Past exam of the physics B course of the University of Cambridge / 2015 / Paper 2 / Section C

A liquid-gas thermodynamic critical point is the endpoint of the coexistence curve where distinct liquid and gas densities merge. In the van der Waals illustration, write the original constants as $a_v,b_v$ to distinguish them from the later polynomial coefficients. The critical isotherm has a horizontal inflection in $p(V)$; setting its first two volume derivatives to zero givesBelow this temperature the homogeneous van der Waals loop contains an unstable region; physical equilibrium instead separates into liquid and gas.^[ref]

For the critical point of a cubic density equation of state, assume $b,c>0$. Its density derivatives areBecause $n=N/V$ is a smooth monotone function of $V$ at finite positive density, a stationary inflection obeys $p_n=p_{nn}=0$ as well. Solving these equations and substituting back givesThe symbols $b,c$ here have different units from the van der Waals excluded-volume parameter. With $x=n/n_c$, $t=T/T_c$, the isotherms becomeThe spinodal densities of a cubic gas delimit the two-extrema regime. The curves are monotone for $t>1$, have a stationary inflection at $(x,p/p_c)=(1,1)$ for $t=1$, and have two extrema $x=1\pm \sqrt{1-t}$ for $0<t<1$.

For the critical-density compressibility of a cubic gas, since ${\kappa }_T=(1/n)(\partial n/\partial p{)}_T$, evaluation at the critical density givesIt diverges as $T\to T_c^{+}$ in this mean-field model. For $T<T_c$ it is negative on the homogeneous branch at $n_c$, violating mechanical stability: a pressure increase would expand rather than compress that branch.

The actual equilibrium below criticality is coexistence of two phases, with equal pressure and chemical potential and different densities. A sample's mean density can change by changing phase proportions. The unstable homogeneous branch is not a realizable equilibrium isotherm. The Maxwell construction in density coordinates requires care: the equal-area construction is performed in $p$ versus volume, not by using unweighted area in $p$ versus density. The truncated polynomial is an approximate mean-field equation, so its predictions need not give quantitative critical behaviour for a real gas.

The Boltzmann distribution assigns probability $P_i=e^{-E_i/(k_{B}T)}/Z$ to each microstate. For the equilibrium height distribution in a uniform gravitational field, take a dilute gas above a floor in a constant gravitational field. Uniform cross-section and no relevant upper boundary give the normalized height densityThusThe gravitational potential is linear in height, so its mean is not a quadratic-mode contribution of $k_{B}T/2$.

The equipartition states that each independent quadratic term in a classical equilibrium Hamiltonian contributes mean energy $k_{B}T/2$, provided the relevant degrees of freedom are thermally accessible. Equivalently, $⟨x_i\partial H/\partial x_i⟩=k_{B}T$ when the phase-space integration by parts has vanishing boundary terms. Three translational velocity components therefore contribute $3k_{B}T/2$.

For rotational equipartition in a rough-ball gas, collisions exchange both translational and rotational energy. In the intended three-dimensional classical equilibrium model there are three translational and three rotational quadratic terms, giving $3k_{B}T$ in total kinetic energy per ball. Adding $⟨U_g⟩=k_{B}T$, and conserving the initial energy $mgh$ per ball, givesRoughness is relevant because it allows rotation to thermalize. The effective mechanical temperature of released rough balls describes the balls' mechanical degrees of freedom, not their internal material temperature; finite ball size and excluded-volume effects are neglected.^[ref]

The height tail isThis can be nonzero because collisions redistribute energy between balls; an individual ball is not restricted to its original energy. The exponential distribution is the large-number statistical approximation under the stated effective-equilibrium assumption.

For the angularly resolved molecular arrival flux, let $f(v)$ be normalized by ${\int }_0^{\infty }f(v)dv=1$. Isotropic velocity directions have probability $d\Omega /(4\pi )$. Integrating over azimuth gives $(1/2)\sin \theta d\theta$, while the speed normal to a stationary piston is $v\cos \theta$. For incident directions $0\le \theta \le \pi /2$, the arrival flux is thereforeAn elastic collision transfers normal momentum $2mv\cos \theta$ to the piston. Integrating the momentum flux givesNo specific Maxwellian speed distribution was needed for this isotropic kinetic-pressure identity.

Use the figure's $+x$ direction toward the piston and let its inward speed $u>0$ mean piston velocity $-u\hat{\mathbf{x}}$. For reflection from a translating planar piston, in the piston frame the incoming normal velocity is $v_x+u$. Elastic reflection reverses it, and transforming back givesThe exact kinetic-energy change in this ideal massive-piston model is $\Delta K=2mu{v}_x+2m{u}^2$. Thus, to first order in the small piston speed,The first-order collision-flux approximation for a slow piston applies: the incident collision flux changes at order $u$, but its product with the order-$u$ energy increment changes the energy rate only at order $u^2$. Hence one may use the stationary flux at leading order. For piston area $S$,Here $dV/dt=-Su$. This compression work from molecular impacts, with no heat flow, is the first law for work done on the gas by compression. The derivation assumes $u$ small compared with thermal speeds and an approximately isotropic distribution at the piston; rapid compression can produce nonequilibrium pressure and higher-order corrections.

The Second law of thermodynamics restricts which energy-conserving processes can occur. In the Kelvin-Planck statement of the second law, no cyclic engine can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work. In the Clausius statement of the second law, no process can have as its sole effect transferring heat from a colder reservoir to a hotter one. An engine or refrigerator combined with the reverse of a reversible device shows that a violation of either statement would violate the other.

A heat engine takes heat $Q_h>0$ from a reservoir at $T_h$, rejects $Q_c>0$ to one at $T_c<T_h$, and returns its working substance to the initial state. Energy conservation gives $W=Q_h-Q_c$ and efficiency $\eta =W/Q_h=1-Q_c/Q_h$. Carnot's theorem states that no engine between these reservoirs is more efficient than a reversible one, and that all reversible engines between them have the same efficiency. Otherwise, drive a reversible engine backwards with the work of a more efficient engine; the composite cycle transfers heat from cold to hot without external work, contradicting the Clausius statement.

A reversible Carnot cycle consists of two isotherms joined by two reversible adiabats. For an ideal gas, $Q_h=N{k}_B{T}_h\log (V_2/V_1)$ and $Q_c=N{k}_B{T}_c\log (V_3/V_4)$. The adiabats give $V_2/V_1=V_3/V_4$, soCarnot's theorem makes this independent of the working substance and defines the thermodynamic absolute-temperature scale consistently with ideal-gas temperature.^[ref]

For any reversible cycle, $\oint \delta Q_{\mathrm{rev}}/T=0$. This makes the integral between two equilibrium states independent of the reversible path, defining the state function entropy:Heat itself is path dependent; entropy is not. An irreversible change can be evaluated using any imagined reversible path between its equilibrium endpoints. For example, a fixed amount of ideal gas with constant $C_V$ has $\Delta S=C_V\log (T_B/T_A)+N{k}_B\log (V_B/V_A)$.

For a general cycle the Clausius inequality is $\oint \delta Q/T_{\mathrm{boundary}}\le 0$. The entropy balance for a process is $\Delta S\ge \int \delta Q/T_{\mathrm{boundary}}$, with equality for reversible heat exchange and no entropy production. The surroundings' entropy must be included, giving $\Delta S_{\mathrm{universe}}\ge 0$. A system's own entropy can decrease by exporting entropy; an isolated system's cannot.

The engine's reservoirs have net entropy change $Q_c/T_c-Q_h/T_h\ge 0$, sowith equality for a reversible engine. Friction, heat flow through a finite temperature difference and unrestrained expansion produce entropy and reduce usable work. Thus energy conservation fixes the balance of heat and work, while the second law fixes their direction and the attainable efficiency.^[ref]