Pre-U STPM Physics Term 1 CC039242a

244 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 10.3 First Law of Thermodynamics Students should be able to: • state and apply the first law of thermodynamics Q = ∆U + W • deduce the relationship ∆U = nCV,m∆T from the first law of thermodynamics • derive and use the equation Cp,m – CV,m = R • relate CV,m and Cp,m to the degrees of freedom • use the relationship γ = Cp.m CV,m to identify the types of molecules Learning Outcomes 1. The first law of thermodynamics is based on the principle of conservation of energy. 2. The first law of thermodynamics states that the heat supplied to a system equals to the sum of the increase in internal energy of the system and the external work done by the system, Q = ∆U + W Heat supplied= Increase in internal energy + Work done by gas 3. Table 10.1 shows the sign convention for Q, ∆U and W Table 10.1 + (positive) – (negative) Q Heat supplied to system Heat loss by system ∆U Increase in internal energy Decrease in internal energy W Work done by system Work done on system 4. The first law of thermodynamics can also be stated as Increase in internal energy, ∆U = Heat supplied, Q + Work done on the system, W ∆U = Q + W Table 10.2 Sign Convention + (positive) – (negative) ∆U Increase in U Decrease in U Q Heat supplied Heat lost W Work done on system Work done by system 5. The two formats of the first law of thermodynamics are consistent. From the format Q = ∆U + W Increase in internal energy, ∆U = Q – (W) W = work done by system = Q + (–W) = Q – (–W) –W = work done on system Increase in = Heat supplied + Work done internal energy on system It is the second format: ∆U = Q + W Exam Tips Do not be confused by the two formats of the first law of thermodynamics. 2008/P1/Q20, 2013/P1/Q14, 2015/P1/Q13 VIDEO First Law of Thermodynamics

245 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Example 5 A fixed mass of gas undergoes the changes of pressure and volume as shown in the figure. When the gas is taken from the state P to R by the stages PQ and QR, 40 J of heat is absorbed and 15 J of work is done by it. When the same resultant change is achieved by the stages PS and SR, 5 J of work is done by the gas. Calculate (a) the change in the internal energy when the state of the gas changes from P to R. (b) the heat supplied to the gas when it undergoes the changes PS and SR. Solution: (a) For PQ and QR, Q = +40 J, W = +15 J Using Q = ∆U + W ∆U = (40 – 15) J = +25 J Internal energy increased by 25 J. (b) For PS and SR, W = +5 J ∆U = +25 J since the initial state is P and final state in R as is PQR. Using Q = ∆U + W = 25 + 5 Heat supplied = +30 J Quick Check 3 Volume Pressure P S Q R 0 1. The first law of thermodynamics states that ∆U, the change of internal energy of a system is related to Q, the heat supplied to it, and W, the work done by the gas by the equation Q = ∆U + W Which of the following quantities is zero for an ideal gas which undergoes change at constant volume? A Q only B ∆U only C W only D Q and ∆U only 2. The first law of thermodynamics states that the change of internal energy of a system, ∆U is related to the heat supply, Q and the work done on the gas W by the equation ∆U = Q + W What are ∆U, Q and W for a fixed mass of gas which undergoes an expansion by obeying Boyle’s law? ∆U Q W A Positive Negative Negative B Zero Negative Negative C Negative Zero Positive D Zero Positive Negative 3. A fixed mass of an ideal gas absorbs 500 J of heat and it expands at constant pressure of 1.0 × 105 Pa from a volume of 5.0 × 10–3 m3 to a volume of 7.5 × 10–3 m3 . The internal energy of the gas A remains unchanged B increased by 250 J C decreased by 250 J D decreased by 750 J

246 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 4. A fixed mass of ideal gas undergoes the change X → Y → Z as shown in the diagram. Y Z X If the increase in internal energy is 600 J, what is the heat absorbed by the gas? A 200 J B 400 J C 600 J D 1 000 J 5. A fixed mass of an ideal gas absorbs 100 J of heat and its pressure increases under constant volume of 2.0 × 10–3 m3 from 1.0 × 105 Pa to 1.5 × 105 Pa. What is the change of internal energy of the gas? A –50 J B –100 J C +100 J D +150 J 6. (a) Write an equation to represent the first law of thermodynamics which relates Q the heat supplied to a system, ∆U the change of internal energy, and W the work done by the system. Explain the sign convention you use. (b) For a system consisting of a real gas, (i) what forms the internal energy is? (ii) give an example of external work done on the system. 7. State and explain in molecular terms, what happens to the internal energy of a fixed mass of an ideal gas when (a) temperature of the gas decreases. (b) the gas expands at constant temperature. (c) the gas which is inside a cylinder in a plane flying at a certain speed. 8. Use the first law of thermodynamics to calculate the difference between the internal energy of 1.00 kg of water at 100°C and 1.00 kg of steam at 100°C at a pressure of 1.01 × 105 Pa. (Volume of 1.00 kg of steam under that condition is 1.67 m3 ; specific latent heat of vaporisation of water is 2.26 × 106 J kg–1). Relation between CV,m and Cp,m 2014/P1/Q17 B constant (a) p V constant p p, Vm T p, Vm + ∆V T + ∆T p + ∆p Vm T + ∆T A C T V T + ∆T (c) (b) p, Vm T Figure 10.5 1. Figure 10.5(a) shows one mole of an ideal gas at pressure p, temperature T and volume Vm. This state of the gas is represented by the point A on the p-V graph (Figure 10.5). 2. When the temperature of the gas is increased to T + ∆T at constant volume, the pressure increases to p + ∆p. This change of state is represented by AB on the p-V graph.

247 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 3. Heat supplied, Q = CV,m ∆T Work done by gas, W = p ∆V = 0 (∆V = 0) Using the first law of thermodynamics, Q = ∆U + W Q = ∆U + 0 Hence ∆U = Q = CV,m ∆T .............................. a For n moles, change in internal, ∆U = nCV,m ∆T. 4. When the temperature of gas is increased from T to T + ∆T at constant pressure p, the volume of the gas increases from Vm to Vm + ∆V. This is represented by the change AC on the p–V graph. 5. Since the gas is heated at constant p, the heat supplied, Q = Cp,m ∆T Work done by gas, W = p ∆V pVm = RT = R ∆T p ∆V = R ∆T Using the first law of thermodynamics, Q = ∆U + W Cp,m ∆T = CV,m ∆T + R ∆T Cp,m = CV,m + R or Cp,m – CV,m = R Example 6 The specific heat capacity of an ideal gas at constant volume is 6.0 × 102 J kg–1 K–1. Calculate the internal energy of 6.0 × 10–3 kg of the gas at 0°C. Solution: 0°C = 273 K. The internal energy of an ideal gas at 0 K = 0. When the temperature is increases from 0 K to 273 K, internal energy of gas = ∆U from T = 0 K to 273 K = mcV ∆T = (6.0 × 10–3)(6.0 × 102 )(273 – 0) = 980 J Example 7 U/ J a b T/ K 0 The graph shows the variation of internal energy U for n moles of ideal gas with temperature T. What is the molar heat capacity of the gas at constant volume in terms of a, b and n? Exam Tips V = constant p = constant Q = nCV,m (∆T) Q′ = nCp,m (∆T) W = 0 W ′ = p(∆V) ∆U = nCV,m (∆T) ∆U = nCV,m (∆T)

248 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Solution: Using ∆U = nCV,m ∆T CV,m = 1 n ∆U ∆T ∆U ∆T = Gradient of graph = a nb Example 8 The molar heat capacity of an ideal gas at constant volume CV,m is 20.5 J mol–1 K–1. (a) Find the heat required to increase the temperature of 2.0 mole of the ideal gas from 300 K to 350 K at constant volume. (b) 2.0 moles of the ideal gas is heated at constant pressure from 300 K using the heat you calculated in (a). What is the final temperature of the gas? Solution: (a) Heat required, Q = nCV,m ∆T = 2.0 (20.5)(350 – 300) = 2 050 J (b) When the gas is heated at constant pressure, Q = nCp,m ∆T Cp,m = CV,m + R 2 050 = 2.0(20.5 + 8.31)(T – 300) T – 300 = 36 T = 336 K Quick Check 4 1. When one mole of an ideal gas is heated at constant pressure p, the volume increases by ∆V when the increase in temperature is ∆T. What is the heat supplied? A CV,m ∆T B (CV,m + R) ∆T C (CV,m – R) ∆T D CV,m ∆T – p ∆V 2. What is meant by the internal energy, and the molar heat capacity at constant volume of an ideal gas? State the relation between the two quantities. 3. (a) Write an equation to relate the principal molar heat capacities of an ideal gas. (b) A cylinder fitted with a light frictionless piston contains one mole of an ideal gas. When 63 J of heat is supplied to the gas, the gas expands at constant pressure and its temperature increases by 3.0 K. What is the heat required to produce the same increase in temperature if the gas is heated and the piston is prevented from moving? 4. A heat pump changes the state of a gas through a complete cycle represented by ABCDA in figure. The heat absorbed by the gas, and the work done on the gas for each stage are as shown in the table. A B C D 0 Pressure Volume

249 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Section of cycle Heat absorbed by gas/J Work done on gas/J A B 280 0 B C 0 190 C D –400 0 D A 0 –70 The cycle is completed 20 times per second. (a) What is the minimum average power of the motor required to run the pump? (b) What is the net average rate of heat supplied by the pump? 5. A fixed mass of an ideal gas changes from the state X to Y as shown by the p-V graph below. p/ Pa Y X V/ m3 0 100 2 4 6 8 10 12 200 300 400 500 (a) Determine whether the thermodynamics quantities W and ∆U for the process XY is positive, negative or zero. Explain your answer. (b) State the condition required for the thermodynamics quantity Q to be negative for the process XY. 6. The gas in the cylinder of a diesel engine undergoes a cycle of changes of pressure, volume and temperature as shown in figure. The temperature of the gas at A and B are 300 K and 660 K respectively. B C D A (a) Use the ideal gas equation and the data from the graph to find the temperature at C and D. (b) For each of the four sections of the cycle, changes occur to the internal energy of the gas. The heat supplied to the gas and the work done by the gas are as shown in the table below. Section of cycle Heat supplied to gas/J Work done by gas/J Increase in internal energy of gas/J A to B 0 –300 B to C 2 580 +740 C to D 0 +440 D to A –1 700 (i) Explain why the work done by the gas is sometimes negative, and find the work done by the gas in the section D to A. (ii) Deduce the value of the increase in internal energy of the gas for each section of the cycle. (c) Explain why the total change in the internal energy of the gas during a complete cycle must be zero. (d) Find the net work done by the gas during a complete cycle. (e) If the efficiency of a heat engine is net work output heat input , calculate the efficiency of this engine. 7. The figure gives the data in graphical form about the pressure, volume and temperature of a fixed mass of ideal gas.

250 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 0.001 0.002 0.003 0.004 0.005 0.006 1 Pressure / fi105 Pa 0 Volume / m3 2 3 4 5 6 7 8 D A C B T = 1 200 K T = 900 K T = 600 K T = 300 K (a) Use the ideal gas equation to find the number of moles of gas. (b) Show that the data of the graph are consistent with the gas behaving as an ideal gas over the range of temperature 300 K to 1 200 K. (c) The heat capacity of the gas at constant volume is 3.33 J K–1, and at constant pressure is 4.66 J K–1. Calculate the heat required to take the gas (i) from A to B along AB, (ii)from A to C along AC. (d) If the internal energy of the gas at A is 2 000 J, find the internal energy at B, and at C. (e) Calculate the work done by the gas when it expands at constant pressure from A to C. (f) Explain why the internal energy of the gas at D is also 2 000 J. 8. (a) Use the first law of thermodynamics to derive a relationship between the molar heat capacity Cp,m at constant pressure to the molar heat capacity CV,m at constant volume for an ideal gas. (b) Explain why Cp,m is greater than CV,m. Ratio of Cp,m CV,m 2014/P1/Q13, 2016/P1/Q13 1. When an ideal gas is heated at constant volume, the heat supplied is used to increase the internal energy of the gas. No external work is done by the gas because there is no change in volume. 2. For one mole of an ideal gas at a temperature T, internal energy, U = f 2 RT where f = degree of freedom. 3. The molar heat capacity at constant volume CV,m = dU dT CV,m = f 2 R Hence, the molar heat capacity at constant pressure Cp,m = CV,m + R = f 2 R + R Cp,m = (f + 2 2 ) R 4. The ratio of Cp,m to CV,m is denoted by the symbol γ. γ = Cp,m CV,m = (f + 2)R 2 f 2 R γ = f + 2 f

251 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 5. At normal temperatures: For monatomic gas, f = 3, γ = 5 3 = 1.67 For diatomic gas, f = 5, γ = 7 5 = 1.40 For polyatomic gas, f = 6, γ = 8 6 = 1.33 6. Whether a gas is monatomic, diatomic or polyatomic can be determined from its value of γ = Cp,m CV,m . 7. Table 10.3 below summarised the above discussion. Table 10.3 Monatomic Diatomic Polyatomic Degrees of freedom f = 3 3 translational f = 5 3 translational 2 rotational f = 6 3 translational 3 rotational Mean kinetic energy of one molecule 3 2 kT 5 2 kT 3kT Internal energy of 1 mole, U 3 2 RT 5 2 RT 3RT CV,m = dU dT 3 2 R 5 2 R 3R Cp,m = CV,m + R 5 2 R 7 2 R 4R γ = Cp,m CV,m 5 3 7 5 4 3 Example 9 The ratio of the principal molar heat capacities γ of an ideal gas is 1.40. If 29.1 J of heat is required to increase the temperature of 1 mole of the gas by 1 K at constant pressure, what is the molar specific heat capacity of the gas at constant volume? Solution: Cp,m = 29.1 J mol–1 K–1 Cp,m CV,m = 1.40 CV,m = Cp,m 1.40 = 29.1 1.40 = 20.8 J mol–1 K–1

252 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Example 10 The values of the specific heat capacity at constant volume, cV of the diatomic gas hydrogen at various temperatures are as given below. T/K cV/J kg–1 K–1 50 6.2 × 103 500 10.3 × 103 2 000 11.4 × 103 If the relative molecular mass of hydrogen is 2, calculate the molar heat capacity a constant volume of hydrogen at the temperatures given. Comment on your answers. Assume that hydrogen gas remains diatomic for all the temperatures. Solution: Mass of 1 mole of hydrogen gas, M = 0.002 kg Molar heat capacity at constant volume, CV,m = McV At 50 K, CV,m = 0.002 × (6.2 × 103 ) = 12.4 J mol–1 K–1 f 2 R = 12.4 Degrees of freedom, f = 12.4 × 2 8.31 = 3 At 500 K, CV,m = 0.002 × (10.3 × 103 ) = 20.6 J mol–1 K–1 f 2 R = 20.6 Degrees of freedom, f = 20.6 × 2 8.31 = 5 At 2 000 K, CV,m = 0.02 × (11.4 × 103 ) = 22.8 mol–1 K–1 f 2 R = 22.8 Degrees of freedom, f = 22.8 × 2 8.31 = 6 At 50 K (low temperature), f = 3. Only the translational kinetic energy of the hydrogen molecules is significant. The rotational kinetic energy is negligible. At 500 K (normal temperature), f = 5. Three for translational kinetic energy and two for rotational kinetic energy. At 2 000 K (high temperature), f = 6. Besides the translational and rotational kinetic energy, the vibrational kinetic energy is also significant.

253 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Example 11 The speed of sound in a gas is given by c = γp ρ . In an experiment, the speed of sound in a gas is 0.68 times the r.m.s. speed of the gas molecules. (a) Using the above equation and the kinetic theory equation for the pressure p of an ideal gas p = 1 3 ρ < c 2 >, determine the value of γ for the gas. (b) What deduction can you make about the molecules in the gas? Solution: (a) Using p = 1 3 ρ < c2 > cr.m.s = <c2 > = 3p ρ Given speed of sound in the gas = 0.68 cr.m.s. γp ρ = 0.68 3p ρ γ = (0.68)2 × 3 = 1.39 = 1.4 to 2 significant figures (b) Using γ = f + 2 f 1.4 = f + 2 f Degrees of freedom, f = 5 Hence, the gas molecules are diatomic. Example 12 The molar heat capacities at constant volume of diatomic hydrogen gas and oxygen gas are both 5 2 R. Two moles of hydrogen gas are mixed with one mole of oxygen gas. (a) What is the heat capacity at constant volume for 0.018 kg of the mixture? (b) When the mixture above is ignited to form 1 mole of steam, the molar heat capacity of steam is 3R. Explain why the molar heat capacity of the compound is different from the heat capacity of 0.018 kg of the mixture. (Relative molecular masses : hydrogen = 2, oxygen = 32) Solution: (a) Total mass of 2 mole of hydrogen and 1 mole oxygen = 2(0.002) + 1(0.032) kg = 0.036 kg Heat required to increase the temperature of 0.036 kg of the mixture by 1 K = 2 (5R 2 ) + (5R 2 ) = 15R 2

254 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 1. The molar heat capacity of an ideal monatomic gas at constant pressure is A 3 2 R C 3R B 5 2 R D 7 2 R 2. The specific heat capacity of a gas at constant volume is 0.410 kJ kg–1 K–1 and the gas molecules have five degrees of freedom. The gas constant is A 0.117 kJ kg–1 K–1 B 0.158 kJ kg–1 K–1 C 0.164 kJ kg–1 K–1 D 0.273 kJ kg–1 K–1 3. The molar heat capacity Cp,m at constant pressure for an ideal gas is 5 2 R. The number of degrees of freedom of the gas molecule is A 3 B 5 C 6 D 7 4. The value of the ratio Cp,m CV,m for hydrogen gas is 1.67 at 30 K but is 1.40 at 300 K. When the temperature increases, A Cp,m remains constant, but CV,m increases. B Cp,m decreases, but CV,m increases. C Cp,m decreases, but CV,m remains constant. D both Cp,m and CV,m increase by the same value. 5. The molar heat capacity at constant volume of an ideal gas is aR where a is a constant and R is the molar gas constant. The ratio of the principal molar heat capacities of the gas is A a – 1 C a a – 1 B a + 1 D a + 1 a 6. The table below shows the values of the molar heat capacity at constant volume, Cv,m at 50 K and 500 K for hydrogen gas which is diatomic. T/K CV,m/J mol–1 K–1 50 500 12.4 20.6 If the molar gas constant is 8.31 J mol–1 K–1, which of the following gives correctly the number of degrees of freedom for translational (T), rotational (R) and vibrational (V) energy which are active? At 50 K At 500 K T R V T R V A B C D 3 3 3 0 0 2 0 3 0 0 0 0 3 3 3 3 2 2 0 2 0 2 2 0 Hence, the heat capacity of 0.018 kg of the gas mixture at constant volume = ( 0.018 0.036 ) × (15R 2 ) = 15 4 R (b) When 0.018 kg of the mixture is ignited, one mole of steam (H2O) is formed. Mass of 1 mole of H2O = (2 + 16) × 10–3 kg = 0.018 kg H2O is polyatomic, hence at normal temperature (100°C), it has 6 degrees of freedom, f = 6, compared to f = 5 for hydrogen and oxygen molecules. This causes the molar heat capacity of H2O to be different from the heat capacity of 0.018 kg of the mixture. For 0.018 kg (1 mole) of H2O, CV,m = f 2 R = 6 2 R = 3R Quick Check 5

255 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 7. (a) (i) What is meant by degrees of freedom of a gas molecule? (ii) State the principle of equipartition of energy. Use the principle of equipartition of energy to derive an expression of γ, the ratio of the principal molar heat capacities of a gas in terms of f, the number of degrees of freedom. (b) (i) Explain why the value of CV,m the molar heat capacity at constant volume of a diatomic gas is greater than monatomic gas. (ii) The value of CV,m for a diatomic gas is different at low temperature (50 K), room temperature (300 K), and at high temperature (5 000 K). Assuming that the molecules of the gas remain diatomic for all the temperatures. Explain the different values of CV,m. Give the numerical values of CV,m at the various temperatures mentioned above. (c) Calculate the heat required to raise the temperature of 1.20 dm3 of hydrogen gas at constant pressure of 1.01 × 105 Pa from 300 K to 450 K. (Volume of 1 mole of hydrogen gas at 1.01 × 105 Pa pressure and temperature 273 K is 22.4 dm3 ) 8. (a) The speed of sound in a gas is given by v = ( γp ρ ) 1 2 where γ is the ratio of the principal molar heat capacities of the gas, p its pressure, and ρ its density. Use the real gas equation pVm = RT, derive an expression for v in terms of T and M the mass of one mole of gas. (b) Is it necessary to mention (i) the temperature (ii) the pressure in the results obtained to determine the speed of sound in a gas? 10.4 Isothermal and Adiabatic Changes Students should be able to: • describe the isothermal process of a gas • use the equation pV = constant for isothermal changes • describe the adiabatic process of a gas • use the equation pV γ = constant and TV γ – 1 = constant for adiabatic changes • illustrate thermodynamic processes with p–V graphs • derive and use the expression for work done in the thermodynamic processes Learning Outcomes Reversible Change 1. The state of a gas is determined by the three variables, pressure p, volume V and temperature T. 2. A gas is said to be in equilibrium if its state does not change with time and satisfies the equation of state of a real gas, pV T = constant. 3. A reversible process or reversible change in the state of a gas is a process where the gas is taken from an equilibrium state to another equilibrium state through a number of small steps, where the gas is always in equilibrium at every intermediate step (Figure 10.6(a)). The process can be reversed to its original state via the same small steps, but in the opposite direction (Figure 10.6(b)) 2009/P1/Q20, 2011/P1/Q20, 2013/P1/Q20, 2014/P1/Q14, 2016/P1/Q14

256 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 1 2 3 4 p 0 V Final state Original state 1 2 3 4 p 0 V Final state Original state (a) Change from original state to final state (b) Process is reversed via the same steps via small steps 1, 2, 3, … . but in the opposite direction. Figure 10.6 4. A reversible change can be represented by a smooth continuous curve as shown in Figure 10.6. 5. A change in the state of a gas becomes irreversible if the gas is not in equilibrium at any of the intermediate steps. An irreversible change cannot be represented by a smooth curve. Isothermal Changes 1. An isothermal change is the change in the condition of a gas which takes place at constant temperature. 2. A reversible isothermal change is the change in the condition of a gas which takes place at constant temperature where the gas is taken from its initial equilibrium state to a final equilibrium state via small steps. The gas remains in equilibrium through all the steps. The change can be reversed via the same small steps but in the opposite direction. 3. A reversible isothermal change obeys Boyle’s law and can be represented by the equation pV = constant where temperature T = constant 4. The Figure 10.7 shows the variation of p with V at different temperatures T1 < T2 < T3. These curves are known as isotherms. 5. In practice, a reversible isothermal compression, or expansion is carried out as described below. Gas Thin conducting wall Light frictionless piston Heat escapes Gas Thin conducting wall Light frictionless piston Heat absorbs (a) Isothermal compression (b) Isothermal expansion Figure 10.8 (i) The cylinder used to contain the gas should be thin-walled and made of good conducting material, so that heat can easily flow out or into the gas. Figure 10.7 Isotherms

257 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases (ii) A light frictionless piston is fitted to the cylinder. (iii) The cylinder is surrounded by a constant temperature bath. (iv) The compression or expansion is done by moving the piston through small steps. Each step is done very slowly. 6. Perfectly reversible isothermal change is not possible because when a gas is compressed, work done on the gas causes the gas temperature to rise before heat starts to flow out from the gas. 7. Conversely, when a gas expands, it does external work. Its temperature drops before heat flows into the gas. Work Done in an Isothermal Change 1. Figure 10.9 shows a fixed mass of an ideal gas at a temperature T expands isothermally from volume V1 to V2. 2. The work done by the gas W is represented by the shaded area under the p–V graph. Work done by gas, W = ∫ V V 1 2 p dV Using the ideal gas equation, pV = nRT p = nRT V W = ∫ V V 1 2 nRT V dV = nRT [1n V] V2 V1 = nRT [1n V2 – 1n V1] W = nRT 1n V2 V1 3. Using the first law of thermodynamics. Q = ∆U + W ∆U = 0 because temperature T is constant. Hence, work done W = Q, the heat absorbed by the gas. The energy required by the gas to do external work is absorbed from the surroundings. 4. When the gas is compressed isothermally at temperature T from (V1′, P1′) to (V2′, P2′) work done on gas, W = ∫ V V2 1 ′ ′ p dV = ∫ V V2 1 ′ ′ nRT V dV = nRT 1n V2 ′ V1 ′ Since V2′ < V1′ the work done is negative, consistent with the sign convention, that work done on gas is negative. 5. Using the first law of thermodynamics. Q = ∆U + W, ∆U = 0 Q = W Since W is negative, Q is negative. This means that the energy from the work done on the gas escapes as heat. V1 V2 p V W (V1, p1) (V2, p2) 0 Figure 10.9 Exam Tips For an isothermal change, • Temperature T = constant • No change in internal energy ∆U = 0 • Work done W = Q Figure 10.10

258 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Example 13 A cylinder fitted with a light frictionless piston contains 2.0 moles of an ideal gas at temperature of 300 K. What is the work done by the gas when it expands isothermally to twice of its initial volume? Solution: Using the ideal gas equation pV = nRT p = nRT V Work done by gas, W = ∫ 2V V1 1 p dV = ∫ 2V V1 1 nRT V dT = nRT [1n V] 2V V1 1 = nRT (1n 2V1 – 1n V1) = nRT 1n 2V1 V1 = 2.0 × 8.31 × 300 × 1n 2 = 864 J Quick Check 6 1. The first law of thermodynamics is represented by Q = ∆U + W. Where Q is the heat supplied, ∆U is change in internal energy, and W is work done by gas. During an isothermal expansion of an ideal gas, which of these quantities Q, ∆U and W, is zero? A Q C W B ∆U D Q and ∆U 2. An ideal gas in a cylinder is compressed at constant temperature from volume V1 at pressure p1 to volume V2 at pressure p2. Which of the following is true? A Heat is lost by the gas. B Internal energy of the gas increases. C Work is done by the gas. D Work done on the gas is p1V1 – p2V2. 3. When an ideal gas undergoes an isothermal change, A the number of degrees of freedom changes. B temperature of the gas changes. C there is no heat transfer with the surrounding. D the internal energy of the gas remains constant. 4. One mole of an ideal gas expands isothermally at temperature T until its volume is 3 times of its initial volume. If R is the molar gas constant, work done by the gas is A 3RT B 3RT ln V C RT 1n 3 D 3 2 RT 5. A fixed mass of an ideal gas is compressed isothermally. Which of the following is true for the heat supplied Q, change in internal energy ∆U, and work done by the gas W? Q ∆U W A Positive Positive Negative B Positive Negative Positive C Negative Zero Negative D Negative Zero Positive

259 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 6. A fixed mass of an ideal gas is first heated at constant volume. It is then cooled to its initial temperature at constant pressure before returning to its initial state through an isothermal process. Which of the following graphs shows correctly the three processes? A C p 0 V p 0 V B D p 0 V p 0 V 7. One mole of an ideal gas is in a cylinder fitted with a frictionless piston. The initial temperature of the gas is T. The gas is heated at constant pressure until its volume is doubled. If R is the molar gas constant, what is the work done by the gas? A RT C RT 1n 2 B 2RT D 3 2 RT 8. Two moles of an ideal gas is compressed isothermally at 400 K from volume V to 1 4 V. What is the work done on the gas? A 4.0 kJ C 8.4 kJ B 6.6 kJ D 9.2 kJ 9.What is the work done on one mole of an ideal gas which is compressed at constant temperature of 400 K from a volume of 15 dm3 to 10 dm3 ? A 0.3 kJ C 1.3 kJ B 0.6 kJ D 1.7 kJ Adiabatic Changes 2008/P1/Q21, 2009/P1/Q21, 2010/P1/Q22, 2015/P1/Q20, 2016/P1/Q17, 2017/P1/Q13 1. An adiabatic change is the change in the state of a gas where no heat enters or leaves the gas. 2. A reversible adiabatic change is the change in the state of a gas when no heat enters or leaves the gas. The gas is taken from an initial equilibrium state to a final equilibrium state via small steps. The gas is in equilibrium through all the steps. The process can be reversed via the same steps, but in the opposite direction. 3. Using the first law of thermodynamics, Q = ∆U + W For an adiabatic change Q = 0 0 = ∆U + W W = –∆U 4. (a) During an adiabatic expansion, work done by the gas W is equal to –∆U, the decrease in internal energy. (b) This implies that the gas used part of its internal energy to do external work as no heat flows into the gas. (c) When the internal energy decreases, the temperature of the gas decreases. 5. (a) During an adiabatic compression, work is done on the gas. W is negative. (b) Hence, change in internal energy ∆U = –W is positive. Internal energy of the gas increases. (c) As no heat leaves the gas. The temperature of the gas increases. 6. In practice, a reversible adiabatic change is carried out as follows. Exam Tips For an adiabatic change, • Q = 0, no heat enters or leaves the gas. • Work done, W = –∆U, change in internal energy.

260 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Gas Thick, insulating wall Light frictionless piston Gas Thick, insulating wall Light frictionless piston (a) Adiabatic compression (b) Adiabatic expansion Figure 10.11 Adiabatic changes (i) The gas cylinder should have a thick wall made of insulating material, to prevent heat from entering or leaving the gas. (ii) The cylinder is fitted with light frictionless piston so that no work is done against friction. (iii) The compression (or expansion) is carried out by moving the piston in (or out) quickly but in small steps, so that the heat leaving (or entering) the gas in the very short time for each stage is negligible. 7. Perfectly adiabatic changes are not possible in practice, as there is no perfect insulator to prevent the flow of heat into or out of the gas. 8. Examples of adiabatic changes are (a) the compression of air in a diesel engine. (b) the compression and rarefaction of air during the propagation of sound waves in a gas. 9. When a gas is compressed adiabatically, the increase in pressure is greater than that produced by a similar isothermal compression because • the temperature of the gas increases • the r.m.s. speed of the gas increases • the density of the gas increases • from the equation p = 1 3 ρ< c2 >, the pressure of the gas increases. Adiabatic Equations 1. Consider a mole of ideal gas at pressure p, volume V and temperature T, then pV = RT p = RT V 2. When the volume of the gas changes by ∆V, its temperature changes by ∆T. Change in internal energy, ∆U = CV,m ∆T Work done W = p ∆V = ( RT V ) ∆V 3. Using the first law of thermodynamics, Q= ∆U + W Q = 0 (adiabatic) 0 = CV,m ∆T + ( RT V ) ∆V R = Cp,m – CV,m Hence CV,m ∆T + (Cp,m – CV,m) T ∆V V = 0 VIDEO Isothermal and Adiabatic Process

261 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases ∆V T + ( Cp,m CV,m – 1) ∆V V = 0 Cp,m CV,m = γ dT T + (γ – 1) dV V = 0 Integrating, ∫ dT T + ∫ (γ – 1) dV V = ∫0 ln T + (γ – 1) ln V = constant ln (TV γ–1) = constant TV γ–1 = constant From pV = RT Substituting, T = pV R into TV γ–1 = constant ( pV R ) V γ – 1 = constant pVγ = constant 4. Figure 10.12 shows the variation of pressure p with volume V for an adiabatic compression (AB), and an adiabatic expansion (AD). 5. The magnitude of the gradient of the adiabatic curve at a point such as A is greater than that of the isothermal curve. 6. For an isothermal change, using pV = constant where T = constant p = constant V Gradient = dp dV = – constant V2 = – pV V2 (constant = pV) = – p V 7. For an adiabatic change, differentiating pVγ= constant V γ dp dV + p (γV γ–1) = 0 dp dV = –p(γV γ–1) V γ Gradient = –γ p V (γ > 1) Hence, magnitude of gradient of the adiabatic curve is greater than that of the isothermal. Exam Tips Adiabatic equations pV γ = constant TV γ–1 = constant pV = nRT p B A D 0 V Adiabatic expansion (V1, p1) T1 T2 T0 (V2, p2) Adiabatic compression (V0, p0) Figure 10.12

262 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Example 14 In the figure, AD and BC represent two isotherms at temperatures of T1 and T2 respectively, of a fixed mass of an ideal gas. The isotherms are intersected by the adiabatics AB and DC of the same gas. Using the notations in the figure, find an expression for (a) the ratio VA VB in terms of VC and VD, (b) the ratio pA pB in terms of pC and pD. Solution: (a) For the adiabatic AB : T1 VA γ–1 = T2 VB γ–1 For the adiabatic DC : T1 VD γ–1 = T2 VC γ–1 Hence (VA VD) γ–1 = (VB VC ) γ–1 VA VB = VD VC (b) For the isotherm T1 : pAVA = pDVD For the isotherm T2 : pBVB = pCVC ( pA pB ) (VA VB ) = ( pD pC ) (VD VC ) ( pA pB )(VD VC ) = ( pD pC ) (VD VC ) VA VB = VD VC pA pB = pD pC Example 15 The temperature of oxygen gas in a cylinder is 30°C. Assuming that oxygen behaves as an ideal gas with relative molecular mass 32.0, calculate the r.m.s. speed of the oxygen molecules (a) after the gas expands slowly under isothermal conditions to twice of its initial volume. (b) after the gas expands adiabatically to twice of its initial volume. Explain why the answers to (a) and (b) are different. (γ = 1.40 for oxygen gas) Solution: (a) When the gas expands isothermally, the temperature remains unchanged, and the r.m.s. speed cr.m.s is given by 1 2 m <c2 > = 1 3 kT cr.m.s = <c2 > = 3kT m (m = M NA ) = 3kTNA M = [ 3 (1.38 × 10–23) (273 + 30) (6.02 × 1023) 32 × 10–3 ] 1 2 = 486 m s–1 p A D B C pA pD pB pC VA VB VD VC 0 V T2 T1

263 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases (b) T1 = 30°C = (283 + 30) K = 303 K, V1 = V, V2 = 2V Using T1V1 γ–1 = T2V2 γ–1 T2 = ( V1 V2 ) γ–1 T1 = ( 1 2 ) 1.4 – 1 × T1 = (0.5)0.4 × 303 = 230 K cr.m.s ∝ T At 230 K, cr.m.s = 230 303 × 486 = 423 m s–1 When the gas expands isothermally, the energy required for the gas to do external work comes from the absorbed heat. The internal energy remains unchanged. When the gas expands adiabatically, the energy required for the gas to do external work comes from its internal energy. The internal energy decreases. Hence, the r.m.s. speed decreases. Work Done during Adiabatic Expansion 1. Consider a fixed mass of a gas at an initial pressure of p1, volume V1 and temperature T1. When the gas expands adiabatically to volume V2, its pressure changes to p2 and its temperature drops to T2. 2. Using pV γ = c, constant Pressure p = c Vγ Work done by gas,W = ∫ V V 1 2 p dV = ∫ V V 1 2 c Vγ dV = c 1 – γ [ 1 V γ–1 ] V2 V1 = c 1 – γ [ 1 V2 γ–1 – 1 V1 γ–1 ] c = p1V1 γ = p2V2 γ = 1 γ – 1 [ p1V1 γ V1 γ–1 – p2V2 γ V2 γ–1 ] = 1 γ – 1 [p1V1 – p2V2] p1V1 = nRT1, p2V2 = nRT2 = nR γ – 1 [T1 – T2] γ = f + 2 f , where f = number of degrees of freedom W = nR f + 2 f – 1 [T1 – T2] = – nfR 2 (∆T) where ∆T = T2 – T1 W (V1, p1) T1 T2 (V2, p2) p V1 V2 0 V Figure 10.13

264 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases From U = nfR 2 T ∆U = nfR 2 (∆T) W = –∆U, change in internal energy 3. The relationship W = –∆U can also be established using the first law of thermodynamics, Q = ∆U + W. For adiabatic change, Q = 0, W = –∆U Example 16 In a diesel engine, diesel is injected into a cylinder where the temperature of the air has been increased by an adiabatic compression to exceed the ignition temperature of diesel, 630°C. Air enters the cylinder at a pressure of 1.0 × 105 Pa and temperature of 28°C. The initial volume of the air is 5.0 × 10–4 m3 . (a) Calculate the ratio initial volume final volume for the air in the cylinder so that its final temperature is the ignition temperature of diesel. (b) What is the work done to compress the air? (For air, γ = 1.40) Solution: (a) T1 = 28°C = 301 K V1 = 5.0 × 10–4 m3 T2 = 630°C = 903 K V2 = final volume Using T1 V1 γ–1= T2 V2 γ–1 (301) V1 1.40–1 = (903) V2 1.40– 1 ( V1 V2 ) 0.40 = 903 301 = 3 ∴ V1 V2 = 35 2 = 15.6 (b) Work done = nR γ – 1 (T1 – T2) n = pV RT = (1.0 × 105 ) (5 × 10–4) (8.31 × 301) × 8.31 (1.40 – 1) × (301 – 903) = –250 J Example 17 (a) (i) What is meant by an isothermal compression and an adiabatic compression? (ii) On the same axes, sketch the p-V graphs for both processes for an ideal gas where the initial volume is V0, and initial pressure is p0. (b) The volume of an ideal gas at pressure p is V. When the gas undergoes an adiabatic change, its volume increases by ∆V. Derive an expression for the corresponding change in pressure ∆p of the gas in terms of p, V, ∆V and γ the ratio of the principal molar heat capacities of the gas.

265 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases (c) The volume of one mole of monatomic ideal gas at 0°C and pressure 1.01 × 105 Pa is 22.4 litre. Calculate the work done to compress the gas to a volume of 10.0 litre if the compression is carried out (i) isothermally, (ii) adiabatically. Explain how the difference in the answers to (i) and (ii) can be deduced from the p-V graphs you sketched in (a). Solution: (a) (i) Isothermal compression – compression of a gas at constant temperature. Adiabatic compression – compression of a gas such that no heat enters or leaves the gas. (ii) Adiabatic compression Isothermal compression p p0 V0 0 V (b) Differentiate the equation pVγ = constant V γ dp dV + p γ V γ–1 = 0 dp dV = – pγVγ–1 Vγ = – pγ V ∴ ∆p= – pγ V (∆V) (c) (i) V1 = 22.4 liter and V2 = 10.0 liter For isothermal compression Work done W = ∫ V V 1 2 p dV pV = RT = ∫ V V 1 2 RT V dV = RT [1n V]V2 V1 = 8.31 × 273 × 1n 10.0 22.4 = –1.83 × 103 J (ii) Using the first law of thermodynamics, for adiabatic change Q = 0. Q = ∆U + W 0 = ∆U + W Work done W = –∆U U = 3 2 RT = – 3 2 R ∆T T1 = 0°C = 273 K

266 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Using T1V1 γ–1 = T2V2 γ–1 γ = 1.67 T2 = 273 ( 22.4 10.0) 1.67–1 = 469 K Hence, work done, W = – 3 2 R(∆T) = – 3 2 × 8.31 × (469 – 273) = –2.44 × 103 J The area under the adiabatic curve is greater than the area under the isotherm. Hence, work done on the gas is greater for adiabatic compression. Example 18 A cylinder of volume 8.0 × 10–3 m3 contains an ideal gas at a pressure of 1.14 × 105 Pa. The tap of the cylinder is opened and the gas expands adiabatically, and the pressure in the cylinder drops to atmospheric pressure (1.01 × 105 Pa), and some of the gas escapes. The tap is then closed and the gas in the cylinder is allowed to return to its initial temperature, the equilibrium pressure of the gas is 1.06 × 105 Pa. (a) Explain why the temperature of the gas changes during an adiabatic expansion. (b) Calculate the final volume of the gas that remains in the cylinder under the initial conditions. (c) Sketch a graph to show the variation of the pressure with the volume for the mass of gas that remains in the vessel during the two stages. (d) Calculate the value of γ, the ratio of the principal heat capacities of the gas. (e) What can you deduce about the nature of the gas molecules. Show how you make the deduction. Solution: (a) When a gas expands adiabatically, work is done by the gas, (W > 0). Since no heat enters the gas (Q = 0), using the first law of thermodynamics Q = ∆U + W ∆U = –W Internal energy of the gas decreases and its temperature decreases. (b) p1 = 1.14 × 105 Pa, V1 = ? p2 = 1.06 × 105 Pa, V2 = 8.0 × 10–3 m3 Since the initial and final temperature of the gas are the same, using Boyle’s law p1V1 = p2V2 V1= (1.06 × 105 )(8.0 × 10–3) (1.14 × 105) = 7.44 × 10–3 m3 (c) Isotherm C B A p/fi 105 Pa V/fi 10–3 m3 0 1.01 1.06 1.14 7.44 8.00

267 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases (d) When the tap is opened, the gas expands adiabatically from A to B. (see graph in (c)). p1 = 1.14 × 105 Pa, V1 = 7.44 × 10–3 m3 p2 = 1.01 × 105 Pa, V2 = 8.00 × 10–3 m3 Using p1V1 γ = p2V2 γ ( V2 V1 ) γ = p1 p2 ( 8.00 × 10–3 7.44 × 10–3 ) γ = 1.14 × 105 1.01 × 105 ∴ γ = 1.67 (e) Using γ = f + 2 f , 1.67 = 5 3 = f + 2 f f = 3 Hence, the molecules of the gas are monatomic. Example 19 (a) Distinguish between internal work and external work done by a real gas when the gas expands. (b) The diagram shows a real gas in vessel A which is connected to evacuate vessel B by a narrow tube fitted with a tap which is closed. When the tap is opened the gas expands adiabatically into B. It is found that the equilibrium temperature of the water bath around the vessels remains unchanged, but the temperature TB of the gas that flows into B is higher than the initial temperature T of the gas, and the final temperature of the gas that remains in A, TA is lower than T. (i) Explain why TB > T > TA. (ii) Based on these observations, discuss whether any external work is done during the expansion. (iii) What conclusion can be made about any internal work done during the expansion? Solution: (a) When a real gas expands, internal work is done by the gas against the force of attraction between gas molecules when the separation between gas molecules increases. External work is done by the gas against the external pressure when the volume of the gas increases. (b) (i) For the gas that flows into B, the temperature increases, TB > T, because the kinetic energy of the gas molecules increases. The temperature of the gas that remains in A decreases because the internal energy of the gas decreases. Part of the internal energy of the gas that remains in A is used to do internal work against molecular attraction between molecules that flows into B. (ii) No external work is done as the initial pressure in B is zero. Work done = p ∆V, p = 0 = 0 (iii) Since the temperature of the water bath remains unchanged, there is no resultant heat flow between the gas and the water bath. The internal work done by the gas comes from the decrease in the internal energy of the gas that remains in A. A B Water bath

268 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 1. An adiabatic change of an ideal gas is one A in which the internal energy remains constant. B in which no heat enters or leaves the gas. C in which no external work is done by or on the gas. D which is carried out very rapidly in a thinwalled conducting cylinder. 2. A fixed mass of ideal gas initially of volume V1 in state X undergoes a reversible adiabatic expansion to state Y, and then undergoes a reversible isothermal compression back to its volume V1 in state Z. Which of the following graphs best represents the changes in the state of the gas? A X Y Z C X Z Y B X Z Y D Z X Y 3. An ideal monatomic gas has a volume of 8.0 m3 at a pressure of 2.43 × 105 Pa. It expands adiabatically to 27.0 m3 . Which is its final pressure? A 0.32 × 105 Pa C 1.08 × 105 Pa B 0.72 × 105 Pa D 1.17 × 105 Pa 4. Two samples of an ideal gas, P and Q are initial at the same state. P is compressed adiabatically, and Q is compressed isothermally until their final pressure is three times of their initial pressure. Which of the following about the sample of gas P and Q in the final state is correct? A Final temperature of P less than that of Q. B Work done on P and Q are the same. C Heat lost from P greater than that of Q. D Final volume of P greater than that of Q. 5. A vessel fitted with a tap contain an ideal gas at a pressure several times of the atmospheric pressure. The tap is opened and then immediately closed. The vessel is set aside until it attains it equilibrium state. Which of the following graphs shows correctly the variation of pressure p and volume V of a unit mass of gas in the vessel? A C B D 6. (a) Write an expression for the first law of thermodynamics to represent an adiabatic change in the state of an ideal gas where 160 J of work is done by the gas. (b) Deduce the change in internal energy of the gas. (c) What can be deduced about the temperature of the gas? 7. A fixed mass of an ideal gas expands isothermally and is then compressed adiabatically back to its original pressure. (a) Sketch a p-V graph to represent the changes in the state of the gas. (b) Describe briefly how the gas may be returned to its original state in a single process. 8. An ideal gas of 0.025 mole has a pressure of 5.0 × 105 Pa at a temperature of 400 K. (a) What is its volume? (b) The gas then undergoes an adiabatic expansion until its volume is doubled. What is its final pressure? (γ = 1.67) Quick Check 7

269 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 9. The pressure of an ideal gas of mass m in an insulated cylinder increases from p0 to p1 when its volume decreases from V0 to V1. The temperature increases from T0 to T1. Write equations to relate (a) p0, V0, and T0. (b) p0, p1, V0 and V1. Identify any additional symbol in your equation. 10. One mole of the hydrogen gas has a volume of 0.060 m3 is compressed adiabatically until its final volume is half of its original volume. The initial mean kinetic energy per molecule of the gas is 1.38 × 10–20 J. Calculate (a) the initial pressure, (b) the final pressure of the gas. Assume the hydrogen gas behaves as an ideal diatomic gas. 11. Initially the volume of 48 g of oxygen gas is 0.037 m3 when its pressure is 1.013 × 105 Pa. The gas then undergoes the following processes. I : Expands adiabatically to twice of its original volume. II : Is compressed at constant pressure, so that the gas returns to its initial volume of 0.037 m3 . III : Is heated at constant volume, so that the gas returns to its original state. Calculate (a) the initial temperature of the gas, and its temperatures at the end of stage I and II. (b) the work done by the gas for each stage I, II and III. (c) the heat transferred during stage I, II and III. [CV,m = 20.8 J mol–1 K–1, Cp,m = 29.1 J mol–1 K–1] Important Formulae 1. Work done by a gas, W = ∫ V V 1 2 p dV = p ∆V when the pressure p = constant 2. CV,m = 1 n dU dT Cp,m – CV,m = R γ = Cp,m CV,m = f + 2 f 3. Molar heat capacities: CV,m Cp,m γ Monatomic gas 3 2 R 5 2 R 5 3 Diatomic gas 5 2 R 7 2 R 7 5 Polyatomic gas 3R 4R 4 3 4. Internal energy, U = nCV,m T Change in internal energy, ∆U = nCV,m (∆T) 5. First law of thermodynamics: Q = ∆U + W 6. Isothermal changes: ∆T = 0 and pV = constant, W = nRT ln ( V2 V1 ) 7. Adiabatic changes: Q = 0, pV γ = constant, TV γ–1 = constant, W = –∆U

270 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases STPM PRACTICE 10 1. A gas undergoes the change from P to Q along the path shown in the p-V graph below. The changes to the thermodynamics quantities, Q, heat transfer, U, internal energy, and W, work done are correctly described in Q U W A Enters the gas Increases by the gas B Enters the gas Decreases on the gas C Leaves the gas Increases by the gas D Leaves the gas Decreases on the gas 2. A monatomic gas expands adiabatically from the state X to the state Y as shown in the figure below. p / 105Pa V / m3 7.2 X 2.0 Y 1.5 0 3.2 What is the change in the internal energy of the gas? A 220 kJ B 340 kJ C 660 kJ D 1.10 MJ 3. When a gas expands isothermally, it does 50 J of work. What is the change in its internal energy ∆U, and the heat transferred Q? ∆U Q A 0 +50 J B 0 –50 J C +50 J 0 D –50 J 0 4. T h e v a r i a t i o n o f i n t e r n a l e n e r g y per mole U with temperature of a gas is as shown in the graph. What is the ratio of the principal molar heat capacities of the gas? A 1.33 C 1.40 B 1.20 D 1.67 5. A vessel contains 5.0 mole of a diatomic gas at an initial temperature of 60o C. How much heat is lost by the gas when its temperature drops to 35o C at constant volume? A 500 J C 2600 J B 1600 J D 3100 J 6. An ideal monatomic gas has initial pressure p, volume V and temperature T. When the gas expands isothermally to twice its initial volume, its pressure is p1. If the gas expands adiabatically to a volume of 2V, its pressure becomes p2. What is the ratio of p1/p2? A 0.40 C 1.60 B 0.67 D 1.67 7. A 2.0 mol sample of an ideal gas is heated at constant pressure and the temperature of the gas increases by θ°C. If the molar heat capacities of the gas at constant volume and at constant pressure are Cv and a Cp respectively, what is the work done by the gas? A 2Cpθ B 2Cp(273 + θ) C 2(Cp – Cv)θ D 2(Cp – Cv)(273 + θ) 8. Heat is absorbed by a gas undergoing an isothermal expansion. Which statement explains why the internal energy of the gas remains unchanged? A The gas is an ideal gas. B There is no change of state. C The pressure of the gas remains constant. D The heat absorbed is totally converted into work done by the gas. 9. What is the molar heat capacity at constant volume of an ideal monatomic gas? A R C 5 2 R B 3 2 R D 3R p P Q V U / J mol–1 T / K 6250 2500 200 500

271 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 10. An ideal gas undergoes the change WXY as shown by the p-V graph below. The gas has the same temperature at W and Y. p W X Y V 0 Which row about the net heat transferred and work done for the change W X Y is correct? Net heat transferred Work done A Absorbed by the gas by the gas B Lost by the gas by the gas C Absorbed by the gas on the gas D Lost by the gas on the gas 11. The molar heat capacity of a gas at constant pressure is Cp,m. The molar gas constant is R. The number of degree of freedom of the gas is given by A Cp,m R B Cp,m R – 1 C  Cp,m R – ! D 2 Cp,m R – ! 12. An ideal gas undergoes the changes PQ and QR as shown in the p-V graph below. V p Q P R Isotherms Which statement is correct? A In the process PQ, the temperature of the gas remain constant. B In the process QR, the internal energy of the gas decreases. C The final temperature at R is lower than the temperature at P. D In the whole process, work is done on the gas 13. A fixed mass of ideal gas undergoes changes of pressure and volume as shown in the graph. Which of the following graphs shows correctly the relation between thermodynamic temperature T and the volume V? A C B D 14. A fixed mass of an ideal gas undergoes the changes represented by XYZ as shown below. Y X Z Which of the following sets describes the changes? XY YZ ZX A Isothermal Adiabatic Compression at compression expansion constant pressure B Isothermal Adiabatic Pressure expansion compression reduction at constant volume C Adiabatic Isothermal Compression at compression expansion constant pressure D Adiabatic Isothermal Pressure expansion compression reduction at constant volume

272 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 15. The initial internal energy of an ideal gas is U1. It expands adiabatically and the work done is W. Heat Q is then supplied to increase its pressure at constant volume. The final internal energy of the gas is U2. The increase in internal energy U2 – U1 is equal to A W C Q – W B Q D W – Q 16. A fixed mass of ideal gas at a temperature of 300 K has pressure and volume as represented by the point X on the graph shown. The gas undergoes changes XYZ as show. What is the temperature of the gas at Z? Volume / cm3 Pressure / fi 105 Pa 0 0.25 0.50 0.75 1.00 5 10 15 20 Z Y X A 200 K C 450 K B 400 K D 600 K 17. (a) Write the equation representing the first law of thermodynamics. Hence deduce the change in the internal energy of a gas that undergoes an adiabatic process in terms of the relevant quantity in the equation. (b) A diatomic gas at 250 K and pressure of 300 kPa has a volume of 1.18 × 10–2 m3 . It expands adiabatically to a volume of 2.00 × 10–2 m3 . Determine (i) the final temperature of the gas, (ii) the work done by the gas. 18. Two moles of hydrogen gas in a cylinder is at a temperature of 300 K. The gas is heated at constant volume until its temperature is 350 K. Assume that hydrogen is an ideal gas. (a) What is the molar heat capacity at constant volume of hydrogen gas? (b) What is the heat supplied to the gas? (c) If the gas is heated at constant pressure from 300 K to 350 K, what is the heat required? 19. (a) Differentiate between CV,m molar heat capacity at constant volume and Cp,m molar heat capacity at constant pressure for a gas. (b) Show that for an ideal gas, γ = cp,m cv,m = f + 2 f where f is the number of degree of freedom of the gas. (c) An ideal monatomic gas at a temperature of 30°C, pressure 1.01 × 105 Pa has a volume of 3.2 × 10–3 m3. The gas is compressed adiabatically to one third of its original volume. (i) State qualitatively, the change, if any, to the internal energy, heat transfer, and work done. (ii) Calculate the final pressure of the gas. (iii) Determine the work done. 20. B A D C When a system changes from A to D via the stages AB and BD, the heat supplied is 5.0 J and the work done by the system is W = 2.0 J. For the stages AC and CD, the heat supplied is 4.0 J. (a) What is the work done by the system via the stages AC and CD? (b) If W = –1.5 J for the stage DA, what is the value of the heat supplied for this stage? (c) If at A, the internal energy U1 = 1.0 J, what is the internal energy of the system at D? (d) If the internal energy U = 2.0 J at C, calculate the heat supplied for the (i) stage AC, (ii) stage CD. 21. (a) State the first law of thermodynamics. (b) Give a practical example of each of the following. (i) A process where heat is supplied to a system without causing any increases in temperature.

273 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases (ii) A process where the temperature of the gas increases even though the gas is not heated. (c) A cylinder fitted with a light frictionless piston contains 1.00 × 1022 molecules of argon gas at a temperature of 30°C. The gas is heated at constant pressure to 50°C. Assuming that argon gas behaves as an ideal gas, calculate the heat supplied to the gas. 22. (a) (i) What is meant by an adiabatic process? (ii) Write an equation to relate the pressure p and the volume V of an ideal gas that undergoes adiabatic change. (b) A vessel of volume 2.0 × 10–2 m3 contains an ideal gas at a pressure of 1.50 × 105 Pa. A tap attached to the vessel is opened and the gas expands adiabatically until its pressure equals the atmospheric pressure of 1.01 × 105 Pa. The tap is then closed. The gas is allowed to return to its original temperature. In the final equilibrium state, the pressure of the gas that remains in the vessel is 1.13 × 105 Pa. (i) Calculate the volume of the gas that remains in the vessel under the initial pressure of 1.50 × 105 Pa and temperature. (ii) Calculate the ratio of the principal heat capacities of the gas. (iii) Sketch a p-V graph to show the variation of p with V for the mass of gas which remains in the vessel for the changes described above. 23. p 0 V X Y Z When an ideal gas is taken from the state X to Z along XYZ, the heat supplied for the stage XY and YZ are 182 J and 65 J respectively. The temperature of the gas at X, Y and Z are respectively 300 K, 400 K and 450 K. (a) What is the increase in internal energy of the gas during the process XYZ? Explain why the increase in internal energy is different from the value of the total heat supplied. (b) Determine the value of γ, the ratio of the principal heat capacities of the gas. Then, state whether the molecule of the gas is monatomic, diatomic or polyatomic. (c) Calculate the number of moles for the gas. 24. (a) The internal energy U of a mole of ideal gas at various temperatures θ°C is given by the table below. θ/°C 0 18 35 45 U/J mol–1 5 730 6 110 6 470 6 680 (i) Use the data in the table, deduce a value for the molar heat capacity at constant volume of the gas. (ii) Find the value of γ, the ratio of the principal heat capacities of the gas. (iii) What conclusion can you make about the gas molecules? Give your reasons. (b) A sample of ideal gas with an initial volume of 3.0 × 10–3 m3 and temperature 17°C undergoes the following changes. I : Its temperature is increased to 27°C at constant pressure. The heat supplied is 20.5 J. II : Its temperature then changes to 17°C at constant volume. The heat lost is 14.6 J. III : The gas undergoes an adiabatic compression until its final volume is 2.0 × 10–3 m3 . (i) Sketch a p-V graph to show the above processes. (ii) Find the ratio of the principal heat capacities of the gas. (iii) What is the volume of the gas at the end of process I? (iv) What is the temperature of the gas at the end of process III? 25. (a) Explain in terms of molecular theory of gas, the change in the temperature of an ideal gas which expands adiabatically and does work. (b) Describe how a gas can be compressed almost adiabatically in practice.

274 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases (c) A fixed mass of an ideal gas is compressed adiabatically from A to B, and then the pressure is increased from B to C as shown in the figure. 20.0 1.4 1.0 16.0 10.0 C B A 0 p / fi 105 Pa V / fi 10–4 m3 Use the data shown on the graph to calculate (i) the value of γ. Then, state whether the gas is monatomic, diatomic or polyatomic, (ii) the temperature at B, if the temperature at A is 450 K, (iii) the work done to compress the gas from A to B, (iv) the temperature at C. How is the process BC carried out? 26. (a) (i) Write an equation to represent the first law of thermodynamics. Identify the symbols in your equation. (ii) Use the equation above to explain the energy used by an ideal gas when it expands isothermally and adiabatically. (b) The figure below shows a pump used to compress air initially at atmospheric pressure 1.01 × 105 Pa and temperature 300 K. The pump consists of a uniform insulated cylinder of length 0.300 m. The valve attached to the pump opens when the pressure of the air in the pump reaches 6.25 × 105 Pa. Insulator 0.300 m Valve Air Piston If air behaves as an ideal gas with molar heat capacity CV,m at constant value of 20.8 J mol–1 K–1, calculate (i) distance move by the piston before the valve opens, (ii) the temperature of the compressed air, (iii) the work done to pump 50.0 mole of air into a vessel using the pump. 27. The figure below shows the p-V graph of 2.0 mole of an ideal monatomic gas in a cylinder. In the initial state X, the temperature of the gas is 360 K and the volume is 1.20 × 10–3 m3 . The gas is compressed at constant pressure until its volume is 0.30 × 10–3 m3 to state Y. It is then heated at constant volume until its temperature returns to 360 K, state Z. Finally it returns to its initial state X isothermally. p / Pa V / 10–3 m3 p2 p1 0 X Z Y 0.30 1.20 Calculate (a) the pressure p1 at X, (b) the temperature at Y, (c) pressure p2 at Z, (d) the heat absorbed for the process Y to Z (e) the nett work done in the cycle XYZX. 28. (a) Two moles of a diatomic ideal gas is heated at constant pressure of 1.01 × 105 Pa from 30o C to 45o C. Calculate (i) the heat supplied to the gas, (ii) the increase in the volume of the gas, (iii)the work done on the gas, (iv) the increase in internal energy of the gas. (b) The gas with the same initial conditions as in (a) is undergoes adiabatic compression until its volume become one third. (i) What is meant by adiabatic compression? (ii) What is the final temperature of the gas? (iii) Calculate the work done on the gas.

275 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 1 1. D 2. No. When heated, increase in volume of liquid is negligible. 3. (a) CV = 7500 (350 – 320) J K–1 = 250 J K–1 (b) cV = CV m = 3130 J kg–1 K–1 (c) CV,m = McV = 12.5 J mol–1 K–1 2 1. B : Nett work = WWX – WY = (4.0 × 105 )(3.0 – 1.0)10–4 – (2.0 × 105 )(3.0 – 1.0)104 J = 40 J 2. B 3. B 4. C 5. B: pV = nRT, p∆V = nR ∆T W = p∆V = nR ∆T = 1.99 kJ 6. (a) 2.26 × 106 J Q = mL (b) 1.69 × 105 J Q = p(Vsteam – Vwater) (c) 2.09 × 106 J DU = Q – W 7. 600 K V ∝ T, if p = constant 0.05 J W = p(V2 < V1) 8. (a) 1.31 × 1025 N = ( m M ) NA (b) 63.1 kJ W = p ∆V (c) 6.01 × 10–20 J ∆U = ml – p ∆V 3 1. C 2. D 3. B 4. D 5. C 6. (a) Q = ∆U + W (b) (i) U = (Kinetic energy + Potential energy) of gas molecules (ii) work done on the gas when it is compressed. 7. (a) U decreases, molecules move slower. (b) U = constant, mean kinetic energy of molecules ∝ T. (c) U = constant. U depends on random motion of molecules. 8. 2.09 × 106 J DU = Q – pDV 4 1. B 2. CV,m = 1 n dU dT 3. (a) Cp,m – CV,m = R (b) 38.07 J Cp,m = 63 3.0 J mol–1 K–1 Cp,m = CV,m + R 4. (a) 3 800 W P = (190)(20)W (b) 2 400 W Net heat per cycle = (400 – 280) J 5. (a) W negative, volume decreases. ∆U positive, value of pV increases, T increases (b) |W| > |∆U| 6. (a) TC = 2 829 K pV = nRT TD = 2 340 K A and D : p ∝ T (b) (i) W negative : work done on the gas W = 0 for DA (ii) ∆U = +300 J, 1 840 J, –440 J, –1 700 J (c) Gas returns to its original state (original temperatures). U ∝ T, ∆U = 0 (d) 880 J (e) 34.1% 880 2 580 × 100% 7. (a) n = 0.160 pV = nRT (b) Calculate value of n at 300 K, 600 K, 900 K, and 1 200 K. (c) (i) 1 998 J Q = nCv ∆T (ii) 2 796 J Q = nCp ∆T (d) UB = 4 000 J U ∝ T UC = 4 000 J TC = TB (e) W = 800 J W = p∆V (f) TD = TA U ∝ T 8. (a) Cp,m = CV,m + R (b) Refer to page 247 5 1. B 2. C 3. A 4. D 5. D 6. A 7. (a) (i) Degree of freedom : independent mode of motion. (ii) Derive γ = (f + 2) f . Refer to page 250 (b) (i) f of diatomic gas, 5 > f of monatomic gas, 3. (ii) 50 K : f = 3, CV,m = 3 2 R = 12.5 J mol–1 K–1 300 K : f = 5, CV,m = 5 2 R = 20.8 J mol–1 K–1 5 000 K : f = 6, CV,m = 3R = 25.0 J mol–1 K–1 (c) 213 J Q = nCp,m ∆T 8. (a) v = ( γRT M ) 1 2 pVm = RT and ρ = M Vm (b) (i) Temperature T, yes (ii) Pressure p, no 6 1. B 2. A 3. D 4. C 5. C 6. C 7. A 8. D 9. C 7 1. B 2. C 3. A 4. D 5. A 6. (a) Q = ∆U + W Q = 0, ∆U = –W (b) ∆U = –W = –160 J (c) Temperature decreases ANSWERS

276 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 7. (a) 0 p V (b) Cool gas at constant pressure. 8. (a) 1.66 × 10–4 m3 pV = nRT (b) 1.57 × 105 Pa p1V1 γ = p2V2 γ 9. (a) p0 V0 = ( m M ) RT0 m : mass of gas, M : mass of 1 mole of gas (b) p0V0 γ = p1V1 γ γ = Cp,m CV,m 10. (a) 5.54 × 104 Pa 5 2 kT = 1.38 × 10–20 J, pVm = RT (b) 1.46 × 105 Pa p1V1 γ = p2V2 γ 11. (a) 301 K, 228 K, 114 K (b) I : 2 271 J W = –∆U = nCV,m ∆T II : 1 420 J W = p∆V III : 0 ∆V = 0 (c) I : 0 adiabatic, Q = 0 II : –4 974 J Q = ∆U + W III : 5 825 J Q = nCV,m ∆T STPM Practice 10 1. A : V increases W by gas. Temperature increases, hence U increases Q supplied to increase U and for the gas to do work. 2. C : pV = nRT Monatomic: ∆U = 3 2 nR(∆T) ∆U = 3 2 [(pV)Y – (pV)X] = 3 2 [(2.0 × 105 )(3.2) – (7.2 × 105 )(1.5)] J = 660 kJ 3. A : Isothermal, T = constant, ∆U = 0 Q = ∆U + W 4. D : For 1 mole, U = f 2 RT Gradient = dU dT = f 2 R f = 2 8.31 ( 6250 – 2500 500 – 200 ) = 3 γ = f + 2 f = 3 + 2 3 = 1.67 5. C : Q = ∆U = 5 2 nR(∆T) = 5 2 (5.0)(8.31)(25)J = 2600 J 6. C : Isothermal expansion: p1(2V) = pV, p1 = p 2 Adiabatic expansion: p2(2V)1.67 = pV1.67, p2 = p 21.67 p1 p2 = 20.67 = 1.60 7. C : Cp > CV because gas expands when heated at constant pressure. Work is done by the gas. 8. D : Q = ∆U + W, ∆U = 0 9. B 10. A 11. D : Cp,m CV,m = γ = f + 2 f Cp,m fR/2 = 1 + 2 f 2 f (Cp,m R – 1) = 1 f = 2( Cp,m R – 1) 12. B : SQ: Temperature at R is lower than the temperature at Q. Hence internal energy decreases. 13. D : When p = constant, V ∝T. When p decreases, T decreases 14. C : Gradient of adiabatic curve greater than isothermal curve. 15. C : Use first law of thermodynamics 16. C : At X: pV = (1.00 × 105 Pa)(10 cm3 ) ∝ 300 K At Z, pV = (0.75 × 105 Pa)(20 cm3 ) ∝ TZ TZ = 1.5(300 K) = 450 K 17. (a) Q = ∆U + W, Q = 0 ∆U = –W (b) (i) T2V2 γ – 1 = T1V1 γ – 1, γ = 1.40 T2 = 202 K (ii) U = 5 2 nRT and pV = nRT W = –∆U = 1.70 × 103 J 18. (a) CV,m = 5 2 R = 20.8 J K–1 mol–1 (b) Q = (2)(20.8)(350 – 300) J = 2080 J (c) Cp,m = CV,m + R = 29.1 J K–1 mol–1 Q = (2)(29.1)(350 – 300) J = 2910 J 19. (a) Refer to page 238. (b) Refer to page 250. (c) (i) U increases, Q = 0, work is done on the gas. (ii) p1V1 γ = p0V0 γ and (V1 = V0 3 , V0 V1 = 3) p1 = ( V0 V1 ) γ p0 = 3.01.40(1.013 × 105 ) Pa = 4.72 × 105 Pa (iii) U = 5 2 nRT = 5 2 pV Q = 0, W= –∆U = – 5 2 (p1V1 – p0V0) = – 5 2 [(4.72 × 105 )( 3.50 × 10–3 2 ) – (1.013 × 105 )( 3.50 × 10–3)] J = –490 J

277 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 20. (a) 1.0 J ∆U = 3.0 J (b) –4.5 J Q = ∆U + W (c) 4.0 J UD = UA + ∆U (d) (i) 2.0 J ∆U = UC – UA, Q = ∆U + W (ii) 2.0 J W = 0 21. (a) Refer to page 244. (b) (i) Isothermal expansion. (ii) Adiabatic compression. (c) 6.90 J Q = nCp,m ∆T, Cp,m = 5 2 R 22. (a) (i) Refer to page 259 – 260 (ii) pVγ = constant, γ = Cp,m CV,m (b) (i) 1.51 × 10–2 m3 (ii) γ = 1.41 (iii) 0 p V 23. (a) (i) 195 J ∆U = 65 J when ∆T = (450 – 400) K X → Z, ∆T = (450 – 300) K Work is done by the gas from X to Y. Heat supply Q = ∆U + W (b) γ = 1.40 182 J = n Cp,m × 100 65 J = n CV,m × 50 Hence, diatomic (c) n = 0.0626 mol 65 J = nCV,m × 50, CV,m = 5 2 R 24. (a) (i) 21.1 J mol–1 K–1 CV,m = dU dT = gradient of graph (ii) γ = 1.40 (iii) Diatomic gas, γ = (f + 2) f , f = 5 (b) (i) 0 p V (ii) γ = 1.41 (iii) 3.1 × 10–3 m3 V ∝ T (iv) 346 K T1V1 γ–1 = T2V2 γ–1 25. (a) Refer to page 259 (b) Refer to page 260 (c) (i) γ = 1.41 p1V1 γ = p2V2 γ (ii) 1008 K T1V1 γ–1 = T2V2 γ–1 (iii) –310 J W = –∆U = –n CV,m ∆T CV, m = 5 2 R, n = pV RT (iv) 1 260 K p ∝ T, when V = constant Heat the gas at constant volume. 26. (a) (i) Q = DU + W (ii) Isothermal : ∆U = 0 Work done by gas, W = Q, heat supplied. Adiabatic, Q = 0 W = –∆U, decrease in internal energy. (b) (i) 0.218 m p1V1 γ = p2V2 γ , V2 = 0.0816 (ii) 505 K T1V1 γ–1 = T2V2 γ–1 (iii) 2.13 × 105 J W = –∆U = – n CV,m ∆T 27. (a) pV = nRT p1 = nRT V = (2.0)(8.31)(360) 1.20 × 10–3 Pa = 4.99 × 106 Pa (b) Along XY, p = constant, V ∝ T TY = (0.30 1.20 )(360 K) = 90 K (c) Along YZ, V = constant, p ∝ T p2 = ( 360 90 )(4.99 × 106 Pa) = 2.00 × 107 Pa (d) Nett work done = WZX –WXY = nRT ln ( V2 V1 ) – p1(1.30 – 0.30) × 10-3 J = (2.0)(8.31)(360)ln( 1.20 × 10–3 0.30 × 10–3) – (4.99 × 106 ) (1.20 – 0.30) × 10-3 J = 3.80 kJ 28. (a) (i) Diatomic gas, CV,m = 5 2 R, Cp,m = 5 2 R + R = 7 2 R Heat supplied, Q = n Cp,m∆T = (2.0)( 7 2 R) (15) J = 873 J (ii) pV = nRT ∆V = nR p (T2 – T1) = 2.0(8.31)(15) 1.01 × 105 m3 = 2.45 × 10–3 m3 (iii) W = p(∆V)= (1.01 × 105 Pa)(2.45 × 10–3 m3 ) = 247 J (iv) ∆U = Q – W= (873 – 247) J = 626 J (b) (i) Adiabatic compression: Gas is compressed without heat entering or leaving the gas. (ii) TVγ– 1 = constant, γ = 7 5 T2 = (273 + 30)( V V/3)7/5 – 1 K = 470 K (iii) W = – ∆U = 5 2 nR(∆T) = – 5 2 (2.0)(8.31)(470 – 303) J = –6940 J

278 CHAPTER Concept Map HEAT TRANSFER 11 Bilingual Keywords 278 Black body: Jasad hitam Conduction: Konduksi Convection: Perolakan Diffusion: Resapan Greenhouse effect: Kesan rumah hijau Insulated: Ditebat Radiation: Pancaran Rate of heat flow: Kadar aliran haba Temperature gradient: Kecerunan suhu Thermal conductivity: Kekonduksian terma Thermal resistance: Rintangan terma 278 Conduction • Mechanism of thermal conduction • Thermal conductivity dQ dt = –kA dθ dx Convection • Natural and forced convection Global Warming • Greenhouse effect • Thermal pollution Radiation • dQ dt = eσAT4 • Black body radiation Heat Transfer

279 11 Physics Term 1 STPM Chapter 11 Heat Transfer INTRODUCTION 1. There are three main methods of heat transfer: conduction, convection and radiation. 2. Conduction is the transfer of heat through a material medium without the medium moving. 3. Convection is the transfer of heat when the material medium moves and carries the heat along. 4. Thermal radiation is the transfer of heat from a hot body to the surrounding with negligible heating of the space round the body. 5. Thermal radiation does not require a material medium. Thermal energy is transferred by infrared radiations. 6. Heat from the Sun reaches the Earth by thermal radiation. 11.1 Conduction 2008/P1/Q22, 2008/P1/Q23, 2009/P1/Q22, 2010/P1/Q23, 2013/P1/Q15,Q17, 2014/P1/Q15, 2015/P1/Q14,Q15, 2016/P1/Q20 Students should be able to: • explain the mechanism of heat conduction through solids, and hence, distinguish between conduction through metals and nonmetals • define thermal conductivity • use the equation dQ dt = – kA dθ dx for heat • describe and calculate heat conduction through a cross-sectional area of layers of different materials • compare heat conduction through insulated and non-insulated rods Learning Outcomes 1. In a solid, atoms are closely packed with strong bonds. 2. When one end of a solid is heated, atoms will vibrate more vigorously. 3. Vibrations of atoms set up waves which are propagated across the solid. 4. The waves are dispersed by dislocations or impurities atoms. Hence, thermal energy is transferred. 5. Conduction of thermal energy by the vibrations of atoms is the sole method of thermal conduction in non-metals. 6. In metals, besidesthe method discussed above, thermal conduction also occursthrough the dispersion of free electrons. The latter method is a much effective method. 7. Metals contain many free electrons. When one end of a metal rod is heated, the free electrons at the hot end gain energy and diffuse from the hot end to the other parts of the rod. 8. Heat is transferred to ions in the lattice when these free electrons collide with the ions. 9. Since electrical conduction is also through the drift of free electrons, a good conductor of electricity is also a good thermal conductor. Thermal Conductivity 1. Figure 11.1 shows heat flowing through two parallel cross-sections X and Y of area A in a solid. The distance between X and Y is Δx. VIDEO Conduction

280 11 Physics Term 1 STPM Chapter 11 Heat Transfer 2. The temperatures of X and Y are θ and (θ – Δθ) respectively. The temperature gradient along the solid = Temperature difference Separation between 2 surfaces = (θ – Δθ) – θ Δx = – Δθ Δx = – dθ dx 3. The negative sign shows that the temperature decreases as the distance x increases. 4. The rate of heat flow through a cross-sectional area A normal to the flow of heat is directly proportional to (a) the cross-sectional area A. (b) the temperature gradient, – dθ dx Hence, rate of heat flow, dQ dt ∝ – A dθ dx dQ dt = –kA dθ dx where k is a constant, known as the thermal conductivity of the material. 5. Hence, thermal conductivity k of a material may be defined using the equation. Thermal conductivity, k = – dQ dt A dθ dx = Rate of heat flow (Cross-sectional area) (Temperature gradient) The thermal conductivity of a material is the rate of heat flow through a unit cross-sectional area of the material in a region of unit temperature gradient. 6. The unit for k is given by W m2 . K m = W m–1 K–1 θ°C (θ – 1)°C Direction of heat flow Area = 1 m2 1 m 1 m 1 m Figure 11.2 7. The meaning of thermal conductivity is illustrated in Figure 11.2, as the rate of heat flow normal across opposite faces of a cube of length 1 m when there is a temperature difference of 1 K (or 1°C) between the two faces of the cube. 8. Table 11.1 gives the values of the thermal conductivity k of a few common materials at 20°C. The value of k is affected by the temperature. For most metals, the value of k decreases slightly with increase of temperature. Figure 11.1 X A Y (θ – ∆θ) ∆x θ Heat Heat Info Physics Aluminium engine block Engine blocks made of aluminium are used in some cars. The aluminum block is both lighter and almost as strong as cast iron engine blocks. Besides aluminium block is 40% lighter, has thermal conductivity 150 W m–1 K–1 compared to cast iron thermal conductivity of 50 W m–1 K–1. Being better thermal conductor, aluminium engine blocks give high cooling performance.

281 11 Physics Term 1 STPM Chapter 11 Heat Transfer Table 11.1 Thermal conductivity State Material k/W m–1 K–1 Solids Silver Copper Carbon Glass Brick Paper Wood 418 380 2 1.0 \0.6 0.06 0.08 Liquids Mercury Water Alcohol 8.0 0.6 0.2 Gases Air Hydrogen Steam 0.03 0.17 0.02 9. Good conductors, such as silver and copper have high values of thermal conductivity. 10. Gases such as air and hydrogen have low thermal conductivity because • gas molecules are far apart. Energy is transferred between gas molecules when they collide with each other. • the number density (number of molecules per unit volume) for gas is low. Thermal Conduction in an Insulated Rod Insulator (a) Insulated rod (b) Temperatue – distance graph Magnitude of temperature gradient, (b) Temperatue gradient – distance graph Rod Heat flow θ2 θ1 θ1 Temperatue, θ x x 0 x 0 θ2 < θ1 dθ dx Figure 11.3 1. Figure 11.3 shows a rod of length l, and cross-sectional area A which is perfectly insulated. Heat flows fromone end to the other end.Whenthe steady state is attained,the temperatures atthe ends of θ1 and θ2 (θ1 > θ2 ) are constants.

282 11 Physics Term 1 STPM Chapter 11 Heat Transfer 2. Since the rod is perfectly insulated, there is no heat loss from the sides of the rod. There is parallel heat flow along the rod as shown in Figure 11.3(a). 3. The rate of heat flow along the rod is the same and constant. From the equation, dQ dt = –kA dθ dx . Since dQ dt , k and A are constant, then temperature gradient dθ dx = constant (Figure 11.3(c)). The graph of temperature θ against distance x from the hot end is a straight line with gradient = dθ dx = constant = – (θ2 – θ1 ) l = (θ1 – θ2 ) l Hence, the rate of heat flow, dQ dt = kA (θ1 – θ2 ) l Example 1 The thermal conductivities of copper and glass at 20°C are 380 W m–1 K–1 and 1.0 W m–1 K–1 respectively. Explain the large differences of the values in terms of the mechanisms of thermal conduction. Solution: Copper is a metal with many free electrons. The main mechanism of thermal conduction isthrough the diffusion of free electrons from the hot end of the copper to the other regions. Fast moving free electrons collide with copper ions in the atomic lattice. Heat is transferred at a fast rate. Glass is a non-metal. There are no free electrons in glass. Mechanism of thermal conduction is through the vibrations of molecules which generate waves. The waves are then dispersed through the glass. Heat is transferred at a slower rate. Example 2 A uniform aluminium rod which is perfectly insulated has a cross-sectional area of 2.50 cm2 and length 20.0 cm. Heat is conducted by the aluminium rod. When the steady state is attained, the temperatures at the ends of the rod are 120°C and 30°C. Calculate (a) the temperature gradient along the rod. (b) the rate of heat flow in the rod. (c) the temperature 15.0 cm from the hot end. (Thermal conductivity of aluminium = 210 W m–1 K–1) Solution: (a) Temperature gradient, dθ dx = (30 – 120)°C 0.20 m = –450 °C m–1 (b) Rate of heat flow, dQ dt = –kA dθ dx = – (210)(2.5 × 10–4)(–450) = 23.6 W (c) Temperature at a distance of 15.0 cm from the hot end = 120°C + (0.15) × (–450)°C = 52.5°C 15.0 cm 50.0 cm 120°C 30°C

283 11 Physics Term 1 STPM Chapter 11 Heat Transfer Thermal Conduction in a Non-insulated Rod 1. Figure 11.4 shows a uniform rod BD of cross-sectional area A and length l in a steady state when heat flows from one end to the other. The temperatures at the ends of the rod are θ1 and θ2 respectively. B D (a) Non-insulated rod Magnitude of temperature gradient, (b) Temperatue – distance graph (b) Magniture of temperature gradient – distance graph Room temperature Rate of heat flow θ2 θ1 θ1 Temperatue, θ x x 0 θ2 (θ2 < θ1) dθ dx Figure 11.4 2. Since the rod is not insulated, heat is lost from the sides of the rod. Figure 11.4(a) shows the rate of heat flow at the end D is smaller than the rate of heat flow at the end B. 3. The rate of heat flow dQ dt decreases as the distance x from the hot end increases. Using dQ dt = –kA dθ dx as dQ dt decreases, the temperature gradient dθ dx also decreases as the distance x increases as shown in Figure 11.4(c). 4. The temperature-distance graph is a curve instead of a straight line as shown in figure 11.4(b). Example 3 A uniform copper rod has cross-sectional area of 80 mm2 . One end of the rod is kept at a high constant temperature. When the steady state is achieved, the temperature gradient at two different cross-sections are 4.6 × 102 K m–1 and 2.4 × 102 K m–1. Calculate the rate of heat loss from the sides of the rod between the two cross-sections. (Thermal conductivity of copper = 380 W m–1 K–1)

284 11 Physics Term 1 STPM Chapter 11 Heat Transfer Solution: The rate of heat flow through the first cross-section ( dQ dt )1 = –kA( dθ dx )1 The rate of heat flow through the second cross-section ( dQ dt )2 = –kA( dθ dx )2 Rate of heat loss from the sides of the rod,( dQ dt )1 – ( dQ dt )2 = [–kA( dθ dx )1 ] – [–kA( dθ dx )2 ] = 380 × (80 × 10–6)(4.6 – 2.4) 102 W = 6.69 W Quick Check 1 1. The equation for rate of heat flow, dQ dt = –kA dθ dx can only be applied when A the conductor is perfectly insulated. B the conductoris a good thermal conductor. C the mechanism of thermal conduction is diffusion of electrons. D the steady state has been achieved. 2. The end of a long metal bar of uniform crosssectional area is kept at a temperature T1 > T0 , the room temperature. If the rod is unlagged, which one of the following graph shows the variation of temperature T with the distance x from the hot end when the steady state has been achieved? A C B D 3. A copper bar is unlagged for a length l of its total length L. The ends are maintained at temperatures T1 and T2 (T1 > T2 ). Which one of the following graph shows the variation of the steady state temperature T with distance x from the hot end? A C B D

285 11 Physics Term 1 STPM Chapter 11 Heat Transfer 4. Heat is conducted along a perfectly lagged metal bar of uniform cross-sectional area A at a steady rate R. The temperature difference Δθ between two points with distance d apart on the axis of the bar is A Δθ = RdA k C Δθ = Rk dA B Δθ = Rd kA D Δθ = RA dk 5. The ends of a well-lagged metal rod of length L is kept at a steady temperatures of T1 and T2 (T1 > T2 ). Which graph best represents the relation between rate of heat flow dQ dt and x, the distance from the hot end? A x L 0 dQ dt C x L 0 dQ dt B x L 0 dQ dt D x L 0 dQ dt 6. The thermal conductivity of a clay brick is 0.60 W m–1 K–1. If the rate of heat flow per unit area is 200 W m–2, what is the temperature difference between two points in the brick distanced 1.5 cm apart along the direction of heat flow? A 1.25°C C 5.00°C B 1.60°C D 8.00°C 7. A lagged metal rod has length l and crosssectional area A. The ends are kept at temperatures T1 and T2 (T2 > T1 ) respectively. At steady state, the rate of heat flow is p. The thermal conductivity of the rod can be expressed as A pA l(T2 – T1 ) C Al p(T2 – T1 ) B pl A(T2 – T1 ) D pA(T2 – T1 ) l 8. Two conducting blocks are kept at temperatures which differ by 20 K. The two blocks are connected by two cylinders of length 0.40 m with thermal conductivity 25 W m–1 K–1. The cross-sectional area of the cylinders are 1.0 × 10–4 m2 and 4.0 × 10–4 m2 respectively. The space between the two blocks is filled with an insulating material as shown below. Lagging Cylinders Cold block Hot block What is the rate of heat flow between the two blocks? A 0.13 W C 0.31 W B 0.25 W D 0.63 W 9. A lagged metal rod of uniform cross-section has its ends kept at different temperatures. At steady state, the temperature difference between two points P and Q along the rod is greater if A the thermal conductivity of the metal is higher. B the cross-sectional area is bigger. C the lagging is removed. D the points P and Q are closer. 10. The thermal conductivity of a metal decreases as the temperature increases because at higher temperatures, A there are fewer conduction electrons. B lattice vibration is slower. C rate of collision between conduction electrons and metal atoms increase. D the distance between metal atoms increases.

286 11 Physics Term 1 STPM Chapter 11 Heat Transfer Determination of Thermal Conductivity of Good Conductors – Searle’s Method M N T1 T2 T4 T3 θ2 θ4 θ3 θ1 L Steam chest Thick metal rod Lagging Mass m Beaker Water Water Constant pressure apparatus Steam Steam Figure 11.5 1. Figure 11.5 shows the apparatus used in Searle’s method which is used to determine the thermal conductivity of a good conductor, such as a metal. 2. The special feature of the apparatus is that the sample is in the form of a thick, long insulated rod. 3. Although the rod is insulated, a little heat is lost from the sides of the rod. The rod possesses a large cross-sectional area, so that the rate of heat lost from the sides is negligible compared to the rate of heat flow along the rod. 4. A long rod is used, so that the larger temperature drop across the long length of the rod can be measured accurately. A bigger temperature difference reduces the percentage error in the measurement of the temperature difference. 5. Since the rod is insulated, the rate of heat flow and the temperature gradient is constant along the rod. Hence, the temperature gradient can be measured along one section of the rod, and the rate of heat flow along another section. 6. The end M of the rod is heated in a steam chest. Water from a constant pressure apparatus flows in a coil around the end N of the rod. 7. When steady state is attained, the readings θ1 , θ2 , θ3 and θ4 of the thermometers T1 , T2 , T3 and T4 respectively are noted. 8. The temperature gradient, dθ dx = (θ2 – θ1 ) l where l = distance between the thermometers T1 and T2 . 9. If m = mass of water collected in time interval t, then the rate of heat flow dQ dt = mc (θ4 – θ3 ) t where c = specific heat capacity of water. 10. Using dQ dt = –kA dθ dx where A = πd 2 4 , d = diameter of rod measured using vernier callipers, mc (θ4 – θ3 ) t = – k πd 2 4 (θ2 – θ1 ) l Thermal conductivity, k = 4 mc (θ4 – θ3 )l πd 2 t (θ2 – θ1 )

287 11 Physics Term 1 STPM Chapter 11 Heat Transfer Example 4 The graph below shows the variation of temperature with distance x along a metal rod obtained from three different experiments A, B and C. For each experiment, the end M is maintained at 100°C and the end N at 60°C. The room temperature is 30°C. Discuss the conditions of the rod during the three experiments. M C N B A Solution: In experiment A : The rod is perfectly lagged. There is no heat loss from the sides of the rod. In experiment B : The rod is not well-lagged. Heat escapes from the sides of the rod. In experiment C : The rod is not lagged. Since the temperature of both ends are higher than room temperature (30°C), heat flows from both ends. Example 5 In a heating system of a house, a room is heated using a radiator where water flows at a rate of 0.12 kg s–1. At steady state, the difference in temperature between the water flowing into the radiator and out of it is 6.0 K. The radiator is made of copper with thermal conductivity 380 W m–1 K–1 having a total effective surface area of 1.5 m2 and thickness 2.0 mm. (a) What is the rate of heat supplied to the room by the radiator? (b) What is the mean temperature difference between the inner and outer walls of the radiator? (Specific heat capacity of water = 4 200 J kg–1 K–1) Solution: (a) From the equation Q = mcθ Rate of heat loss by hot water in the radiator, dQ dt = dm dt c θ Rate of heat flow from the radiator to the room = Rate of heat loss by hot water = (0.12) × (4 200)(6.0) = 3.02 × 103 W (b) Using dQ dt = –kA dθ dx Temperature difference across walls of radiator, Δθ = dQ dt ∆x kA = 3.02 × 103 × (2.0 × 10–3) 380 × 1.5 = 0.0106°C

288 11 Physics Term 1 STPM Chapter 11 Heat Transfer 1. In an experiment to determine the thermal conductivity of a good conductor, the sample used is long so that A theheatflow is parallel along the conductor B a sufficiently large temperature difference is obtained between the ends of the conductor C the heat loss from the sides of the rod can be corrected accurately D the temperature gradient along the conductor is constant 2. The figure shows a non-uniform metal rod PQ which is well-lagged. The end P is at a higher temperature than the end Q. At steady state, the rate of heat flow through the crosssection at X and Y is Rx and Ry respectively. The magnitudes of the temperature gradient at X and Y is mx and my respectively. P X Lagging Y Q Which of the following pair in relation is correct? A Rx = Ry and mx < my B Rx = Ry and mx > my C Rx > Ry and mx > my D Rx < Ry and mx < my 3. The temperature gradient at a distance x from the cold end of an unlagged metal rod is directly proportional to x2 when steady state is achieved. Which of the following shows the correct rate of heat flow dQ dt with the distance x? A C B D 4. The figure shows a uniform metal rod PQRS of length l and thermal conductivity k. The ends of the rods P and S are kept at temperature θ1 and θ2 respectively (θ2 < θ1 ). The rod is well-lagged except for the section QR. P Q Metal rod Lagging R S The cross-sectional area of the rod is A and the temperature gradient across PQ and RS is m1 and m2 respectively. The rate of heat loss from the unlagged section QR is given by A kA (m1 – m2 ) C kA l (m1 – m2 ) B kA (m1 + m2 ) D kA l (m1 + m2 ) 5. The ends of an unlagged metal rod is kept at temperatures T1 and T2 . When steady state is achieved, the variation of temperature along the rod is as shown by the graph below. Which of the following about the rod and the room temperature T3 is correct? A Rod lagged, T3 > T2 B Rod unlagged, T3 < T2 C Rod lagged, T3 < T2 D Rod unlagged, T3 > T2 6. In experiments to determine the thermal conductivity of a good conductor, explain why the sample used is usually long and is lagged, and the temperature difference between two points separated at least 100 mm along the sample is measured. Quick Check 2

289 11 Physics Term 1 STPM Chapter 11 Heat Transfer Determination of Thermal Conductivity of a Poor Conductor – Lee’s Method 1. The apparatus used to determine the thermal conductivity of a poor conductor is as shown in Figure 11.6. 2. A thick brass disc B with a hole holding a thermometer T2 is hung using thin suspension wires. 3. The sample, in the form of a thin circular disc with the same diameter as the disc B, is placed on B. 4. No lagging is required because the rate of heat loss from the sides of the thin sample is negligible. The surface area of the sides of the sample is small compared to the large cross-sectional area of the sample. 5. Since the sample is a poor conductor, for steady state to be achieved in a short time, the sample must be thin. Heat will require a very long time to flow through a thick sample. 6. On top of the sample is placed a steam chest which has a thick brass base with a hole holding a thermometer T1 . 7. Steam is allowed into the steam chest. When the readings of the thermometers T1 and T2 do not increase further but stay constant, steady state has been achieved. 8. Heat flows from the steam in the steam chest through the brass base C, the sample, the thick brass disc B and out from the base of disc B. 9. Brass is a good conductor of heat. Hence, the temperature θ1 at the upper surface of the circular sample is recorded by the thermometer T1 . Thermometer T2 records the temperature θ2 of the lower surface of the sample. 10. The thickness, x of the sample is measured using a micrometer screw gauge. Hence, temperature gradient across the sample dθ dx = (θ2 – θ1 ) x 0 θ2 t/s θ/°C B Sample Brass disc T2 θ (a) (b) Figure 11.7 11. The second part of the experiment is to measure the rate of heat flow dQ dt when the steady state is achieved. (a) The steam chest and sample is removed. (b) The brass disc B is slowly heated using a Bunsen burner until its temperature is a few degrees above θ2 recorded in the first part of the experiment. (c) The sample is placed on top of disc B and the reading θ of the thermometer T2 is recorded every 20 seconds until the reading is a few degrees below θ2 (Figure 11.7(a)). C x Steam chest Steam Steam Suspension wire Sample Brass disc Thermometer T2 Thermometer T1 θ1 θ2 Steam B Figure 11.6

290 11 Physics Term 1 STPM Chapter 11 Heat Transfer (d) A graph of temperature θ against time t is plotted (Figure 11.7(b)). (e) The gradient of the graph when θ = θ2 ,( dθ dt )θ2 is calculated. 12. Using the equation, Q = mcθ dQ dt = mc dθ dt When the lower surface temperature is θ2 , the rate of heat flow through the sample is dQ dt = mc ( dθ dt )θ2 where m = mass of disc B c = specific heat capacity of brass 13. Using the equation, dQ dt = –kA dθ dx A = πD2 4 , D = diameter of sample mc ( dθ dt )θ2 = –k ( πD2 4 )(θ2 – θ1 x ) Thermal conductivity, k = mc ( dθ dt )θ2 ( πD2 4 )(θ2 – θ1 x ) Example 6 The table below showsthe approximate values of the thermal conductivity λ of a number of materials at room temperature. The values may be taken as typical of metals, insulators, liquids and gases. Material λ/W m–1 K–1 Aluminium (metal) 200 Rubber (insulator) 0.2 Methanol (liquid) 0.2 Air (gas) 0.02 (a) Explain why metals are good conductors of both heat and electricity, whereas a thermal insulator is also electrically insulating. (b) Suggest a reason why thermal conductivity of typical liquids and typical insulating solids are similar. (c) Although gases have very low thermal conductivities, they do not make good practical thermal insulators unless contained in cellular structures, such as expanded polystyrene. Comment on this. (d) Give the reason why thermal conductivity of gases are generally rather less than that of liquids. Solution: (a) Metals have many free electrons. Thermal and electrical conduction are due to the translational motion of free electrons. Hence, metals are good conductors of heat and electricity. Thermal insulators have no free electrons. Hence, thermal insulators are also electrical insulators. (b) Thermal conductivity of typical liquids and insulating solids are similar because • the mechanisms are the same, the vibrations of atoms or molecules. • separations between atoms or molecules are almost the same.

291 11 Physics Term 1 STPM Chapter 11 Heat Transfer (c) The main method of heat transfer in gases is convection, not conduction. In cellular structures, convection is limited to each cell. Heat is not transferred to other regions by convection. (d) The number of molecules per unit volume in gas is less than that in liquids. Thermal energy is transferred when molecules collide. Frequency of gas molecules colliding is less than that in liquids. Example 7 A vacuum flask contains 0.40 kg water at initial temperature of 75°C is covered with a cork of cross-sectional area 4.5 × 10–3 m2 and thickness of 25 mm. After 1.0 hour, the temperature of the water is 72°C. The mean temperatures of the outer and inner surface of the cork is 50°C and 22°C. Calculate the fraction of the total heat lost through the cork. Suggest other methods of heat loss. (Specific heat capacity of water = 4 200 J kg–1 K–1. Thermal conductivity of cork = 5.3 × 10–2 Wm–1 K–1) Solution: Total heat loss by water Q = mc (θ2 – θ1 ) = 0.40 × 4 200 × (75 – 72) = 5.04 × 103 J Rate of heat flow through the cork dQ dt = –kA dθ dx = (5.3 × 10–2)(4.5 × 10–3)(50 – 22) (25 × 10–3) = 0.2671 W Heat loss through the cork in 1.0 hour = 0.267 × (60 × 60) = 0.962 × 103 J Fraction of heat loss through cork = 0.962 × 103 5.04 × 103 = 0.191 The balance of the heat loss is through convection and radiation from the flask. Example 8 (a) Explain why electric insulators are poor thermal conductors. (b) Figure (a) shows Lee’s apparatus for the determination of the thermal conductivity of glass in the form of a disc of thickness x. At steady state, the temperatures recorded by the thermometers is θ1 and θ2 . Steam Steam θ1 θ2 Glass disc Brass base x θ Brass base Glass disc (a) (b)

292 11 Physics Term 1 STPM Chapter 11 Heat Transfer (i) Why must the glass disc used be thin? (ii) Referring to figure (a), explain what is meant by steady state. (c) The following data is obtained in an experiment. Temperature θ1 = 99.4°C Temperature θ2 = 73.5°C Diameter of glass disc, d = 12.00 cm Thickness of glass disc, x = 2.00 mm Mass of brass base, M = 1.600 kg Specific heat capacity of brass = 375 J kg–1 °C–1 The variation of the temperature of the brass base with time in the arrangement shown in figure (b) is as shown in the table below: Time t/s θ/°C 0 82.0 10 78.4 20 75.2 30 72.2 40 69.8 50 67.6 60 65.6 70 63.8 80 62.2 (i) Derive an expression for the thermal conductivity k of the glass disc in terms of the quantities mentioned above and identify any other symbols in your expression. (ii) Plot a graph of temperature θ against time t. Use your graph to calculate a value for the thermal conductivity of glass. (iii) Sketch a graph to show the variation of the temperature between the upper surface and lower surface of the glass disc. Explain the shape of your graph. (iv) On the same axes, sketch the same graph for a copper disc of the same dimensions. Mark clearly the temperatures θ1 , and θ2 and x, the thickness of the disc. Explain any difference in the two graphs. Solution: (a) Electric insulators have no free electrons. Thermal conduction in electric insulators is by the vibrations of atoms in the lattice which is a slower process. (b) (i) The glass disc must be thin so that heat can flow through it in a short time and a long period is not required for steady state to be achieved. (ii) Steady state is achieved when the readings of the thermometers θ1 and θ2 are steady. (c) (i) From the equation, Q = Mcθ Rate of heat flow through the glass disc at steady state dQ dt = Mc ( dθ dt )θ2 Using dQ dt = – kA dθ dx A = πd 2 4 , dθ dx = (θ2 – θ1 ) x Mc( dθ dt )θ2 = –k ( πd 2 4 )(θ2 – θ1 x )

293 11 Physics Term 1 STPM Chapter 11 Heat Transfer Thermal conductivity, k = Mc ( dθ dt )θ2 ( πd2 4 )(θ1 – θ2 x ) (ii) 0 80 Temperature, θ/°C Time, t / s 75 73.5 70 65 60 20 40 60 80 From the graph, ( dθ dt )θ2 = magnitude of gradient of graph at θ = 73.5 °C = 15.0 50 = 0.30 °C s–1 Using the equation, k = Mc ( dθ dt )θ2 ( πd2 4 )(θ1 – θ2 x ) = 1.600 × 375 × 0.3 π 0.122 4 (99.4 – 73.5 2 × 10–3 ) = 1.23 W m–1 K–1 (iii) 0 Temperature /°C Distance / mm Copper (iv) Glass 73.5 99.4 2.00 Since the glass disc is rather thin (2.0 mm) and its diameter is large (12.0 cm), the rate of heat loss from the sides of the disc is negligible compared to the rate of heat flow through the glass disc. Hence, the temperature gradient is constant, and the graph is a straight line. (iv) Copper is a good thermal conductor. The temperature gradient is smaller.