CND-22

Dulong–Petit law

C = 3 n R. Classical high-temperature heat capacity of a solid.

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PhononsDulong–Petit

Governing equation

C=3nR

$n \in \left[0.1, 10\right]$

where

n: Amount (mol)
C: Heat capacity (J/K)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

C=3nR

$n \in \left[0.1, 10\right]$

$n = 0.1$ $C = 2.494$

Along the curve

②

Domain of the variable — blue

n_{\min} = 0.1,\quad n_{\max} = 10

$n \in \left[0.1, 10\right]$

$n = 0.1$ $n = 0.1$

Along the curve

③

Response curve — green

C=3nR

$n \in \left[0.1, 10\right]$

$n = 0.1$ $C = 2.494$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}C}{\mathrm{d}n}

$n \in \left[0.1, 10\right]$

$n = 0.162$ $C^{\prime} = 24.94$

Along the curve

Lecture brief

Historical brief

Drude electrons, Bloch waves, BCS pairing (1957) and Wiedemann–Franz heat are the first solids-and-metals laws. The lab is conductivity, gap and phonon heat in closed form. This sheet (CND-22 — Dulong–Petit law) is the form associated with Dulong–Petit. Working symbols: $n$ $\rightarrow$ $C$ . Equipartition: 3 oscillators per atom, ½ kT kinetic + ½ kT potential each.

Purpose

Purpose: compute

C

from

n

in Condensed matter via

C=3nR

C = 3 n R. Classical high-temperature heat capacity of a solid. Use it when a real condensed matter question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

n = 1.000\,\mathrm{mol}

, the governing relation

C=3nR

yields

C = 24.94\,\mathrm{J/K}

. A saturating C bar at 3R. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Amount n1 mol

Outputs

Heat capacity C24.94 J/K

Reading speed

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Narration of this film

A saturating C bar at 3R.

Equipartition: 3 oscillators per atom, ½ kT kinetic + ½ kT potential each.

Reading speed

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