Thermal Physics IBDP - Lecture Notes Dr. Belal Al Qassem Ibn Rushd National Academy Topic 3: Thermal Physics Syllabus Outline 3.1 – Thermal concepts • Molecular theory of solids, liquids and gases • Temperature and absolute temperature • Internal energy • Specific heat capacity • Phase change • Specific latent heat 3.2 – Modelling a gas • Pressure • Equation of state for an ideal gas • Kinetic model of an ideal gas • Mole, molar mass and the Avogadro constant • Differences between real and ideal gases Recommended Teaching Hours: 11 Hours References: 1. Kirk, Tim. Physics: for the IB Diploma. Oxford University Press, 2014. 2. Homer, David and Bowen-Jones, Michael. Physics: course companion. Oxford University Press, 2014. 3. Tsokos, K. A. Physics: for the IB Diploma. Cambridge University Press, 2014. 2|Page Dr. Belal Al Qassem Thermal Physics Thermal Concepts Solid • • • • Solids have a fixed volume and fixed shape. The molecules in a solid are held close together by forces (the intermolecular bonds). The bonds are not absolutely rigid, so the molecules vibrate/oscillate around a mean (average) position. The higher the temperature, the greater the vibrations. The molecules in a solid are held close together by the intermolecular bonds. Model for Solid Material The springs represent the bonds. Liquid • • • Liquids have a fixed volume but its shape can change (it takes the shape of the container). In liquids the forces between the particles are weaker. The particles are able to move around the volume of the liquid and the liquid will take the shape of the container in which it is placed. However, the inter-particle forces between the particles in a liquid are sufficiently strong that the particles cannot move far from each other. Model for Liquid Material • Gas • In gases the inter-particle forces are very weak so as to be almost negligible. The only time significant forces exist between the particles is during collisions. Model for gas Molecules are free to move randomly. 3|Page Dr. Belal Al Qassem Temperature • Temperature usually described as the “degree of hotness” of an object. • Temperature is a scalar quantity and is measured in units of degrees Celsius (°C) or kelvin (K) using a thermometer. • The absolute Temperature scale (Kelvin) is defined as: π(πΎ ) = π( ππΆ) + 273 Notes: 1. βπ»(π²) = βπ»( ππͺ) 2. The lowest possible temperature on the absolute scale is zero kelvin, 0 K, on the Celsius scale this is equivalent to -273 oC. • Temperature is a measure of the average kinetic energy of the molecules in a substance. Heat and Temperature Thermal energy naturally flows from hot to cold. Eventually, the two objects would be expected to reach the same temperature. When this happens, they are said to be in thermal equilibrium. • Heat is energy that is transferred from one body to another as a result of a difference in temperature. • The unit of heat energy is Joules (J). • It is a scalar quantity. 4|Page Dr. Belal Al Qassem Internal Energy • Internal energy is the total random kinetic energy of the particles of a substance, plus the total inter-particle potential energy of the particles. Internal Energy = Total random KE + Total inter-particle PE • The internal energy of a system can changes as a result of heat added or taken out and as a result of work performed. Heat capacity (C) The energy required to increase the temperature of a body by one kelvin. πͺ= πΈ βπ» or πΈ = πͺβπ» Where C is the heat capacity (in J K-1) Q is the heat energy (in J) οT is the change in temperature (in K or oC) 5|Page Dr. Belal Al Qassem Specific Heat capacity (c) The heat energy needed to raise the temperature of a unit mass (1 kg) of the body by one kelvin (or degree) π= πΈ πβπ» or πΈ = ππβπ» Where c is the specific heat capacity (in Jkg-1K-1) Q is the heat energy (in J) m is the mass of the substance (in kg) οT is the change in temperature (in K or oC) Relationship between Heat Capacity and Specific Heat Capacity Simply πΆ = ππ or π= 6|Page πΆ π Dr. Belal Al Qassem Measuring Specific Heat Capacity 1. Electrical Method The experiment would be set up as follows (Figure 1 (a) for solid material and Figure 1 (b) for liquid). The electric heater will produce heat energy and we will assume that The heat energy (Q) = The electrical heat energy Or π = π. π‘ = ππΌπ‘ So π= π½π°π π(π»π − π»π ) Sources of Errors • Loss of thermal energy from the apparatus. • It will take some time for the energy to be shared uniformly through the substance. • The container for the substance and the heater will also be warmed up. In this case, we can take into consideration the heat energy absorbed by the container. π = ππΌπ‘ = ππ ππ (π2 − π1 ) + ππ ππ (π2 − π1 ) 7|Page Dr. Belal Al Qassem 1. Method of Mixtures The experiment would be set up as follows: • • • • • measure the masses of the liquids mA and mB measure the two starting temperatures TA and TB mix the two liquids together. record the maximum temperature of the mixture Tmax If no energy is lost from the system then, energy lost by hot substance cooling down = energy gained by cold substance heating up ππ΄ ππ΄ (ππ΄ − ππππ₯ ) = ππ΅ ππ΅ (ππππ₯ − ππ΅ ) Sources of Errors • The main source of experimental error is the loss of thermal energy from the apparatus; particularly while the liquids are being transferred. • The changes of temperature of the container also need to be taken into consideration for a more accurate result. In minimize the error, we can take into consideration the heat energy absorbed by the container. ππ΄ ππ΄ (ππ΄ − ππππ₯ ) = ππ΅ ππ΅ (ππππ₯ − ππ΅ ) + ππΆ ππΆ (ππππ₯ − ππΆ ) 8|Page Dr. Belal Al Qassem Change of Phase When heat is provided to a body or removed from it, the body may not necessarily change its temperature. The body may change phase instead. Changes of phase happen at constant temperature and include: • melting – when a solid changes to a liquid (heat must be provided to the solid) • freezing – when a liquid changes into a solid (heat must be taken out of the liquid) • vaporisation (or boiling) – when a liquid changes into vapour (by giving heat to the liquid) • condensation – when a vapour changes into a liquid (by taking heat out of the vapour). Specific Latent Heat (L) The amount of energy required to change the phase of a unit mass at constant temperature. or π πΏ= π π = πΏπ 9|Page 1kg water at 100oC 1kg Ice at 0 oC Heat Heat 1kg steam at 100oC 1kg water at 0 oC Dr. Belal Al Qassem Notes: 1. we use the term fusion if the change is from solid to liquid, and we use the term vaporization for a change from liquid to gas. 2. During the change of phase, the kinetic energy of the molecules stays the same (so the temperature is constant). 3. The amount of heat given to the molecules will increase the potential energy of the molecules by breaking the intermolecular bonds. 4. The specific latent heat of vaporization is greater than specific latent heat of fusion. π³(ππππππππππππ) > π³(ππππππ) Why? This is because the increase in separation of the molecules is much larger when going from the liquid to the vapour phase than when going from the solid to the liquid phase. More work is needed to achieve the greater separation, and so more energy is required. 1kg Ice at -20oC 1kg Ice at 0oC Heat π = ππβπ Heat π = πΏπ 10 | P a g e 1kg ice at 0oC 1kg water at 0 oC 1kg water at 0 oC 1kg water at 100oC Heat π = ππβπ Heat π = πΏπ 1kg water at 100oC 1kg steam at 100oC Dr. Belal Al Qassem Example Calculate the heat energy needed to convert 5 kg of ice at -20 oC into steam at 100 oC. • • • • The specific heat capacity of ice is 2100 J/kg.oC. The specific heat capacity of water is 4200 J/kg.oC. The specific latent heat of fusion of ice is 3.3 × 105 J kg–1 The specific latent heat of vaporization of water is 2.3 × 106 Jkg–1 Solution 5kg Ice at -20oC 5kg Ice at 0oC 5kg water at 0 oC 5kg water at 100oC Heat π1 = ππβπ Heat π2 = πΏπ Heat π3 = ππβπ Heat π4 = πΏπ 5kg ice at 0 oC 5kg water at 0oC π1 = ππβπ = 5 × 2100 × 20 = 210000 π½ π2 = πΏπ = 5 × 3.3 × 105 = 16.5 × 105 π½ 5kg water at 100oC π3 = ππβπ = 5 × 4200 × 100 = 2100000π½ 5kg steam at 100oC π4 = πΏπ = 5 × 2.3 × 106 = 11.5 × 106 π½ πππ‘ππ βπππ‘ ππππππ¦ = π1 + π2 + π3 + π4 = 1.55 × 107 π½ 11 | P a g e Dr. Belal Al Qassem Measuring the Specific latent heat 1. A method for measuring the specific latent heat of vaporization of water a. b. c. d. e. Measure the mass of water. Connect the circuit as shown in the diagram. Once the water starts boiling, turn on the stopwatch to measure the time. After 5 min, turn off the circuit and measure the final mass of water. Use the following equation to measure the specific latent heat of fusion: π½. π°. π π. π π³= = βπ βπ 2. A method for measuring the specific latent heat of fusion of water π»ππ ππππ ππππ ππ πππ πππππ = π»ππ ππππ ππππ ππ ππππ πππ πππ + πππ ππππ ππππ ππ ππππππππ πππ πππππππππππ ππ πππ πππππππ. ππππππ ππππππ (π» − π»πππ ) = ππππ π³ππππππ + ππππ ππππππ (π»πππ − π) 12 | P a g e Dr. Belal Al Qassem Practice Problems (P1) 1. 2. The Kelvin temperature of an ideal gas is a measure of the A. average speed of the molecules. B. average momentum of the molecules. C. average kinetic energy of the molecules. D. average potential energy of the molecules. Two ideal gases X and Y, are contained in a cylinder at constant temperature. The mass of the atoms of X is m and of Y is 4m. Which one of the following is the correct value of the ratio average kinetic energy of the atoms of Y ? average kinetic energy of the atoms of X 3. 4. A. 1 B. 2 C. 4 D. 16 A substance changes from solid to liquid at its normal melting temperature. What change, if any, occurs in the average kinetic energy and the average potential energy of its molecules? Average kinetic energy Average potential energy A. constant constant B. increases constant C. increases decreases D. constant increases The specific latent heat of vaporization of a substance is greater than its specific latent heat of fusion because A. boiling takes place at a higher temperature than melting. B. thermal energy is required to raise the temperature from the melting point to the boiling point. C. the volume of the substance decreases on freezing but increases when boiling. D. the increase in potential energy of the molecules is greater on boiling than on melting. 13 | P a g e Dr. Belal Al Qassem 5. 6. Which of the following is the internal energy of a system? A. The total thermal energy gained by the system during melting and boiling. B. The sum of the potential and the kinetic energies of the particles of the system. C. The total external work done on the system during melting and boiling. D. The change in the potential energy of the system that occurs during melting and boiling. A hot liquid X has specific heat capacity SH. It is mixed with an equal mass of a cold liquid Y of specific heat capacity SC. The best estimate of the ratio temperature fall of liquid X temperature rise of liquid Y is 7. A. SC SH B. SH SC C. 2 SC SH D. 2 SH SC Heat lost by hot = heat gained by cold πππ» . βππ = π. ππΆ . βππ βππ ππΆ = βππ ππ» A copper block and a steel block each have the same mass. The copper block is at a higher temperature than the steel block. The blocks are placed in thermal contact and they then reach thermal equlibrium. There is no energy exchange with the surroundings. How do the magnitude of the change in temperature βT and the magnitude of the change in internal energy βU of the two blocks compare? βT βU A. same same B. same different C. different same D. different different 14 | P a g e Dr. Belal Al Qassem 8. A container holds 20 g of neon (mass number 20) and also 8 g of helium (mass number 4). What is the ratio number of atoms of neon ? number of atoms of helium A. 0.4 1 mole of Ne → the mass of it = 20 g → # of atoms = 6 x 1023 B. 0.5 1 mole of He → the mass of it = 4 g → # of atoms = 6 x 1023 C. 2.0 D. 2.5 6 x 1023 1 = 23 2 × 6 × 10 2 9. A temperature of 23 K is equivalent to a temperature of A. °C. B. °C. C. °C. D. °C. 10. The specific latent heat is the energy required to change the phase of A. one kilogram of a substance. B. a substance at constant temperature. C. a liquid at constant temperature. D. one kilogram of a substance at constant temperature. 11. The internal energy of any substance is made up of the A. total random kinetic and potential energy of its molecules. B. total potential energy of its molecules. C. total random kinetic energy of its molecules. D. total vibrational energy of its molecules. 12. A sealed container contains water at 5 °C and ice at 0 °C. This system is thermally isolated from its surroundings. What happens to the total internal energy of the system? A. It remains the same. B. It decreases. C. It increases until the ice melts and then remains the same. D. It increases. 15 | P a g e Dr. Belal Al Qassem 13. A 1.0 kW heater supplies energy to a liquid of mass 0.50 kg. The temperature of the liquid changes by 80 K in a time of 200 s. The specific heat capacity of the liquid is 4.0 kJ kg–1 K–1. What is the average power lost by the liquid? A. 0 B. 200 W C. 800 W D. 1600 W π = 1000π π = ππβπ = 0.5 × 80 × 4000 = πππππππ± π = 0.5ππ π΅π’π‘ βπ = 80πΎ π‘ = 200 π π = 4000 π½ππ−1 πΎ −1 πβπππ‘ππ = π. π‘ = 1000 × 200 = πππππππ± ππππ π‘ = 200000 − 160000 = ππππππ± ππππ π‘ 40000 ππππ π‘ = = = 200 π π‘ 200 14. The graph shows the variation with time t of the temperature T of two samples, X and Y. X and Y have the same mass and are initially in the solid phase. Thermal energy is being provided to X and Y at the same constant rate. What is the correct comparison of the specific latent heats LX and LY and specific heat capacities in the liquid phase cX and cY of X and Y? οΎ 16 | P a g e Dr. Belal Al Qassem 15. A mass m of ice at a temperature of –5 °C is changed into water at a temperature of 50 °C. Specific heat capacity of ice = ci Specific heat capacity of water = cw Specific latent heat of fusion of ice = L Which expression gives the energy needed for this change to occur? A. 55 m cw + m L B. 55 m ci + 5 m L C. 5 m ci + 50 m cw + m L D. 5 m ci + 50 m cw + 5 m L 16. A container with 0.60kg of a liquid substance is placed on a heater at time t=0. The specific latent heat of vaporization of the substance is 200kJkg–1. The graph shows the variation of the temperature T of the substance with time t. π = πΏπ = 0.6 × 200 = 120ππ½ π= π 120000 = = 3000 π π‘ 40 What is the power of the heater? A. 1200 W B. 3000 W C. 4800 W D. 13 300 W 17 | P a g e Dr. Belal Al Qassem 17. Two objects are in thermal contact and are at different temperatures. What is/are determined by the temperatures of the two objects? I. The direction of thermal energy transfer between the objects II. The quantity of internal energy stored by each object III. The process by which energy is transferred between the objects A. I only B. II only C. I and II only D. I, II and III 18. Molecules leave a boiling liquid to form a vapour. The vapour and the liquid have the same temperature. What is the change of the average potential energy and the change of the average random kinetic energy of these molecules when they move from the liquid to the vapour? οΎ 19. Equal masses of water at 80°C and paraffin at 20°C are mixed in a container of negligible thermal capacity. The specific heat capacity of water is twice that of paraffin. What is the final temperature of the mixture? π2π (80 − π) = ππ(π − 20) A. 30°C B. 40°C 2(80 − π) = π − 20 160 − 2π = π − 20 C. 50°C D. 60°C 18 | P a g e Dr. Belal Al Qassem 20. A block of iron of mass 10 kg and temperature 10°C is brought into contact with a block of iron of mass 20 kg and temperature 70°C. No energy transfer takes place except between the two blocks. What will be the final temperature of both blocks? A. 30°C B. 40°C C. 50°C D. 60°C 21. A sample of solid copper is heated beyond its melting point. The graph shows the variation of temperature with time. During which stage(s) is/are there an increase in the internal energy of the copper? A. P, Q and R B. Q only C. P and R only D. Q and R only 19 | P a g e Dr. Belal Al Qassem Practice Problems (P2) 1. This question is about an experiment to measure the temperature of a flame. (a) Define heat (thermal) capacity. ..................................................................................................................................... ..................................................................................................................................... (1) A piece of metal is held in the flame of a Bunsen burner for several minutes. The metal is then quickly transferred to a known mass of water contained in a calorimeter. flame water calorimeter container Bunsen burner lagging (insulation) The water into which the metal has been placed is stirred until it reaches a steady temperature. (b) Explain why (i) the metal is transferred as quickly as possible from the flame to the water; .......................................................................................................................... . (1) (ii) the water is stirred. .......................................................................................................................... (1) The following data are available: heat capacity of metal = 82.7 J K–1 heat capacity of the water in the calorimeter = 5.46 × 102 J K–1 heat capacity of the calorimeter = 54.6 J K–1 initial temperature of the water = 288 K final temperature of the water = 353 K (c) Assuming negligible energy losses in the processes involved, use the data to calculate the temperature T of the Bunsen flame. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (4) (Total 7 marks) 20 | P a g e Dr. Belal Al Qassem 2. This question is about thermal physics. (a) Explain why, when a liquid evaporates, the liquid cools unless thermal energy is supplied to it. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (3) (b) State two factors that cause an increase in the rate of evaporation of a liquid. 1. ................................................................................................................................. 2. ................................................................................................................................. (2) (c) Some data for ice and for water are given below. Specific heat capacity of ice Specific heat capacity of water Specific latent heat of fusion of ice = 2.1 × 103 J kg–1 K–1 = 4.2 × 103 J kg–1 K–1 = 3.3 × 105 J kg–1 A mass of 350 g of water at a temperature of 25°C is placed in a refrigerator that extracts thermal energy from the water at a rate of 86 W. Calculate the time taken for the water to become ice at –5.0°C. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (4) (Total 9 marks) 21 | P a g e Dr. Belal Al Qassem 3. This question is about specific heat capacity and specific latent heat. (a) Define specific heat capacity. ..................................................................................................................................... ..................................................................................................................................... (1) (b) Explain briefly why the specific heat capacity of different substances such as aluminum and water are not equal in value. ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... (2) A quantity of water at temperature θ is placed in a pan and heated at a constant rate until some of the water has turned into steam. The boiling point of the water is 100°C. (c) (i) Using the axes below, draw a sketch-graph to show the variation with time t of the temperature θ of the water. (Note: this is a sketch-graph; you do not need to add any values to the axes.) (1) 100°C °C 0 time at which heating starts 22 | P a g e t time at which water starts to boil Dr. Belal Al Qassem (ii) Describe in terms of energy changes, the molecular behavior of water and steam during the heating process. .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... (5) Thermal energy is supplied to the water in the pan for 10 minutes at a constant rate of 400 W. The thermal capacity of the pan is negligible. (d) (i) Deduce that the total energy supplied in 10 minutes is 2.4 × 105 J. .......................................................................................................................... . (1) (ii) Using the data below, estimate the mass of water turned into steam as a result of this heating process. initial mass of water = 0.30 kg initial temperature of the water θ = 20°C specific heat capacity of water = 4.2 × 103 J kg–1 K–1 specific latent heat of vaporization of water = 2.3 × 106 Jkg–1 .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... (3) (iii) Suggest one reason why this mass is an estimate. .......................................................................................................................... .......................................................................................................................... (1) (Total 14 marks) 23 | P a g e Dr. Belal Al Qassem Gas Laws There are 4 factors to describe the gas: • • • • Pressure (P) Temperature (T) Volume (V) Amount of the gas Boyle’s Law: For a fixed mas of a gas, the volume of the gas is inversely proportional to its pressure provided that the temperature is constant. π π·∝( ) π½ π· = ππππππππ × π π½ π·π½ = ππππππππ Or π·π π½π = π·π π½π Example Pressure Volume 1 20 1 X 20 = 20 24 | P a g e 2 10 2 X 10 = 20 4 5 4 X 5 = 20 5 4 10 2 20 20 1 20 20 Dr. Belal Al Qassem 25 20 20 Pressure Pressure 25 15 10 5 15 10 5 0 0 0 5 10 15 20 0.00 25 0.50 1.00 1.50 1/Volume Volume 25 20 PV 15 10 5 0 0 5 10 15 20 25 Pressure PV = Constant 25 | P a g e Dr. Belal Al Qassem Pressure Pressure is defined as Force per unit Area π·πππππππ = π·= πππππ π¨πππ π π¨ The unit of Pressure is N/m2 = Pascal (Pa). 1 N/m2 = Pascal (Pa). There are other units for Pressure such as: 1 atm = 1.013 ο΄ 105 Pa ο» 1 ο΄ 105 Pa ο» 101 ο΄ kPa 1 atm = 76 cm.Hg = 760 mm.Hg π·= But π π¨ A ππ π·= π¨ π= π π h π = ππ = πβπ΄ π·= πβπ΄π π¨ π· = πππ Pressure depends on density and Depth. 26 | P a g e Liquid of density ο² Dr. Belal Al Qassem Atmospheric Pressure π· = πππ π· = ππ. π × πππ × π. π × π. ππ = π. πππ × πππ π·π If we used water instead of Mercury π· = π × πππ × π. π × π = π. πππ × πππ π·π π ≈ ππ π A h h A The weight of the water in the Cylindar is Greater. F = mg So if we used the equation P=F/A It seems that the pressure at the bottom of the cylindar is greater !!!!!!!!! BUT If we used the equation π = ππβ Then both will have the SAME Pressure at the bottom. WHO CAN SOLVE THIS PARADOX. 27 | P a g e Dr. Belal Al Qassem Why Pressure inversely proportional to Volume? π·= π π¨ πππππ = πΉπππ ππ πͺπππππ ππ π΄πππππππ • Pressure depends on the number of collisions per second between the molecules and the wall of the container. • • • Increasing the Volume will decrease the rate of collisions (pressure). As a result, will decrease the pressure of the gas. Since the temperature is constant, then the speed of the molecules stays the same. 28 | P a g e Dr. Belal Al Qassem Charles’ Law The volume of a fixed mass of a gas is directly proportional to its absolute temperature (Temperature in Kelvin) if the pressure is constant. π½∝π» π½ = ππππππππ × π» π½ = πͺπππππππ π» π½π π½π = π»π π»π V V 0 -273 T/K T/oC T (K) = T(oC) + 273 So 0 K = -273 oC Using Charles’ Law, explain why -273 oC (Zero Kelvin) is the lowest possible temperature? Since at -273 oC, the volume is zero, and below that temperature the volume becomes negative which is impossible. 29 | P a g e Dr. Belal Al Qassem Why Volume is directly proportional to Temperature at constant pressure? (explanation) A higher temperature means faster moving molecules. • • • • Faster moving molecules hit the walls with a greater microscopic force. If the volume of the gas increases, then the rate at which these collisions take place on a unit area of the wall must go down. The average force on a unit area of the wall can thus be the same. Thus the pressure remains the same. 30 | P a g e Dr. Belal Al Qassem The Pressure Law The pressure of a fixed mass of a gas is directly proportional to its absolute temperature (Temperature in Kelvin) if the volume is constant. π·∝π» π· = ππππππππ × π» π· = πͺπππππππ π» π·π π·π = π»π π»π P P 0 T/K -273 T/oC Using the Pressure Law, explain why -273 oC (Zero Kelvin) is the lowest possible temperature? Since at -273 oC, the pressure is zero, and below that temperature the pressure becomes negative which is impossible. 31 | P a g e Dr. Belal Al Qassem Why Pressure is directly proportional to Temperature? (explanation) If the temperature of a gas goes up, the molecules have more average kinetic energy – they are moving faster on average. • • • • • Fast moving molecules will have a greater change of momentum when they hit the walls of the container. Thus the microscopic force from each molecule will be greater. The molecules are moving faster so they hit the walls more often. For both these reasons, the total force on the wall goes up. Thus the pressure goes up. 32 | P a g e Dr. Belal Al Qassem 33 | P a g e Dr. Belal Al Qassem Ideal Gas Law (The equation of state) π·π½ = ππππππππ π½ = πͺπππππππ π» π·π½ = ππππππππ π» π· = πͺπππππππ π» π·π π½π = π·π π½π π½π π½π = π»π π»π π·π π½π π·π π½π = π»π π»π π·π π·π = π»π π»π π·π½ = ππππππππ = ππΉ π» n = The number of moles. R = The molar gas constant = 8.31 J mol-1 K-1. π·π½ = ππΉπ» 34 | P a g e Dr. Belal Al Qassem Definitions Ideal gas: An ideal gas is one that follows the gas laws (PV = nRT) for all values of P, V and T. Mole: The mole is the basic SI unit for ‘amount of substance’. One mole of any substance is equal to the amount of that substance that contains the same number of particles as 0.012 kg of carbon–12 (12C). When writing the unit it is (slightly) shortened to the mol. • The mass of 1 atom of C-12 = 12 u (atomic mass unit) • The mass of 1 mole of C-12 = 12 g = 0.012 kg • There are 6 ο΄ 1023 atoms in 1 mole. The unified atomic mass unit One unified mass unit is defined as exactly one twelfth the mass of a carbon-12 atom. i.e. π ππ = × ππππ ππ ππππͺ ππππ ππ Note: 1 u = 1.661 × 10−27 kg Avogadro This is the number of atoms in 0.012 kg, of carbon–12 (12C). It is constant, NA 6.02 × 1023. Molar mass The mass of one mole of a substance is called the molar mass. A simple rule applies. If an element has a certain mass number, A, then the molar mass will be A grams. π΅ π= π΅π¨ ππππππ ππ πππππ = 1 πππ π ππππ 35 | P a g e π΅πππππ ππ πππππ π¨πππππ ππ ππππππππ → 6 × 1023 ππ‘πππ → π ππ‘πππ Dr. Belal Al Qassem Example a) What volume will be occupied by 8 g of helium (mass number 4) at room temperature (20 °C) and atmospheric pressure (1.0 × 105 Pa). ANS: π = 20π πΆ = 20 + 273 = πππ π², π = 1 × 105 ππ, π =? 1 πππ ππ π»π → 4π ? πππ ππ π»π → 8π ? (ππ’ππππ ππ πππππ ) = 8 = 2 πππ. 4 π·π½ = ππΉπ» 1.0 × 105 × V = 2 × 8.3 × 293 V = 0.049 m3 b) How many atoms are there in 8 g of helium (mass number 4)? ANS: ππππππ ππ πππππ = 2= π΅πππππ ππ πππππ π¨πππππ ππ ππππππππ π 6.02 × 1023 N = 12 × 1023 = 1.2 × 1024 36 | P a g e Dr. Belal Al Qassem Gas Laws – Practice Questions 1. A container holds 20 g of neon (mass number 20) and also 8 g of helium (mass number 4). What is the ratio 2. A. 0.4 B. 0.5 C. 2.0 D. 2.5 number of atoms of neon ? number of atoms of helium The equation of state of an ideal gas is pV = nRT. In this equation, the constant n is the number of 3. A. atoms in the gas. B. molecules in the gas. C. particles in the gas. D. moles of the gas. A fixed mass of an ideal gas is heated at constant volume. Which one of the following graphs best shows the variation with Celsius temperature t with pressure p of the gas? A. B. p οΎ 0 p 0 C. 0 t /°C 0 t /°C D. p 0 37 | P a g e 0 t /°C p 0 t /°C 0 Dr. Belal Al Qassem 4. The graph below shows the variation with absolute temperature T of the pressure p of one mole of an ideal gas having a volume V. R is the molar gas constant. p ππ = ππ π πΊπππππππ‘ = π ππ = π π T Which of the following is the best interpretation of the intercept on the temperature axis and the gradient of the graph? Intercept on temperature axis / K Gradient of graph – 273 R V 0 R V 0 V R – 273 V R A. B. C. D. 5. When a gas in a cylinder is compressed at constant temperature by a piston, the pressure of the gas increases. Consider the following three statements. I. The rate at which the molecules collide with the piston increases. II. The average speed of the molecules increases. III. The molecules collide with each other more often. Which statement(s) correctly explain the increase in pressure? A. I only B. II only C. I and II only D. I and III only 38 | P a g e Dr. Belal Al Qassem 6. 7. The equation of state for an ideal gas, pV = nRT, describes the behaviour of real gases A. only at low pressures and large volumes. B. only at high temperatures. C. only at large volumes and large pressures. D. at all pressures and volumes. Two identical boxes X and Y each contain an ideal gas. Box X Box Y n moles 2n moles temperature T temperature pressure PX pressure PY T 3 In box X there are n moles of the gas at temperature T and pressure PX. In box Y there T are 2n moles of the gas at temperature and pressure PY. 3 The ratio A. 8. 2 . 3 B. 3 . 2 C. 2. D. 3. PX PY ππ = ππ π is For Box X; ππ₯ = ππ π π ππ¦ = 2ππ π 3π For Box Y; ππ π ππ₯ 1 3 = π = = 2ππ π 2 2 ππ¦ 3π 3 The temperature of an ideal gas is reduced. Which one of the following statements is true? A. The molecules collide with the walls of the container less frequently. B. The molecules collide with each other more frequently. C. The time of contact between the molecules and the wall is reduced. D. The time of contact between molecules is increased. 39 | P a g e Dr. Belal Al Qassem 9. 10. The Kelvin temperature of an ideal gas is a measure of the A. average speed of the molecules. B. average momentum of the molecules. C. average kinetic energy of the molecules. D. average potential energy of the molecules. Two ideal gases X and Y, are contained in a cylinder at constant temperature. The mass of the atoms of X is m and of Y is 4m. Which one of the following is the correct value of the ratio average kinetic energy of the atoms of Y ? average kinetic energy of the atoms of X 11. A. 1 Since the Temperature is constant B. 2 Then they have SAME average KE C. 4 Ratio ….. = 1 D. 16 An ideal gas is kept in a container of fixed volume at a temperature of 30ο°C and a pressure of 6.0 atm. The gas is heated at constant volume to a temperature of 330ο°C. pressure 6.0 atm new pressure temperature 30ο° C temperature 330ο° C gas gas π· π π·π = π»π π»π The new pressure of the gas is about A. 0.60 atm. B. 3.0 atm. C. 12 atm. D. 40 | P a g e 66 atm. π × πππ π·π × πππ = ππ + πππ πππ + πππ π πππ = π·π πππ → ππ = ππ Dr. Belal Al Qassem 12. Container X below has volume V and holds n moles of an ideal gas at kelvin temperature T. Container Y has volume 2V and holds 3n moles of an ideal gas also at kelvin temperature T. The pressure of the gas in X is PX and in Y is PY. The ratio A. 13. PX is PY 2 . 3 B. 3 . 2 C. 5. D. 6. ππ = ππ π For Box X; ππ₯ = ππ π π ππ¦ = 3ππ π 2π For Box Y; ππ π ππ₯ 1 2 = π = = 3 3 ππ¦ 3ππ π 2π 2 When the volume of a fixed mass of an ideal gas is reduced at constant temperature, the pressure of the gas increases. This pressure increase occurs because the atoms of the gas 14. A. collide more frequently with each other. B. collide more frequently with the walls of the containing vessel. C. are spending more time in contact with the walls of the containing vessel. D. are moving with a higher mean speed. A fixed quantity of an ideal gas is compressed at constant temperature. The best explanation for the increase in pressure is that the molecules A. are moving faster. B. are colliding more frequently with the container walls. C. exert greater forces on each other. D. are colliding more frequently with each other. 41 | P a g e Dr. Belal Al Qassem 16. Gas leaks slowly out of a cylinder of constant volume. The temperature of the gas in the cylinder does not change. Which of the following is constant for the gas molecules in the cylinder? A. The number striking unit area of surface in unit time B. The number of the collisions between molecules per unit time n : decreases (leaks) C. The number per unit volume p: decreases, so less number of collisions. D. The average speed π·π½ = ππΉπ» T: Constant, so Same Speed. 17. Which of the following is not an assumption on which the kinetic model of an ideal gas is based? A. All molecules behave as if they are perfectly elastic spheres. B. The mean-square speed of the molecules is proportional to the kelvin temperature. C. Unless in contact, the forces between molecules are negligible. D. The molecules are in continuous random motion. 42 | P a g e Dr. Belal Al Qassem Kinetic Model of an Ideal Gas The kinetic theory of gases is a statistical treatment of the movement of gas molecules in which macroscopic properties such as pressure are interpreted by considering molecular movement. The key assumptions of the kinetic theory are: • A gas consists of a large number of identical tiny particles called molecules; they are in constant random motion. • The number of molecules is large enough for statistical averages to be made. • Each molecule has negligible volume when compared with the volume of the gas as a whole. • At any instant as many molecules are moving in one direction as in any other direction. • The molecules undergo perfectly elastic collisions between themselves and also with the walls of their containing vessel; during collisions each momentum of each molecule is reversed. • There are no intermolecular forces between the molecules between collisions (energy is entirely kinetic). • The duration of a collision is negligible compared with the time between collisions. • Each molecule produces a force on the wall of the container. • The forces of individual molecules will average out to produce a uniform pressure throughout the gas (ignoring the effect of gravity). 43 | P a g e Dr. Belal Al Qassem y m L cx -cx m x L z L The initial momentum of the molecule = mcx The final momentum of the molecule = -mcx The change in momentum of the molecule = -mcx-mcx = -2mcx π= βπ −ππππ = π π Newton’s 2nd Law From Newton’s 3rd Law, the force by the wall of the container on the molecule equals the force by the molecule on the wall of the container but in the opposite direction, i.e., βπ ππππ π= = Newton’s 3rd Law π π t=? π= ππ³ ππ → ππππ πππ π π= = π π³ 44 | P a g e The force on the wall of the container from ONE molecule. Dr. Belal Al Qassem For N molecules The total force exerted on the shaded face of the container: NOTE: ππππ π ππππ π ππππ π ππππ΅ π π= + + + β―+ π³ π³ π³ π³ π π = (πππ π + πππ π + β― + πππ΅ π ) π³ All molecules have the same mass but not the same speed. Where cx1 : the x component of the velocity of the 1st molecule. cx2 : the x component of the velocity of the 2nd molecule. NOTE: Μ Μ Μ Μ π ≠ πΜ 2 π π π₯ The mean value of the square of the velocities (πππ π + πππ π + β― + πππ΅ π ) Μ Μ Μ Μ π ππ = π΅ Μ Μ Μ Μ π = (π π + π π + β― + π π ) π΅. π π ππ ππ ππ΅ So we can write the force F as follows: ππ = π΅π Μ Μ Μ Μ π . ππ π³ In 3D, we can use Pythagoras theorem, and write π Μ Μ Μ Μ Μ Μ Μ Μ π Μ Μ Μ Μ π πΜ π = Μ Μ Μ Μ ππ π + π π + ππ = π. ππ π πΜ Μ Μ Μ π = πΜ π π 2 = ππ₯ 2 + ππ¦ 2 + ππ§ 2 cy c π π΅π πΜ π ππ = . π³ π cz cx The pressure on the wall of the container: ππ ππ π΅π πΜ π ππ = = π= π . π¨ π³ π³ π 45 | P a g e Dr. Belal Al Qassem ππ = π΅π π . πΜ ππ½ In Fluids (liquids and gases), the pressure at a point acts equally in all directions: π= π΅π π . πΜ ππ½ But π»ππ πππππ ππππ ππ πππππππππ = π΅π and π«ππππππ = π= So we can write π= ππππ ππππππ π΅π π½ π π. πΜ π π It links a mAcroscopic quantity (such as pressure) with a mIcroscopic quantity (such as the speed of the molecule). Molecular Interpretation of Temperature π·= π π΅π Μ Μ Μ π π Μ Μ Μ π . π = π. π π π½ π π·π½ = π Μ Μ Μ π π΅π. π π Multiply each side by 3/2 π π π Μ Μ Μ π π·π½ = × π΅π. π π π π π π Μ Μ Μ π π·π½ = π΅π. π π π Compare it with pV = nRT 46 | P a g e Dr. Belal Al Qassem π π Μ Μ Μ π ππΉπ» = π΅. π. π π π ππ π Μ Μ Μ π πΉπ» = π. π ππ΅ π π= π΅ π΅π¨ π π = π΅ π΅π¨ π πΉ π Μ Μ Μ π π» = π. π π π΅π¨ π BUT ππ© = πΉ = π. ππ × ππ−ππ π±π²−π , π΅π¨ π©ππππππππ ππππππππ π π Μ Μ Μ π ππ© π» = π. π = Μ Μ Μ Μ π¬π π π It links the mIcroscopic quantity (Kinetic energy of the molecules) with the mAcroscopic quantity (the Temperature). Ideal Gases and Real Gases An ideal gas is a one that follows the gas laws for all values of p, V and T and thus ideal gases cannot be liquefied. Real gases, however, can approximate to ideal behaviour providing that the intermolecular forces are small enough to be ignored. For this to apply, the pressure/density of the gas must be low and the temperature must be moderate. 47 | P a g e Dr. Belal Al Qassem