Chapter 6 Practice Problems [Thermal & Statistical Mechanics I Studied #12]

Ex 6.2

What volume does 1 mol of ideal gas occupy at STP (standard temperature & pressure)?

Ideal gas law at STP: pV = n_m RT with 1 atm, 1 mol, 0.082 atm·L/(mol·K) and 273 K gives V = 22.386 L = 0.022386 m³

Ex 6.3

What is the relation between pressure and kinetic energy density?

The sum of the kinetic energies of the $n$ particles contained in unit volume can be called the kinetic energy density ($u$).

When we derived pressure $p$ (in the previous post Ideal Gas Equation of State and Pressure (Thermal & Statistical Mechanics I Studied #11)),

The end of the pressure derivation from the previous post: the integral splits into a v part, which is ⟨v²⟩ = 3k_BT/m, and a θ part, which is 1/3, so p = nk_BT

we showed that $p = \frac13 nm\langle v^2\rangle$ . Therefore,

Prob 6.2

Calculate the kinetic energy density ($u$) for air at atmospheric pressure.

Air at atmospheric pressure is about 10⁵ Pa, so u = (3/2) × 10⁵ Pa = 1.5 × 10⁵ J/m³

Prob 6.3

Mr. Fourier is sitting in a living room at 18 degrees Celsius. He thinks it’s a bit cold, so he adjusts the heating so the temperature becomes 25 degrees Celsius.

What happens to the total energy of the air in the living room? (hint: what controls the pressure in the room?)

The total energy U = uV = (3/2)pV = (3/2) n k_B V T

Prob 6.5

a) If the number of molecules hitting per unit area, per unit time, with the angle made with the direction perpendicular to the surface between $\theta$ and $\theta+d\theta$ and the speed between $v$ and $v+dv$ is given by the equation below,

show that the mean value of $\cos\theta$ for these molecules is $2/3$.

— that was the count. The count! Let’s multiply each of those numbers by the $\cos\theta$ for each of their respective angles (think of it as a weight), sum them all, and then divide by the total number of hits to get the average.

The weighted average of cosθ over the hits works out to (1/3)/(1/2) = 2/3

b) Using the result above again, show that for a gas satisfying the Maxwell distribution, the average energy of all the molecules is $\frac32 k_BT$ but the average energy of molecules hitting the surface is 2k_BT .

The average of $E$ for the ones that hit…… hmm. Multiply each of those numbers by the KE for each of their respective motions (think of it as multiplying by a weight), sum them all, and divide it by the total number of hits.

The average kinetic energy of the molecules that hit: (1/4)mn⟨v³⟩ / ((1/2)n⟨v⟩)

Working out ⟨v³⟩ with the Maxwell speed distribution and the formula ∫x⁵e^(−αx²)dx = 1/α³

(proof of integration formula Proofs of Integrals Commonly Used in Thermal and Statistical Mechanics (Reference Links) (Thermal & Statistical Mechanics I Studied #9))

The average energy of the molecules that hit works out to 2k_BT

(proof of $\langle v\rangle$ Expected Speed, Expected Energy, and Statistical Analysis of Ideal Gas Speed (Thermal & Statistical Mechanics I Studied #10))

Prob 6.6

Molecules within some gas move with different velocities from one another. A specific molecule moves with velocity $\mathbf{v}$, with speed $|\mathbf{v}|$, at an angle $\theta$ with respect to some fixed axis.

In some gas, it has already been shown that the number of molecules whose speed is between $v$ and $v+dv$ and whose angle with the specified axis is between $\theta$ and $\theta+d\theta$ is

The number of molecules with speed between v and v+dv and angle between θ and θ+dθ: ½ n f(v) dv sinθ dθ

Here $n$ is the number per unit volume, and $f(v)$ is a function of $v$. ($f(v)$ could be the Maxwell distribution given above, but do not assume so and calculate for the general case.)

Integrate and obtain the following. ($u$ is one rectangular coordinate component of $\mathbf{v}$.)

Let me interpret the problem first.

Our

The Maxwell speed distribution f(v) = (4/√π)(m/2k_BT)^(3/2) v² e^(−mv²/2k_BT)

how this was derived,

How the speed distribution f(v) comes out of the Maxwell-Boltzmann distribution of one velocity component

something like this was how we derived it, but here it’s saying to also consider the case when it’s not the Maxwell-Boltzmann distribution….. $f(v)$

The Maxwell speed distribution f(v) again

I think it’s saying not to rewrite it as this exponential term, but just go with $f(v)$ as is.

And the idea for solving the problem, we have to use the idea from Prob 6.5!!!! So let’s solve the problem gogo

a) $\langle u\rangle = 0$

Since $u$ was said to be one of the rectangular coordinate components, we can just flat-out set $u = v\cos\theta$!

Because the number that $\frac12 nf(v)\,dv\sin\theta\,d\theta$ represents is the number of things with the same $\cos\theta$ value for the direction, so we can think of that direction as one of the coordinate axes!!! Then the average of $v\cos\theta$ will be obtained by multiplying each number having $v$ and $\theta$ by the weight $v\cos\theta$, adding them all~~~ up, and then dividing by the total number again to get the average!

The average of v cosθ works out to ⟨u⟩ = 0

An obvious result. For a component in any one direction, it’s distributed evenly in all directions, so it’s obvious that the average in that component direction comes out to 0!!!

b) $\langle u^2\rangle = \frac13\langle v^2\rangle$

Let’s also take a look at the square of that component. Calculating the average by the same principle,

The average of u² = v²cos²θ works out to ⟨u²⟩ = ⅓⟨v²⟩

Without assuming the Maxwell-Boltzmann form, all we know is ∫f(v)dv = 1, so ⟨v²⟩ cannot be tied to T and m

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

In the ideal-gas picture, the middle value $22.3486\ \mathrm{L}$ is a typo for $22.386\ \mathrm{L}$ ($0.082 \times 273 = 22.386$); the result $0.022386\ \mathrm{m^3}$ is right.
In the screenshot of the pressure integral from the previous post, $d\theta$ is written twice; there is only one, inside $\frac{1}{2}\sin\theta\,d\theta$.
The last line of the $\langle v^3\rangle$ calculation should be $\frac{4}{\sqrt{\pi}}\left(\frac{2k_BT}{m}\right)^{3/2}$; the picture turns the fraction upside down. The next picture repeats it and also drops the factor $m$: the ratio is $\frac{m}{2}\cdot\frac{\langle v^3\rangle}{\langle v\rangle}$. Its result, $2k_BT$, is right.
In the picture with the Maxwell-Boltzmann distribution, the velocity distribution should read $g(v_x) = \sqrt{\frac{m}{2\pi k_BT}}\,e^{-mv_x^2/2k_BT}$; the $\pi$ under the root is missing.