IB Physics DP Thermal Physics Notes PDF

Head to savemyexams.co.uk for more awesome resources YOUR NOTES IB Physics DP  3. Thermal Physi...

Head to savemyexams.co.uk for more awesome resources YOUR NOTES IB Physics DP  3. Thermal Physics CONTENTS 3.1 Thermal Concepts 3.1.1 Solids, Liquids & Gases 3.1.2 Temperature 3.1.3 Internal Energy 3.1.4 Specific Heat Capacity 3.1.5 Specific Latent Heat 3.1.6 Phase Change 3.1.7 Investigating Thermal Energy 3.2 Modelling a Gas 3.2.1 Ideal Gas Laws 3.2.2 Ideal Gas Equation 3.2.3 Kinetic Model of an Ideal Gas 3.2.4 Mole Calculations 3.2.5 Investigating Gas Laws Page 1 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.1 Thermal Concepts YOUR NOTES  3.1.1 Solids, Liquids & Gases Solids, Liquids & Gases The three states of matter are solid, liquid and gas The kinetic theory of matter is a model that attempts to explain the properties of the three states of matter In this model, particles are assumed to be small solid spheres Water has three states of matter; solid ice, liquid water and gaseous steam. The difference between each state is the arrangement of the particles Solids Particles in solids: Are held together by strong intermolecular forces Are closely packed Are arranged in a fixed pattern (lattice structure) Can only vibrate about their fixed positions Have low energies compared to particles in liquids and gases Page 2 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  In a solid, particles are arranged in a fixed pattern, with no spaces between them, and are only able to vibrate about their fixed positions As a result of the arrangement and behaviour of their particles, solids: Have a fixed shape (although some solids can be deformed when forces are applied) Have a fixed volume Are very difficult to compress Have higher densities than liquids and gases Liquids Particles in liquids: Are held together by weaker intermolecular forces compared to the forces between particles in solids Are closely packed Are randomly arranged (i.e. there is no fixed pattern) Can flow past each other Have higher energies than particles in solids, but lower energies than gas particles In a liquid, particles are arranged randomly and are able to flow past one another As a result of the arrangement and behaviour of their particles, liquids: Do not have a fixed shape and take the shape of the container they are held in Have a fixed volume Are difficult to compress Have lower densities than solids, but higher densities than gases Gases Page 3 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Particles in gases: YOUR NOTES Have negligible intermolecular forces between them  Are far apart (the average distance between the particles is ∼10 times greater than the distance between the particles in solids and liquids) Are randomly arranged Move around in all directions at a variety of speeds, occasionally colliding with each other and with the walls of the container they are in Are negligible in size compared to the volume occupied by the gas Have higher energies than particles in solids and liquids In a gas, particles can move around freely in all directions (shown by the arrows). As a result of the arrangement and behaviour of their particles, gases: Do not have a fixed shape and take the shape of the container they are held in Do not have a fixed volume and expand to completely fill the available volume Can be compressed Have the lowest densities (∼1000 times smaller than the densities of solids and liquids) Page 4 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES   Worked Example Liquids are about 1000 times denser than gases. Let d be the diameter of a molecule. Estimate the average intermolecular distance in a gas. Give your answer in terms of d. Step 1: Recall the equation for density m ρ= v Step 2: Write down the relationship between the density of a gas ρgas and the density of a liquid ρliquid ρ liquid = 1000 × ρ gas Step 3: Substitute into the density equation to show the relationship between the masses and volumes of a liquid a gas Page 5 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources m liquid ⎛⎜ mgas ⎞⎟ YOUR NOTES = 1000 × ⎜⎜⎜ ⎟⎟ v liquid v ⎟  ⎝ gas ⎠ Step 4: Since the mass stays the same, the relationship between the densities translates into a relationship between volumes as mass cancels out m liquid ⎛⎜ mgas ⎞⎟ = 1000 × ⎜⎜⎜ ⎟⎟ v liquid v ⎟ ⎝ gas ⎠ v gas = 1000 × v liquid Step 4: Relate the volume to the average distance between the molecules, x The average distance x between the molecules is related to the cube root of the volume x = ∛ 1000 v liquid = 10 × d Page 6 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.1.2 Temperature YOUR NOTES  Temperature Temperature is a measure of how hot or cold objects are Temperature also determines the direction in which thermal energy will flow between two objects (or between an object and its surroundings) When thermal energy is exchanged, the objects (or systems) involved are said to have a thermal interaction The thermal energy exchanged during a thermal interaction is referred to as heat During a thermal interaction: Thermal energy always flows from the hotter object to the colder object The energy transfer continues until the two objects are in thermal equilibrium (i.e. they both have the same temperature) Thermal energy can be transferred via conduction, convection or radiation Temperature is a scalar quantity and it is measured using a thermometer It is measured in degrees Celsius (°C) or kelvin (K) The kelvin is the SI base unit for temperature The temperature of an object is a macroscopic measure of the average kinetic energy of the particles (atoms or molecules) that make up the object Absolute temperature Absolute temperature is temperature measured in kelvin (K) Absolute zero is a temperature of zero kelvin (0 K) and corresponds to the temperature at which the average kinetic energy of the molecules is at its minimum The conversion between the Kelvin and the Celsius scale is given by: T(K) = T(°C) + 273.15 It is important to notice that differences in absolute temperatures correspond to differences in Celsius temperatures ΔT(K) = ΔT(°C) Where ΔT stands for temperature change The absolute temperature of a body is directly proportional to the average kinetic energy of the molecules within the body Page 7 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  The ice point is determined by placing a thermometer in a beaker containing melting ice, while the steam point is determined by placing the thermometer in a beaker with boiling water  Worked Example Give an estimate of room temperature in kelvin (K). Step 1: State a reasonable value for room temperature in degree Celsius (°C) room temperature (°C) ~ 20°C Step 2: Write down the conversion between Celsius scale and Kelvin scale T(K) = T(°C) + 273.15 Step 3: Convert the room temperature value and express it in kelvin (K) room temperature (K) ~ 293 K  Exam Tip Remember that the lowest possible temperature on the Kelvin scale is absolute zero (0 K). Therefore, if you are calculating temperature in kelvin and you end up with a negative number, you need to check your work, since negative numbers do not exist on the Kelvin scale. Page 8 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.1.3 Internal Energy YOUR NOTES  Internal Energy When a substance gains or loses thermal energy, its internal energy increases or decreases The internal energy of a substance is defined as: The sum of the total kinetic energy and the total intermolecular potential energy of the particles within the substance As thermal energy is transferred to a substance, two things can happen: An increase in the average kinetic energy of the molecules within the substance - i.e. the molecules vibrate and move at higher speeds An increase in the potential energy of the molecules within the substance - i.e. the particles get further away from each other or move closer to each other Since temperature is a measure of the average kinetic energy of the molecules, only an increase in the average kinetic energy of the molecules will result in an increase in temperature of the substance Due to thermal expansion, when the temperature of a substance increases, the potential energy of the molecules also increases When only the potential energy of the molecules changes, the temperature of the substance does not change This is the case for all state changes (e.g. melting, boiling) Page 9 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Exam Tip YOUR NOTES  Remember that a change in internal energy does not necessarily corresponds to a  change in temperature. A change in the average kinetic energy of the molecules corresponds to a change in temperature A change in the average potential energy of the molecules does not affect temperature Page 10 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.1.4 Specific Heat Capacity YOUR NOTES  Specific Heat Capacity The amount of thermal energy needed to change the temperature of an object depends on: The change in temperature required ΔT - i.e. the larger the change in temperature the more energy is needed The mass of the object m - i.e. the greater the mass the more energy is needed The specific heat capacity c of the given substance - i.e. the higher the specific heat capacity the more energy is needed The equation for the thermal energy transferred, Q, is then given by: Q = mcΔT Where: m = mass of the substance in kilograms (kg) ΔT = change in temperature in kelvin (K) or degrees Celsius (°C) c = specific heat capacity of the substance (J kg–1 K–1) The specific heat capacity of a substance is defined as: The amount of energy required to change the temperature of 1 kg of a substance by 1 K (or 1°C) This definition can be explained when the above equation is rearranged for c: Q c= m ∆T This means that, the higher the specific heat capacity of a substance the longer it takes for the substance to warm up or cool down Note that the specific heat capacity is measured in J kg–1 K–1 Page 11 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES   Worked Example A 2 kg piece of copper is kept inside a freezer at a temperature of –10°C. The copper is taken out of the freezer and placed into 5 litres of water at 20°C. A thermometer is placed into the water. After some time, the thermometer indicates that the water has cooled to 18°C.Determine the temperature of the copper at this time. Give your answer in degrees Celsius (°C). The specific heat capacity of water is 4200 J kg–1 K–1 The specific heat capacity of copper is 390 J kg–1 K–1 Step 1: Write down the known quantities Mass of copper = 2 kg Mass of water = 5 L = 5 kg Initial temperature of copper = –10°C Initial temperature of water = 20°C Final temperature of water = 18°C Change in temperature of water = 18°C – 20°C = –2°C Specific heat capacity of water = 4200 J kg–1 K–1 Specific heat capacity of copper = 390 J kg–1 K–1 Step 2: Write down the equation for thermal energy Q = mcΔT Step 3: Determine the energy transferred from the water to the copper The water is at a higher temperature than copper, hence thermal energy will flow from the water to the copper To quantify this energy, substitute numbers into the above equation In this case, the mass m is that of the water The specific heat capacity is that of water Since this is the energy lost by the water, it will be negative Q = 5 kg × 4200 J kg–1 K–1 × (–2°C) = – 42000 J Page 12 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Step 3: Determine the change in temperature ΔT of the copper YOUR NOTES The energy lost by the water is the same as the energy gained by the copper  Since this is the energy gained by the copper, it is positive The equation for thermal energy can be rearranged to calculate the change in temperature ΔT of the copper In this case, the mass m is that of the copper The specific heat capacity is that of copper Q 42000 ∆T = = = 54°C mc 2 × 390 Step 4: Determine the final temperature of the copper Since the copper gains thermal energy, its final temperature will be higher than its initial temperature final temperature of copper = ΔT + initial temperature of copper = 54°C – 10°C final temperature of copper = 44°C  Exam Tip You should notice that changes in temperature ΔT can usually be written in degrees Celsius (although this is not the SI base unit for temperature) and do not need to be converted into kelvin (K). This is because differences in absolute temperatures always correspond to differences in Celsius temperature.If the question asks to determine the initial or final temperature of a substance, make sure you always check the unit of measure (°C or K) in which you are required to give your final answer. Page 13 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.1.5 Specific Latent Heat YOUR NOTES  Specific Latent Heat During a phase change (i.e. a change of state) thermal energy is transferred to a substance or removed from it, while the temperature of the substance does not change In this case, the thermal energy is calculated as follows: Q = mL Where: Q = heat energy transferred (J) m = mass of the substance in kilograms (kg) L = specific latent heat of the substance in J kg–1 The specific latent heat of a substance is defined as: The amount of energy required to change the state of 1 kg of a substance without changing its temperature This definition can be explained when the above equation is rearranged for L: Q L= m This means that the higher the specific latent heat of a substance, the greater the energy needed to change its state Note that the specific latent heat is measured in J kg–1 The amount of energy required to melt (or solidify) a substance is not the same as the amount of energy required to evaporate (or condense) the same substance Hence, there are two types of specific heat: Specific latent heat of fusion, Lf Specific latent heat of vaporisation, Lv Specific latent heat of fusion is defined as: The energy released when 1 kg of liquid freezes to become solid at constant temperature This applies to the following phase changes: Solid to liquid Liquid to solid Therefore, the definition for specific latent heat of fusion could also be: The energy absorbed when 1 kg of solid melts to become liquid at constant temperature Specific latent heat of vaporisation is defined as: The energy released when 1 kg of gas condenses to become liquid at constant temperature Page 14 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources This applies to the following phase changes: YOUR NOTES Liquid to gas  Gas to liquid Therefore, the definition for specific latent heat of vaporisation could also be: The energy absorbed when 1 kg of liquid evaporates to become gas at constant temperature For the same substance, the value of the specific latent heat of vaporisation is always much higher than the value of the specific latent heat of fusion In other words, Lv > Lf This is because much more energy is needed to evaporate (or condense) a substance than it is needed to melt it (or solidify it) In melting, the intermolecular bonds only need to be weakened to turn from a solid to a liquid When evaporating, the intermolecular bonds need to be completely broken to turn from liquid to gas. This requires a lot more energy.  Worked Example Determine the energy needed to melt 200 g of ice at 0°C. The specific latent heat of fusion of water is 3.3 × 105 J kg–1 The specific latent heat of vaporisation of water is 2.3 × 106 J kg–1 Step 1: Determine whether to use latent heat of fusion or vaporisation We need to use the specific latent heat of fusion because the phase change occurring is from solid to liquid Step 2: List the known quantities Mass of the ice, m = 200 g = 0.2 kg Specific latent heat of fusion of water, Lf = 3.3 × 105 J kg–1 Page 15 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Step 3: Write down the equation for the thermal energy YOUR NOTES Q = mLf  Step 4: Substitute numbers into the equation Q = 0.2 kg × (3.3 × 105) J kg–1 Q = 6.6 × 104 J = 66 kJ  Worked Example Energy is supplied to a heater at a rate of 2500 W.Determine the time taken to boil 0.50 kg of water at 100°C. Ignore energy losses. The specific latent heat of fusion of water is 3.3 × 105 J kg–1 The specific latent heat of vaporisation of water is 2.3 × 106 J kg–1 Step 1: Determine whether to use latent heat of fusion or vaporisation We need to use the specific latent heat of vaporisation because the phase change occurring is from liquid to gas Step 2: Write down the known quantities Power, P = 2500 W Mass, m = 0.50 kg Specific latent heat of vaporisation of water, Lv = 2.3 × 106 J kg–1 Step 3: Recall the equation linking power P, energy E and time t E = Pt Step 4: Write down the equation for the thermal energy E The energy E in the previous equation is the thermal energy Q transferred by the heater to the water Q= mLf Step 5: Equate the two expressions for energy Pt = mLf Step 6: Solve for the time t t = 460 s Page 16 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.1.6 Phase Change YOUR NOTES  Phase Change A phase change happens whenever matter changes its state During a phase change, thermal energy is transferred to or from a substance This energy transfer does not change the temperature of the substance undergoing the phase change This means: The thermal energy provided (or removed) does not affect the kinetic energy of the molecules within the substance Only the potential energy (i.e. the spacing between the atoms or molecules) is affected The four main phase changes are: Melting - i.e. when a substance changes from solid to liquid as it absorbs thermal energy Freezing - i.e. when a substance changes from liquid to solid as it releases thermal energy Vaporisation (or boiling) - i.e. when a substance changes from liquid to gas as it absorbs thermal energy Condensation - i.e. when a substance changes from gas to liquid as it releases thermal energy Water Each substance has its own melting (or freezing) and boiling points For example, the freezing point of water is 0°C and its boiling point is 100°C Possible phase changes of water include: Solid ice melting into liquid water at 0°C Liquid water boiling and changing into gaseous water vapour at 100°C Both these changes happen when thermal energy is absorbed If thermal energy is released from water vapour at 100°C, it condenses back into water If water continues to release thermal energy, it cools down until it reaches 0°C and freezes into ice Phase changes for water Page 17 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Melting and freezing happen at the melting / freezing point of a substance YOUR NOTES Vaporisation and condensation happen at the boiling point of a substance  Page 18 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Phase Change Graphs YOUR NOTES A heating or cooling curve shows how the temperature of a substance changes with time  The 'flat' sections of the graph indicate that there is no change in temperature over time, hence the substance is undergoing a phase change The thermal energy supplied to or removed from the substance only affects the potential energy of the particles The regions of the graph that are not flat indicate that the substance is being heated or cooled down The thermal energy supplied to or removed from the substance changes the average kinetic energy of the particles, hence resulting in an overall change in temperature of the substance Heating Curves As energy is being supplied to a solid substance, its temperature increases until it reaches its melting point The temperature remains constant until the substance has melted completely If energy continues to be supplied, the liquid substance warms up until the boiling point is reached, and the substance vaporises Then, the temperature of the gas increases Cooling Curves As energy is being removed from a gaseous substance, its temperature decreases until the boiling point is reached The temperature remains constant until the substance has condensed completely Page 19 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources If energy continues to be removed, the liquid substance cools down until its freezing point YOUR NOTES and changes into a solid  Then, the temperature of the solid decreases Heating or cooling curves can also display how the temperature of a substance changes with energy In the following worked example, energy (in J) is plotted on the x-axis instead of time Page 20 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  The graph below is the heating curve for a 25 g cube of ice being heated at a constant rate. Calculate: The specific heat capacity of water in its liquid phase The specific latent heat of vaporisation Step 1: Write down the mass m of ice in kilograms (kg) m = 25 g = 0.025 kg Step 2: Read from the graph the amount of energy E1 being supplied to the water in its liquid phase as it warms up The first flat section of the graph indicates the change of phase of ice into water The non-flat region that follows is the one relating to water in its liquid phase being heated up E1 = 20 kJ – 10 kJ = 10 kJ Step 3: Convert this energy from kilojoules into joules E1 = 10 kJ = 10000 J Step 4: Read from the graph the change in temperature ΔT1 of the water as it warms up ΔT1 = 100°C – 0°C = 100°C Step 5: Write down the equation linking thermal energy E1 to mass m, specific heat capacity c and change in temperature ΔT1 E1 = mcΔT1 Step 6: Solve for the specific heat capacity c Page 21 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  c = 4000 J kg–1 °C–1 Step 7: Read from the graph the amount of energy E2 being supplied to the water as it changes into a gas at 100°C The second flat section of the graph indicates the change of phase of water into water vapour This energy must be converted from kilojoules (kJ) into joules (J) E2 = 56 kJ = 56000 J Step 8: Write down the equation linking thermal energy E2 to mass m and specific latent heat of vaporisation Lv E2 = mLv Step 9: Solve for the specific latent heat of fusion Lv Lv = 2.2 × 106 J kg–1 Page 22 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.1.7 Investigating Thermal Energy YOUR NOTES  Investigating Thermal Energy Estimating the Specific Heat Capacity of a Metal Aim of the Experiment The aim of the experiment is to determine the specific heat capacity of a metal block Variables Independent variable = Time, t (s) Dependent variable = Temperature, T (°C) Control variables: Mass of the metal block Voltage of the power supply Equipment List Resolution of measuring equipment: Page 23 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Digital scale = 0.1 g YOUR NOTES Liquid-in-glass thermometer = 1°C  Stop-clock = 0.01 s Voltmeter = 1 mV Ammeter = 1 mA Method Part 1: Measuring the temperature change Apparatus used to investigate the specific heat capacity of a metal block 1. Use the digital scale to measure the mass m of the metal block. Record this value with its uncertainty Δm (this is simply the smallest division on the digital scale - e.g. ± 0.0001 kg) 2. Wrap the insulating material (e.g. cotton wool) around the block of metal to minimise energy transfers with the surroundings 3. Insert the immersion heater into the central hole of the metal block 4. Place the thermometer into the smaller hole in the metal block 5. Put a few drops of oil or water into the hole where the thermometer is, to seal the air gap between the metal and the thermometer 6. Measure the temperature T of the metal block and record this value in the first row of the table (t = 0), together with its uncertainty ΔT 7. Connect the immersion heater to the power supply 8. Turn on the power supply and start the stop-clock at the same time 9. Measure the temperature of the block every minute (i.e. 60 seconds) and record the temperature values in the table (for a total of 10 readings) 10. Turn the immersion heater off An example of a results table might look like this: Page 24 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Part 2: Calculating the energy transferred Apparatus used to measure the energy transferred to the metal block Page 25 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 1. Connect the ammeter in series with the immersion heater and the voltmeter across the YOUR NOTES immersion heater (i.e. in parallel)  2. Turn on the power supply using the same value of voltage used to take the readings of temperature 3. Read the value of the current I from the ammeter and record this value with its uncertainty ΔI 4. Read the value of the voltage V from the voltmeter (not from the power supply) and record this value with its uncertainty ΔV An example of a results table might look like this: Analysis of Results The equation linking the thermal energy E transferred to the metal block, the mass m of the block, the specific heat capacity c of the block and the temperature change ΔT is: E = mcΔT Rearrange this to calculate the specific heat capacity c: Write the energy E in terms of power P and time Δt: E = PΔt Substitute this expression for E into the equation for the specific heat capacity: Write the power P as the product of voltage V and current I: Plot a graph of temperature T (°C) against time t (s) Calculate the gradient of the linear portion of the graph Gradient = ΔT/Δt Page 26 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Calculate the specific heat capacity c of the metal (in J kg–1 K–1 or, equivalently, in J kg–1 °C–1) as follows: Compare this result with the accepted value of the specific heat capacity for the metal used Evaluating the Experiment Systematic errors: Some energy is lost to the surroundings The layer of insulation around the metal block and the drops of oil or water in the hole where the thermometer is placed all help reducing energy losses, but some energy is inevitably transferred to the surroundings, causing a systematic error The thermometer might not be calibrated correctly Place the thermometer in a beaker with a mixture of ice and water for 30 s, of the thermometer reads 0°C, then it is reading correctly and can be used Random errors: There might be parallax error when reading the values of temperature from the thermometer Make sure you hold the thermometer at eye level when taking the temperature readings to reduce random error The temperature readings might not all be taken exactly 60 s apart from each other Make sure you work with a partner, so you can read out the temperature values to them and they can write or type these in the table, while you are only looking at the stop- clock and thermometer Page 27 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Delays in the responsiveness of the thermometer might still cause random errors, YOUR NOTES using a thermometer with a smaller bulb and thinner glass walls will improve its  responsiveness Safety Considerations Position all equipment away from the edge of the desk Do not touch the metal block or the immersion heater when the power supply is switched on Allow time to cool before touching hot parts of the apparatus  Worked Example Use error analysis to obtain expressions for the relative error δc and the absolute error Δc on the specific heat capacity. Step 1: Determine the relative errors on voltage V, current I, temperature T, mass m and time t δV = ΔV/V δI = ΔI/I δT = ΔT/T (where T is the increase in temperature for the straight portion of the graph) δm = Δm/m δt = Δt/t (where t is the increase in time for the straight portion of the graph) Step 2: Sum all relative errors in Step 1 to obtain the relative error δc on the specific heat capacity δc = δV + δI + δT + δm + δt Step 3: Recall the relationship between absolute and relative errors to determine the absolute error on the specific heat capacity Δc Remember that absolute errors have units (in this case, J kg–1 K–1 or in J kg–1 °C–1) c in the equation below is the experimental value of the specific heat capacity Δc = c × δc Page 28 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.2 Modelling a Gas YOUR NOTES  3.2.1 Ideal Gas Laws Ideal Gas Laws Boyle's Law Boyle's Law states: For a fixed mass of gas at a constant temperature, the pressure p is inversely proportional to the volume V This can be expressed in equation form as: Plotting the pressure against the volume for a gas at a constant temperature on a graph (i.e. p-V graph) gives the so-called isothermal curve for the gas Graph of pressure against volume for a fixed mass of gas at three different temperatures, with T1 < T2 < T3. The curves are called isotherms (i.e. the temperature along each curve is constant) Plotting the pressure against the reciprocal of the volume (i.e. 1/V) for a gas at constant temperature still gives an isothermal, but this time, the line is straight Page 29 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Graph of pressure against reciprocal of the volume for a fixed mass of gas at three different temperatures, with T1 < T2 < T3. In this case, the isotherms are straight lines Boyle's law can be rewritten as follows: pV = constant Which means that: p1V1 = p2V2 Where: p1 = initial pressure in pascals (Pa) or atmospheres (atm) V1 = initial volume in metres cubed (m3) or litres (L) p2 = the final pressure in pascals (Pa) or atmospheres (atm) V2 = the final volume in metres cubed (m3) or litres (L) Charles's Law Charle's Law states: For a fixed mass of gas at constant pressure, the volume V is directly proportional to the absolute temperature T This can be expressed in equation form as: The direct proportionality relationship is only valid if the gas temperature is measured in kelvin (K) Plotting the volume against the temperature for a gas at constant pressure gives a straight line along which the gas pressure does not change Page 30 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Graph of volume against temperature for a fixed mass of gas at three different pressures, with p1 < p2 < p3 Charles's law can be rewritten as follows: Which means that: Where: V1 = initial volume in metres cubed (m3) or litres (L) T1 = initial temperature in kelvin (K) V2 = final volume in metres cubed (m3) or litres (L) T2 = final temperature in kelvin (K) Gay Lussac's Law Gay Lussac's Law states: For a fixed mass of gas at constant volume, the pressure p is directly proportional to the absolute temperature T This can be expressed in equation form as: The direct proportionality relationship is only valid if the gas temperature is measured in kelvin (K) Plotting the pressure against the temperature for a gas at constant volume gives a straight line along which the gas volume is the same Page 31 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Graph of pressure against temperature for a fixed mass of gas at three different volumes, with V1 < V2 < V3 Gay Lussac's law can be rewritten as follows: Which means that: Where: p1 = initial pressure in pascals (Pa) or atmospheres (atm) T1 = initial temperature in kelvin (K) p2 = final pressure in pascals (Pa) or atmospheres (atm) T2 = final temperature in kelvin (K) Gas Laws Combined The three gas laws can be combined into one For a fixed mass of gas, the following holds: Which means that: Where: p1 = initial pressure in pascals (Pa) or atmospheres (atm) V1 = initial volume in metres cubed (m3) or litres (L) T1 = initial temperature in kelvin (K) p2 = final pressure in pascals (Pa) or atmospheres (atm) Page 32 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources V2 = final volume in metres cubed (m3) or litres (L) YOUR NOTES T2 = final temperature in kelvin (K)   Worked Example An ideal gas occupies a volume equal to 5.0 × 10–4 m3. Its pressure is 2.0 × 106 Pa and its temperature is 40°C. The gas is then heated and reaches a temperature of 80°C. It also expands to a new volume of 6.0 × 10–4 m3.Determine the new pressure of the gas. Step 1: Write down the given quantities Initial volume, V1 = 5.0 × 10–4 m3 Initial pressure, p1 = 2.0 × 106 Pa Initial temperature, T1 = 40°C = 313 K Final volume, V2 = 6.0 × 10–4 m3 Final temperature, T2 = 80°C = 353 K Remember to: Use the appropriate subscripts for initial (i.e. 1) and final (i.e. 2) values Convert the temperature from degrees Celsius into Kelvin (K) Step 2: Write down the equation for the three gas laws combined Step 3: Rearrange the equation to calculate the unknown final pressure p2 Step 4: Substitute numbers into the equation p2 = 1.9 × 106 Pa  Exam Tip When dealing with gas laws problems, always remember to convert temperatures from degrees Celsius (°C) to kelvin (K).After you solve a problem using any of the gas laws (or all of them combined), always check whether your final result makes physically sense - e.g. if you are asked to calculate the final pressure of a fixed mass of gas being heated at constant volume, your result must be greater than the initial pressure given in the problem (since Gay- Lussac's law states that pressure and absolute temperature are directly proportional at constant volume). Page 33 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.2.2 Ideal Gas Equation YOUR NOTES  Ideal Gas Equation Avogadro's Law Avogadro's Law states: For a gas at constant temperature and pressure, the number of moles n is directly proportional to the volume V of the gas This can be expressed in equation form as: This means that two different gases of equal temperatures, pressures and volumes have the same number of particles N Note that the number of particles N is directly proportional to the number of moles n Equation of State for an Ideal Gas Boyle's Law, Charles's Law and Gay-Lussac's law can be combined with Avogadro's law to give a single constant, known as the ideal gas constant, R Combining the four equations leads to the equation of state of an ideal gas Where: p = pressure in pascals (Pa) V = volume in metres cubed (m3) T = temperature in kelvin (K) n = number of moles in the gas (mol) R = 8.31 J K–1 mol–1 (ideal gas constant)  Worked Example A gas has a temperature of –55°C and a pressure of 0.5 MPa. It occupies a volume of 0.02 m3.Calculate the number of gas particles. Step 1: Write down the known quantities Temperature, T = –55°C = 218 K Pressure, p = 0.5 MPa = 0.5 × 106 Pa Volume, V = 0.02 m3 Note the conversions: The pressure p must be converted from megapascals (MPa) into pascals (Pa) The temperature must be converted from degrees Celsius (°C) into kelvin (K) Page 34 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Step 2: Write down the equation of state of ideal gases YOUR NOTES  Step 3: Rearrange the above equation to calculate the number of moles n Step 4: Substitute numbers into the equation From the data booklet, R = 8.31 J K–1 mol–1 n = 5.5 mol Step 5: Calculate the number of particles N Write down the relationship between number of particles and number of moles N = nNA From the data booklet, NA = 6.02 × 1023 mol–1 (Avogadro constant) N = 5.5 mol × (6.02 × 1023) mol–1 N = 3.3 × 1024  Exam Tip When using the equation of state of ideal gases, always remember to convert temperatures from degrees Celsius (°C) to kelvin (K).Note that the number of moles n is not the same as the number of particles N: When a question asks to calculate the number of particles in a sample of gas, you should first use the equation of state to determine the number of moles n of the gas, and then calculate the number of particles using N = nNA If a question gives the number of particles in a sample of gas instead of the number of moles, you should first use n = N/NA to calculate the number of moles of the gas, and then use the equation of state to perform any further calculation (e.g. volume, pressure, etc.) Page 35 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.2.3 Kinetic Model of an Ideal Gas YOUR NOTES  Kinetic Model of an Ideal Gas Gas Pressure A gas is made of a large number of particles Gas particles have mass and move randomly at high speeds Pressure in a gas is due to the collisions of the gas particles with the walls of the container that holds the gas When a gas particle hits a wall of the container, it undergoes a change in momentum due to the force exerted by the wall on the particle (as stated by Newton's Second Law) Final momentum = –mv Initial momentum = mv Therefore, the change in momentum Δp can be written as: Δp = final momentum – initial momentum Δp = –mv – mv = –2mv ∆p 2mv −F = =− ∆t ∆t According to Newton's Third Law, there is an equal and opposite force exerted by the particle on the wall (i.e. F = 2mv/Δt) A particle hitting a wall of the container in which the gas is held experiences a force from the wall and a change in momentum. The particle exerts an equal and opposite force on the wall Since there is a large number of particles, their collisions with the walls of the container give rise to gas pressure, which is calculated as follows: Page 36 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources p= F YOUR NOTES A  Where: p = pressure in pascals (Pa) F = force in newtons (N) A = area in metres squared (m2) Average Random Kinetic Energy of Gas Particles Particles in gases have a variety of different speeds The average random kinetic energy of the particles EK, which can be written as follows: Where: EK = average random kinetic energy of the particles in joules (J) kB = 1.38 × 10–23 J K–1 (Boltzmann's constant) T = absolute temperature in kelvin (K) kB is known as Boltzmann's constant, and it can be written as follows: Where: R = 8.31 J K–1 mol–1 (ideal gas constant) NA = 6.02 × 1023 mol–1 (Avogadro constant) Internal Energy of the Gas Using the equation of state of ideal gases, the internal energy can be written as follows: Where: U = internal energy of the gas in joules (J) p = gas pressure in pascals (Pa) V = gas volume in metres cubed (m3)  Worked Example 2 mol of gas is sealed in a container, at a temperature of 47°C. Determine: a. The average random kinetic energy of the particles in the gas b. The internal energy of the gas Part (a) Step 1: Write down the temperature T of the gas in kelvin (K) Page 37 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources T = 47°C = 320 K YOUR NOTES Step 2: Write down the equation linking the absolute temperature T of the gas to the  average random kinetic energy EK of the gas particles Step 3: Substitute numbers into the equation From the data booklet, kB = 1.38 × 10–23 J K–1 EK = 6.6 × 10–21 J Part (b) Step 1: Write down the equation linking the internal energy U of the gas to the number of moles n and the absolute temperature T Step 2: Substitute numbers into the equation From the data booklet, R = 8.31 J K–1 mol–1 U = 8000 J = 8kJ Note that, alternatively, the internal energy can be calculated using the following equation: U = NEK = 8000 J = 8kJ N = nNA = 2 mol × (6.02 × 1023) mol–1 = 1.2 × 1024 EK = 6.6 × 10–21 J (calculated in Step 3)  Exam Tip Momentum is a Mechanics topic that should have been covered in a previous unit. The above derivation of change in momentum and resultant force should have already been studied - if you're not comfortable with it then make sure you go back to revise this! Page 38 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Real & Ideal Gases YOUR NOTES The equation of state of ideal gases and the equations for the kinetic energy of the particles  and the internal energy of the gas derived previously, only apply to ideal gases An ideal gas is defined as one to which all the statements below apply: 1. It contains a very large number of identical particles (atoms or molecules) 2. Each gas particle occupies a negligible volume compared to the volume of the gas This means they can be considered as point particles 3. The gas particles move randomly at high speeds 4. The gas particles obey Newton's laws of motion 5. There are no intermolecular forces between the gas particles Therefore, the internal energy of the gas is equal to the total kinetic energy of the particles (U = NEK) 6. The gas particles undergo elastic collisions with each other and with the walls of the container in which the gas is held Hence, the total kinetic energy of the particles and the temperature of the gas do not change as a result of these collisions 7. The duration of the collisions is negligible compared with the time interval between collisions 8. Each particle exerts a force on the wall of the container with which it collides This means the average of the forces produced by all gas particles results in a uniform gas pressure Real gases are not ideal gases, however, under certain conditions, they can be considered as ideal gases. An ideal gas is a good approximation of a real gas when: Pressure and density are low Temperature is moderate Page 39 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.2.4 Mole Calculations YOUR NOTES  Mole Calculations The Mole The mole is one of the seven SI base units It is used to measure the amount of substance One mole is defined as follows: The amount of substance that contains as many elementary entities as the number of atoms in 12 g of carbon-12 This amount of substance is exactly 6.02214076 × 1023 elementary entities (i.e. particles, atoms, molecules) At IB level, this number can be rounded to 6.02 × 1023 One mole of gas contains a number of particles (atoms or molecules) equal to the Avogadro constant Therefore, to calculate the number of particles N in a gas, knowing the number of moles n, the following relationship must be used: N = nNA Where: N = number of gas particles n = number of moles of gas (mol) NA = 6.02 × 1023 mol–1 (Avogadro constant) Molar Mass The molar mass M of a substance is defined as the mass m of the substance divided by the amount (in moles) of that substance The molar mass is calculated as follows: Where: M = molar mass in g mol–1 m = mass in grams (g) n = number of moles (mol)  Worked Example Nitrogen is normally found in nature as a diatomic molecule, N2. The molar mass of nitrogen is 28.02 g mol–1. Determine the mass of a nitrogen atom. Step 1: Write down the relationship between the molar mass M and the mass m Page 40 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Step 2: Rearrange the equation to find the mass m of n = 1 mol of nitrogen YOUR NOTES m = Mn  m = 28.02 g mol–1 × 1 mol m = 28.02 g Step 3: Identify the number of nitrogen atoms in 1 mol of nitrogen n = 1 mol of nitrogen contains a number of molecules equal to the Avogadro constant From the data booklet, NA = 6.02 × 1023 Nmolecules = nNA Nmolecules = 6.02 × 1023 Each nitrogen molecule contains 2 atoms of nitrogen n = 1 mol of nitrogen contains a number of atoms equal to twice the number of molecules Natoms = 2 × Nmolecules = 2 × (6.02 × 1023) Natoms = 1.2 × 1024 Step 4: Divide the mass of 1 mol of nitrogen m by the number of atoms in 1 mol of nitrogen Natoms in order to find the mass of a nitrogen atom Mass of a nitrogen atom = 2.3 × 10–23 g Page 41 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 3.2.5 Investigating Gas Laws YOUR NOTES  Investigating Gas Laws Investigating Gay-Lussac's Law Aim of the Experiment The aim of the experiment is to investigate the relationship between the pressure and the temperature of a gas at constant volume Variables Independent variable = pressure, p (Pa) Dependent variable = temperature, T (°C) Control variables: Volume of the gas Temperature of the room Equipment List Resolution of measuring equipment: Beaker = 50 ml Liquid-in-glass thermometer = 1°C Pressure meter = 0.001 × 105 Pa Page 42 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Method YOUR NOTES  Apparatus used to investigate the relationship between the pressure and the temperature of a gas at constant volume 1. Set up the apparatus as shown in the diagram 2. Pour 500 ml of ice and water into the beaker, and stir it with the stirring rod 3. Measure the temperature T of the water and record this value in the table, together with its uncertainty ΔT 4. Measure the pressure p of the gas and record this value in the table, together with its uncertainty Δp 5. Light the Bunsen burner and start to heat the water 6. Wait until the temperature reading on the thermometer reaches 10°C and measure the pressure of the gas. Record these new values of temperature and pressure in the table 7. Keep heating the water and measure the pressure of the gas at increments of 10°C between 0°C and 100°C An example of a results table might look like this: Page 43 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Analysis of Results Plot a graph of pressure p (Pa) against temperature T (°C) The y-intercept is the value of the pressure of the gas when the water and ice are not heated (T = 0°C) Extrapolate the graph backwards The x-intercept is the value of the temperature of the gas when the pressure is zero This refers to the (theoretical) temperature of the gas when the particles would stop moving, corresponding to the absolute zero on the kelvin scale: T = 0 K = –273°C Verify that the x-intercept corresponds to this value Page 44 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Calculate the gradient of the graph Gradient = Δp/ΔT Recall the equation of state of ideal gases Use this to write the gradient as follows: Where: n = number of moles of gas (mol) V = volume of the gas (m3) R = 8.31 J K–1 mol–1(ideal gas constant) Using the volume V of the gas sample, and the number of moles n, the value of the ideal gas constant R can be calculated Evaluation Parallax error might affect the temperature readings Make sure to hold the thermometer at eye level when taking the temperature readings to reduce random error There might be a delay in the responsiveness of the thermometer Use a thermometer with a smaller bulb and thinner glass walls to improve its responsiveness In this experiment, the temperature of the gas inside the flask is assumed to be the same as the temperature of the water bath, but this might not always be true, as sometimes the reading of pressure might be taken when the gas is not yet in thermal equilibrium with the Page 45 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources water in the beaker. The two improvements below ensure the temperature of the gas is YOUR NOTES much closer to that of the water:  Cover the beaker with a lid and wrap it in tin foil to reduce thermal energy transfers with the surroundings due to evaporation and convection Use a flask with thinner glass walls to improve conduction The volume of the gas inside the flask might not be exactly constant due to gaps in the flask's aperture Make sure there is no gap between the stopper used to seal the flask and the pressure sensor inserted into it. Seal any small gap with Blu Tack if necessary Safety Considerations Position all equipment away from the edge of the desk Tie any long hair back to avoid it catching fire Make sure not to spill hot water onto your skin Make sure you turn off the gas tap as soon as you finish collecting your data Allow time to cool before touching hot parts of the apparatus  Worked Example A student used an online simulation to investigate Gay-Lussac's law for an ideal gas made of 100 particles, kept at a constant volume. She obtained the following graph of pressure (kPa) against absolute temperature (K). The gradient of the graph is 4.239 kPa K–1 Determine the constant volume of the gas. Step 1: Write down the known quantities Gradient = 4.239 kPa K–1 = 4239 Pa K–1 Number of gas particles, N = 100 Note that you must convert the gradient from kPa K–1 into Pa K–1 Page 46 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Step 2: Calculate the number of moles n YOUR NOTES Write down the relationship between the number of particles N and the number of  moles, N = nNA Rearrange this to calculate the number of moles n Substitute the numbers into the equation for n From the data booklet, NA = 6.02 × 1023 mol–1 n = 1.7 × 10–22 mol Step 3: Recall the relationship between the gradient of the graph and the volume of the gas V Step 4: Rearrange the above equation to calculate the volume V Step 5: Substitute the numbers into the above equation From the data booklet, R = 8.31 J K–1 mol–1 V = 3.3 × 10–25 m3 Page 47 of 47 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers