The Gaseous State — Part 2: Mixtures, real gases and liquefaction: visual lesson, CBSE Class 11 Chemistry

Lesson

Mixtures and real gases

Part 2 — gas mixtures, why real gases are not ideal, and how they liquefy. As in Part 1, this chapter is formative assessment only and is taught at textbook depth for school tests.

In this lesson you will

Use Dalton’s law, including the vapour pressure of water for gases collected over water
Explain deviations from ideal behaviour and read the compressibility factor curves
Explain the van der Waals equation qualitatively and the idea of critical temperature

Table contrasting ideal and real gases
Ideal gas	Real gas
Molecules have no volume	Molecules have a small but real volume
No attractions between molecules	Weak attractions exist and cause liquefaction
pV = nRT at all p and T	Obeys it best at low p and high T
Z = 1 always	Z differs from 1

Concept

Dalton’s law of partial pressures

For non-reacting gases in a container, the total pressure is the sum of the partial pressures: $p_\text{total} = p_1 + p_2 + \dots$ . The partial pressure of a gas is $p_i = x_i\,p_\text{total}$ , where $x_i$ is its mole fraction; each gas also obeys $p_iV = n_iRT$ .

p_\text{total} = p_1 + p_2 + p_3

p_i = x_i\,p_\text{total}

Table of partial pressure, mole fraction and Dalton law
Term	Meaning	Formula
Partial pressure	Pressure the gas would exert alone in the same volume at the same T	pᵢ = nᵢRT/V
Mole fraction	Fraction of the molecules that are gas i	xᵢ = nᵢ/n total
Dalton’s law	Total pressure of a mixture	p total = Σ pᵢ
Use	Breathing mixtures, gas analysis, collection over water	—

Worked example

Worked example: oxygen and nitrogen in a flask

Problem. 3.2 g of oxygen and 2.8 g of nitrogen are placed in a 4 L vessel at 300 K. Find the total pressure and the partial pressure of each gas. (O = 16, N = 14; R = 0.08314 L bar K⁻¹ mol⁻¹)

Step 1. $n(\ce{O2}) = 3.2 \div 32 = 0.10$ mol and $n(\ce{N2}) = 2.8 \div 28 = 0.10$ mol, so $n = 0.20$ mol.

Step 2. $p = \dfrac{nRT}{V} = \dfrac{0.20 \times 0.08314 \times 300}{4} = 1.25$ bar.

Step 3. The mole fractions are 0.50 each, so each partial pressure is 0.624 bar.

Table of moles, mole fraction and partial pressure of oxygen and nitrogen in the vessel
Gas	n / mol	Mole fraction	Partial pressure / bar
O₂	0.10	0.50	0.624
N₂	0.10	0.50	0.624
Total	0.20	1.00	1.25

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