NJC 2023 Gases Lecture Notes Student Version

National Junior College SH1 H2 Chemistry 1 The Gaseous State Content • Ideal gas behaviour and deviations from it • pV = nRT and its use in determining a value for Mr • Dalton’s Law and its use in determining the partial pressures of gases in a mixture Learning Outcomes required for H2 (9729) Chemistry: Candidates should be able to • state and use the general gas equation pV = nRT in calculations, including the determination of Mr • use Dalton’s Law to determine the partial pressure of gases in a mixture • state basic assumptions of the kinetic theory as applied to an ideal gas • explain qualitatively in terms of intermolecular forces and molecular size: ➢ the conditions necessary for a real gas to approach ideal behaviour; ➢ the limitations of ideality at very high pressures and very low temperatures References: ➢ Chemistry in Context by G Hill & J Holman ➢ Chemistry for Advanced Level by Peter Cann & Peter Hughes Instructions to Students Period Activities 13 March – 9 April Self-Directed Learning Access: 1) Google classroom for Checkpoint suggested answers 2) SLS lesson - “Ideal and Real Gases” 3) SLS lesson - “Ideal Gas Laws” 10 April – 14 April 1 Tutorial for topic discussion 17 April – 21 April Class Test Copyright © 2023 National Junior College All Rights Reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording or any other information storage and retrieval system, without prior permission in writing from the copyright owner.

National Junior College SH1 H2 Chemistry 2 1 Introduction 1.1 The three states of matter Physical State Solid Liquid Gas Attractive forces between particles Strong Intermediate Weak Motion of particles Vibrate and rotate about fixed positions (no translation motion) Move freely throughout liquid (vibrate, rotate and translate) Move freely within the container (vibrate, rotate and translate) Space between particles Negligible Small Large Order of Packing of Particles Regular packing in crystals Loose clusters of particles No order; far apart Comparing Properties of Solids, Liquids and Gases Volume Fixed Fixed Particles will spread evenly throughout any container, thus volume is indefinite and shape is that of the container. Shape Fixed Takes the shape of the container but may not occupy it completely Relative Density High Moderate Low Relative Compressibility Negligible or almost nil Very slight Very easily compressed 1.2 Relationship between the three states of matter

National Junior College SH1 H2 Chemistry 3 2 The Gas Laws The properties of gases were studied as their behaviour deviated very much from that of solids and liquids. Scientists conducted experiments and derived gas laws from their results. The gas laws were formulated by measuring how the volume of a sample of gas varies with amount of gas, pressure and temperature. Success Criteria: • Be able to state and use the general gas equation pV = nRT in calculations, including the determination of Mr Note: SI units should be used consistently in calculations involving gas equations. 2.1 Parameters of Gas Laws Four parameters are needed to define the state of a gas. Parameter Symbol SI unit Conversion Amount of the gas n mol - Temperature of the gas T K (Kelvin) x C = (x + 273) K Volume of the gas V m3 1m3 = 1 × 103 dm3 1m3 = 1 × 106 cm3 Pressure of the gas p Pa (Pascal) 1 Pa = 1 Nm–2 1 atm = 101325 Pa 1 bar = 105 Pa 2.2 Avogadro’s Law At constant temperature and pressure, the volume of an ideal gas is proportional to the amount of gas. V ∝ n At standard room temperature and pressure (s.t.p): 105 Pa (1 bar) and 273 K (0 C) • 1 mol of gas occupies a volume of 22.7 × 10−3 m3 (22.7 dm3) • 2 mol of gas occupies a volume of 45.4 × 10−3 m3 (45.4 dm3) At room temperature and pressure (r.t.p): 101325 Pa (1 atm) and 293 K (20 C) • 1 mol of gas occupies a volume of 24 × 10−3 m3 (24.0 dm3) • 2 mol of gas occupies a volume of 48 × 10−3 m3 (48.0 dm3)

National Junior College SH1 H2 Chemistry 4 2.3 Boyle’s Law At constant temperature, the volume of a given mass of ideal gas is inversely proportional to the applied pressure. V ∝ 1 p Note: If the temperature data is given in degree Celsius, you must convert the value into Kelvin. Please refer to the table in 2.1 for the unit conversion. 2.4 Charles’ Law The volume of a given mass of ideal gas at constant pressure is directly proportional to its temperature expressed in Kelvin. V ∝ T 2.5 The Combined Gas Law Combining Avogadro’s Law ( V ∝ n), Boyle’s Law ( V ∝ 1 p) and Charles’ Law (V ∝ T), we have: V ∝ nT p V = constant × nT p pV nT = constant The exact value of the constant depends on the amount of gas (n). Thus for 1 mol of an ideal gas, the constant (i) is denoted by the symbol R (ii) is known as the molar gas constant (iii) has a value of 8.31 J K−1 mol−1 (in SI unit) Note: The value of R is taken to be 8.31 J K−1 mol−1 in the ideal gas equation, the following units must be used. Parameter Units p Pa V m3 T K 2.6 The Ideal Gas Equation Since for 1 mol of gas, pV T = R Therefore for n mol of gas, pV T = nR Rearranging, we have the ideal gas equation pV = nRT

National Junior College SH1 H2 Chemistry 5 Worked Example 1 At 150 × 103 Pa, a fixed mass of an ideal gas occupies 300 × 10–6 m3 at temperature T. What will be its volume at 250 × 103 Pa at the same temperature? Using the ideal gas equation, p1V1 = nRT – (1) and p2V2 = nRT – (2) Given a fixed amount of ideal gas at constant temperature T, (1) = (2) p1V1 = p2V2 (150 × 103)(300 × 10–6) = (250 × 103)(V2) V2 = 1.80 × 10–4 m3 Note: You must always convert the given temperature in degree Celsius to Kelvin scale. Worked Example 2 A sample of neon gas occupies a volume of 4.0 × 10–3 m3 at 50 C. Determine the new volume occupied by the neon gas when the sample is heated to 80 C at constant pressure. Using the ideal gas equation, pV1 = nRT1 – (1) and pV2 = nRT2 – (2) Given a fixed amount of Neon (g) at constant pressure, (1) (2) V1 V2 = T1 T2 4.0 × 10-3 V2 = 50 + 273 80 + 273 V2 = 4.37 × 10–3 m3 Note: We can also use the volume in dm3 for calculations p1V1 p2V2 = T1 T2 (1.0 × 105)(6 dm3) (0.45 × 105)(V2) = (22 + 273) (–21 + 273) V2 = 11.4 dm3 Worked Example 3 An inflated balloon has a volume of 6 dm 3 at sea level (1.0 × 105 Pa) and is allowed to ascend in altitude until the pressure is 0.45 × 105 Pa. During ascent, the temperature falls from 22 C to –21 C. Calculate the volume of the balloon at its final altitude. Using the ideal gas equation, p1V1 = nRT1 – (1) and p2V2 = nRT2 – (2) Given a fixed amount of gas inside the balloon, (1) (2) p1V1 p2V2 = T1 T2 (1.0 × 105)(6 × 10−3) (0.45 × 105)(V2) = (22 + 273) (–21 + 273) V2 = 11.4 × 10–3 m3

National Junior College SH1 H2 Chemistry