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For a fixed-composition simple compressible system, $dG=-SdT+Vdp$ gives $S=-(\partial G/\partial T{)}_p$ and $V=(\partial G/\partial p{)}_T$. Equality of mixed derivatives yields $(\partial S/\partial p{)}_T=-(\partial V/\partial T{)}_p$. ThereforeFor one mole of ideal gas, $V_m=RT/p$, so $(\partial V_m/\partial T{)}_p=R/p$. Reversible heat at fixed pressure gives $(\partial S_m/\partial T{)}_p=C_{mp}/T$, and consequentlyWith constant positive molar heat capacity, integrate from a reference state $(p_0,T_0)$:If heat capacity varies, the last term must instead be ${\int }_{T_0}^T{C}_{mp}(T^{\prime })d{T}^{\prime }/T^{\prime }$. The reference ratios make the logarithms dimensionless. The following classical ideal-gas entropy limits test where the constant-heat-capacity expression remains applicable.

The zeroth law of thermodynamics states the transitivity of thermal equilibrium: if two systems each equilibrate thermally with a third, they equilibrate with one another. A reproducible thermometric property can therefore label their temperature. Its empirical numerical scale is not itself the complete thermodynamic content; reversible heat-engine relations establish an absolute temperature scale.

The first law of thermodynamics expresses energy conservation. For a closed system, $dU=\delta Q-\delta W$ when $\delta W$ is work done by the system. internal energy is a state function, while heat and work depend on the path. In reversible simple compressible motion, $\delta W=pdV$ and $dU=TdS-pdV$; an open system additionally needs the energies carried by transferred matter. An ideal gas can expand freely with $Q=W=0$ and unchanged internal energy. The first law alone does not decide whether the reverse spontaneous process occurs.

The Second law of thermodynamics introduces entropy and restricts direction. For a reversible path, $dS=\delta Q_{\mathrm{rev}}/T$; for an irreversible process the entropy balance includes nonnegative production. The entropy of an isolated system cannot decrease. In a cycle, the Clausius inequality is $\oint \delta Q/T\le 0$, using the appropriate boundary/reservoir temperatures. The Kelvin–Planck statement forbids a cyclic device whose sole effect is extracting heat from one reservoir and turning all of it into work; the Clausius statement forbids moving heat from cold to hot with no compensating effect. A refrigerator is allowed because it consumes work. A two-reservoir engine has $\eta \le 1-T_c/T_h$, with equality in the reversible limit. Work may be fully dissipated as heat without contradicting this one-way restriction.

The third law of thermodynamics constrains low-temperature entropy. An equilibrium perfect crystal with a nondegenerate ground state approaches zero entropy as $T\to 0$, fixing the conventional absolute entropy reference. Degenerate ground states or frozen nonequilibrium disorder require care when discussing residual entropy. Under regular equilibrium conditions the heat capacity must decrease sufficiently for $\int C/TdT$ to remain finite; a constant positive classical heat capacity cannot be extrapolated to zero temperature. The unattainability formulation states that absolute zero cannot be reached by a finite sequence of admissible cooling operations. These statements concern the limiting thermodynamic regime, rather than predicting that every substance stays in one classical phase during cooling.

The laws work together: conservation sets a device's energy budget, entropy production restricts the attainable budget, and constitutive equations give the properties of its working material. For fixed temperature and volume, Helmholtz free energy decreases toward equilibrium; at fixed temperature and pressure the corresponding potential is Gibbs free energy. Stable equilibrium, phase coexistence and Maxwell relations follow from these potentials with appropriate constraints. Calorimetry, mechanical-work measurements, thermometry and reversible-path comparisons test different parts of this framework. None of the laws by itself supplies a particular gas equation of state or microscopic heat capacity.

For a reversible isothermal expansion of one mole of an ideal gas, internal energy depends only on temperature, so $\Delta U=0$. The first law of thermodynamics givesThis uses quasistatic expansion against the gas pressure. Specifying only the initial and final volumes and constant temperature would not fix heat for an arbitrary irreversible process: an insulated ideal-gas free expansion has $Q=W=0$.

Use $c_p,c_V$ for the heat capacities of the entire $n$-mole sample. For a stable simple compressible gas, a Maxwell relation giveswhere ${\alpha }_V=V^{-1}(\partial V/\partial T{)}_p$ and ${\kappa }_T=-V^{-1}(\partial V/\partial p{)}_T>0$. This heat-capacity difference and mechanical stability explains the sign without assuming the internal energy depends only on temperature. It is strictly positive for an ordinary expanding gas; the unrestricted word “always” needs the qualification that equality is possible when thermal expansion vanishes.

For an ideal gas, thermodynamic consistency givesConsequently $U=U(T)$, while $H=U+pV=U+nRT$. Differentiating gives $c_p-c_V=nR$: at fixed pressure extra supplied heat permits expansion work in addition to the internal-energy increase.

Differentiate the supplied entropy at fixed $n$:Thus $U=5nRT/2$, consistently recovering $pV=nRT$, andThe coefficient in this entropy corresponds to five active quadratic thermal degrees of freedom per molecule. It should not be replaced by the monatomic value. Reference scales needed to make logarithms dimensionless are absorbed into the entropy constant and do not affect these derivatives.

With open windows the house equilibrates mechanically at approximately the same atmospheric pressure before and after heating. It is an open system: its mole number is not fixed. The isobaric ideal-gas internal-energy density isAt a fixed house volume, $n_{300}/n_{273}=273/300=0.91$: fewer molecules occupy the house, each with greater mean energy. Heating energy can enter while enthalpy leaves with escaping air. Treating the room as a sealed fixed-$n$ gas would give the wrong energy-density comparison.

For a monatomic ideal gas, $\gamma =C_P/C_V=5/3$. If the expansion is reversible adiabatic, $T{V}^{\gamma -1}$ is constant, soAdiabatic isolation alone does not imply this temperature law. For example, free expansion into vacuum has $Q=W=0$, so an ideal gas remains at $300\mathrm{K}$. Thus $75\mathrm{K}$ is the usual intended answer with the additional isentropic assumption; the literal adjective “adiabatic” does not uniquely select it.^[ref]

A heat engine's thermal efficiency is net work out divided by heat absorbed,where both heat quantities are positive magnitudes. The last equality follows from the first law over a complete cycle.

Use $n$ moles of an ideal gas with constant molar $C_V,C_p$ and $\gamma =C_p/C_V>1$. For the Otto cycle, reversible adiabats give $T_2=T_1{r}^{\gamma -1}$ and $T_3/T_4=r^{\gamma -1}$. Isochoric heat input and rejection are $Q_{\mathrm{in}}=n{C}_V(T_3-T_2)$ and $Q_{\mathrm{out}}=n{C}_V(T_4-T_1)$. ThereforeThe denominator's temperature difference is $r^{\gamma -1}$ times the numerator's difference.

For the ideal Atkinson cycle, heating at fixed volume gives $T_3/T_2=r_p$. The longer expansion ends when $p_{4^{\prime }}=p_1$. Combining the adiabatic pressure–temperature relation with $p_2/p_1=r^{\gamma }$ yieldsAlso $V_{4^{\prime }}=V_1{r}_p^{1/\gamma }>V_1$ for $r_p>1$. The PDF asks explicitly for $T_3,T_{4^{\prime }}$; the transcription additionally names $T_2$, which is supplied above too.

Heat input remains $n{C}_V{T}_1{r}^{\gamma -1}(r_p-1)$, while the isobaric return rejects $n{C}_p{T}_1(r_p^{1/\gamma }-1)$. Hence the Atkinson efficiency advantage from extended expansion is quantified byCompare cycles with the same initial state, compression ratio and heat input. The extra expansion and constant-pressure return enclose positive extra work,It uses heat that the Otto cycle would reject, while $Q_{\mathrm{in}}$ is unchanged. Equivalently, concavity gives $\gamma (r_p^{1/\gamma }-1)<r_p-1$, so ${\eta }_A>{\eta }_O$ for $r_p>1$ under this matched comparison. This is an ideal-cycle result; the question's named real-engine application is not needed to establish it.

For a simple compressible one-component equilibrium system, the fundamental thermodynamic relation isThe internal energy has natural variables $(S,V,N)$: its derivatives give $T$, $-p$ and chemical potential $\mu$. The natural variables of a thermodynamic potential identify the independent arguments whose conjugates appear directly in a potential's differential. A Legendre transform exchanges a variable for its conjugate, giving potentials suited to different controlled conditions:

For fixed composition and the appropriate closed-system constraints, equilibrium minimizes $U$ at fixed $S,V$, $H$ at fixed $S,p$, $F$ at fixed $T,V$, or $G$ at fixed $T,p$, with other allowable internal variations understood. Enthalpy is useful for constant-pressure heat exchange with only pressure–volume work. Helmholtz free energy is suited to fixed-temperature, fixed-volume systems, and Gibbs free energy to fixed-temperature, fixed-pressure phase or chemical equilibrium. Coexisting phases of one substance have equal chemical potentials.

Where a potential is twice continuously differentiable, equality of mixed derivatives gives Maxwell relations. At fixed $N$, the four usual forms areThe fixed subscripts and signs follow from the chosen natural variables. Derivatives through a nonsmooth phase boundary require separate phase descriptions rather than automatic use of the smooth single-phase formulas.

As an application, the Helmholtz relation turns pressure data into an isothermal entropy change. For an ideal gas, $(\partial p/\partial T{)}_V=nR/V$, givingCombining the same relation with $dU=TdS-pdV$ yields $(\partial U/\partial V{)}_T=T(\partial p/\partial T{)}_V-p$, which vanishes for an ideal gas and reveals attractive-energy effects for a van der Waals gas. Thermodynamic potentials organize equilibrium constraints, while Maxwell relations convert difficult entropy derivatives into measurable response derivatives.

Treat the vapour as an ideal gas, neglect the liquid's molar volume, and take the molar latent heat as constant across this small interval. Integrating the Clausius-Clapeyron equation givesOnly the pressure ratio enters, so conversion from atmospheres to pascals is unnecessary.

Put $P_{\ast }=p_1{V}_1=nR{T}_1$. For the specified ideal gas, $C_V=3nR/2$, $C_p=5nR/2$ and $\gamma =5/3$. Isobaric expansion doubles temperature, so $T_2=2T_1$. Reversible adiabatic closure gives $T{V}^{\gamma -1}=\mathrm{constant}$, henceThere is no heat input on that compression. The other stages giveTherefore $W_{\mathrm{net}}^{\mathrm{by}}=P_{\ast }-W_{31}^{\mathrm{on}}=0.44494P_{\ast }$ andThe hot reservoir changes entropy by $-Q_{12}/(2.5T_1)=-P_{\ast }/T_1$. The cold reservoir changes by $-Q_{23}/(0.5T_1)=4.11012P_{\ast }/T_1$. Since the gas completes a cycle,Although the mechanical paths can be quasistatic, heat flows across finite temperature differences in both contacts. This entropy generation explains why the efficiency is well below the Carnot value $1-0.5/2.5=80\%$ between these reservoirs.

For an isothermal final compression at $T_1$, the preceding constant-volume cooling must instead end at $T_3=T_1$; keeping the old state 3 would not close the cycle. Compression then requires work $P_{\ast }\ln 2$, while the heat input remains $5P_{\ast }/2$. Thuswhich is lower: this isothermal compression occurs at a higher temperature throughout than the original adiabat and consumes more work. It also rejects heat during compression.

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.

Label the initial state 1, the end of compression 2, the end of constant-pressure heating 3, and the end of adiabatic expansion 4. The volume sequence is $V_1\to V_2\to V_3\to V_1\to V_1$, with $V_2<V_3<V_1$. The Diesel cycle has a compression adiabat $1\to 2$, a horizontal constant-pressure heat-input segment $2\to 3$, an expansion adiabat $3\to 4$, and a vertical constant-volume cooling segment $4\to 1$.

For an ideal gas, a reversible adiabat obeys $p{V}^{\gamma }=\mathrm{constant}$, so it falls more steeply than an isotherm. The expansion adiabat lies above the compression adiabat at equal volume between $V_3$ and $V_1$, giving a clockwise cycle with positive work output.

A Carnot cycle consists of four reversible stages: isothermal expansion at the hot-reservoir temperature $T_h$, adiabatic expansion to $T_c$, isothermal compression at $T_c$, and adiabatic compression back to the starting state. The adiabatic stages exchange no heat. For one mole of an ideal gas, the two isothermal stages have heats $Q_h=R{T}_h\ln (V_B/V_A)$ and $Q_c=R{T}_c\ln (V_C/V_D)$ in magnitude. The reversible adiabatic relations ensure $V_B/V_A=V_C/V_D$.

More generally, the total entropy change over any reversible cycle is zero. The two heat exchanges therefore satisfyWith work output $W=Q_h-Q_c$, its thermal efficiency isThe temperatures must be absolute. The second law makes this the maximum efficiency for an engine exchanging heat only with those two reservoirs; irreversible entropy production lowers the efficiency.

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy, $e^{-m{v}^2/(2k_{B}T)}$. Velocities with speed between $v$ and $v+dv$ fill a spherical shell of velocity-space volume $4\pi v^{2}dv$. The normalized Maxwell speed distribution is thereforeThe factor $v^2$ counts states, rather than representing an extra Boltzmann energy factor.

Set $\alpha =m/(2k_{B}T)$ and use $I_j={\int }_0^{\infty }{v}^j{e}^{-\alpha v^2}dv$. Normalization cancels in ratios, so the given recurrence givesConsequentlyThe root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

Model air as a dilute ideal gas of molecules described by the hard sphere model. The number density and mean free path areThe factor $\sqrt{2}$ accounts for the relative motion of the colliding molecules. A molecular collision diameter has to be estimated because none is supplied. Taking $d\simeq 0.37\mathrm{nm}$, appropriate as an order-of-magnitude size for the main constituents of air, givesThus the mean free path is about $7\times 10^{-8}\mathrm{m}$, or $70\mathrm{nm}$. A slightly smaller assumed diameter $0.35\mathrm{nm}$ gives about $76\mathrm{nm}$, illustrating the intended precision of the estimate.

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

For a reversible adiabatic process, an ideal gas obeys $T{V}^{\gamma -1}=\text{constant}$. A monatomic gas has heat capacity ratio $\gamma =5/3$. Taking the stated expansion factor to mean $V_f=9V_i$,This assumes the usual quasistatic isentropic process. Heat insulation alone does not determine the final temperature: an ideal-gas free expansion is also adiabatic but leaves the temperature unchanged. The numerical answer uses nine times the original volume, the conventional intended meaning of the source's “nine times larger”.

Pressure and its physical terms. Solving the molar Van der Waals equation for pressure givesThe excluded-volume parameter $b$ reduces the volume available for translational motion, raising the kinetic pressure relative to $RT/V$. The attraction parameter $a$ lowers the pressure because attractive interactions reduce the momentum transferred to the boundary. This mean-field model treats $a,b$ as temperature-independent material constants. For one mole, $V$ is the sample volume with the molar amount already fixed; $b$ has units of that volume and $a$ of pressure times volume squared.

volume dependence of entropy. The Helmholtz free energy has differential $dF=-SdT-PdV$. Equality of mixed derivatives yields the Maxwell relationFor $a=0$ this derivative is $R/(V-b)$, givingEquivalently, $S=R\ln (V-b)+f(T)$ with a common reference volume understood inside the logarithm. The equation of state fixes this volume dependence, not the otherwise unspecified function of temperature.

Excess entropy at fixed volume and temperature. The ideal gas has $(\partial S_{\mathrm{id}}/\partial V{)}_T=R/V$. ThereforeIntegrate from infinite volume, where the entropy difference is specified to vanish:It is negative because excluding configurations reduces the accessible state count. It tends to zero as $V\to \infty$ and has leading dilute limit $-Rb/V$. This is the entropy excess of a van der Waals gas.

Effect of attraction. Restoring a temperature-independent $a$ changes the pressure by $-a/V^2$, but not its derivative with respect to temperature. With the same infinite-volume entropy reference, it therefore does not change $\Delta S^{\mathrm{ex}}$ at fixed $V,T$. This can also be seen fromThe attraction affects internal energy and the equilibrium volume, so it can affect entropy indirectly when different state points are compared. If $a$ were temperature dependent, its derivative would contribute directly to the entropy; that is outside the stipulated constant-parameter model.

Coexistence isotherm. Below the Van der Waals critical point, the formal homogeneous isotherm has a local minimum and maximum. Its middle branch has $\partial P/\partial V>0$, hence negative isothermal compressibility, and is mechanically unstable. Portions of the negative-slope branches can be metastable rather than globally stable. The equilibrium isotherm replaces the loop between liquid volume $V_l$ and vapour volume $V_g$ by a constant saturation pressure $P_s$. The two volume are fixed byThe Maxwell equal-area construction expresses equality of the two phases' chemical potentials. Intermediate total volume are mixtures: changing the vapour fraction changes total volume at the same pressure, rather than following the unstable homogeneous curve. The spinodals are the turning points of the formal isotherm; they are distinct from the coexistence endpoints.

Dense-phase volume at low temperature. At a general subcritical temperature, the two coexistence equations above must be solved together. For this equation of state the area condition is explicitlyThis supplies an exact implicit determination of the liquid volume even when a simple closed form is unavailable.

For the asymptotically low-temperature limit, put $\tau =RTb/a\ll 1$. The liquid is dense, $V_l\simeq b$, while the vapour is dilute, $P_s{V}_g\simeq RT$ and $V_g\gg V_l$. The area condition then givesThus the vapour volume grows exponentially in $1/\tau$, and the saturation pressure is exponentially small. It is consequently consistent to neglect $P_s$ in the liquid equation to algebraic order in $\tau$:Choose the root near $b$, not the large-volume zero-pressure root:Expanding gives the leading requested dense-phase behavior,This is the low-temperature liquid volume of a van der Waals gas. Setting $P_s=0$ is an asymptotic approximation, not an exact coexistence condition at every temperature below $T_c$. The figure illustrates generic subcritical coexistence; its finite-temperature liquid volume is determined by the full equal-area equations. The van der Waals liquid also ceases to be a realistic description at sufficiently low physical temperature, where other phases or quantum effects can matter.

For the intended reversible adiabatic process, an ideal gas obeys $T{V}^{\gamma -1}=\text{constant}$. A monatomic gas has heat capacity ratio $\gamma =5/3$, soThe required extra assumption is that “adiabatic” denotes the usual quasistatic isentropic process. Absence of heat transfer alone would not fix the final temperature: an ideal gas undergoing free expansion into vacuum is also adiabatic but leaves its temperature unchanged.

The expression is the canonical partition function for a system in thermal equilibrium with a large heat bath at fixed temperature $T$. The system exchanges energy with the bath, while its particle number and external mechanical parameters, such as volume, are fixed. The bath is large enough that its temperature is not appreciably changed by this exchange.

The system has allowed energy levels ${ϵ}_j$. The index $j$ labels distinct energy values, and the integer degeneracy factor $g_j$ counts distinct quantum states with that same energy. $k_B$ is the Boltzmann constant, and $\exp [-{ϵ}_j/(k_{B}T)]$ is a state's Boltzmann factor. The dimensionless sum $Z$ normalizes the probabilities:Thus the degeneracy multiplies the probability of occupying the level, not the probability assigned to each individual state within it.

The canonical model is not restricted to a classical ideal gas. It applies to a quantum rotor, for example, provided the sum counts all of that system's allowed states with their proper degeneracies. For a many-particle system, ${ϵ}_j$ would be the energy of a many-particle state, with any quantum-statistical restrictions already built into the state count.

Use an ideal gas to illustrate the Carnot cycle, with hot and cold heat baths at $T_h=T_2$ and $T_c=T_1$. The four processes are all reversible thermodynamic processes. Lettered states below belong only to this illustration; the engine later in the question uses numbered states.

Starting at $A$, perform the following four stages:

The clockwise enclosed area in the pressure-volume diagram is the positive net work per cycle, $W=\oint pdV=Q_h-Q_c$.