At $(x,0,0)$ the field is perpendicular to the loop plane. Its Biot-Savart numerator isThe leading denominator variation must be kept:The integral of $-xd{y}^{\prime }$ is zero, but its product with $3x^{\prime }/x$ is not. Keeping all terms of the leading nonzero order givesFor the counterclockwise rectangle, $\oint x^{\prime }d{y}^{\prime }=\Delta x\Delta y$ and $\oint y^{\prime }d{x}^{\prime }=-\Delta x\Delta y$. HenceThis transverse field is opposite to the magnetic dipole moment. Replacing every denominator by a constant before examining the term proportional to $x$ would give the wrong answer.

Use source coordinates $(x^{\prime },y^{\prime },0)$ and observation point $(0,0,z)$. The axial numerator in the Biot-Savart law isFor $|z|$ much larger than the loop dimensions, replace the denominator $[z^2+x^{\prime 2}+y^{\prime 2}{]}^{3/2}$ by $|z{|}^3$ to leading order. Transverse components cancel between opposite sides. The loop integral is twice its oriented area, soThereforeIt points along $\mathbf{m}$ on both halves of the dipole axis, not along the signed radial unit vector.

The largest gap field is at the apex: $|E{|}_{\max }=3V/d$. If the fixed breakdown threshold is $E_{\mathrm{bd}}$, a smooth parallel-plate gap permits $V_{\mathrm{flat}}=E_{\mathrm{bd}}d$, whereas this small hemispherical boss permitsThus the maximum voltage is reduced by a factor of three, within the stated local-field approximation. This comparison holds the material dielectric strength and gap $d$ fixed.

On the lower plane outside the boss, write $s$ for distance from the boss centre along the plane, so $s>a$. Differentiate the potential normally before putting $z=0$:Thus the normal field is $E_z=-E_0(1-a^3/s^3)$ andThe source uses $r$ for this distance on the plane. Near the rim it tends to zero, while far from the boss it recovers the uniform parallel-plate surface charge.

At the hemispherical surface $r=a$, the outward normal from the conductor into the gap is $\hat{\mathbf{r}}$. The tangential field is zero there andThe electrostatic interface boundary conditions give surface charge density $\sigma ={\epsilon }_0{E}_n$, because the interior electric field is zero. HenceIt is negative for a positive upper-plate voltage, greatest in magnitude at the apex, and zero at the equator. Its units are coulombs per square metre.

Choose a flat wall and let $\theta$ measure the angle from the direction towards it. Particles of speed $v$ in solid angle $d\Omega$ occupy fraction $f(v)dvd\Omega /(4\pi )$ of an isotropic gas. In time $dt$, those within normal distance $v\cos \theta dt$ reach the wall. Their number per area per time is therefore $nf(v)dvv\cos \theta d\Omega /(4\pi )$.

Only the incoming hemisphere contributes. Integrating gives the one-way collision rate per unit area,Its units are ${\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. This is not $n\overline{v}/2$: selecting the incoming half of the particles must also account for their mean normal velocity. For effusion through a thin hole whose diameter is much less than the mean free path, the same one-way rate applies because collisions within the opening can be neglected.

Denote the heated and cold vessels by subscripts $h,c$. For argon, the opposing effusive particle rates are equal in the steady state:Because $\overline{v}\propto \sqrt{T}$ and $p=n{k}_{B}T$,This is thermal transpiration in the free-molecular limit. It fixes a ratio; the initial conditions determine the absolute pressures.

For the stated two identical vessels, take their fixed volume to be $V$ and the combined system sealed. Initially it contains $2p_{0}V/(k_B{T}_0)$ atoms. HenceLet $q=\sqrt{T_1/T_0}$. Combining this inventory with $p_h=q{p}_c$ gives the closed-vessel thermal transpiration resultHeating makes the hot side less dense but at higher pressure. Both expressions reduce to $p_0$ when $T_1=T_0$. For a small imposed temperature rise $\Delta T$, the pressure difference starts as $p_0\Delta T/(2T_0)$.

If instead the cold vessel is interpreted as an externally maintained pressure reservoir with $p_c=p_0$, the answer becomes $\Delta p=p_0(q-1)$. That alternative requires the fixed-pressure reservoir assumption; it does not conserve the initial particle inventory of two sealed equal-volume vessels. Merely describing the vessels as large does not keep their pressures fixed through the redistribution. The pressures in the boxed answer assume the natural closed-vessel interpretation.

Let L be the vessel initially containing argon and R the vessel initially containing helium. Both have volume $V$ and fixed temperature $T_0$, and the aperture has area $A_h$. The incoming and outgoing rates of each species must be considered separately. For species $s$,Thus its density difference relaxes exponentially with time constantThe two-vessel effusive mixing solution for the four densities isUsing atomic masses of approximately $40$ and $4$ atomic mass units gives ${\tau }_{\mathrm{He}}/{\tau }_{\mathrm{Ar}}\simeq 1/\sqrt{10}\simeq 0.32$. Helium therefore transfers faster. Each species eventually reaches $n_0/2$ in each vessel, but its transient is different. In particular, the total density temporarily rises in the initially argon vessel and falls in the initially helium vessel, before both return to $n_0$.

For unequal maintained temperatures $T_L,T_R$, replace the rate equation byWith each species' total density sum still $n_0$, its final densities and relaxation time areEvery species has the same final density ratio $\sqrt{T_R/T_L}$ and partial-pressure ratio $\sqrt{T_L/T_R}$. Thus there is no permanent composition separation in this ideal steady state, although the rates remain mass dependent. A maintained hot side has lower density and higher pressure. Equal particle rates do not imply thermal equilibrium: atoms arriving from the hotter side carry more energy, so external heating and cooling are required to hold different temperatures. The model assumes thin-hole effusion, local equilibrium within each vessel and negligible aperture volume.

Wall arrivals are not an unbiased sample of the gas: faster atoms arrive more frequently. The normalized speed distribution among incident atoms isThe angular factor in the arrival rate cancels when calculating the mean kinetic energy. ThusUsing $I_5=2I_3/\alpha$ givesBy contrast, an atom chosen randomly from the bulk has mean kinetic energy $3k_{B}T/2$. The arrival weighting raises the normal-direction contribution to $k_{B}T$ while each tangential direction still contributes $k_{B}T/2$. This is also the mean energy carried by atoms effusing through a thin aperture from an equilibrated reservoir.

At every $T>0$, $V>b$,There is no constant-pressure coexistence interval, unstable loop or finite-temperature critical inflection. The isothermal compressibility is positive:Thus this equation of state predicts no liquid-gas phase transition in its physical domain. Its divergence at the excluded-volume boundary is a repulsive constraint, not a liquid-gas coexistence mechanism. This conclusion concerns the stated model and cannot rule out an ordering transition requiring additional physics or an additional thermodynamic variable. Extrapolation to absolute zero is also outside the classical gas model's reliable regime.

For the given one-mole model,At fixed positive temperature its isotherm is an ideal-gas hyperbola shifted right by $b$: pressure diverges as $V\to b^{+}$ and tends to zero as $V\to \infty$. Higher temperatures give higher pressure at any fixed allowed volume.

For positive $b$, the volume $V-b$ is the effective volume available to molecular motion. Thus $b$ describes excluded-volume repulsion or an effective incompressible molecular core. It is a molar volume in this one-mole equation, not necessarily the literal sum of molecular geometric volume. Setting $b=0$ recovers the ideal gas law. No attractive-energy term is included.

An equation of state relates equilibrium state variables such as pressure, volume, temperature and material amount. At fixed composition, it reduces the number of independent thermodynamic variables. It does not by itself specify all caloric properties: information such as heat capacity or internal energy is also needed.

At a liquid-gas first-order transition, an equilibrium isotherm can contain a constant-pressure interval in which changing total volume changes the phase fractions. A metastable single-phase extrapolation can instead show a region with $(\partial p/\partial V{)}_T>0$; that region has negative isothermal compressibility and is mechanically unstable. Thermodynamic phase coexistence requires equal temperature, pressure and chemical potential, and a consistent free-energy description replaces the unstable region by coexistence. For an appropriate analytic fluid equation of state, a liquid-gas critical point occurs where $(\partial p/\partial V{)}_T=({\partial }^{2}p/\partial V^2{)}_T=0$.

These tests diagnose the phase behaviour represented by that equation of state. They do not guarantee detection of every possible transition involving an additional order parameter which the mechanical equation omits.

For a reversible adiabatic change, $dS=0$, equivalently $C_{V}dT=-pdV$. Substitute $p=RT/(V-b)$:Integration, with fixed positive reference values $T_{\ast },W_{\ast }$, givesSince $T=p(V-b)/R$ and $\gamma =1+R/C_V$,The relation applies to the quasistatic reversible adiabat. Adiabatic heat isolation alone does not enforce it for an irreversible process which generates entropy. The shifted volume $V-b$, rather than $V$ itself, is the appropriate coordinate in this model.

Use the enthalpy $H=U+pV$. The equation of state gives $pV=RT+pb$, soAt fixed pressure, $pb$ is constant. ThereforeSince $C_V$ and $R$ are constants, this ratio is independent of both temperature and pressure. Although the gas is not mechanically ideal when $b\ne 0$, its constant-pressure/constant-volume heat-capacity difference remains $R$ in this model.

Start with the fundamental relation at fixed material amount, $dU=TdS-pdV$, and expand entropy in independent variables $T,V$:The last equality uses the supplied Maxwell relation. ConsequentlyFor the excluded-volume ideal gas, $(\partial p/\partial T{)}_V=R/(V-b)=p/T$, so the bracket vanishes:Once $U$ depends only on temperature, $C_V=dU/dT$ cannot depend on volume either. Its stipulated temperature independence therefore makes it a constant, giving $U=C_{V}T+U_{\ast }$ with arbitrary energy reference $U_{\ast }$. Excluded volume changes the pressure without adding the volume-dependent attractive internal energy that a more general interacting-gas model might have.

For one mole, $pV=RT$, and constant heat capacities satisfy $C_p-C_V=R$. The specified diatomic-gas model hasA reversible adiabatic process satisfies $T{V}^{\gamma -1}=\mathrm{constant}$. Constant-pressure expansion satisfies $T\propto V$. Applying these in order givesEvery volume ratio is dimensionless, so the expressions have units of temperature. For the intended volume ordering, the minimum temperature is $T_1$ and the maximum is $T_3$.

For $\rho \to 1^{+}$, Taylor expansion gives ${\rho }^{\gamma }-1=\gamma (\rho -1)+O((\rho -1{)}^2)$. The vanishing-cutoff limit of the Diesel cycle isThe maximum temperature tends to $T_2=T_1{r}^{\gamma -1}$, while the minimum stays at $T_1$. A Carnot engine between those limiting extrema hasso the limiting efficiencies coincide. At exactly $V_3=V_2$, however, both heat input and net work vanish, and the cycle retraces its compression adiabat. The efficiency there is an undefined $0/0$; the boxed value is the continuous limit for neighbouring nonzero cycles, not finite work at zero heat input.

For a finite cutoff ratio $\rho >1$, the Carnot comparison uses $T_{\max }=T_3=T_1{r}^{\gamma -1}\rho$, giving ${\eta }_C=1-1/(r^{\gamma -1}\rho )$. The Diesel cycle has smaller efficiency. For example,which makes its rejected/input heat ratio larger than the Carnot ratio.

Use heat into the gas as positive. The two adiabatic stages have $Q_{12}=Q_{34}=0$. During constant-pressure heating and constant-volume cooling,The first law over a complete cycle gives $W_{\mathrm{out}}=Q_{23}+Q_{41}$, since the internal energy returns to its initial value. ThereforeSubstitute the temperatures:HenceEquivalently, define compression ratio $r=V_1/V_2$ and cutoff ratio $\rho =V_3/V_2$:These formulas use the ideal constant-heat-capacity model and reversible adiabatic strokes. The heat-addition and heat-rejection strokes occur over changing gas temperatures, unlike the isothermal heat exchanges of a Carnot engine.

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.

Archimedes' principle states that the resultant hydrostatic pressure force on a body equals the weight of the fluid displaced, directed opposite to gravity. Let $V_d$ be the displaced fluid region, with outward normal $\hat{\mathbf{n}}$ from the body into the fluid. Since the fluid pressure force is inward,Hydrostatic balance is $\nabla p={\rho }_f\mathbf{g}$, with downward gravitational acceleration $\mathbf{g}$. Thereforefor uniform fluid density and gravity. Uniform atmospheric pressure gives no net force on a closed surface. For a floating body, the same displaced-volume construction includes the free surface; any significant air buoyancy can be included as a second displaced fluid.

A freely floating body of mass $m$ satisfies $mg={\rho }_{f}g{V}_d$, so its displaced volume is $m/{\rho }_f$. A homogeneous body of density ${\rho }_b<{\rho }_f$ floats with submerged fraction ${\rho }_b/{\rho }_f$; if ${\rho }_b>{\rho }_f$, hydrostatic buoyancy alone cannot support it. The buoyant force acts through the centre of buoyancy, the centroid of the displaced volume for constant density, while weight acts through the centre of mass. Their relative positions determine rotational equilibrium and stability. For a fully submerged body, a centre of mass below the centre of buoyancy gives a restoring torque; a floating body's changing displaced shape additionally requires the metacentric height criterion. Surface tension, externally driven flow and contact forces can alter the simple force balance.