Past exam of the physics B course of the University of Cambridge / 2016 / Paper 2 (Physics B)

The Boltzmann entropy isThe thermodynamic definition of temperature givesHolding the other system parameters fixed, $E=(5N/2)k_{B}T$, so its heat capacity is $C=5N{k}_B/2$. The constants ${\Omega }_0$ and $E_0$ set an additive reference for entropy; neither affects this derivative or the resulting heat capacity.

For a lossless transmission line with characteristic impedance $Z_0$ and resistive termination $R_L$, the voltage reflection coefficient isThe reflected pulse is inverted and has half the incident voltage, so $\Gamma =-1/2$. Substituting the termination givesThe echo delay is a round trip from the observation point to the termination and back, not a one-way travel time. Thus the propagation speed isFor a homogeneous, nonmagnetic coaxial cable dielectric, $v=c/\sqrt{{ϵ}_r}$, giving ${ϵ}_r\simeq 2.25$. More generally the measured speed fixes ${ϵ}_r{\mu }_r=(c/v{)}^2$; identifying this with relative permittivity assumes ${\mu }_r\simeq 1$ and negligible dispersion or attenuation of the pulse.

Separate a ball's orbital motion from its spin. For a rolling ball between concentric races, the ball centre lies $2a$ from the axle axis. Let its tangential centre velocity be $v$ in the direction of axle motion and let its signed spin angular velocity be $\Omega$, positive in the same sense as $\omega$. Rolling without slipping at the moving inner contact and stationary outer contact requiresHenceThe spin is opposite to the axle's rotation, while the centre orbits in the same sense. The total angular momentum about the axle is the sum of its orbital angular momentum and its angular momentum about the centre of mass:For a uniform solid spherical ball, $I_{\mathrm{cm}}=2m{a}^2/5$, soThe solid-ball assumption supplies the usual moment of inertia; the preceding expression in terms of $I_{\mathrm{cm}}$ also states the answer for a different internal mass distribution.

In the thin-gap Couette flow approximation, the tangential velocity varies linearly through the gap. At radius $r$, its difference between the two discs is $r\omega$, so a Newtonian fluid exerts shear stressAn annulus has area $2\pi rdr$, and its tangential force acts at lever arm $r$. Integrating its torque givesThe driving torque is in the direction of rotation; the liquid's resisting torque has the opposite sign. This parallel-disc viscometer result assumes laminar shear with no slip, negligible edge effects and negligible secondary circulation. Its fourth-power radius dependence comes from the combined radial dependence of speed, annular area and lever arm.

At fixed temperature, equilibrium blackbody radiation has fixed energy density $u$, with radiation pressure $p=u/3$. Doubling the volume therefore doubles its internal energy, so $\Delta U=U_0$. The work done by the radiation isThe first law of thermodynamics, with positive work done by the radiation, givesThus $1.00\mathrm{mJ}$ is absorbed from the walls. The walls must replace the energy used to do work as well as supply the extra stored energy; using only $\Delta U=750\mu \mathrm{J}$ would omit that work. This is a reversible isothermal expansion of a photon gas, not an adiabatic expansion.

An electric dipole consists of equal and opposite electric charges separated by a displacement $\mathbf{d}$. Its electric dipole moment is $\mathbf{p}=q\mathbf{d}$, directed from negative to positive electric charge. At distances large compared with the separation, its leading electric potential and electric field areThere is no monopole term because the net electric charge is zero. The axial electric field points along $\mathbf{p}$ with magnitude $2p/(4\pi {ϵ}_0{r}^3)$, while the equatorial electric field points oppositely with half that magnitude.

A uniform applied electric field exerts zero net force on an ideal fixed electric dipole, but it exerts torque $\mathbf{\tau }=\mathbf{p}\times \mathbf{E}$. Its orientation-dependent potential energy is $U=-\mathbf{p}\cdot \mathbf{E}$, so alignment with the electric field is stable. A nonuniform applied electric field can also exert a net force, $\mathbf{F}=\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed permanent electric dipole moment.

A permanent electric dipole moment exists without an applied electric field. An induced electric dipole moment instead develops in response to the local electric field; in an isotropic linear model $\mathbf{p}=\alpha \mathbf{E}$. Creating that induced moment requires accounting for the work of polarization, so its mechanical potential energy in a specified applied electric field is $-\alpha E^2/2$, rather than $-\alpha E^2$. In matter, the dipole moment per unit volume is the polarization density $\mathbf{P}$.

Mutual inductance expresses how a changing electric current in one circuit induces an electromotive force in another. Choose winding orientations and define their linked magnetic fluxes byFor a reciprocal linear magnetic system, the same $M$ appears in both expressions. Faraday's law of induction gives ${\mathcal{E}}_2=-Md{I}_1/dt$ when the secondary electric current does not itself change. The sign depends on winding orientation and is consistent with Lenz's law.

The stored magnetic energy isIts nonnegativity for arbitrary electric currents requires $M^2\le L_1{L}_2$. Strong coupling makes $|M|$ approach $\sqrt{L_1{L}_2}$; flux leakage makes it smaller.

An ideal electrical transformer has two windings linked by the same time-varying core magnetic flux. The induced winding voltages are proportional to their turn counts. For appropriately chosen terminal polarities,The electric current convention in the power relation takes primary power into the device and secondary power out to the load. Consequently a load electrical impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_1/N_2{)}^2{Z}_L$. A larger secondary voltage therefore comes with a smaller available secondary electric current, rather than extra energy.

A steady direct electric current produces no sustained transformer action after its switching transient. Practical departures from the ideal include winding electrical resistance, leakage magnetic flux, finite magnetizing electric current, core saturation, eddy currents and magnetic hysteresis losses.

Free charge is charge assigned to externally supplied or mobile carriers in a macroscopic description. Bound charge describes the net charge associated with a material's polarization density $\mathbf{P}$. Its bulk and surface densities arewhere the normal points out of the polarized material. Uniform bulk polarization density has no bulk bound charge, but it can still have surface bound charge.

The electric field is sourced by the total electric charge. Introducing the electric displacement field separates the free source in the macroscopic field equation:This does not mean that bound charges are physically absent from $\mathbf{D}$: they are represented through $\mathbf{P}$. In a linear isotropic dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and $\mathbf{D}=ϵ\mathbf{E}$, with $ϵ={ϵ}_0(1+{\chi }_e)$.

Across a stationary ordinary dielectric interface, electrostatics gives continuity of tangential electric field, while a free sheet charge givesThe normal electric field need not be continuous. For a fully filled parallel-plate capacitor at fixed free plate charge, polarization reduces the internal electric field by ${ϵ}_r$ and increases capacitance by the same factor. If a battery instead holds the voltage fixed, extra free plate charge flows from the battery; the constraint must be stated before comparing stored energies.^[ref]

A permanent magnet retains a substantial magnetization without an applied coil electric current. In a ferromagnetic material this reflects microscopic magnetic moments, magnetic domains and their resistance to rearrangement. The history-dependent relation between applied field and magnetization gives magnetic hysteresis, remanence and coercivity. A soft core used in an electromagnet or electrical transformer is chosen for easy magnetization and small hysteretic loss; a permanent-magnet material must resist demagnetization.^[ref]

The macroscopic fields obeyin magnetostatics. The magnetization can be represented by equivalent bound currents,Thus both permanent magnetic matter and free coil electric currents generate a magnetic field. Its magnetic field lines form continuous loops; apparent north and south poles are not isolated magnetic charges.

In a long ideal solenoid, $H\simeq nI$. A linear core would give $B\simeq {\mu }_0{\mu }_{r}nI$, until saturation or other nonlinear effects invalidate a constant magnetic permeability. In a finite magnetic circuit, air gaps and demagnetizing fields affect the field strongly: because the gap has low magnetic permeability, it often needs most of the available magnetomotive force. A permanent magnet can have a nonzero internal $\mathbf{H}$ even when ${\mathbf{J}}_f=0$, because its boundaries create a demagnetizing field. Hence setting $\mathbf{H}=0$ merely because there is no free coil electric current is generally incorrect.

An electrostatic field calculation begins by specifying the charge information, material response and boundary conditions. In vacuum, prescribed volume and surface electric charge densities can be superposed directly using Coulomb's law. For a volume distribution,Differentiating this electric potential gives the electric field. This is useful for known source distributions, but in the presence of conductors the induced surface electric charge is usually an unknown rather than an independently specified input.

Gauss's law is especially effective when symmetry fixes the direction and magnitude of the electric field on a suitable closed surface. A spherical charge distribution gives a radial field, an infinitely long uniform line gives a cylindrical field, and a uniform infinite sheet gives a planar field. In an arbitrary geometry, the flux law remains true but does not alone determine the pointwise electric field. A multipole expansion is another useful approximation: far from a localized charge distribution, successive monopole, dipole and higher terms separate the effects of total charge and its spatial structure.

For a boundary-value formulation, use the electrostatic equationsThey imply Poisson equation ${\nabla }^2\varphi =-\rho /{ϵ}_0$, or Laplace equation where $\rho =0$. A conductor in electrostatic equilibrium is an equipotential with zero internal electric field. Its exterior normal electric field fixes the surface electric charge through $E_n=\sigma /{ϵ}_0$. One must specify whether each conductor is grounded, maintained at a prescribed voltage, or isolated with prescribed total electric charge.

Uniqueness of Poisson equation justifies methods that construct a convenient candidate solution. With fixed boundary electric potential and suitable behaviour at infinity, a solution satisfying the field equation is the solution. With normal-derivative boundary conditions, the source and flux must be compatible, and the electric potential is determined only up to an additive constant.

For example, a point charge $q$ a distance $h$ above a grounded infinite plane has the same upper-half-space electric potential as $q$ and a mirrored image charge $-q$ below it. Their potentials cancel on the plane. The image is a mathematical source in the excluded region, not a second physical charge. Separation of variables similarly constructs solutions suited to rectangular, cylindrical or spherical boundaries; matching coefficients enforces the prescribed boundary data.

In dielectrics, the distinction between total and free charge is essential. In a linear isotropic material,Only where permittivity is constant may this be reduced to ${\nabla }^2\varphi =-{\rho }_f/ϵ$. At an ordinary interface, $\varphi$ and the tangential electric field are continuous, while the jump in normal electric displacement field is the free surface charge. Applying Gauss's law to total charge with ${ϵ}_0$, or its displacement-field form to free charge, gives consistent descriptions; mixing their sources and permittivities does not.

A useful interface example is the image charge at a dielectric interface. If $q$ lies in medium 1, the source-side potential is represented bywhere $r_1$ is distance to the real charge and $r_2$ to its mirrored position. The other-side potential can be written ${\varphi }_2=q_t/(4\pi {ϵ}_2{r}_1)$, where $q_t=2{ϵ}_{2}q/({ϵ}_1+{ϵ}_2)$. Continuity of $\varphi$ and of normal $\mathbf{D}$ fixes both coefficients. The equal-permittivity limit gives no image; the conductor limit ${ϵ}_2\to \infty$ gives $q^{\prime }\to -q$.

For a sphere of radius $a$ and relative permittivity ${ϵ}_r$ in a uniform external electric field ${\mathbf{E}}_0$, matching the spherical solutions gives a uniform internal electric field and an external induced electric dipole moment:This illustrates how a material boundary modifies an otherwise simple applied electric field.

For complex boundaries or spatially varying materials, finite difference methods or the finite element method solve the same boundary value problem numerically. A sound calculation checks its boundary conditions, charge or flux balance and convergence with mesh refinement. The governing equations, source classification and uniqueness argument remain the organizing principles even when an analytic formula is unavailable.

A horizontal pressure gradient force accelerates air from higher toward lower pressure. In a rotating frame the Coriolis force deflects the flow. Away from strong friction, at large spatial scales and small Rossby number, a slowly changing flow can reach geostrophic balance:where $f=2\Omega \sin \phi$ is the Coriolis parameter. The geostrophic wind runs along atmospheric isobars, with speed $|{\nabla }_{h}p|/(\rho |f|)$. Closely spaced atmospheric isobars therefore indicate stronger geostrophic wind when density and latitude are comparable. In the northern hemisphere lower pressure lies to the left of this wind, producing counterclockwise circulation around a low; the sense reverses in the southern hemisphere.

A curved flow also has centripetal acceleration. For cyclonic circular flow in the northern hemisphere, the gradient wind relation isNear the ground, friction reduces wind speed and weakens its Coriolis contribution, so air crosses atmospheric isobars toward lower pressure. The simple geostrophic formula fails near the equator, where $f$ becomes small, and where rapid changes or tight curvature make neglected acceleration important. Vertically, large-scale motion is often approximated by hydrostatic balance, $dp/dz=-\rho g$.

Bernoulli's principle can relate speed changes to pressure changes along an appropriate inviscid streamline, but its simple dynamic-pressure relation is not a replacement for the rotating force balance governing a broad atmospheric pressure map.

Coupled oscillators have more than one collective pattern of motion. For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and linked to each other by stiffness $\kappa$, small displacements obeyHere equal linear damping $b$ is included to make finite steady-state responses at resonance. The normal mode coordinates $q_{+}=(x_1+x_2)/\sqrt{2}$ and $q_{-}=(x_1-x_2)/\sqrt{2}$ diagonalize these equations. Their natural angular frequencies areThe symmetric normal mode leaves the coupling spring unextended; the antisymmetric normal mode stretches it and therefore has the higher frequency. With $F_{\pm }=(F_1\pm F_2)/\sqrt{2}$, their complex steady-state responses areThus the drive's spatial pattern matters as well as its frequency. Equal in-phase forces drive only the symmetric normal mode; opposite forces drive only the antisymmetric one; driving a single mass generally excites both. Finite damping broadens the resonance peaks and produces a phase lag. In the ideal undamped system, a resonant drive with nonzero projection onto a normal mode produces secular growth rather than a bounded resonant steady state.

The displacement of a particular mass adds contributions from both normal modes. They can cancel at an antiresonance. For $F_2=0$ and no damping, $x_1$ vanishes at ${\Omega }^2=(k+\kappa )/m$, while the partner still moves. This differs from the two nearby resonances, where a mode responds strongly; a small measured response need not mean that no motion or stored energy exists elsewhere.

A Newtonian fluid exerts viscous shear stress proportional to the rate of strain. For an incompressible fluid of constant dynamic viscosity $\eta$,The local momentum balance is the Navier-Stokes equation:The left side is density times material acceleration. A steady velocity field can still have convective acceleration when a moving parcel changes speed or direction; setting $\partial \mathbf{u}/\partial t=0$ does not eliminate it.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscous effects. At low Reynolds number, the fluid's own inertia can often be neglected, giving Stokes flow. This does not say that every immersed body has zero acceleration: its mass and the net pressure, viscous and body forces still enter its own equation of motion.

For a small rigid sphere moving through a slowly varying viscous background, Stokes drag is $-6\pi \eta a(\mathbf{v}-\mathbf{u})$. In still fluid, a simple settling model includes weight, buoyancy and drag:Ignoring fluid added mass and unsteady history forces, this gives relaxation time $m/(6\pi \eta a)$ and terminal velocityAt the terminal velocity the net force and acceleration vanish, while viscous drag and gravity remain individually nonzero. The dissipated power $6\pi \eta a{v}_t^2$ is supplied by the loss of gravitational potential energy relative to buoyant support. The creeping-flow and quasisteady-drag assumptions must be checked before applying this expression at large speed or immediately after a rapid start.

For two bodies interacting through a potential energy that depends only on their separation, introduce the centre of mass coordinate and the two-particle relative coordinate:Their kinetic energy separates exactly:The centre of mass moves uniformly if there is no external force, while the internal two-body problem reduces to one effective particle of reduced mass $\mu$:The original trajectories follow from ${\mathbf{r}}_1=\mathbf{R}+(m_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=\mathbf{R}-(m_1/M)\mathbf{r}$.

For Newtonian gravity, $U=-G{m}_1{m}_2/r$, so the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Conserved relative angular momentum and energy then give the usual conic sections. For a bound orbit of relative semimajor axis $a$,The two bodies share this orbital period and orbit their common centre of mass with semimajor axes in inverse proportion to their masses. If $m_2\gg m_1$, then $\mu \simeq m_1$, recovering the fixed-heavy-centre approximation. Using $M$ as the internal inertial mass would confuse centre-of-mass translation with relative motion.

Elasticity describes recoverable deformation. For small displacements $\mathbf{u}(\mathbf{x})$, the infinitesimal strain tensor is the symmetric part of the displacement gradient,Its diagonal entries describe extension, while its off-diagonal entries describe shear. The engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$; the antisymmetric part of the displacement gradient is a small rigid rotation, not elastic strain.

The stress tensor specifies the force per unit area on each oriented plane: $\mathbf{t}(\hat{\mathbf{n}})=\mathbf{\sigma }\hat{\mathbf{n}}$. Positive normal stress is tensile in the convention used here; tangential traction gives shear stress. In the absence of internal couples, angular-momentum balance makes the stress tensor symmetric. Force balance gives $\nabla \cdot \mathbf{\sigma }+\mathbf{f}=0$ in static equilibrium, where $\mathbf{f}$ is body force per unit volume.

Linear elasticity relates stress and strain. A homogeneous isotropic solid needs two independent elastic constants:Here $G$ is the shear modulus and $\lambda$ is the other Lamé parameter. For uniaxial loading, Young's modulus is $E={\sigma }_{xx}/{ϵ}_{xx}$, with lateral contraction described by Poisson's ratio $\nu =-{ϵ}_{yy}/{ϵ}_{xx}$. The bulk modulus is $K=-p/(\Delta V/V)$ for a small hydrostatic compression. The relations between these constants areStable isotropic elastic strain energy density requires $G>0$ and $K>0$, implying $-1<\nu <1/2$. The incompressible limit is approached as $\nu \to 1/2$. Anisotropic crystals generally need more constants; the isotropic relations should not be imposed on arbitrary directions in such a crystal.

The elastic energy density is $w=\mathbf{\sigma }:\mathbf{ϵ}/2$. In the isotropic linear model this becomes $\lambda (\mathrm{tr}\mathbf{ϵ}{)}^2/2+G\mathbf{ϵ}:\mathbf{ϵ}$. The quadratic form explains why reversible loading stores energy and unloading can return it, unlike plastic deformation or a dissipative hysteresis cycle.

For a slender beam described by the Euler-Bernoulli beam equation, plane sections approximately remain plane. A fibre a distance $y$ from the neutral axis has axial strain $-y\kappa$, where $\kappa$ is curvature, and stress $-Ey\kappa$. Integrating its moment about the neutral axis gives, with consistent signs,$I_A$ is a second moment of area, with units of length to the fourth power; it is not a mass moment of inertia. For a rectangular section of width $b$ and thickness $h$, $I_A=b{h}^3/12$, explaining the strong dependence of bending stiffness on thickness. For small transverse deflection $w(x)$, curvature is approximately $w^{\prime \prime }$, and the Euler-Bernoulli beam equation is $E{I}_A{w}^{\prime \prime \prime \prime }=q(x)$ with a suitable load-sign convention. A cantilever of length $\ell$ under end load $F$ has tip deflection $F{\ell }^3/(3E{I}_A)$ and largest bending stress of magnitude $F\ell c/I_A$, where $c$ is the furthest distance from the neutral axis.

In torsion of a solid circular shaft, a cross-section rotated by angle $\theta (x)$ has shear strain $r{\theta }^{\prime }(x)$. ThereforeHere $J$ is the polar second moment of area. A uniform shaft of length $\ell$ under constant torque twists by $\mathcal{T}\ell /(GJ)$, and its largest shear stress is $\mathcal{T}a/J$. Noncircular sections can warp, so replacing their torsional stiffness by $G$ times their polar area moment is not generally valid. Both bending and torsion formulas presume small, elastic deformations; exceeding the elastic limit can produce permanent deformation or fracture.

A Carnot cycle consists of two reversible isothermal branches joined by two reversible adiabatic branches. Let the hot isothermal expansion run from volume $V_a$ to $V_b$, and the cold isothermal compression from $V_c$ to $V_d$. For $n$ moles of an ideal gas, their absorbed and rejected heat magnitudes areOn a reversible ideal-gas adiabat, $T{V}^{\gamma -1}$ is constant. Applying this to the two joining branches gives $V_b/V_a=V_c/V_d$. ThereforeThe result is independent of the gas's heat-capacity ratio. This derivation presumes the usual constant-heat-capacity ideal-gas model; more generally the equal entropy changes on the two isotherms give the same heat ratio without that restriction.

To establish the upper bound, suppose an engine were more efficient between these same two reservoirs. Use its output work to drive a reversed Carnot cycle, scaled so that the work magnitudes match. For fixed work $W$, the hypothetical engine would take $W/\eta$ from the hot bath, whereas the reversed Carnot cycle would return $W/{\eta }_C>W/\eta$ to it. The combined device would transfer heat from the cold bath to the hot bath without any net external work. That contradicts the Clausius statement of the second law. Hence no such two-reservoir engine can exceed $1-T_2/T_1$.

For the triangular cycle, call the three successive vertices $A,B,C$, and measure energies in units of $p_0{V}_0$. For one mole of a monatomic ideal gas, $U=3pV/2$, and positive work done by the gas is $W=\int pdV$. The net values on the three complete branches areThus the net work is $W_{\mathrm{cycle}}=p_0{V}_0/2$, equal to the triangle's enclosed area.

There is a small but important distinction between the net heat on a branch and its total positive heat input to a cycle. On the sloping branch put $x=V/V_0$, so $p/p_0=3-x$, with $1\le x\le 2$. ThenThere is a heat-flow reversal along a thermodynamic path at $x=15/8$. The first part absorbs heat; the final part rejects it. Integrating the two signed contributions separately givesThe true total absorbed and rejected heats and thermal efficiency are thereforeFor clarity, the positive heat input on $A\to B$ is $3p_0{V}_0/2$, on $B\to C$ it is $49p_0{V}_0/32$, and on $C\to A$ it is zero. Using just the positive net heats of the complete branches would give $Q_{\mathrm{in}}=3p_0{V}_0$ and $\eta =1/6$; that shortcut misses the short heat-rejection interval on $B\to C$.

Finally, put $T_0=p_0{V}_0/R$. The minimum temperature is $T_0$ at $A$. Along $B\to C$,which has its maximum $9/4$ at $x=3/2$, not at either endpoint. Thus the corresponding two-reservoir Carnot efficiency isThe triangular cycle receives heat at a range of temperatures below $T_{\max }$ and rejects it at temperatures above $T_{\min }$. It therefore makes less work per unit input heat than a Carnot cycle using only those extrema. The geometric path alone does not prove irreversibility: it could in principle exchange heat reversibly with a continuum of matched-temperature reservoirs, while still having this lower efficiency. Restricting it to only the two extreme-temperature reservoirs would introduce additional irreversible heat transfers.

The classical equipartition theorem assigns mean energy $k_{B}T/2$ to each independent quadratic term in the Hamiltonian, provided the canonical distribution and its relevant boundary integrals exist. For a quadratic term $a{q}^2$, the Boltzmann distribution is proportional to $e^{-\beta a{q}^2}$, with $\beta =1/(k_{B}T)$. Integrating a total derivative givesAfter dividing by the normalization,The same argument applies to quadratic momentum terms. Other independent canonical coordinates factor out of the ratio. It is a classical statement: widely spaced quantum energy levels can freeze out a contribution at low temperature.

The confining quartic potential energy $U(x)=Q{x}^4$, with $Q>0$, is not quadratic, so the $k_{B}T/2$ rule cannot be assigned to it. This does not invalidate the canonical Boltzmann distribution or its generalized integration-by-parts identity. Indeed,Hence equipartition for a homogeneous potential givesEquivalently, $Z_x=\Gamma (1/4)(\beta Q{)}^{-1/4}/2$ and $-\partial \ln Z_x/\partial \beta =1/(4\beta )$. This is the potential contribution only; an additional standard kinetic term $p^2/(2m)$ would contribute a further $k_{B}T/2$ to energy and $k_B/2$ to heat capacity.

For an intact diatomic molecule, the classical degrees of freedom comprise three translational quadratic terms, two rotational quadratic terms, and two terms for its one bond-stretching vibration: one kinetic and one potential. Thus complete classical equipartition predictsThe heat capacity of a diatomic gas has lower-temperature regimes because rotational and vibrational levels are quantized. In a dilute molecular gas, the expected progression isThe rotational energy scale is normally much smaller than the vibrational scale, so these steps are separated. The high-temperature value treats the vibration as harmonic and excludes dissociation and electronic excitation; the lowest-temperature gas regime is meaningful only while the sample remains dilute and gaseous.^[ref]

The plotted example uses a quantum rigid rotor and harmonic vibration, with an illustrative ratio of characteristic temperatures. It is a model calculation, not measured data for a particular molecular species. For the vibrational part,which tends to zero on freezing out and to one when many vibrational levels are accessible. The classical equipartition theorem is recovered when the relevant level spacings are small compared with $k_{B}T$.

For the two unlabelled follow-up requests, distinguish re-equilibrating effusion from collisionless effusion. The calculations below show that escaping atoms carry mean kinetic energy $2k_{B}T$, whereas a bulk atom has mean kinetic energy $3k_{B}T/2$. If the aperture is small and interatomic collisions maintain near-equilibrium conditions, faster atoms escape preferentially, and the remaining gas relaxes toward a lower-temperature Maxwell-Boltzmann distribution. No heat supplied by the walls is available to replace the lost energy.

For a monatomic gas at fixed container volume, conservation of energy gives a quantitative check on this effusive cooling picture:Since $dN<0$, the temperature falls. Under these idealizations $T\propto N^{1/3}$; the gas is not left with a permanently truncated velocity tail when collisions rapidly restore equilibrium.

If interatomic collisions are absent, the individual surviving atoms do not rethermalize to a new common temperature. Their speed distribution becomes selectively depleted, generally favouring slower survivors, and their angular and spatial distribution can also develop structure determined by which trajectories reach the hole. A single lower temperature does not describe the full nonequilibrium distribution. With a direction-randomizing but energy-preserving wall model, an illustrative survival law is $f(v,t)\propto f(v,0)e^{-\kappa vt}$; specular walls can give more detailed directional selection and even persistent nonescaping trajectories. The question supplies no wall-scattering or container-geometry model from which a unique collisionless loss law could be inferred.

For the labelled wall integrals, choose the polar axis toward the wall so that incoming atoms have $0\le \theta \le \pi /2$, and putThe needed radial integrals are $I_3=1/(2{\alpha }^2)$, $I_4=3\sqrt{\pi }/(8{\alpha }^{5/2})$ and $I_5=1/{\alpha }^3$. They are related to the supplied integrals by rescaling the speed.

The collision flux contains an extra normal velocity $v\cos \theta$: a faster atom sweeps out a longer approach cylinder per unit time. Integrating the Maxwell-Boltzmann distribution over the incoming hemisphere givesThereforeHere $J$ is the number striking unit area per unit time and $\overline{v}=\sqrt{8k_{B}T/(\pi m)}$ is the bulk mean speed. A simple factor of one half would count the incoming hemisphere but miss the projection factor and velocity weighting.

An elastic reflection transfers normal momentum $2mv\cos \theta$ to the wall. Multiplying this impulse by the collision flux and integrating givesUsing the radial integral and $\alpha =m/(2k_{B}T)$ yieldsThis agrees with the ideal-gas equation of state. The calculation assumes a stationary wall and elastic normal momentum reversal, appropriate to a gas in equilibrium with it.

The energy average must be taken over the atoms that actually arrive, not over the bulk population. Weighting the collision flux by $m{v}^2/2$ givesThe wall-impact speed distribution is proportional to $v^3{e}^{-\alpha v^2}$ rather than the bulk $v^2{e}^{-\alpha v^2}$. Faster atoms are overrepresented among arrivals. Equivalently, the mean normal kinetic energy of the arriving flux is $k_{B}T$, while each tangential component still contributes $k_{B}T/2$.

An equation of state relates thermodynamic variables within a material model; a phase diagram records which equilibrium phase or phases are stable under specified controls. For a pure substance at fixed temperature and pressure, stability means minimizing Gibbs free energy. Two phases coexist when their molar chemical potentials agree. The equation of state and the free-energy stability condition must therefore be used together: an algebraic relation between $p,V,T$ alone need not identify the stable branch.

A pressure-temperature phase diagram contains single-phase regions separated by phase coexistence curves. For a one-component system the Gibbs phase rule gives $F=3-P$: a single phase has two independent controls, two-phase coexistence lies on a curve, and three phases coexist at a triple point. The liquid-gas coexistence curve ends at a liquid-gas critical point; beyond it a continuous path connects liquidlike and gaslike states without crossing a first-order coexistence boundary. A fixed $p,T$ point on a coexistence curve does not specify the proportions of the two phases: an additional extensive constraint, such as total volume, fixes the phase fraction.

At a first-order phase transition, first derivatives of Gibbs free energy, such as molar entropy or volume, are discontinuous. The conversion has latent heat $L=T\Delta s$. At a continuous transition the first derivatives match, while response functions associated with higher derivatives can be singular or discontinuous. At an ordinary liquid-gas critical endpoint, the two coexisting densities approach one another and the latent heat tends to zero; the material does not retain a sharp liquid-gas boundary above it.

The coexistence slope follows directly from equality of the two molar chemical potentials. Since $d\mu =-sdT+vdp$, differentiating ${\mu }_{\alpha }={\mu }_{\beta }$ along the boundary givesThis is the exact Clapeyron equation, also commonly called the Clausius-Clapeyron relation. For vaporization, both $L$ and $\Delta v$ are positive, so the saturation pressure rises with temperature. For ordinary ice melting into liquid water, $L>0$ but the molar volume decreases, giving the familiar negative melting-curve slope. The sign follows from the volume change, not from a different law.

For a dilute ideal vapour with $v_g\gg v_l$, $\Delta v\simeq RT/p$. The ideal-vapor Clausius-Clapeyron approximation becomesIf molar latent heat is approximately constant over the interval,These approximations deteriorate close to the critical point, where liquid and gas volumes become comparable, or over a range in which latent heat changes appreciably.^[ref]

The ideal gas law $pV=nRT$ is an equation of state for a dilute gas, but it has no liquid branch and cannot predict condensation. A simple model including repulsion and attraction is the Van der Waals molar equation,where $b>0$ represents excluded molar volume and $a>0$ represents attraction. At its critical point, $\partial p/\partial v={\partial }^{2}p/\partial v^2=0$, givingBelow $T_c$, the homogeneous isotherm has an interval with $\partial p/\partial v>0$. That would have negative isothermal compressibility and is mechanically unstable. Equilibrium liquid-gas coexistence replaces the loop by a constant-pressure segment. Equality of chemical potentials gives the Maxwell construction:The equal-area condition enforces the common tangent to the appropriate free energy; it is a stability construction rather than permission to follow the unstable loop physically. The average molar volume between $v_l$ and $v_g$ varies by changing the phase fraction.