = R(\ln1.005 \times 10^{7} + 2.5) = 154.85\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$. The tabulated standard entropy, obtained from heat capacities measured down to a few kelvins, is $154.846 \pm 0.003$: the formula has no adjustable parameter, and it contains Planck’s constant. A heavier atom has a larger entropy (more translational states at the same energy), and so has a gas at lower pressure (more volume per atom).

Proposition 10.25 (Rotational and vibrational molar entropies)

For a linear molecule with $T \gg \theta_{\mathrm R}$, and a vibration with $x = \theta_{\mathrm V}/T$,

$$S^{\mathrm R}_{\mathrm m} = R\Bigl[\ln\frac{T}{\sigma\theta_{\mathrm R}} + 1\Bigr], \qquad S^{\mathrm V}_{\mathrm m} = R\Bigl[\frac{x}{\eu^x - 1} - \ln\bigl(1 - \eu^{-x}\bigr)\Bigr],$$

and an electronic ground level of degeneracy $g_0$, alone populated, adds $R\ln g_0$.

Method 10.26 (The standard entropy of a gas from its constants)

1. Translation: Sackur–Tetrode with the molar mass, $T = 298.15\,\mathrm{K}$ and $p^\circ = 1\,\mathrm{bar}$.
2. Rotation: $B_0 = B_e - \alpha_e/2$, $\theta_{\mathrm R} = hcB_0/k$, the symmetry number, and $R[\ln(T/\sigma\theta_{\mathrm R}) + 1]$ (or the exact sum if $T < 30\,\theta_{\mathrm R}$).
3. Vibration: one term per normal mode, with the fundamental wavenumbers.
4. Electronic: $R\ln g_0$, or $R[\ln q^{\mathrm E} + \langle\varepsilon\rangle/kT]$ if low levels exist.
5. Add; compare with the table, where nuclear spin is left out in the same way.

gas	$S^{\mathrm T}$	$S^{\mathrm R}$	$S^{\mathrm V}$	$S^{\mathrm E}$	sum	table
$\ce{Ar}$	154.85	0.00	0.00	0.00	154.85	154.85
$\ce{N2}$	150.42	41.18	0.00	0.00	191.60	191.61
$\ce{CO}$	150.42	47.23	0.00	0.00	197.65	197.66
$\ce{HCl}$	153.71	33.16	0.00	0.00	186.87	186.90
$\ce{Cl2}$	162.00	58.65	2.24	0.00	222.89	223.08
$\ce{I2}$	177.91	74.24	8.43	0.00	260.58	260.69
$\ce{Cl}$	153.36	0.00	0.00	11.83	165.19	165.19

The sums agree with the tables to better than 0.2 $\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$, and to a few hundredths for the light molecules. The small residuals of $\ce{Cl2}$ and $\ce{I2}$ come from approximations this chapter made: the constants of the main isotopologue, the harmonic oscillator for a molecule with many vibrational levels populated. The tabulated entropy of $\ce{CO}$ is itself the statistical value: measured calorimetrically, it comes out several $\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$ lower, a discrepancy explained in Chapter 11.

In the lab — Measuring a third-law entropy

The calorimetric route heats a sample from a few kelvins to $298.15\,\mathrm{K}$ in an adiabatic calorimeter, measuring its heat capacity in small steps. The entropy is the sum of $\int C_p\dd T/T$ over each phase, plus $\Delta H/T$ at every transition (solid–solid, fusion, vaporisation), plus a small correction for the non-ideality of the real gas; below the lowest measured temperature, $C_p$ of a non-metallic solid is extrapolated as $aT^3$. The third law fixes the entropy of the perfect crystal at 0 K to zero. Agreement with the statistical value, within the uncertainty of the measurement, both tests the third law and reveals the crystals that are not perfect at 0 K.

History — Boltzmann, 1877

Ludwig Boltzmann showed in 1877 that the entropy of a gas is proportional to the logarithm of the number of ways its molecules can share the energy, and derived the distribution that bears his name by counting those ways. The interpretation was fiercely contested by physicists who doubted the existence of atoms. The formula in the form $S = k\log W$, with the constant $k$, was written by Max Planck, who used the same counting in 1900 for the radiation of a hot body; it is carved on Boltzmann’s tomb in the central cemetery of Vienna.

10.5 Exercises

Exercise 10.1 ★

Four oscillators share four quanta. List the distributions, give the weight of each, and check that they add up to the number of ways of placing four indistinguishable quanta in four oscillators, $\binom{7}{3}$. Which distributions are the most probable?

Solution

Solution of Exercise 10.1.

Write each distribution as the quanta held by the four oscillators, largest first. “4, 0, 0, 0”: $4!/(3!\,1!) = 4$; “3, 1, 0, 0”: $4!/(2!\,1!\,1!) = 12$; “2, 2, 0, 0”: $4!/(2!\,2!) = 6$; “2, 1, 1, 0”: $4!/(1!\,2!\,1!) = 12$; “1, 1, 1, 1”: 1. Total 35 $= \binom73$. The most probable are “3, 1, 0, 0” and “2, 1, 1, 0” (12 each); the second, with populations 1, 2, 1 for 0, 1, 2 quanta, already falls off with energy as the Boltzmann distribution does for large numbers.

Exercise 10.2 ★

Two non-degenerate levels are $2.5\,\mathrm{kJ}/\mathrm{mol}$ apart. Compute the ratio of their populations at $298.15\,\mathrm{K}$ and at $1000\,\mathrm{K}$. What is the limit at very high temperature?

Solution

Solution of Exercise 10.2.

$\eu^{-2500/(8.314 \times 298.15)} = \eu^{-1.009} = 0.365$; at $1000\,\mathrm{K}$, $\eu^{-0.301} = 0.740$. At very high temperature the ratio tends to 1 (equal populations, never an inversion).

Exercise 10.3 ★

Compute the thermal wavelength of an argon atom at $298.15\,\mathrm{K}$ and its translational partition function in a volume of $1\,\mathrm{dm}^{3}$.

Solution

Solution of Exercise 10.3.

$m = 39.95\times10^{-3}/6.022\times10^{23} = 6.634 \times 10^{-26}\,\mathrm{kg}$; $\Lambda = h/\sqrt{2\pi mkT} = 1.600 \times 10^{-11}\,\mathrm{m} = 16.0\,\mathrm{pm}$; $q^{\mathrm T} = 10^{-3}/(1.600\times10^{-11})^3 = 2.44 \times 10^{29}$.

Exercise 10.4 ★

With $B_0 = 1.990\,\mathrm{cm}^{-1}$ ($\ce{N2}$) and $10.440\,\mathrm{cm}^{-1}$ ($\ce{H^{35}Cl}$), compute $\theta_{\mathrm R}$ and the high-temperature rotational partition function of each molecule at $298.15\,\mathrm{K}$.

Solution

Solution of Exercise 10.4.

$hc/k = 1.4388\,\mathrm{cm}\,\mathrm{K}$. $\ce{N2}$: $\theta_{\mathrm R} = 2.863\,\mathrm{K}$, $q^{\mathrm R} = 298.15/(2 \times 2.863) = 52.1$. $\ce{HCl}$: $\theta_{\mathrm R} = 15.02\,\mathrm{K}$, $q^{\mathrm R} = 298.15/15.02 = 19.85$ (exact sum 20.19).

Exercise 10.5 ★★

The fundamental wavenumbers of $\ce{I2}$ and $\ce{N2}$ are $213.27\,\mathrm{cm}^{-1}$ and $2329.92\,\mathrm{cm}^{-1}$. Compute $\theta_{\mathrm V}$, $q^{\mathrm V}$ at $298.15\,\mathrm{K}$, and the fraction of molecules that are not in $v = 0$.

Solution

Solution of Exercise 10.5.

$\ce{I2}$: $\theta_{\mathrm V} = 1.4388 \times 213.27 = 306.9\,\mathrm{K}$, $q^{\mathrm V} = 1/(1 - \eu^{-1.029}) = 1.556$; fraction excited $1 - 1/q^{\mathrm V} = 0.357$. $\ce{N2}$: $\theta_{\mathrm V} = 3352\,\mathrm{K}$, $q^{\mathrm V} = 1.000013$, fraction excited $1.3 \times 10^{-5}$.

Exercise 10.6 ★★

Compute the electronic partition function at $298.15\,\mathrm{K}$ of $\ce{NO}$ (two doubly degenerate levels $119.73\,\mathrm{cm}^{-1}$ apart) and of the chlorine atom (${}^2P_{3/2}$ ground level, ${}^2P_{1/2}$ at $882.35\,\mathrm{cm}^{-1}$). Which fraction of the $\ce{NO}$ molecules is in the upper level? What does $q^{\mathrm E}(\ce{NO})$ tend to at high temperature?

Solution

Solution of Exercise 10.6.

$\ce{NO}$: $q^{\mathrm E} = 2 + 2\eu^{-119.73/207.22} = 2 + 2 \times 0.561 = 3.12$; upper fraction $1.12/3.12 = 0.36$. At high temperature $q^{\mathrm E} \to 4$ (two levels equally populated). $\ce{Cl}$: $q^{\mathrm E} = 4 + 2\eu^{-882.35/207.22} = 4 + 2 \times 0.0142 = 4.03$.

Exercise 10.7 ★★

From $q^{\mathrm T} = V/\Lambda^3$, derive the internal energy and the heat capacity at constant volume of one mole of a monatomic ideal gas, and evaluate $U - U(0)$ at $298.15\,\mathrm{K}$.

Solution

Solution of Exercise 10.7.

$\ln q = \ln V + \frac32\ln T + \text{const}$, so $U - U(0) = NkT^2 \times \frac{3}{2T} = \frac32NkT$; for one mole $\frac32RT = 3.72\,\mathrm{kJ}/\mathrm{mol}$ at $298.15\,\mathrm{K}$, and $C_V = \frac32R = 12.47\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$.

Exercise 10.8 ★★

Use the Sackur–Tetrode equation to compute the standard molar entropy of argon at $298.15\,\mathrm{K}$; compare with the tabulated $154.846\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$. By how much does it change if the pressure is divided by 10?

Solution

Solution of Exercise 10.8.

$kT/p^\circ = 4.116 \times 10^{-26}\,\mathrm{m}^{3}$, $\Lambda^3 = 4.093 \times 10^{-33}\,\mathrm{m}^{3}$, ratio $1.006 \times 10^{7}$: $S_{\mathrm m} = 8.3145 \times (16.124 + 2.5) = 154.85\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$, the tabulated value to the last digit. Dividing $p$ by 10 adds $R\ln10 = 19.14\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$.

Exercise 10.9 ★★

Compute the rotational contribution to the standard molar entropy of $\ce{H^{35}Cl}$ at $298.15\,\mathrm{K}$ ($\theta_{\mathrm R} = 15.02\,\mathrm{K}$).

Solution

Solution of Exercise 10.9.

$S^{\mathrm R}_{\mathrm m} = R[\ln(298.15/15.02) + 1] = 8.3145 \times (2.988 + 1) = 33.16\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$.

Exercise 10.10 ★★★

Treating $J$ as continuous, show that the most populated rotational level of a linear molecule is near $J_{\max} = \sqrt{T/2\theta_{\mathrm R}} - \frac12$. Evaluate it for $\ce{HCl}$ ($\theta_{\mathrm R} = 15.02\,\mathrm{K}$) and $\ce{CO}$ ($\theta_{\mathrm R} = 2.766\,\mathrm{K}$) at $300\,\mathrm{K}$. Why are the two branches of an infrared band strongest a few lines away from the centre?

Solution

Solution of Exercise 10.10.

$\frac{\dd}{\dd J}\bigl[(2J+1)\eu^{-\theta J(J+1)/T}\bigr] = \bigl[2 - (2J+1)^2\theta/T\bigr]\eu^{-\theta J(J+1)/T} = 0$ gives $2J + 1 = \sqrt{2T/\theta}$, i.e. $J_{\max} = \sqrt{T/2\theta} - \frac12$. $\ce{HCl}$: 2.66, so $J = 3$; $\ce{CO}$: 6.86, so $J = 7$. The intensity of a line of the P or R branch follows the population of its lower level, which peaks at $J_{\max}$, a few lines from the gap at the band centre.

Exercise 10.11 ★★★

The Euler–Maclaurin formula gives $\sum_{J\ge0}f(J) \approx \int_0^\infty f(J)\dd J + \frac12f(0) - \frac1{12}f'(0)$. Apply it to $f(J) = (2J+1)\eu^{-\theta J(J+1)/T}$ and show that $q^{\mathrm R} \approx T/\theta + \frac13$. For which temperatures is $T/\theta$ correct to 1 %? Is it acceptable for $\ce{H2}$ ($B_0 = 59.32\,\mathrm{cm}^{-1}$) at $298.15\,\mathrm{K}$?

Solution

Solution of Exercise 10.11.

$\int_0^\infty f\dd J = T/\theta$, $f(0) = 1$, $f'(0) = 2 - \theta/T$. Hence $q^{\mathrm R} \approx T/\theta + \frac12 - \frac16 + \frac{\theta}{12T} \approx T/\theta + \frac13$. The relative error of $T/\theta$ is about $\theta/3T$: below 1 % when $T > 33\,\theta$. For $\ce{H2}$, $\theta = 1.4388 \times 59.32 = 85.3\,\mathrm{K}$: at $298.15\,\mathrm{K}$ the error is 10 %, not acceptable (and the nuclear-spin restriction on $J$ also has to be treated exactly, Chapter 11).

Exercise 10.12 ★★★

Show that the vibrational molar entropy tends to zero when $T \ll \theta_{\mathrm V}$ and to $R[1 + \ln(T/\theta_{\mathrm V})]$ when $T \gg \theta_{\mathrm V}$. Evaluate both the exact expression and this limit for $\ce{I2}$ ($\theta_{\mathrm V} = 306.9\,\mathrm{K}$) at $298.15\,\mathrm{K}$, and comment.

Solution

Solution of Exercise 10.12.

For $x = \theta_{\mathrm V}/T \to \infty$, $x/(\eu^x - 1) \to 0$ and $\ln(1 - \eu^{-x}) \to 0$: $S \to 0$. For $x \to 0$, $x/(\eu^x - 1) \approx 1 - x/2$ and $-\ln(1 - \eu^{-x}) \approx -\ln x + x/2$: $S \to R(1 - \ln x) = R[1 + \ln(T/\theta_{\mathrm V})]$. $\ce{I2}$, $x = 1.029$: exact $R(0.5721 + 0.4420) = 8.43\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$; limit $R(1 - 0.0289) = 8.07\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$. Even at $T \approx \theta_{\mathrm V}$ the classical limit is only 4 % low.

10.6 Problem: Counting the Entropy of Nitrogen

Problem 10.1

Weekend problem — counting the entropy of nitrogen: the translational, rotational, vibrational and electronic contributions to its standard entropy, computed from its spectrum and compared with the table

Nitrogen is the main gas of the air. Its standard molar entropy at $298.15\,\mathrm{K}$ is computed here from its molar mass, $28.014\,\mathrm{g}/\mathrm{mol}$, and from its spectroscopic constants: $B_e = 1.998\,241\,\mathrm{cm}^{-1}$, $\alpha_e = 0.017\,318\,\mathrm{cm}^{-1}$, $\omega_e = 2358.57\,\mathrm{cm}^{-1}$, $\omega_ex_e = 14.324\,\mathrm{cm}^{-1}$; its electronic ground term is ${}^1\Sigma_g^+$. Standard pressure $p^\circ = 1\,\mathrm{bar}$; $k = 1.380\,649 \times 10^{-23}\,\mathrm{J}/\mathrm{K}$, $h = 6.626\,070\,15 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}$.

Part I — Translation.

1. Compute the mass of one molecule.
2. Compute its thermal wavelength at $298.15\,\mathrm{K}$.
3. Compute the volume available per molecule, $kT/p^\circ$.
4. Deduce $q^{\mathrm T}/N$, and comment on its size.
5. Compute the translational molar entropy.
6. By how much would it change at half the pressure?

Part II — Rotation.

7. Compute $B_0$.
8. Compute $\theta_{\mathrm R}$.
9. Give the symmetry number and say what it accounts for.
10. Compute $q^{\mathrm R}$ at $298.15\,\mathrm{K}$.
11. Is the high-temperature form justified?
12. Compute the rotational molar entropy.
13. Ignoring the nuclear-spin weights, which rotational level is the most populated?

Part III — Vibration and electronic states.

14. Compute the vibrational quantum $G(1) - G(0)$.
15. Compute $\theta_{\mathrm V}$.
16. Compute $q^{\mathrm V}$ at $298.15\,\mathrm{K}$.
17. What fraction of the molecules is in $v = 1$?
18. Compute the vibrational molar entropy.
19. What is the electronic contribution? Why can excited electronic states be ignored?

Part IV — Sum and comparison.

20. Add the contributions.
21. Compare with the tabulated $191.609\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$.
22. Which measurements does the calorimetric value of a table rest on?
23. Why can the tables leave nuclear spin out?
24. State the result: the standard molar entropy of nitrogen at $298.15\,\mathrm{K}$ computed from its spectroscopic constants.

Solution

Solution of Problem 10.1.

1. $m = 28.014\times10^{-3}/6.02214\times10^{23} = 4.652 \times 10^{-26}\,\mathrm{kg}$. 2. $\Lambda = h/\sqrt{2\pi mkT} = 1.910 \times 10^{-11}\,\mathrm{m}$ ($19.1\,\mathrm{pm}$). 3. $kT/p^\circ = 1.380649\times10^{-23} \times 298.15/10^5 = 4.116 \times 10^{-26}\,\mathrm{m}^{3}$. 4. $q^{\mathrm T}/N = 4.116\times10^{-26}/(1.910\times10^{-11})^3 = 5.905 \times 10^{6}$: about six million translational states per molecule, so two molecules almost never share one and $Q = q^N/N!$ holds. 5. $S^{\mathrm T}_{\mathrm m} = R(\ln5.905 \times 10^{6} + \frac52) = 8.3145 \times 18.091 = 150.42\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$. 6. $+R\ln2 = 5.76\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$, i.e. $156.18$. 7. $B_0 = 1.998241 - 0.008659 = 1.989\,582\,\mathrm{cm}^{-1}$. 8. $\theta_{\mathrm R} = 1.438777 \times 1.989582 = 2.8626\,\mathrm{K}$. 9. $\sigma = 2$: the two nuclei are identical; turning the molecule end over end gives an indistinguishable orientation, and the exchange symmetry of the nuclei allows, for each nuclear-spin state, only half the rotational levels. 10. $q^{\mathrm R} = 298.15/(2 \times 2.8626) = 52.08$. 11. Yes: $T/\theta_{\mathrm R} = 104$, an error of about 0.3 % on $q^{\mathrm R}$ and much less on the entropy. 12. $S^{\mathrm R}_{\mathrm m} = R(\ln52.08 + 1) = 8.3145 \times 4.9527 = 41.18\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$. 13. $J_{\max} = \sqrt{298.15/5.725} - 0.5 = 6.72$: $J = 7$ (population 0.0839, just above 0.0831 for $J = 6$). 14. $2358.57 - 2 \times 14.324 = 2329.92\,\mathrm{cm}^{-1}$. 15. $\theta_{\mathrm V} = 1.438777 \times 2329.92 = 3352\,\mathrm{K}$. 16. $x = 3352.2/298.15 = 11.24$; $q^{\mathrm V} = 1/(1 - \eu^{-11.24}) = 1.000013$. 17. $\eu^{-11.24}/q^{\mathrm V} = 1.3 \times 10^{-5}$. 18. $S^{\mathrm V}_{\mathrm m} = R[11.24\eu^{-11.24} + \eu^{-11.24}] = 0.0013\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$, negligible. 19. $g_0 = 1$ (${}^1\Sigma$): $R\ln1 = 0$. The excited electronic states lie several electronvolts up, more than a hundred times $kT$: their Boltzmann factors are utterly negligible. 20. $150.42 + 41.18 + 0.00 + 0 = 191.60\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$. 21. The difference is $0.01\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$, at the level of the approximations made (rigid rotor, separation of the motions). 22. Heat capacities of the solid measured from a few kelvins (extrapolated as $T^3$ below), the enthalpies of the solid–solid transition, of fusion and of vaporisation divided by their temperatures, the heat capacity of the gas, and a correction for non-ideality. 23. The nuclei are conserved in every reaction, so their spin entropy is the same on both sides and cancels; calorimetry does not see it either, since the nuclear spins stay disordered in the crystal down to the lowest temperatures measured. 24. $S^\circ(\ce{N2}, 298.15\,\mathrm{K}) = 191.6\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$ from the spectroscopic constants, 150.4 of it from translation and 41.2 from rotation.