Gas Laws

25min Part 2 / Ch3 / Lesson 1

Prerequisites: 1-2-1

Objectives

Explain Boyle's law and Charles's law
Perform calculations using the ideal gas equation PV=nRT
Understand Dalton's law of partial pressures

Three Gas Laws

Boyle’s law (constant T): $PV = k$ (constant) → $P_1V_1 = P_2V_2$

Charles’s law (constant P): $\dfrac{V}{T} = k$ (constant) → $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$

Note: $T$ must be in absolute temperature (K). $T\ \mathrm{(K)} = t\ \mathrm{(°C)} + 273$

Boyle's law and Charles's law

Ideal Gas Equation

$PV = nRT$

Symbol	Meaning	Unit
$P$	Pressure	Pa or atm
$V$	Volume	L
$n$	Amount of substance	mol
$R$	Gas constant	$8.31\ \mathrm{J/(mol \cdot K)}$ or $0.0821\ \mathrm{L \cdot atm/(mol \cdot K)}$
$T$	Absolute temperature	K

At STP ( $0\ \mathrm{°C}$ , $1.013 \times 10^5\ \mathrm{Pa}$ ), $1\ \mathrm{mol}$ of any gas occupies $22.4\ \mathrm{L}$ .

Gas Mixtures and Partial Pressure

Dalton’s law of partial pressures: the total pressure of a gas mixture equals the sum of the partial pressures.

$P_{\text{total}} = P_A + P_B + P_C + \cdots$

Partial pressure: $P_A = x_A \cdot P_{\text{total}}$ ( $x_A$ : mole fraction)

Worked Example

Example: Ideal gas equation

Find the volume of $0.50\ \mathrm{mol}$ of an ideal gas at $27\ \mathrm{°C}$ and $2.0 \times 10^5\ \mathrm{Pa}$ . ( $R = 8.31\ \mathrm{J/(mol \cdot K)}$ )

$T = 27 + 273 = 300\ \mathrm{K}$

$V = \frac{nRT}{P} = \frac{0.50 \times 8.31 \times 300}{2.0 \times 10^5}$

$V = \frac{1246.5}{2.0 \times 10^5} = 6.2 \times 10^{-3}\ \mathrm{m^3} = 6.2\ \mathrm{L}$

Check Your Understanding

Q1 At constant temperature, if gas pressure is doubled, the volume becomes:

Q2 What is 0°C in Kelvin?

Q3 In PV = nRT, what does R represent?

Exercises

Q1. Find the mass of oxygen gas occupying $5.6\ \mathrm{L}$ at STP. ( $\mathrm{O_2} = 32$ )

Solution

$n = \frac{5.6}{22.4} = 0.25\ \mathrm{mol}$

$m = 0.25 \times 32 = 8.0\ \mathrm{g}$

Q2. A mixture of $\mathrm{N_2}$ ( $0.30\ \mathrm{mol}$ ) and $\mathrm{O_2}$ ( $0.20\ \mathrm{mol}$ ) has a total pressure of $1.0 \times 10^5\ \mathrm{Pa}$ . Find the partial pressure of $\mathrm{N_2}$ .

Solution

Mole fraction: $x_{\mathrm{N_2}} = \dfrac{0.30}{0.30 + 0.20} = 0.60$

Partial pressure: $P_{\mathrm{N_2}} = 0.60 \times 1.0 \times 10^5 = 6.0 \times 10^4\ \mathrm{Pa}$