Physics 4311: Thermal Physics – HW Solution 3

due date: Wednesday, Feb 18, 2026

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Problem 1: Root-mean-square velocity of gas molecules (6 points)

\begin{array}{rcl} ⟨ {\vec{v}}^{2} ⟩ & = & ⟨ v_{x}^{2} + v_{y}^{2} + v_{z}^{2} ⟩ = 3 k_{B} T ∕ m \\ r_{𝑟𝑚𝑠} & = & \sqrt{⟨ {\vec{v}}^{2} ⟩} = \sqrt{3 k_{B} T ∕ m} \end{array}

mass of N $_{2}$ molecule: $4.65 \times 1 0^{- 26}$ kg

for $T = 293$ K: $v_{𝑟𝑚𝑠} = 511$ m/s, for $T = 1$ K: $v_{𝑟𝑚𝑠} = 29.8$ m/s

Problem 2: Velocity components (6 points)

$⟨ v_{x} ⟩ = 0$ , $⟨ v_{y}^{2} ⟩ \neq 0$ , $⟨ v_{z}^{3} ⟩ = 0$ , $⟨ v_{x} v_{z} ⟩ = 0$ , $⟨ v_{y}^{2} v_{z} ⟩ = 0$ , $⟨ v_{x}^{2} v_{y}^{4} ⟩ \neq 0$ .

Problem 3: Two-dimensional Maxwell distribution (12 points)

a)

g (v_{x}, v_{y}) = \frac{m}{2 π k_{B} T} e^{- \frac{1}{2} m (v_{x}^{2} + v_{y}^{2}) ∕ k_{B} T}

b)

\begin{array}{rcl} f (u) & = & \int d v_{x} d v_{y} g (v_{x}, v_{y}) δ (u - \sqrt{v_{x}^{2} + v_{y}^{2}}) \\ = & \frac{m}{2 π k_{B} T} \int 𝑣𝑑𝑣𝑑𝜙 e^{- \frac{1}{2} m v^{2} ∕ k_{B} T} δ (u - v) \\ = & \frac{m}{k_{B} T} u e^{- \frac{1}{2} m u^{2} ∕ k_{B} T} \end{array}

Problem 4: Low-speed molecules (16 points)

a)

0.05 k_{B} T = \frac{1}{2} m v_{𝑚𝑎𝑥}^{2} \Rightarrow v_{𝑚𝑎𝑥} = \sqrt{0.1 k_{B} T ∕ m}

b)

g (v_{x}, v_{y}, v_{z}) = {(\frac{m}{2 π k_{B} T})}^{3 ∕ 2} e^{- \frac{1}{2} m (v_{x}^{2} + v_{y}^{2} + v_{z}^{2}) ∕ k_{B} T}

c)

P (v < v_{𝑚𝑎𝑥}) = \int_{-}^{v_{𝑚𝑎𝑥}} v^{2} 𝑑𝑣 \int 𝑑𝜙 sin 𝜃𝑑𝜃 {(\frac{m}{2 π k_{B} T})}^{3 ∕ 2} e^{- \frac{1}{2} m v^{2} ∕ k_{B} T}

d)

\begin{array}{rcl} P (v < v_{𝑚𝑎𝑥}) & \approx & 4 π {(\frac{m}{2 π k_{B} T})}^{3 ∕ 2} \int_{0}^{v_{𝑚𝑎𝑥}} v^{2} 𝑑𝑣 = \frac{4}{3} π {(\frac{m}{2 π k_{B} T})}^{3 ∕ 2} v_{𝑚𝑎𝑥}^{3} \\ use result of part a) \\ P (v < v_{𝑚𝑎𝑥}) & \approx & \frac{4}{3} \frac{1}{\sqrt{π}} 0.0 5^{3 ∕ 2} \approx 8.41 \times 1 0^{- 3} \end{array}