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Acetone reacts with iodine to form iodoacetone, iodide, and a hydrogen ion:In water, ${\mathrm{H}}^{+}$ denotes a solvated hydrogen ion; added hydrochloric acid provides acidic conditions.

Moorpark College's protocol mixes acetone, hydrochloric acid and water from burets in matched test tubes; pipet in iodine, swirl and time colour loss against white paper. Use water, then reacted solution, as references. Duplicate each mixture; keep total volume $5.00\mathrm{mL}$ and run temperatures within $2{}^{\circ }\mathrm{C}$.^[ref]

The stock dilutions give $[\mathrm{acetone}{]}_0=0.800$, $[{\mathrm{H}}^{+}{]}_0=0.200$, and $[{\mathrm{I}}_2{]}_0=0.00100\mathrm{mol}{\mathrm{L}}^{-1}$. B and C separately double acetone and acid; D halves iodine. The manual supplies no measured times; it states zero partial order of reaction in iodine and asks students to determine the other exponents.^[ref]

The endpoint measurement is a time $\tau$ to an operational colour threshold, not direct observation of a reaction intermediate. If the iodine disappearance rate $r$ is approximately constant until that endpoint, $r\approx [{\mathrm{I}}_2{]}_0/\tau$. Halving $[{\mathrm{I}}_2{]}_0$ at fixed other concentrations then approximately halves $\tau$ without changing $r$. A longer or shorter endpoint time alone is not a comparison of reaction rates unless the starting concentration is included. For an observed power-form rate law, doubling only acetone gives a rate ratio $2^m$ and hence $m=\ln (r_B/r_A)/\ln 2$; doubling only acid similarly estimates its own partial order of reaction. Those are inference formulas, not asserted experimental results.

A useful proposed reaction mechanism would explain how iodine is consumed after an earlier transformation sets the throughput, making the observed reaction rate insensitive to iodine over a tested range. The observed zero exponent would contradict treating this net chemical equation as a single elementary reaction with directional rate law $k[\mathrm{acetone}][{\mathrm{I}}_2]$. It would not uniquely prove which reaction intermediate or earlier step is involved. Independently detecting a candidate reaction intermediate would provide another kind of evidence.

The excess-reactant approximation has an explicit scale: the baseline iodine inventory is only $0.00100/0.800=0.125\%$ of the initial acetone inventory, and its net hydrogen ion production is $0.00100/0.200=0.5\%$ of the initial acid inventory, assuming only the stated one-to-one chemical reaction. These are calculated bounds on composition change, not measured reaction orders. Mixing delay, a subjective colour threshold, and temperature drift limit the inference. Repeats and a fixed viewing geometry help; a calibrated absorbance trace could test linearity during conversion instead of relying on an endpoint alone. That extension would require a reference blank and a justified Beer-Lambert law calibration, rather than treating visible colour as an exact concentration. A zero-order reaction approximation in iodine must fail at depletion; it cannot imply consumption after no iodine remains.

A model in which each excitation changes volume by a fixed increment $v$. If the all-ground configuration has volume $V_0$, then $V=V_0+v⟨N_{\mathrm{ex}}⟩$. Energies and statistical weights must be specified under the relevant mechanical constraint; volume endpoints alone do not define arbitrary intermediate configurations.

For a homogeneous cosmic fluid, the physical volume bounded by fixed comoving coordinates obeys $V\propto a^3$. The cosmological continuity equation therefore gives $d(\rho c^{2}V)=-PdV$. This is the first law of thermodynamics with conserved entropy and no additional chemical work. The internal energy of that volume need not be constant: a component with negative pressure gains internal energy as the universe expands. Local stress-energy conservation does not imply a global conserved energy for an arbitrary expanding spacetime.

The affine breathing mode changes all three principal axes by the same fraction and therefore changes the stellar volume and density.

An affine quadrupole mode has axis perturbations whose fractional sum vanishes. It changes the ellipsoidal shape without changing volume to first order.

A device that transfers air into or out of a vessel. Historical vacuum experiments used a receiver, valves and a mechanical pump; leakage limited the attainable pressure.

A statistician and professor at the London School of Economics.

A scaling model of overlapping end-grafted polymers assigns each chain a column of lateral width $D\sim s^{-1/2}$ at grafting density $s$. Each concentration blob contains $g\sim (D/a{)}^{1/\nu }$ statistical segments, and stacking $N/g$ blobs gives polymer brush height $H\sim ND/g$. For assumed segment volume $a^3$, the mean volume fraction is $\varphi \sim N{a}^{3}s/H$. It assumes a long many-segment-blob brush and supplies scaling powers rather than a precise chain-end distribution.

Scientist associated with Avogadro's law. Authentic institutional portrait: Science History Institute portrait record.

For an amphiphile, let $v$ be its hydrophobic-tail volume, $a_0$ its effective head-group area, and $\ell$ its maximum tail extension. The dimensionless packing parameter $P=v/(a_0\ell )$ estimates compatible aggregate curvature. For a spherical micelle of core radius $r$ containing $n$ molecules, $nv=4\pi r^3/3$ and $n{a}_0=4\pi r^2$, giving $v/a_0=r/3$. Since $r\le \ell$, spherical packing requires $P\le 1/3$. The analogous long-cylinder calculation gives $v/a_0=r/2$ and the limiting value $1/2$; a planar monolayer has $v/a_0$ equal to its thickness and limiting value one. Thus small $P$ favors spherical micelles, intermediate values favor cylindrical aggregates, and values near one favor lipid bilayers. This is a geometric guide, not a full thermodynamic prediction of every phase boundary.

Side losses need not destroy the method. Let $\theta$ be the small temperature disturbance about a steady background in surroundings at fixed temperature. For a rod of cross-sectional area $S$ and perimeter $\mathcal{P}$, a uniform real surface heat transfer coefficient $h_s$ produces a loss per unit volume of $H\theta$, with $H=h_s\mathcal{P}/S$. Assume constant $k,\rho ,c_p,H$, a cross-section nearly uniform in temperature, and no distributed heat source. The balance becomesHere $H$ has units $\mathrm{W}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, while $\lambda$ has units ${\mathrm{s}}^{-1}$. This is a distributed version of Newton's law of cooling; a constant linearized contribution from thermal radiation may be included over a small temperature range. It does not represent arbitrary time-dependent coupling to a surrounding material with its own thermal storage.

For the decaying exponential branch, the phasor equation givesso thatReal positive side loss increases $a$ and decreases $b$ at fixed angular frequency and thermal diffusivity. Their product eliminates $\lambda$, yielding the combined inferenceThis result requires that attenuation and lag belong to the same exponential complex wavenumber at the same frequency, and that the loss term is real. It does not generally survive a superposition caused by a nearby finite end, spatially varying coefficients, transverse temperature gradients, or unequal sensor delays. The inferred $\lambda =\alpha (a^2-b^2)$ supplies an additional model check; a negative value signals inconsistency with this passive-loss model rather than negative physical cooling.

The bulk photon mode density per volume per angular frequency is ${\omega }^2/({\pi }^2{c}^3)$, including two transverse polarizations. Multiply it by the thermal oscillator energy $\hslash \omega /(e^{\hslash \omega /(k_{B}T)}-1)$ to obtain the displayed spectrum. Its integral is $u={\pi }^2{k}_B^4{T}^4/(15{\hslash }^3{c}^3)$. This is volume energy density rather than radiance or surface exitance; isotropic equilibrium radiation has outward black-surface flux $cu/4$. zero-point energy is excluded from this thermal spectrum.

For a Gaussian kernel $K{q}_{\parallel }^2+L{q}_{\perp }^4+t$, keeping both gradient coefficients fixed gives $q_{\parallel }^{\prime }=b{q}_{\parallel }$ and $q_{\perp }^{\prime }=b^{1/2}{q}_{\perp }$. The volume rescales by $b^{(d+1)/2}$. Fourier-field normalization is $m^{\prime }(q^{\prime })=m(q)/b^{(d+5)/4}$; the real-space normalization instead divides by $b^{(3-d)/4}$. The thermal coefficient scales by $b^2$ and the uniform source by $b^{(d+5)/4}$. Thus ${\xi }_{\parallel }\sim t^{-1/2}$, ${\xi }_{\perp }\sim t^{-1/4}$ and the formal Gaussian free-energy powers are $\alpha =(7-d)/4$, $\Delta =(d+5)/8$.

For the continuum dispersion ${\epsilon }_{\mathbf{k}}={\hslash }^2(k_x^2+k_y^2+k_z^2/\gamma )/(2m)$ with $\gamma >0$, substituting $q_z=k_z/\sqrt{\gamma }$ makes the energy isotropic and gives Jacobian $\sqrt{\gamma }$. Every momentum-state counting integral therefore acquires that factor. Per internal state and unit volume, $N=\sqrt{\gamma }(2m/{\hslash }^2{)}^{3/2}\sqrt{\epsilon }/(4{\pi }^2)$. The result assumes an unrestricted three-dimensional continuum; increasing this mass anisotropy is not the same operation as imposing a two-dimensional Brillouin zone and a finite interlayer band width.

Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.

A lumped arterial reservoir stores blood by elastic expansion. Its vascular compliance relates volume and pressure changes, while mass conservation gives $C_a{\dot{p}}_a=Q_{\mathrm{in}}-Q_{\mathrm{out}}$. This storage smooths intermittent ventricular input into continuing peripheral flow.

An astrophysical cooling function packages radiative losses of a plasma into a temperature-dependent coefficient. In the hydrogen-density-squared convention, the volume energy loss is $n_H^2\Lambda (T)$, so the coefficient has units $\mathrm{erg}\mathrm{c}{\mathrm{m}}^3{\mathrm{s}}^{-1}$. Its dependence on ionization state, chemical abundances and radiation field must be specified; a low-density collisional-equilibrium curve is not universal.

The ideal Atkinson cycle considered here has reversible adiabatic compression, isochoric heat addition, a longer reversible adiabatic expansion and isobaric heat rejection. Its expansion ends at the initial pressure rather than at the initial volume. With constant ideal-gas heat capacities, $r=V_1/V_2$ and $r_p=p_3/p_2$, the temperatures are $T_2=T_1{r}^{\gamma -1}$, $T_3=r_p{T}_2$, and $T_{4^{\prime }}=T_1{r}_p^{1/\gamma }$. This ideal cycle is distinguished from the detailed valve timing and losses of a practical engine.

For the same initial state, compression ratio and constant-volume heat input, the ideal Atkinson cycle adds net work during the extended expansion and subsequent constant-pressure return. The extra work is the integral of $p_{\mathrm{expansion}}-p_1$ over the added volume, which is positive. Equivalently, concavity of $x^{1/\gamma }$ gives $\gamma (r_p^{1/\gamma }-1)<r_p-1$ when $r_p>1$ and $\gamma >1$, so its efficiency exceeds the corresponding Otto cycle efficiency. The comparison fixes these shared conditions rather than asserting universal superiority under arbitrary engine constraints.