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The internal energy is the population-weighted sum of the molecular energy levels:At fixed volume, assume the energy levels themselves are temperature-independent. Differentiating the molecular partition function givesChoosing the zero of energy changes $U$ by the corresponding constant per molecule, so the ground-level convention must be consistent throughout.

For identical, non-interacting molecules, the translational canonical partition function is $Q_N=q_{\mathrm{trans}}^N/N!$. Stirling's approximation gives the Helmholtz free energyAt fixed $V$, $q_{\mathrm{trans}}\propto T^{3/2}$, so differentiation gives $S=Nk[\log (q_{\mathrm{trans}}/N)+5/2]$. For one mole at the standard-state molar volume,Here the quoted molar translational quantity is per mole, and its ratio to $N_A$ is dimensionless. The derivative defining entropy is taken at fixed volume before inserting the standard-state volume; differentiating along a constant-pressure path would give the wrong additive constant.

The primed $f_{\mathrm{TS}}^{\prime }$ is the corresponding transition state partition function per unit volume with the unstable reaction-coordinate degree of freedom omitted. Its replacement by the crossing-frequency factor $kT/h$ is already outside the partition-function ratio. For a diatomic atom-pair transition state, the omitted coordinate is the bond stretch, leaving translation and rotation.

The quantities $f_A$ and $f_B$ are molecular partition functions per unit volume for the two reactants, with their internal rotational, vibrational and electronic factors included. Thus their translational factors are $(2\pi m_{j}kT/h^2{)}^{3/2}$, rather than those factors multiplied by $V$. Their energy zeros must match the separately displayed barrier convention.

For identical atoms, use $\mu =m_A/2$ and the homonuclear rotational symmetry number $\sigma =2$. With a rate defined as reaction events per volume and time, this halves the geometric prefactor relative to distinguishable labelled pairs. Do not independently count both ordered pairs and then add another unrelated factor of two. If the rate is instead defined by disappearance of A, each event consumes two atoms, so $-d{n}_A/dt=2r$; that rate coefficient is twice the event-rate coefficient.

Write $M=m_A+m_B$ and $\mu =m_A{m}_B/M$. An atom has no rotational or vibrational internal factor under the stated electronic assumption. The transition-state moment of inertia is $I=\mu R^2$, so the high-temperature rigid rotor result isThe translational factors therefore giveSubstituting into the transition state theory expression and simplifying,For distinct atoms, $\sigma =1$. This is a number-density rate coefficient, with units of volume per time. A rate law in molar concentrations uses $N_A{k}_{2nd}$. The derivation assumes a classical high-temperature rotor at the dividing surface and a transmission coefficient of one.

The Fermi sphere volume in reciprocal space isThe primitive real-space cell has volume $a^3/2$, so the first Brillouin zone has volumeEach band accommodates two electrons per primitive cell when spin is counted. Six electrons therefore require three such zone-volumes of occupied reciprocal-space states in the free-electron counting picture. The extended-zone sphere is not confined to the first zone; folding it produces multiple bands. It does not imply that the actual tungsten Fermi surface is spherical.

At fixed particle number and pressure, the joint isothermal-isobaric ensemble weight for a microstate of energy $E$ and volume $V$ is $e^{-\beta (E+PV)}$. Write its enthalpy as $H=E+PV$. Differentiating the isothermal-isobaric partition function with a temperature-independent volume measure givesAfter division by $Q_P$ this is the normalized ensemble average, soThe differentiation assumes that the microscopic Hamiltonian and the integration measure have no additional explicit temperature dependence. An extra temperature-dependent normalization factor must be differentiated separately.

Differentiate the normalized average, including the derivative of its denominator. The enthalpy fluctuation formula follows fromSince $d\beta /dT=-1/(k_B{T}^2)$,Thus a larger equilibrium spread of enthalpy corresponds to a larger response to heating at fixed pressure. The variance is nonnegative, so this finite-system equilibrium description has $C_P\ge 0$. The relation connects a response coefficient to fluctuations of the appropriate exchanged quantity; at fixed volume the corresponding quantity is internal energy.

The volume integral is a gamma function integral,For the prescribed bridge relation, $\beta P{Q}_P=(\beta P{\Lambda }^3{)}^{-N}$, so $G=N{k}_{B}T\ln (\beta P{\Lambda }^3)$. Since $\Lambda \propto T^{-1/2}$, this gives the macroscopic ideal-gas resultHeating at fixed pressure requires both an increase in internal energy and expansion work; at fixed volume the latter contribution is absent.

The finite-size normalization of the isothermal-isobaric ensemble matters here. Applying part (a) literally to the unprefactored $Q_P$ instead gives $⟨H{⟩}_P=(5N/2+1)k_{B}T$ and $C_P=(5N/2+1)k_B$. The two answers differ because $-{\partial }_{\beta }\ln (\beta P{Q}_P)=-{\partial }_{\beta }\ln Q_P-k_{B}T$. The normalized continuous-volume distribution has $⟨PV⟩=(N+1)k_{B}T$; its variance agrees with the unprefactored result in part (b). The bridge convention subtracts an order-one contribution and recovers the usual extensive thermodynamic answer. Both agree to leading order in the thermodynamic limit, but their exact finite-$N$ derivatives should not be mixed.

Excited $\mathrm{O}({}^{1}D)$ has two competing fates. Collisional deactivation returns ground-state oxygen, which reforms ozone rapidly; reaction with water instead gives net ozone loss and two OH radicals. The chemical branching fraction for the latter isThus only about 4.92% of these ozone photolyses directly lead to the specified net chemical loss; treating every photodissociation as permanent removal would overestimate the sink. The corresponding first-order rate is $J_1{f}_w=9.8361\times 10^{-7}{\mathrm{s}}^{-1}$. Converting the given hydroperoxyl mixing ratio to a concentration gives $[\mathrm{H}{\mathrm{O}}_2]=20\times 10^{-12}[\mathrm{M}]=4.4\times 10^8\mathrm{c}{\mathrm{m}}^{-3}$, so $k_4[\mathrm{H}{\mathrm{O}}_2]=9.6800\times 10^{-7}{\mathrm{s}}^{-1}$.

For a well-mixed marine atmospheric boundary layer of depth $H=2\times 10^5\mathrm{cm}$, the downward surface flux is $v_{\mathrm{dep}}[{\mathrm{O}}_3]$. Dividing by the layer volume per unit area gives a first-order dry deposition rate $v_{\mathrm{dep}}/H=7.5\times 10^{-7}{\mathrm{s}}^{-1}$. The air density cancels on expressing the entire balance in ozone mixing ratio $\chi$:At steady state, with $P_{\chi }=0.25/3600\mathrm{ppb}{\mathrm{s}}^{-1}$,The effective removal time is about 4.28 days. This calculation assumes a uniform layer of the diagnosed depth, fixed chemical conditions and a local time-independent balance.

The high-pressure lattice constant follows from its first $(100)$ reflection:Conserve the number of NaCl formula units. The low-pressure conventional unit cell contains four, whereas the high-pressure cell contains one. The volume per formula unit changes from $a_0^3/4$ to $a_P^3$, soThe sample contracts by about $39.5\%$. Comparing $a_P^3$ directly with $a_0^3$ would compare different amounts of material.

The Debye model treats a three-dimensional solid as an isotropic elastic continuum with three independent acoustic phonon polarizations. It assumes harmonic normal modes, a linear dispersion relation $\omega =vq$, and equilibrium Bose-Einstein statistics with zero phonon chemical potential. Real anisotropy and nonlinear dispersion are replaced by one effective Debye sound speed. A cutoff restores the finite number of atomic vibrational degrees of freedom; the continuum itself would otherwise have infinitely many modes.

For sample volume $V$, the wavevector state density is $V/(2\pi {)}^3$ per polarization. Counting a sphere givesThere are $3N$ modes for $N$ atoms. HenceThe associated phonon density of states isIf longitudinal and transverse sound speeds differ, use $3/v^3=1/v_L^3+2/v_T^3$ for the low-frequency count. The Debye frequency is an effective mode-counting cutoff, not a claim that every actual branch terminates at exactly that frequency.

Assume thermal equilibrium, nondegenerate carriers, and isotropic parabolic band dispersions, with the supplied effective masses interpreted as density-of-states masses. Take one effective valley with two electron spin states for each band; any additional valley multiplicity must instead be absorbed into the masses or explicit degeneracy factors.

Let the conduction band edge be $E_c$, the valence band edge $E_v$, and $E_g=E_c-E_v$. The spin-summed conduction density of states per volume isUsing the nondegenerate limit of the Fermi-Dirac distribution and the substitution $x=(E-E_c)/(k_{B}T)$ givesThe factor $\sqrt{\pi }/2$ in the supplied integral produces the normalization. Counting missing valence band electrons similarly givesMultiplication cancels the chemical potential, establishing the semiconductor mass-action law:The result does not require an intrinsic semiconductor: semiconductor doping changes $n$ and $p$ separately while preserving this product under the stated equilibrium and nondegeneracy assumptions.

A Bravais lattice is an infinite periodic set of equivalent points $\mathbf{R}=n_1{\mathbf{a}}_1+n_2{\mathbf{a}}_2+n_3{\mathbf{a}}_3$, with integer $n_i$ and independent primitive translations ${\mathbf{a}}_i$. It describes translation symmetry, without assigning a chemical species to a point. An atomic basis of a crystal is the motif of atoms, with their species and relative positions, repeated identically at every lattice point. The resulting crystal structure consists of the lattice decorated by that basis; the basis can contain several atoms even when the lattice is primitive.

A primitive unit cell tiles space under lattice translations and contains exactly one lattice point, counting shared boundary points fractionally. Its volume is $|{\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)|$. A conventional unit cell is chosen to display useful crystal symmetry and may contain several lattice points. Thus a centered conventional cell is not primitive, but the crystal still has a primitive cell. Neither cell size nor the number of dots drawn on its boundary should be confused with the number of atoms in the crystal basis.

After compression along $z$, let the original conventional edges be $a,a,c$ with $c=a/\sqrt{2}$. This gives exactly the specified volume reduction. The compressed FCC points admit the alternative translation vectorsThey are mutually perpendicular, have equal length $b=a/\sqrt{2}$, and form a right-handed cube. Besides its corners, the lattice has the pointwhich is the compressed image of an original FCC face centre. All compressed FCC sites are generated by the cube's integer translations and this body-centre coset. ThereforeThe old compressed tetragonal conventional cell has volume $a^3/\sqrt{2}$ and four sites; the new BCC cube has half that volume and two sites. This verifies the same physical site density rather than silently losing atoms during the change of cell. The continuous uniaxial deformation is the Bain path, named for Edgar Bain.

In the spin-degenerate free electron model, a monovalent metal has conduction-electron number density $n$. The density of wave-vector states is $1/(2\pi {)}^3$ per unit real-space volume, with two spin states at each wave vector. Filling a sphere at zero temperature givesFor the Debye model, there are three acoustic polarizations and $3n$ modes per unit volume in total. Replacing their dispersions by $\omega =ck$ with the effective Debye sound speed, mode counting givesThe factor two in the electron count is spin, whereas the factor three in the phonon count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.

Dividing gives $T_F/{\theta }_D=\lambda (3{\pi }^{2}n/2{)}^{1/3}$, where $\lambda =\hslash /(2m_{e}c)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. ThereforeHere and below $c$ is the speed of sound, not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the Brillouin zone; it is not an electron-state cutoff.

The SI volume magnetic susceptibility is the weak-field response ${\chi }_M=\partial M/\partial H$, or $M/H$ for a linear isotropic sample with no spontaneous moment. It is dimensionless; a molar susceptibility would be a different quantity. For weak magnetization, $B\simeq {\mu }_{0}H$. Neglect local-field and demagnetizing corrections in this independent-spin model.

A spin-$1/2$ ion has moment projections $\pm g{\mu }_B/2$. The requested coefficient assumes a spin-only $g\simeq 2$, giving projections $\pm {\mu }_B$ and Zeeman energies $\mp {\mu }_{B}B$. With $x={\mu }_{B}B/(k_{B}T)$, the one-ion partition function and magnetization areFor $|x|\ll 1$, $\tanh x\simeq x$, soHere $N$ is a number density, as specified, rather than total ion number. The small-field condition ismeaning the Zeeman splitting $2{\mu }_{B}B$ is small compared with the thermal scale up to an irrelevant order-one convention. It is not a temperature-independent absolute upper bound on $B$. For a different $g$ value the coefficient contains $(g{\mu }_B/2{)}^2$ instead of ${\mu }_B^2$.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

Pauli exclusion principle prevents most filled electronic states from changing occupation at $T\ll T_F$. Only a shell of width of order $k_{B}T$ around the Fermi energy participates. Since the density of states near that energy is of order $n/E_F$ per volume and energy, the fraction of active electrons is of order $k_{B}T/E_F=T/T_F$. Their typical change of energy is of order $k_{B}T$, so the thermal energy increment per volume scales asDifferentiating with respect to temperature gives the electronic heat capacityThe simple shell argument fixes the linear law, not its coefficient. A Sommerfeld expansion using the spin-summed free-electron density of states $g(E_F)=3n/(2E_F)$ gives $C_e=({\pi }^2/3)g(E_F)k_B^{2}T$, hence $A={\pi }^2/2$, consistent with the supplied value. The result excludes strong band-structure or interaction corrections and requires a degenerate electron population.