Chapter 7 Practice Problems [Thermal & Statistical Mechanics I Studied #14]

Ex 7.3

Find the particle flux of nitrogen gas in the STP (standard Temperature & Pressure) state.

The way we computed the flux was: we wrote the number of gas molecules hitting a unit area per unit time as

and then, thinking of that number as being equal to the number passing through a unit area per unit time, if we add up~~ all of them for all $v$ and all $\theta$,

Adding it all up over v and θ: the θ integral of sinθ cosθ from 0 to π/2 gives ½, so the flux is ¼ n⟨v⟩, and with ⟨v⟩ = √(8k_BT/πm) and n = p/k_BT it becomes p/√(2πm k_BT)

The numbers for nitrogen at STP: 28 g per mol gives 0.028/(6.022×10²³) kg per molecule, k_B = 1.3807×10⁻²³ J/K, and 1 atm = 1.01325×10⁵ Pa = N/m²

Ex 7.5

What is the average kinetic energy of the gas molecules effusing through a small hole?

We have to solve the problem with the same idea as in Chapter 6. Multiply the gas molecules at each $v$ and $\theta$ by a weight KE, and add them all up, then just divide it all by the total count, and we can call that the average energy.

The number of molecules hitting the wall per unit area, per unit time = the number of molecules going out per unit area, per unit time, since they’re the same, so the number of molecules going out per unit area, per unit time =

That is,

The average energy of the effusing molecules: ⟨E⟩ = (m/2)⟨v³⟩/⟨v⟩ = 2k_BT

Because effusion preferentially selects molecules with higher energy, it ends up a bit bigger than the overall average kinetic energy $\frac32 k_BT$ , they say.

Prob 7.1

Inside a vacuum chamber designed for surface science experiments, in order to keep the surface clean, the pressure of the remaining gas is made as low as possible. Covering the surface with just a single layer requires $10^{19}$ atoms per square meter.

What pressure is needed so that, from the remaining gas, less than one layer is covered per hour (/h)?

You may assume that once a molecule strikes the surface, it sticks to it.

A monolayer per hour means at most 10¹⁹ molecules per m² per hour, so Φ = p/√(2πm k_BT) ≤ 2.77×10¹⁵ m⁻²s⁻¹ and p ≤ 2.77×10¹⁵ √(2πm k_BT)

Prob 7.3

A closed vessel is partially filled with liquid mercury, and above the liquid surface there is a hole of area $10^{-7}$ square meters. This vessel is placed in a high-vacuum region at a temperature of 273 K, and after 30 days have passed it was found to have lost $2.5\times10^{-5}$ kg in weight.

Estimate roughly the vapor pressure of mercury at 273 K (the relative molecular mass of mercury is 200.59).

The mass of one mercury molecule: 200.59 g per mol and 6.022×10²³ molecules per mol give about 0.033×10⁻²³ kg

The number of mercury molecules lost: 2.5×10⁻⁵ kg divided by 0.033×10⁻²³ kg gives 75.75×10¹⁸

The number leaving per second through 1 m² is Φ = p/√(2πm k_BT); through the 10⁻⁷ m² hole it is Φ′ = 75.75×10¹⁸ = p/√(2πm k_BT) × 10⁻⁷

Solving for p with m = 0.033×10⁻²³ kg, k_B = 1.3807×10⁻²³ J/K and T = 273 K gives p = 0.66963×10⁵

Prob 7.5

Gas is effusing into a vacuum through a small hole of area $A$. These particles are adjusted nicely so that they pass through a very small circular hole of radius $a$ on a screen located a distance $d$ away from the first hole.

Show that the rate at which particles come out from the second hole is $\frac14 nA\langle v\rangle\left(\frac ad\right)^2$ .

(Assume that after the gas has passed through the second hole no collisions occur, and that the relation $d \gg a$ holds.)

Hand-drawn sketch: gas comes through the small hole A, travels a distance d, and reaches a tiny circular hole of radius a on the screen

So we have to work out the number of gas molecules that go through $A$ and then through $a$ like this, meaning we need to count the gas molecules passing through $A$ at al~most $\theta\approx 0$ , but we don’t know that $\theta$ value…..

Let’s set it as an unknown and go go go

(It’ll soon turn out $\theta = a/d$, no worries no worries)

The number of gas molecules passing through a unit area per unit time, and through an area A per unit time: A v cosθ · ½ n f(v) dv sinθ dθ

Let’s count this number for all~~~ the kinds.

Adding up all speeds and the angles from 0 to θ′: A(n/2) ∫v f(v)dv ∫sinθ cosθ dθ = A(n/4)⟨v⟩ (½ − ½ cos 2θ)

For small θ, use the Maclaurin series: cos 2θ ≈ 1 − 2θ²

So the count comes out as A(n/4)⟨v⟩ θ′²

For small θ, θ ≈ sinθ ≈ tanθ = a/d, so the count is A(n/4)⟨v⟩ (a/d)²

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

In the effusion-rate integral, $dv$ appears twice in the first line (there is only one), and the antiderivative should be $\left[-\frac{1}{4}\cos 2\theta\right]$, not $\sin 2\theta$; from $0$ to $\pi/2$ it gives $\frac{1}{2}$, as used.
Under the square root of $\langle v^3\rangle$ the numerator should be $2^7 k_B^3T^3$ ($=128\,k_B^3T^3$), not $3^3 k_B^3T^3$, since $\langle v^3\rangle = \frac{4}{\sqrt\pi}\left(\frac{2k_BT}{m}\right)^{3/2}$. With it, $\frac{m}{2}\frac{\langle v^3\rangle}{\langle v\rangle} = 2k_BT$, as shown.
In the vacuum problem, the unit area is $1\ \mathrm{m^2}$, not $1\ \mathrm{m^3}$; the formula below it uses $\mathrm{m^2}$.
In the mercury problem, $75.75\times10^{18}$ is the number of atoms lost in the whole 30 days, not per second. Per second it is $75.75\times10^{18}/(30\times24\times3600\ \mathrm{s}) \approx 2.92\times10^{13}$, so the vapor pressure comes out at about $0.026\ \mathrm{Pa}$, not $0.66963\times10^{5}\ \mathrm{Pa}$.

Ex 7.3

Ex 7.5

Prob 7.1

Prob 7.3

Prob 7.5

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