气体摩尔体积 Molar volume of a gas — 公式推导、符号与局限

\[n = \frac{V}{V_m}, \qquad V_m \approx 22.4\ \mathrm{L\,mol^{-1}}\ (0\,{}^\circ\mathrm{C},\ 101.325\,\mathrm{kPa})\]

由理想气体状态方程from the ideal-gas law

\[V_m = \frac{RT}{p}\]

气体密度与相对密度gas density and relative density

\[\rho = \frac{M}{V_m},\qquad D = \frac{\rho_1}{\rho_2} = \frac{M_1}{M_2}\]

含义Meaning

同温同压下，1 mol 任何气体占有的体积相同，标准状况下约 22.4 L。

At the same temperature and pressure one mole of any gas occupies the same volume — about 22.4 L at STP.

由理想气体状态方程 \(pV = nRT\) 得 \(V/n = RT/p\)，与气体种类无关（阿伏加德罗定律）。
From \(pV = nRT\), \(V/n = RT/p\), independent of the gas (Avogadro’s law).
代入 T = 273.15 K、p = 101.325 kPa：
Insert T = 273.15 K, p = 101.325 kPa:
\[V_m = \frac{8.314 \times 273.15}{101325}\ \mathrm{m^3/mol} = 0.022414\ \mathrm{m^3/mol} = 22.4\ \mathrm{L/mol}\]
25 °C、101.325 kPa 时 \(V_m\) = 24.5 L/mol；IUPAC 现行“标准状况”为 0 °C、100 kPa，\(V_m\) = 22.7 L/mol。
At 25 °C and 101.325 kPa, \(V_m\) = 24.5 L/mol; with the current IUPAC STP (0 °C, 100 kPa) it is 22.7 L/mol.
密度 \(\rho = m/V = nM/(nV_m) = M/V_m\)，所以同温同压下气体密度之比等于摩尔质量之比。
Density \(\rho = m/V = M/V_m\), so gas densities at equal T, p are in the ratio of molar masses.

只适用于气体（含混合气体），不适用于液体和固体（标况下的水、SO₃、CCl₄ 不是气体！）。
Gases only — not liquids or solids (water, SO₃, CCl₄ are not gases at STP).
必须注明温度和压强；22.4 L/mol 只在 0 °C、1 atm 下成立。
Always state T and p; 22.4 L/mol holds only at 0 °C and 1 atm.
基于理想气体近似；高压、低温或强分子间作用的气体（NH₃、H₂O 蒸气）偏差可达数个百分点。
Ideal-gas approximation; deviations of a few per cent at high pressure, low temperature or for strongly interacting gases.

标况下 5.6 L O₂ 的物质的量为 5.6/22.4 = 0.25 mol，质量 8.0 g。

5.6 L of O₂ at STP is 0.25 mol, i.e. 8.0 g.