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Use $v$ for molar volume to distinguish it from the fluctuating parcel volume below. The equation is $p=RT/(v-b)-a/v^2$. The parameter $b>0$ models excluded volume, while $a>0$ models the reduction in pressure from attractive interactions.

Below the critical temperature, a raw isotherm has a loop. The positive-slope segment is mechanically unstable and surrounding metastable portions do not give the stable equilibrium mixture. Replace the loop by a constant coexistence pressure $p_s$ satisfying the Maxwell equal-area condition,The endpoints are liquid and vapour molar volumes; intermediate mean volumes correspond to phase mixtures with changing proportions. Above the critical temperature, liquid-like and gas-like states connect continuously without a coexistence plateau.

In the pressure-temperature diagram, the liquid-vapour coexistence line ends at the critical point. Below it, the low-pressure side is vapour and the high-pressure side liquid. “Gas” conventionally denotes the region above the critical temperature, where compression alone cannot cross a liquid-vapour transition. The model has no solid phase.

The critical isotherm has an inflection: $p_v=p_{vv}=0$. Thuswhich giveFor fluctuations about a stable equilibrium value $x_0$, the system plus reservoir has probability proportional to $e^{-A(x)/(k_{B}T)}$. Expanding availability,gives a Gaussian distribution and, to this order, $⟨x⟩=x_0$ withThis requires positive curvature and sufficiently small fluctuations; it is not exact when higher-order terms matter.

The whole cylinder's volume is fixed. For the volume fluctuations of a small fluid parcel, interpret the small quantity as a material parcel containing $s$ moles, whose volume can fluctuate against the much larger remainder. Its mean volume at the critical density is $\bar{V}=s{v}_c=3sb$. At fixed temperature its availability is $A(V)=F_s(V,T)+p_{\mathrm{bath}}V$, soOn the critical isochore, approached from $T>T_c$,ConsequentlyEquivalently, $⟨\delta V^2⟩=k_{B}T{\kappa }_T\bar{V}$, with ${\kappa }_T=4b/[3R(T-T_c)]$. The mean-field variance diverges as $(T-T_c{)}^{-1}$, so its critical exponent is $1$; the root-mean-square fluctuation has exponent $1/2$. A fixed spatial observation box instead has number fluctuations, not fluctuating geometric volume. Finite size and non-Gaussian terms cut off the literal divergence; the approximation fails exactly at the critical point.

Let $E$ denote the fluctuating microscopic energy and $U=⟨E⟩$ its mean, at fixed $N,V$. In a canonical ensemble with $\beta =1/(k_{B}T)$,Since $dT/d\beta =-k_B{T}^2$, this givesHere $\Delta U$ denotes an energy fluctuation about the mean, as in the question. The relation is not restricted to ideal gases when the Hamiltonian operator has no explicit temperature dependence. Alternatively, the stated equilibrium free-energy curvature gives a Gaussian energy weight proportional to $\exp [-(\Delta U{)}^2/(2k_B{T}^2{C}_V)]$, with the same variance.

For the relative energy fluctuation of a monatomic ideal gas, $U=3N{k}_{B}T/2$ and $C_V=3N{k}_B/2$, henceTaking room conditions as approximately $T=300$ K and $p=10^5$ Pa, with $V=10^{-6}$ m${}^3$, givesThe mean energy is about $0.15$ J and its standard deviation about $2.49\times 10^{-11}$ J. The tiny relative fluctuation explains why macroscopic equilibrium internal energy behaves as a sharply defined thermodynamic quantity. The estimate uses a thermal reservoir, not a particle reservoir; allowing $N$ itself to fluctuate changes the relevant fluctuation problem.

Availability and total entropy. The reservoir changes by $d{U}_R=-dU$ and $d{V}_R=-dV$. Its particle number and the system's particle number are fixed. The reservoir's first law of thermodynamics therefore givesAdd the system's entropy change. Because $T_R,p_R$ are constant,Thus define the availabilityAn additive constant in $A$ is immaterial; its value above the equilibrium reference is the exergy. The Second law of thermodynamics gives $\Delta S_{\mathrm{tot}}\ge 0$, hence $\Delta A\le 0$. A stable equilibrium maximizes total entropy over the allowed variations and therefore minimizes $A$. The environmental $T_R,p_R$ enter this definition even when the system has not yet reached their values. Because both thermal and mechanical exchanges are allowed, equilibrium here includes mechanical balance; equality of temperatures alone does not ensure the minimum.

Probability and curvature. For a fixed value of a macroscopic coordinate $x$, let $S(x)$ count the compatible system microstates. The product of the system and reservoir multiplicities isAll factors independent of $x$ enter a normalization constant, givingHere $A(x)$ is the constrained or effective availability for this coordinate, including the multiplicity of any eliminated variables. A nonuniform coordinate measure must likewise be retained or incorporated in that effective availability.

At a stable interior minimum, $A^{\prime }(x_0)=0$ and $A^{\prime \prime }(x_0)>0$. Its Taylor expansion givesKeeping the quadratic term yields a normal distribution,This is availability curvature and equilibrium fluctuations. The quoted fluctuation formula holds at quadratic order in the Gaussian fluctuation approximation; it is not an exact identity for an arbitrary nonlinear $A(x)$. Higher terms can shift the mean away from its most probable value $x_0$.

Bubble equilibrium. Keep the number of air molecules fixed and include the surface energy in the system's total internal energy. For constant surface tension $\Gamma$, this contribution is $\Gamma \Omega =\Gamma \alpha V^{2/3}$, with no surface entropy under the stated approximation. Its differential isHenceIndependent thermal and volume variations give $T=T_R$ andThe air pressure is evaluated at $V_0$; since the bubble is closed it generally depends on volume, so this equation is implicit until an air equation of state is supplied. For a sphere, $\alpha =(36\pi {)}^{1/3}$, $\Omega =4\pi R^2$, and the pressure jump becomes $2\Gamma /R$, the Young–Laplace equation for a single gas–liquid interface. There is no second surface as in a thin soap film.

Isothermal volume fluctuations. With the air maintained at $T_R$, eliminate its entropy in favor of its Helmholtz free energy:The identity $(\partial F_{\mathrm{air}}/\partial V{)}_{T,N}=-p_{\mathrm{air}}$ givesUsing the curvature result,These are the volume fluctuations of a closed gas bubble. Stability requires the denominator to be positive. The negative surface contribution reflects the concavity of $V^{2/3}$; gas compression supplies the positive restoring curvature. For ideal air, ${\kappa }_T^{-1}=p_{\mathrm{air}}$, and mechanical equilibrium reduces the denominator toThe units are consistent: $D$ is pressure, so $k_B{T}_R{V}_0/D$ has units ${\mathrm{m}}^6$, the units of a volume variance. Holding $p_{\mathrm{air}}$ externally fixed while varying $V$ would remove the gas-compression term and change this stability problem; that is not the fixed-particle-number situation here.

Numerical size of the fluctuations. The volume conversion is important:Use the specified $\alpha =4.84$, $\Gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $p_R=10^5\mathrm{Pa}$. ThenWith the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$,Thus the fractional root mean square fluctuation and fractional variance are respectivelyThe absolute root mean square fluctuation is $2.01\times 10^{-19}{\mathrm{m}}^3$. The very small fractional value supports the local Gaussian fluctuation approximation. This estimate neglects exchange of dissolved air, variations of surface tension, shape fluctuations and temperature fluctuations, consistently with the prescribed fixed-number, isothermal bubble model.

Occupation and the two-dimensional Fermi energy. For an independent single-particle state, the Fermi-Dirac distribution isThe chemical potential enforces the chosen mean particle number through $N={\sum }_{j}n({\epsilon }_j)$, including spin states in the sum. It is common to all states in particle-exchange equilibrium and must generally change with temperature or field when the total mean number is held fixed. At zero temperature, the distribution becomes a Heaviside step function at the Fermi energy.

Let the sample area be $\mathcal{S}$ and let $g_s$ be its internal spin spin degeneracy. In the continuum limit, filling the disk $k<k_F$ givesThereforeFor spin-half fermions with both spin states available, $g_s=2$ and ${\epsilon }_F=\pi {\hslash }^{2}N/(m\mathcal{S})$; for a spin-polarized gas $g_s=1$. The question does not specify the area or spin degeneracy, so these conventions must be stated rather than extracting an energy from $N$ alone. Although the original text calls $\mathbf{k}$ momentum, the supplied ${\hslash }^2{k}^2/(2m)$ dispersion makes it a wave vector; physical momentum is $\mathbf{p}=\hslash \mathbf{k}$. The result agrees with the constant two-dimensional free-electron density of states. Discrete level spacings in a finite sample would require a specified boundary spectrum.

One level occupied on average. Write $\delta ={\mu }_{B}B$ and $z=e^{\beta (\epsilon -\mu )}>0$. The two mean spin occupations areImposing $N_{+}+N_{-}=1$ gives $2+z(e^{\beta \delta }+e^{-\beta \delta })=1+z(e^{\beta \delta }+e^{-\beta \delta })+z^2$, hence $z^2=1$. Thus, for every positive temperature,The total magnetic moment is $\mathcal{M}=-{\mu }_B(N_{+}-N_{-})$. ConsequentlyThis is the grand-canonical spin susceptibility of a half-filled single level. Empty, singly occupied and doubly occupied orbital states are allowed in the grand canonical ensemble. At zero field, the singly occupied configurations together have probability $1/2$, explaining the factor of two relative to one permanent spin-half moment. If exactly one electron were required at all times, a canonical two-state partition function would instead give $\mathcal{M}={\mu }_B\tanh (\beta {\mu }_{B}B)$ and $\chi ={\mu }_B^2/(k_{B}T)$. That different answer does not implement the stated average-number constraint.

Density-of-states formula for the susceptibility. Here $g$ is the zero-field single-spin density of states. Regard its argument $E$ as the full spin-shifted energy. The shifted functions are $g_{\pm }(E)=g(E\mp {\mu }_{B}B)$, soThe total number is even in $B$; thus $\mu (B)=\mu (0)+O(B^2)$ at fixed total number on a smooth branch. More directly, terms from ${\mu }^{\prime }(0)$ cancel in the spin difference because the two densities coincide at zero field. To first order,It follows thatFor a physical energy band, take $g$ to vanish below its lower edge. Provided the endpoint term vanishes, integration by parts also givesThis form displays the positive response and the narrow thermal window around $\mu$. A sharp band edge is included by the corresponding distributional derivative in the first expression. The susceptibility defined here is the derivative of total magnetic moment with respect to $B$, with units $\mathrm{J}{\mathrm{T}}^{-2}$; it is not automatically a dimensionless SI volume susceptibility. A density-of-states convention per unit volume likewise gives a response per unit volume. These are the field conventions for magnetic susceptibility.

Chemical potential at fixed number. Define the integrated single-spin state countThe probability kernel $K_T$ has unit integral, mean $\mu$, and variance $a(k_{B}T{)}^2$. Using integration by parts,Expand $G$ about $\mu$. The linear term integrates to zero, while $G^{\prime \prime }=g^{\prime }$ givesThis is the number form of the Sommerfeld expansion. Since $N/2=G({\epsilon }_F)$ at zero temperature, set $\mu ={\epsilon }_F+\delta \mu$ with $\delta \mu =O(T^2)$:ThereforeThe coefficient can be obtained directly. With $y=(E-\mu )/(k_{B}T)$,For example, the sum follows by expanding $e^{-y}/(1+e^{-y}{)}^2$ on $y>0$, using the evenness of the kernel and integrating each exponential. Thus $a/2={\pi }^2/6$. A density of states increasing with energy gives a downward quadratic chemical-potential shift. For the constant two-dimensional density of states that shift vanishes to this order; a finite lower band edge still allows exponentially small corrections.

Low-temperature susceptibility at fixed number. Apply the same kernel expansion to $g$ itself:The shift of $\mu$ must be retained in the first term. Writing $g_F=g({\epsilon }_F)$ and similarly for its derivatives,ConsequentlyThis is the fixed-number low-temperature Pauli susceptibility. The $g_F^{\prime \prime }$ contribution is thermal broadening at fixed chemical potential; the $-g_F^{\prime 2}/g_F$ contribution enforces fixed total number. Both are needed. If the density of states were instead defined with both spin directions included, the prefactor $2$ would be absorbed into that definition. Only the spin response has been calculated; no orbital magnetic response is included.

The expansion assumes $g_F>0$, sufficient smoothness near the Fermi energy, and a temperature small compared with the energy scale over which $g$ changes or encounters a band edge. It does not apply unmodified at a singular density of states. For the ideal two-dimensional parabolic band, $g_F^{\prime }=g_F^{\prime \prime }=0$, so the leading Pauli response is temperature independent, consistently with the earlier state count.

The one-step master equation for the biased random walk isExpand in a Taylor series and retain the first time derivative and the first two space derivatives. Since $w_{+}+w_{-}=1$ and $w_{-}-w_{+}=2ϵ$,Thus the usual continuum truncation gives the requested Fokker-Planck equation,The diffusion coefficient has units $\mathrm{lengt}{\mathrm{h}}^2/time$, while $C$ has units of velocity. The physical drift velocity is $v=-C=a(w_{+}-w_{-})/\tau$, because the standard advection term is $-v{\partial }_{x}P$. A rightward bias therefore means $ϵ<0$, $C<0$, and $v>0$.

The approximation must be stated carefully. The truncation is a controlled diffusion limit of a weakly biased random walk when $a\to 0$, $\tau \propto a^2$, and $ϵ\propto a$, keeping $D_0$ and $C$ finite. It is also the spatial diffusion approximation for a continuous-time jump process with exponentially distributed waiting times of mean $\tau$.

If the original fixed-step-time walk has a finite bias, its exact variance from (a) instead gives the diffusion coefficient of the Gaussian envelope about the drifting mean:One obtains the same correction by retaining $\frac{1}{2}{\tau }^2{\partial }_t^{2}P$ in the time expansion and using the leading advection equation to replace ${\partial }_t^{2}P$ by $v^2{\partial }_x^{2}P$. The result is $D_{\mathrm{fixed}}=D_0-v^2\tau /2$. In particular, a deterministic one-way walk has zero variance, as this expression requires. Thus $D_0$ is the usual intended first-time-derivative approximation, not an exact variance-growth coefficient at arbitrary bias. The later continuum estimates use that convention, with the fixed-time alternative identified where relevant.

For the binomial distribution, the binomial theorem givesDifferentiation of the probability generating function recovers the first expected value and second factorial moment:Since $X^2=X(X-1)+X$, the variance isThusThe formulas also hold for $n=0,1$ and the degenerate choices $p=0,1$, with derivatives read directly from the resulting polynomial.

For the discrete uniform distribution, each value has probability $1/n$. Using the given finite sums,Subtracting the square of the expected value gives the variance:Hence

Assume the variances are finite. Expand the square and apply linearity of expectation:Part (i), applied to the identity functions, makes the bracketed covariance zero by independence of random variables. Hence

Part (a) with $n=6$ gives $\mathbb{E}[S_1]=\mathbb{E}[S_2]=7/2$ and $\mathrm{var}(S_1)=\mathrm{var}(S_2)=35/12$. By linearity of expectation,Independence of random variables and $\mathrm{var}(-S_2)=\mathrm{var}(S_2)$ give the same variance for the sum and difference:

Put $S=X+Y$. Given $X$, its conditional expected value is $3X/2$ and its conditional variance is $X/4$, since $X$ is then fixed. The law of total expectation and law of total variance giveThe two scores are dependent: the law of total covariance gives $\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,X/2)=q/(2p^2)$. Adding their two variances without that covariance term would give an incorrect total.

Given $X=k$, the score $Y$ has a binomial distribution with success probability $1/2$. Its conditional expected value and variance are $k/2$ and $k/4$. ThusThis is binomial thinning of the first-round count; the second formula uses the identity $\mathbb{E}[Y^2\mid X]=\mathrm{Var}(Y\mid X)+\mathbb{E}[Y\mid X{]}^2$.

Each $X_i$ has the same probability distribution as $-X_i$. By independence, the entire vector and its negative have the same joint law, so $S_m$ has a symmetric distribution about zero. The two disjoint tail events consequently have equal probabilities:Apply Chebyshev's inequality with the mean and variance from (i):Using only containment in the two-sided event would give $8/(3m)$; the symmetry supplies the necessary extra factor $1/2$.

For each random variable, the probability mass function givesLet $S_m={\sum }_{i=1}^m{X}_i$. By linearity of expectation and variance additivity for independent random variables,The independence is needed for the variance calculation, but not for the mean.

Since $X^2=X(X-1)+X$, the probability generating function identities giveSubtracting the squared mean yields the variance

For the Poisson distribution with parameter $1$, the exponential series givesEvery derivative is $G^{(k)}(s)=e^{s-1}$, so the factorial moments areIn particular $G^{\prime }(1)=G^{\prime \prime }(1)=1$, and the variance is

For a continuous random variable with probability density function $f_X$, its mean isprovided ${\int }_{-\infty }^{\infty }|x|f_X(x)dx<\infty$. Its variance is the expectation of the squared deviation from its mean:This is finite when $\mathbb{E}[X^2]<\infty$; otherwise it may be $+\infty$ if the finite mean exists.

The second moment isSubtracting the squared mean gives the variance

The standard normal distribution has mean zero and variance one, so $\mathbb{E}[Z^2]=\mathrm{var}(Z)+(\mathbb{E}[Z]{)}^2=1$. Part (iv) therefore givesUsing the second moment from part (iii),

By linearity of expectation, $\mathbb{E}(S_n)={\sum }_{i=1}^{n}p=np$. Part (ii) gives $\mathrm{Var}(X_i)=\mathbb{E}(X_i^2)-[\mathbb{E}(X_i){]}^2=p-p^2$. Independent random variables have zero pairwise covariance, so the variance of their sum is the sum of their variances. Therefore

Write $F_X(x)=\mathbb{P}(X\le x)$ and $F_Y(x)=\mathbb{P}(Y\le x)$. For $0\le x\le \mu$, the factor $t/\mu$ is at most one throughout $0\le t\le x$, soFor $x>\mu$, compare the tails instead: $t/\mu \ge 1$ throughout $t\ge x$, givingRearranging provesThis is stochastic domination of $X$ by its size-biased distribution. The proof needs only the finite positive mean, not the finite variance assumed in (ii).