Practice Problems: Chapter 19 [Thermal & Statistical Mechanics I Studied #74]

Prob 19.1

He atom, Xe atom, Ar atom, Kr atom: what is the average energy of the gas at room temperature, in eV?

Hint: do we even have to calculate all four?

Prob 19.1 by hand: all four are called atoms, so 3 degrees of freedom and ⟨E⟩ = (3/2)k_BT for each; R = N_A k_B = 0.0821 L·atm/(mol·K), k_B = 1.3806 × 10⁻²³ J/K, so ⟨E⟩ = (3/2)(1.3806 × 10⁻²³ J/K)(25 + 273.15 K)

At 25°C, that is about $0.0385\,\mathrm{eV}$ per atom for all four.

Prob 19.3

A particle at position $r$ is inside the following potential well $V(r)$.

The potential well V(r) = A/rⁿ − B/r

Here $A$ and $B$ are positive constants and $n \gt 2$. Show that the bottom of the well can be approximated by a quadratic term. So, assuming the equipartition theorem is valid in this situation, find the average energy of the particle above the bottom of the well at temperature $T$.

Taylor series of V(r): V′(r) = −nA r^−(n+1) + B r^−2 and V″(r) = n(n+1)A r^−(n+2) − 2B r^−3, so near r₀, V(r) ≈ V(r₀) + V′(r₀)(r − r₀) + ½V″(r₀)(r − r₀)²

The bottom of the well is where the slope is 0: V′(r₀) = 0 gives r₀ = (nA/B)^(1/(n−1)); with V(r₀) and V″(r₀) written out, V(r) ≈ V(r₀) + ½V″(r₀)(r − r₀)², a quadratic term, so the equipartition theorem holds

In the local classical approximation, the mean potential energy above the minimum is $k_BT/2$. If r is the single coordinate of a one-dimensional oscillator and we include its kinetic energy too, the total mean energy above the bottom is $k_BT$.

Prob 19.4

In Example 19.1, show the following:

To show: ⟨x_i²⟩ = k_BT/(2α_i)

Three quadratic terms: the probability of each x_i goes as e^(−α_i x_i²/k_BT); they are independent, so P(x₁, x₂, x₃) is the product e^(−(α₁x₁² + α₂x₂² + α₃x₃²)/k_BT) divided by its triple integral

Averaging: ⟨E⟩ = ∭E·P(x₁, x₂, x₃) dx₁dx₂dx₃ splits into three one-variable ratios, each ½k_BT, so α_i⟨x_i²⟩ = ½k_BT and ⟨x_i²⟩ = k_BT/(2α_i)

Prob 19.5

If the energy $E$ of the system is not a quadratic term but is given like The energy E = α|x| with α positive , show that the average energy is The result to show, ⟨E⟩ = k_BT .

For E = αx with x from 0 to ∞: p(x) = e^(−αx/k_BT) / ∫₀^∞ e^(−αx/k_BT) dx, so ⟨E⟩ = α ∫₀^∞ x e^(−αx/k_BT) dx / ∫₀^∞ e^(−αx/k_BT) dx

Doing the integral: ∫₀^∞ e^(−αx/k_BT) dx = k_BT/α; d/dα of both sides gives ∫ x e^(−αx/k_BT) dx = (k_BT)²/α², so ⟨E⟩ = ⟨αx⟩ = k_BT

The general case E = α|x|ⁿ with n = 1, 2, 3, …: the substitution t = αxⁿ/k_BT, and ⟨E⟩ = −αk_BT times d/dα of ∫e^(−αxⁿ/k_BT) dx, over ∫e^(−αxⁿ/k_BT) dx; the sheet stops here

(For the very last problem, I got this far but I can’t get to the conclusion. I’m putting it up here because I’d love for someone to point out what I’m missing.)

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

In the last line of the general case $E = \alpha|x|^n$, the minus sign of the line above stays: $\langle E\rangle = -\alpha k_BT\,\frac{\frac{d}{d\alpha}\int e^{-\alpha x^n/k_BT}dx}{\int e^{-\alpha x^n/k_BT}dx}$. The integral is proportional to $\alpha^{-1/n}$, so this finishes at $\langle E\rangle = \frac{k_BT}{n}$ ($\frac{1}{2}k_BT$ for $n = 2$, as equipartition says, and $k_BT$ for $n = 1$, as in the problem before).
A quadratic approximation does not by itself guarantee classical equipartition: the thermal motion must stay near the minimum and quantum level spacing must be negligible on the temperature scale.

Prob 19.1

Prob 19.3

Prob 19.4

Prob 19.5

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