Consider Again Perrin s Data on the Number of Mastic Particles as a Function of Height

Previous answers are correct but incomplete, in that I suppose you practice non much about Statistical Mechanics so maybe information technology's bettere to elaborate a trivial.

On a historical note, this is the same question that led Einstein to codify is theory of Brownian motion and the same system I will describe now has been used past Perrin to measure out Avogadro'southward number with a Nobel-winning experiment.

I will endeavour to derive everything from uncomplicated formulas to make the answer articulate.

EDIT: the author mentions the gas having costant PVT at equilibrium, whereas I Will assume that we are dealing with a system of N particles bars in a volume Five and in contact with a thermostat (or with the environment) and then that the temperature is always T. This situation I think amend describes the everyday situation (eastward.g. sedimentation of colloids at room temperature) and it is easier to handle. The concepts about equilibrium in the respond are Full general and I Remember the writer did not cull those boundary conditions for a item reason.

We beginning past defining our system: a cylinder of base of operations $A$ and volume $5$ with inside $Northward$ molecules with mass $m$ at a temperature $T$. Gravity is acting parallel to the sides of the cylinder. The gas is ideal and then that $P(z) V(z) = North(z) k_B T$. Notice that while temperature is defined (as the mean kinetic energy is not afflicted by gravity), force per unit area and number of particles are a function of the tiptop $z$ because gravity volition pull moleculues to the bottom and a gradient of density (and thus of pressure) will exist formed as you suppose. Moreover $V(z)$ is the volume considered which contains $N(z)$ molecules. Call up of it as a piece of the cylinder at height $z$.

Why doesn't everything sediment to the bottom and why are we still talking nearly equilibrium?

Molecules are still able to randomly move due to thermal fluctuations i.east. to diffuse. Yet as gravity pulls them to the bottom, a situation is created in which, since in the bottom there are more molecules than at the top, the probability of a molecule jumping "up" is higher than ane jumping "downwardly", and so that as molecules sediment to the bottom they as well manage to "jump upwards" a little. The more molecules at a given height, the more than the probability of "jumping up" (the aforementioned holds for "jumping down", but at the top at that place are less molecules).

Equilibrium means that the two fluxes (one upward due to diffusion and one downward due to sedimentation) compensate and so that there is no net flux of particles. The concluding (equilibrium distribution $N(z)$ (come across image) is the 1 for which, at any $z$ the probability of jumping upwardly/down are perfectly compensated.

Can we find this distribution?

Yes, in 2 ways. Notice that from now on we will call $\rho(z)=N(z)/V$.

1)The thermodynamic (piece of cake) way

We at present that $P(z)V(z)=N(z)k_B T$. This means $\rho(z) = Northward(z)/5(z)=P(z)/k_BT$.

Now, what is the pressure inside a cylinder slice (of base $A$ and height $h$) at a given peak $z$ from the bottom of the cylinder? It is given by the pressure "higher up" it $P(z+h)$, plus the weight of the slice (the number of particles $Northward(z)=\rho(z)V(z)$ inside the slice times $mg$) divided past the cylinder area $A$, so that

$$P(z)=P(z+h)+{\rho(z)Five(z) mg \over A}$$

Using $Five(z)=Ah$ and $\rho(z)=P(z)/k_BT$: $$P(z)=P(z+h)+{P(z) mgh}$$ past dividing by $h$ and taking the limit $h\rightarrow 0$ nosotros become a differential equation $${dP(z)\over dz}=-P(z){mg\over k_B T}$$

which is easily solved as:

$$P(z)=P(0)e^{-mgz\over k_B T}$$

So we got the pression gradient. Equally nosotros are in equilibrium, pressure is perfectly defined (despite beingness different) at whatever height. We only used the fact that there exists a land of the arrangement in which the number of particles at a given height does not modify over time (i. due east. equilibrium or steady state - notice that this two concepts are actually slightly unlike but this is non relevant here).

What is the distribution of particles? Again using $\rho(z)=P(z)/k_BT$ we get

$$\rho(z)={P(0)\over k_B T}e^{-mgz\over k_B T}=\rho(0)e^{-mgz\over k_B T}$$

which as mentioned earlier is Boltzmann's distribution in the instance of gravity.

Finally notice as well that this section is equivalent to putting, equally suggested by Chester Miller, the full force at any height to be vanishing equally $A\left(P(z+h)-P(z)\right)+N(z)mg=0$, where $N(z)mg$ is weight and $A\left(P(z+h)-P(z)\right)$ the difference in force on the two sides of the cylinder, would give the same result. Notice as well that if $one thousand=0$ (no gravity) you observe the normal result of constant pressure.

2)The statistical (hard) way

Just presume this the following formulas if you do not know them. They are quite easy to find anyway (await for mobility, diffusion electric current and migrate current).

Gravity creates a flux of particles $J_g=\rho(z) \mu m g$ where $\mu$ is the particle mobility then that $\mu grand grand$ is the velocity due to the gravitational field. The flux due to particles moving randomly (diffusion) is given by $J_d=-\mu k_B T{d\rho(z)\over dz}$. Past balancing fluxes then that $J_g=J_d$:

$$\rho(z) \mu m g=-\mu k_B T{d\rho(z)\over dz}$$

and once more we find

$${d\rho(z)\over dz} = -\rho(z){mg\over k_B T}$$ which is the aforementioned equation every bit before with solution

$$\rho(z)=\rho(0)e^{-mgz\over k_B T}$$

Annotation that you can use both procedures with any external field, irresolute the strength!

Both ways prove that, even though there is a slope of pressure, not just equilibrium exists (no fluxes of molecules and force balance) but that we can exploit information technology to find relevat results as $\rho(z)$ and $P(z)$ using what we know about equilibrium, due east.g. $PV=Nk_B T$.

P.s. Take a await at Perrin's book about how He exploited this results to win a Nobel prize!

wilfongfroldn.blogspot.com

Source: https://physics.stackexchange.com/questions/291621/could-a-gas-be-in-thermodynamical-equilibrium-even-with-a-gravitational-field

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