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Physics · Glossary

What is Kinetic temperature; equation of state?

Definition 20.7 University Physics — Year 1 · Chapter 20 — Kinetic Theory and the Perfect Gas

The temperature of a perfect gas is defined by the mean kinetic energy of its molecules,

⟨ϵk⟩=12m⟨v2⟩=32kBT,\langle\epsilon_k\rangle = \tfrac12 m\langle v^2\rangle = \tfrac32 k_BT ,

so that, combining with Theorem 20.6,

PV=NkBT=nRT,R=NAkB=8.314 J/(mol K),PV = Nk_BT = nRT, \qquad R = N_Ak_B = 8.314\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K}) ,

the equation of state of the perfect gas. The root-mean-square speed is u=⟨v2⟩=3kBT/m=3RT/Mu = \sqrt{\langle v^2\rangle} = \sqrt{3k_BT/m} = \sqrt{3RT/M}, MM the molar mass.

The Clapeyron diagram of a perfect gas: at each temperature the isotherm is a hyperbola P = nRT/V, higher for higher T. A transformation is a path in this plane; a cycle, a closed loop ().
The Clapeyron diagram of a perfect gas: at each temperature the isotherm is a hyperbola P=nRT/VP = nRT/V, higher for higher TT. A transformation is a path in this plane; a cycle, a closed loop (Chapter 24).

Examples

Example 20.8 (Numbers for air)

Nitrogen at 300 K300\,\mathrm{K}: u=3×8.314×300/0.028=517 m/su = \sqrt{3 \times 8.314 \times 300/0.028} = 517\,\mathrm{m}/\mathrm{s}; hydrogen, 1930 m/s1930\,\mathrm{m}/\mathrm{s}; the heavier the slower, as 1/M1/\sqrt M. At 1 bar1\,\mathrm{bar} and 300 K300\,\mathrm{K}, n∗=P/kBT=2.4×1025 m−3n^* = P/k_BT = 2.4 \times 10^{25}\,\mathrm{m}^{-3}: 24 L24\,\mathrm{L} per mole, 2.7×1022 2.7 \times 10^{22}\, molecules in a liter. A classroom of 200 m3200\,\mathrm{m}^{3}: 8300 mol8300\,\mathrm{mol}, 240 kg240\,\mathrm{kg} of air.

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