Mastering Molar Gas Volume and the Ideal Gas Equation | Edexcel A-Level Chemistry | 掌握气体摩尔体积与理想气体方程

📚 Mastering Molar Gas Volume and the Ideal Gas Equation | Edexcel A-Level Chemistry | 掌握气体摩尔体积与理想气体方程

This article focuses on a core quantitative skill in Edexcel A-Level Chemistry: linking chemical amounts, in moles, to the volumes of gases. It begins with the molar gas volume rule at room temperature and pressure, then extends the idea to non-standard conditions using the ideal gas equation pV = nRT. The foundation for this topic often appears in Edexcel Combined Science as the rule that one mole of gas occupies about 24 dm³ at r.t.p.

本文聚焦 Edexcel A-Level 化学的核心定量技能：将化学物质的量（摩尔）与气体体积联系起来。文章先介绍室温常压下的摩尔气体体积规律，再扩展到非标准条件下使用理想气体方程 pV = nRT。该主题的基础常见于 Edexcel Combined Science 中：1 摩尔气体在室温常压下约占 24 dm³。

1. The Core Rule: One Mole of Gas at r.t.p. | 核心规律：室温常压下一摩尔气体

At room temperature and pressure, usually taken as 20 °C and 1 atmosphere, one mole of any gas occupies approximately 24 dm³. This value is often called the molar gas volume, represented as 24 dm³ mol⁻¹.

在室温常压（通常为 20 °C、1 个大气压）下，1 摩尔任何气体约占据 24 dm³。这个数值常被称为摩尔气体体积，记作 24 dm³ mol⁻¹。

V = n × 24 dm³ mol⁻¹

n = V ÷ 24 dm³ mol⁻¹

The rule works because gas particles are widely spaced. Under the same temperature and pressure, the average separation of particles is much larger than the particle size, so the gas volume is essentially determined by the number of moles, not by the identity of the gas.

这一规律成立的原因是气体粒子间距很大。在相同温度和压强下，粒子的平均间距远大于粒子本身尺寸，因此气体体积基本上由摩尔数决定，而几乎与气体种类无关。

• 1 mol of gas at r.t.p. ≈ 24 dm³ = 24 000 cm³
• 0.5 mol of gas at r.t.p. ≈ 12 dm³
• 3 mol of gas at r.t.p. ≈ 72 dm³

上述列表中，1 摩尔气体在室温常压下约 24 dm³，即 24 000 cm³；0.5 摩尔约 12 dm³；3 摩尔约 72 dm³。

2. Using Molar Volume With Balanced Equations | 在配平方程中使用摩尔体积

Gas volumes can be used directly in reaction calculations when the reaction occurs at constant temperature and pressure. In a balanced equation, the mole ratio between gases is the same as the volume ratio.

当反应在恒温恒压下进行时，气体体积可以直接用于反应计算。在配平方程中，气体之间的摩尔比等于体积比。

Consider the reaction between calcium carbonate and dilute hydrochloric acid:

例如碳酸钙与稀盐酸反应：

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

From the equation, 1 mole of CaCO₃ produces 1 mole of CO₂. At r.t.p., the volume of CO₂ produced by 0.050 mol CaCO₃ is:

由方程可知，1 摩尔 CaCO₃ 产生 1 摩尔 CO₂。在室温常压下，0.050 mol CaCO₃ 产生的 CO₂ 体积为：

V = 0.050 × 24 = 1.2 dm

Published by TutorHao | A-Level Revision Series | aleveler.com

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