What is Kinetic temperature; equation of state? — Physics

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,

\langle\epsilon_k\rangle = \tfrac12 m\langle v^2\rangle = \tfrac32 k_BT ,

so that, combining with Theorem 20.6,

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 $u = \sqrt{\langle v^2\rangle} = \sqrt{3k_BT/m} = \sqrt{3RT/M}$ , $M$ 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 $P = nRT/V$ , higher for higher $T$ . A transformation is a path in this plane; a cycle, a closed loop (Chapter 24).

Examples

Example 20.8 (Numbers for air)

Nitrogen at $300\,\mathrm{K}$ : $u = \sqrt{3 \times 8.314 \times 300/0.028} = 517\,\mathrm{m}/\mathrm{s}$ ; hydrogen, $1930\,\mathrm{m}/\mathrm{s}$ ; the heavier the slower, as $1/\sqrt M$ . At $1\,\mathrm{bar}$ and $300\,\mathrm{K}$ , $n^* = P/k_BT = 2.4 \times 10^{25}\,\mathrm{m}^{-3}$ : $24\,\mathrm{L}$ per mole, $2.7 \times 10^{22}\,$ molecules in a liter. A classroom of $200\,\mathrm{m}^{3}$ : $8300\,\mathrm{mol}$ , $240\,\mathrm{kg}$ of air.

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