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After the load is removed, a reversibly stretched DNA molecule relaxes towards a thermally fluctuating coil, releasing stored free energy as heat through the surrounding fluid. There are many more disordered chain conformations than strongly aligned ones, so entropy favours retraction; viscous drag sets how rapidly it occurs. At long times an unconstrained ideal chain has zero mean end-to-end vector but a nonzero typical size, $R_{\mathrm{rms}}=b\sqrt{N}$; it does not contract to a point.

At finite temperature the chain continues fluctuating, and an anchored or otherwise constrained molecule need not regain an isotropic ensemble. If stretching broke attachments, denatured the DNA or caused irreversible structural changes, recovery can be incomplete or hysteretic. Those outcomes cannot be decided from the ideal FJC alone.

Dense DNA packaging opposes three main physical effects. The negatively charged DNA backbones repel through electrostatic interactions, moderated by electrostatic screening and counterion condensation. At close approach, hydration forces and excluded volume become important. Bending a semiflexible DNA polymer on a radius comparable with its persistence length stores elastic energy. Confinement also removes configurational entropy, and concentrating mobile counterions can impose an additional osmotic pressure and free energy cost. Their relative importance depends on geometry and ionic conditions.

A schematic lytic bacteriophage life cycle is illustrated below.

Attachment recognizes a suitable host; delivery brings the genetic information into the host; host resources support production of components; assembly creates progeny that are released. Some bacteriophages instead have a latent integrated phase, but that is not a necessary stage of this lytic schematic.

In many double-stranded DNA bacteriophages, a portal-associated molecular motor uses ATP hydrolysis to move DNA into a preformed capsid. Chemical free energy pays for bending, repulsion and confinement, so the opposing force rises as filling proceeds. This is active packaging rather than spontaneous equilibrium compaction. Optical tweezers can infer that opposing force from displacement of a trapped bead and monitor packaging through the changing external DNA extension. Such measurements establish the mechanical principle without requiring any virus-handling procedure. The inverse spool model used below is an idealized packing geometry, not an exact structural claim for every bacteriophage.

The bare packed-sphere estimate exceeds $2.1\times 10^5\mathrm{pN}\mathrm{nm}$ by approximately $659$; subtracting the estimated extended-rod self-energy still leaves a factor of about $629$. The dominant missing effect is electrostatic screening: mobile positive counterions accompany the negatively charged DNA, greatly reducing long-range repulsion. Counterion condensation can further reduce its effective electric charge. Treating the genome's bare charge as uncompensated throughout and outside a $20\mathrm{nm}$ sphere therefore grossly overestimates its long-range electrostatic energy.

An optical tweezers measurement integrates an imposed force over a displacement. A reversible extraction or packaging comparison concerns a free energy difference relative to the unconfined DNA and ionic reservoirs. It can include entropy, bending and hydration contributions; dissipative pulling can add further work. Extraction can also be assisted by stored internal free energy, so the sign of externally supplied work requires the experimental convention. Here the supplied positive value is compared as an energy magnitude.

Adding $40000$ monovalent counterions makes the combined genome-plus-ion electric charge nominally zero. It does not make the local electric field or free energy vanish: opposite charges have spatial distributions, and confining mobile ions costs entropy.

A definite screened continuation of b replaces the bare Coulomb potential by $e^{-\kappa r}/(4\pi ϵr)$, with $\kappa =1/{\lambda }_D$ and ${\lambda }_D=1\mathrm{nm}$. In the Debye–Hückel approximation, a uniformly distributed fixed charge density $\rho =Q/V_c$ satisfies $({\nabla }^2-{\kappa }^2)\phi =-\rho /ϵ$ inside, with zero fixed charge outside and matching potential and radial field at $R$. Its interior solution isThe screened charging free energy $U_{\kappa }=\frac{1}{2}\int \rho \phi dV$ is thereforeFor $R\gg {\lambda }_D$, the bracket approaches one: only charges within roughly a screening length contribute appreciably, yielding the local estimate $U_{\kappa }\simeq Q^2{\lambda }_D^2/(2ϵV_c)$. With the same ${ϵ}_r=80$ as before, $x=20$ gives bracket $0.92519$ andabout $173$ times smaller than the unscreened packed-sphere self-energy. This is a specified continuum estimate of packed-state charging free energy; a full packaging free energy also subtracts the screened unpacked reference and includes ion redistribution. It should not be presented as a unique exact packaging work fixed by the problem's few inputs.

The supplied counterion count also permits an independent estimate of the entropic contribution. For $N$ ideal confined ions, $F_{\mathrm{ion}}=-N{k}_{B}T\ln V+\text{constant}$, so compression from an accessible reference volume $V_0$ to $V_f$ costsFor example, approximate the original screening atmosphere as an annulus of thickness ${\lambda }_D$ around a rod. With $a=0.5\mathrm{nm}$, $V_0=\pi [(a+{\lambda }_D{)}^2-a^2]L=4.27\times 10^4\mathrm{n}{\mathrm{m}}^3$. Taking $V_f=V_c-V_D=1.35\times 10^4\mathrm{n}{\mathrm{m}}^3$ gives $\Delta F_{\mathrm{ion}}\simeq 1.91\times 10^5\mathrm{pN}\mathrm{nm}$. This useful order-of-magnitude ion-entropy model is distinct from the screened charging model; the two estimates must not simply be added as independent terms without a consistent ionic free energy functional. The rod cutoff, fixed DNA volume and screening-atmosphere geometry are rough geometric assumptions, not a microscopic reconstruction of the source sketch.

There is a further consistency limit. The nominal ion number density is $n=N/V_c=1.19\mathrm{n}{\mathrm{m}}^{-3}$. For a uniform ideal monovalent screening gas, ${\lambda }_D^2=ϵk_{B}T/(n{e}^2)$ would give ${\lambda }_D\simeq 0.309\mathrm{nm}$ at $300\mathrm{K}$, rather than $1\mathrm{nm}$; added salt would shorten it further. Thus the stated ion inventory and screening length cannot both be interpreted as a self-consistent dilute uniform screening gas with this dielectric. Bound counterions, nonuniform distributions and an effective phenomenological screening length can change that interpretation. Linear screening is itself only an estimate near highly charged DNA.

The specified screened-sphere charging model gives $8.00\times 10^5\mathrm{pN}\mathrm{nm}$, about $3.8$ times the supplied experimental magnitude. It is much closer than the unscreened result. The separate ideal counterion-compression estimate gives $1.91\times 10^5\mathrm{pN}\mathrm{nm}$, within about $9\%$ of that magnitude, but this numerical proximity does not validate its rough geometry.

Remaining differences can arise from the unpacked reference self-energy, counterion condensation, ion exchange with the surrounding reservoir, nonuniform charge density, nonlinear electrostatic screening, finite DNA thickness and the actual capsid shape. DNA bending energy, hydration forces and configurational entropy also contribute. The experiment measures a process-dependent work or inferred free energy, whereas the simple screened sphere computes just one model contribution. Neither a zero-net-charge argument nor agreement of one approximate number establishes an exact total energy.

Each Cartesian component of one link has zero mean and variance $b^2/3$, with zero cross-component covariance. By the multivariate central limit theorem, for large $N$ on the coil scale the vector probability density isIt is normalized with respect to $d^{3}r$; the radial probability density is $4\pi r^{2}p(\mathbf{r})$. The Gaussian limit of the end-to-end distribution of a freely jointed chain is an approximation, not the exact finite-$N$ law near $r=Nb$.

At prescribed vector separation the configurational entropy is $S(\mathbf{r})=k_B\ln p(\mathbf{r})+\text{constant}$. The ideal chain has no extension-dependent internal energy, so its Helmholtz free energy isThe externally required force is ${\nabla }_{\mathbf{r}}A$, while the chain's restoring force has the opposite sign. ThusThe last equation also follows by multiplying $p(\mathbf{r})$ by the Boltzmann factor $e^{fz/(k_{B}T)}$ and completing the square. This entropic spring resists extension because fewer link configurations realize a stretched chain. At the same small applied force, increasing temperature reduces extension; at the same extension it increases the required force. Do not differentiate $-k_{B}T\ln [4\pi r^{2}p(\mathbf{r})]$ to obtain the fixed-vector Hooke law, since that radial measure includes an additional geometric multiplicity. The low-force condition is $fb/(k_{B}T)\ll 1$, equivalently $⟨z⟩\ll Nb$.

In atomic force microscopy, one end of the polyacrylic acid chain is attached to the substrate and the other to a compliant cantilever tip. Controlled separation displaces the cantilever; calibrated cantilever stiffness converts its deflection into force. The molecular extension is the tip–surface separation after correcting for cantilever deflection and instrumental offsets. A quasi-static interpretation requires relaxation to be fast compared with the pulling motion.

At low force a coiled polymer has many available configurations and extends mainly by changing segment orientations. This is the entropic spring regime. As the available contour is aligned, the loss of configurational entropy becomes steep. For an inextensible three-dimensional freely jointed chain,where $\mathcal{L}$ is the Langevin function. The limiting extension is the polymer contour length $Nb$; the ideal model approaches it only as $f\to \infty$. Real polymer backbone stretching permits extension beyond the fitted rigid-link contour, explaining why the experimental curve lies on the more extensible side at high force.

The original trace has a sharp force drop after its peak. Finite contour length explains the steep preceding rise, but does not itself cause that drop: release of an attachment, desorption or mechanical failure ends tensile loading. The graph alone does not distinguish these possibilities. If the intact chain is released from tension, thermal motion restores a coil with characteristic end-to-end distance of a polymer chain of order $b\sqrt{N}$, because this macrostate has many more configurations. Recoil is reversible for an intact equilibrating chain; irreversible chain scission or attachment loss can prevent repetition of the same pulling curve.

For the adsorbed lattice gas, the Helmholtz free energy isUse Stirling's approximation with $M,N,M-N$ large, then differentiate at fixed $M,T,B$. Writing surface coverage as $\theta =N/M$ gives the adsorption chemical potentialThe occupation ratio comes from site-mixing entropy; it is essential near saturation.

Let $N_g$ be the number of free atoms and $n_g=N_g/V$. In the field-free gas, both spin states have zero energy, so the one-atom partition function is $q_g=2V/{\Lambda }^3$, where $\Lambda$ is the thermal de Broglie wavelength. For a dilute ideal gas,ThereforeThe last equality uses $p=n_g{k}_{\mathrm{B}}T$. The bulk spin degeneracy is two even though the bulk magnetic field vanishes: degeneracy is not the absence of spin states.

For fixed degeneracies and ionization energy,Therefore $K_c$ increases with temperature in this model. Ionization costs positive energy, so it is suppressed at low temperature by its Boltzmann factor. At higher temperature that penalty becomes less important.

Ionization also creates an additional independently translating particle and more ways to distribute the particles and their momenta. This favours the ionized mixture through its translational and mixing entropy, consistent with the $T^{3/2}$ prefactor and increasing ionization at high temperature.

Assume a dilute classical ideal gas of $N$ indistinguishable, noninteracting particles, with translational states dense enough to replace their sum by a continuum integral. Separate translation from all internal degrees of freedom. ThenUsing Stirling's approximation in the Helmholtz free energy,Differentiate at fixed $N,V$. Since ${\Lambda }^{-3}\propto T^{3/2}$,Only after this derivative set $N=N_A$ and $V=RT/p^{\circ }$ for one mole at standard pressure:This is the translational Sackur-Tetrode equation. The $N!$ factor is essential for the correct extensive entropy; replacing $V$ by a temperature-dependent standard molar volume before taking the fixed-volume derivative would give the wrong constant term.

For a linear rigid rotor in the high-temperature regime, $q_{\mathrm{rot}}\propto T$. Its molar rotational entropy isThe rotational symmetry number must already be included in $q_{\mathrm{rot}}$.

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 substituted molecule has the larger entropy. Increased mass and moment of inertia reduce translational and rotational spacings and make more states thermally accessible. The largest contribution here is the loss of the twofold rotational indistinguishability of the oxygen ends, contributing $R\log 2$. It is a rotational symmetry number correction for one isotope-defined species, not an entropy of mixing between different molecules.

Use the same standard pressure and temperature, the same bond length, and the same electronic and vibrational-ground-state approximations. The translational partition function scales as mass to the power $3/2$, giving the factor $(47/44{)}^{3/2}$. For the original isotopologue, $I=32m_u{R}^2$ and the rotational symmetry number is two. For the isotope-substituted molecule, the two oxygen ends are distinguishable, so the symmetry number is one. HenceThe change in standard molar entropy issoThis uses a convention omitting a separate nuclear-spin entropy. If nuclear-spin degeneracies are instead included explicitly, the spin-$1/2$ carbon-13 nucleus adds a further $R\log 2$ relative to the spin-zero original isotopologue; that convention must be stated rather than silently mixed with the quoted partition functions.

Under the stated vibrational-ground-state approximation and a nondegenerate electronic ground state, only translation and rotation contribute. ThereforeA temperature-independent ground-state energy, including zero-point energy, contributes no entropy. This estimate deliberately omits thermal vibrational excitation; it need not equal a tabulated value that includes all modes.

As a first estimate, neglect the small mass change and changes in geometry and vibrational populations. Removing an electron creates an unpaired spin, giving at least the spin-only electronic entropy increaseFor a more specific estimate, the actual cation ground term is $X{}^2{\Pi }_g$, rather than an orbitally nondegenerate doublet. Ignoring spin-orbit coupling, its two orbital components and two spin components give electronic degeneracy four andThe ground-state assignment is confirmed by state-resolved photoionization measurements.^[ref] Spin-orbit splitting and changed vibrational/rotational constants refine this estimate; the question's supplied partition functions alone do not determine those corrections. The $R\log 2$ answer is a spin-only estimate, while $R\log 4$ additionally retains the orbital degeneracy.

The strongest donor belongs to the acid, ${\alpha }_1=3.5$, and the strongest acceptor to the aminopyridine, ${\beta }_2=8.5$. The cocrystal pairs the strongest donor with the strongest acceptor, while the reciprocal amino-NH···carbonyl contact closes a complementary two-contact motif.

Using the simple favorable-contact score $\alpha \beta$, one heterodimer givesThe self-associated comparison, normalized to the same two molecules and two hydrogen bonds, is half an acid dimer plus half an aminopyridine dimer:ThusIn a model where bond free energy contains $-\alpha \beta$, this is a favorable energy change. No bonds have been artificially added to one side of the comparison: both alternatives contain two contacts per acid/aminopyridine pair.

The calculation rationalizes the hydrogen-bond preference. It is not a complete crystal free energy: packing, additional contacts, entropy and possible solvent effects also matter. Without a specified calibration, the numerical score should not be promoted to a measured cocrystallization free energy.

In water, separated benzene molecules require separate hydration environments around nonpolar surface. Bringing them together buries part of that surface and reorganizes the surrounding water. The resulting solvent contribution favours association relative to cyclohexane, where benzene was already well solvated by dispersion. This solvent-mediated contribution is the hydrophobic effect, not formation of a new covalent bond or a claim that benzene has become a conventional hydrogen-bond acceptor like oxygen.

The supplied constants imply a thirtyfold enhancement. With matched standard states,at an illustrative 298 K. The water value is still slightly below $1{\mathrm{M}}^{-1}$, so enhancement relative to cyclohexane does not mean a negative standard association free energy for a 1 M standard state, nor does it imply that dilute benzene is all dimer. No universal “entropy only” explanation is needed: solvent enthalpy and entropy both contribute and depend on temperature.

Association in solution includes replacement of solute–solvent contacts, not just formation of the isolated dimer. cyclohexane solvates the benzene surface well through dispersion. Much of the London dispersion gained on bringing two benzene molecules together replaces comparable benzene–cyclohexane contacts, leaving only a modest net gain before the translational and orientational entropy penalty. cyclohexane cannot provide the complementary donor/acceptor interactions available when two waters hydrogen-bond, so water association gains a more distinctive interaction relative to its separated solvated monomers.

Using the same concentration standard state, $K^{\circ }=K_c{c}^{\circ }$ and $\Delta G_{\mathrm{assoc}}^{\circ }=-RT\ln K^{\circ }$. For illustration at 298 K and $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, the ratio of the two supplied association constants corresponds toThus the water-molecule pair can associate more favourably in this solvent even though the stipulated isolated-pair binding energies are similar. Gas-phase potential energy alone omits both differential solvation and the entropy of association.

The clearest comparison is between the two amines, which have the same molecular formula and molar mass. isopropylamine can donate N–H hydrogen bonds to neighbouring nitrogen lone pairs; trimethylamine has no N–H donor. The resulting intermolecular association raises the primary amine's boiling point, allowing it to remain liquid at room temperature, whereas trimethylamine is gaseous. Independent phase-change data place their normal boiling points on opposite sides of room temperature.^[ref]

acetone has a substantial carbonyl dipole and London dispersion forces sufficient for a room-temperature liquid, even though it cannot build a strong donor–acceptor hydrogen bond network by itself. isobutane has mainly London dispersion forces and a compact shape and remains gaseous. acetamide combines carbonyl acceptors with two N–H donors, allowing an extended, directionally bonded crystal lattice; favourable packing as well as intermolecular association stabilizes its solid state.

These compounds have similar molar masses, so mass alone cannot explain the differences. The hydrogen-bond acceptor parameter also cannot predict the phase on its own: a pure substance needs complementary donors to exploit strong acceptor sites, and freezing additionally depends on crystal packing and entropy. A stronger acceptor does not automatically imply a higher melting point for arbitrary molecules.