PHY 103 Complete Slides (Real) PDF

PHY 103: Basic Principle of Physics II Heat and Thermodynamics LECTURE 1 By: Dr. E. Oyeniyi Mr. A. O. Ayoola Course Content Temperature scales. 1st and 2nd laws of thermodynamics. Application of the laws thermodynamics (e.g. to...

PHY 103: Basic Principle of Physics II Heat and Thermodynamics LECTURE 1 By: Dr. E. Oyeniyi Mr. A. O. Ayoola Course Content Temperature scales. 1st and 2nd laws of thermodynamics. Application of the laws thermodynamics (e.g. to calorimetry, gas properties and expansion of liquids). Application of 1st and 2nd laws to heat engines,heat pumps and refrigeration. Third law and absolute zero of temperature. Course Content contd. Thermal conductivity. Types of radiation and energy Understanding of the following is key to this aspect: Temperature Heat Thermal Contact Thermal Equilibrium Zeroth Law Temperature scales Thermometers Temperature and “Zeroth Law” We often associate the concept of temperature with how hot or cold an object feels when we touch it. In this way, our senses is a qualitative indicator of temperature (but it is unreliable and could be misleading). For example, if two objects of different thermal conductivities (e.g a metal tray and a plastic bottle) are removed from the freezer, the two objects are at the same temperature but will feel different when touched. The object with the higher conductivity (the metal tray) will feel colder than the one with lower conductivities if we touch them. This will give a false idea that it is at a lower temperature. Temperature and “Zeroth Law” (cont’d) HEAT is the transfer of energy from one object to another object as a result of a difference in temperature between them. NOTE: (i)The two bodies must be in thermal contact. (ii)Heat always flow from a higher temperature body to a lower temperature body Two objects are said to be in thermal contact with each other if energy can be exchanged between them.  Temperature and “Zeroth Law” (contd) Zeroth Law of thermodynamics This law states that if two objects A and B, which are not in thermal contact, are separately in thermal equilibrium with a third object C, then objects A and B are in thermal equilibrium with each other. This law allows the development of thermometers. Object C is our thermometer in this case. For example, the length of a mercury column (object C in the zeroth law as defined above) may be used as a measure to compare the temperature of the two other objects. Temperature and “Zeroth Law” contd. Thermal equlibrium This occurs when two objects in thermal contact with each other cease to exchange energy by the process of heat or by electromagnetic radiation. It also means that two objects in thermal equilibrium are at the same temperature. Thermometer and temperature scales Temperature of an object depends on measuring devices for its measurements and on the temperature scale adopted. The Most common temperature scales are:  Thermodynamic (Kelvin) Celsius Fahrenheit Though few scales have been used, there are many types of devices for measuring temperature. The devices are called Thermometers.  Thermometer and temperature scales Thermometers are devices used to measure the temperature of a system. All thermometers are based on the principle that some physical properties change as temperature changes. Some of the physical properties that change with temperature are: The volume/length of a liquid The length of a solid Thermometer and temperature scales contd. The pressure of a gas at constant volume (as in constant-volume gas thermometer) The volume of a gas at constant pressure (as in constant-pressure gas thermometer) The electric resistance of a conductor (as in resistance thermometer) The colour of an object The emf of a thermo couple TEMPARATURE SCALES Kelvin (Thermodynamic) scale Characteristics of the Kelvin Scale On this scale, Ice point has temp. of 273.15 K Triple point of water is defined as 273.16 K Triple point of water is the temp. at which saturated water vapour, pure water (distilled water from which dissolved air has been driven out) and melting ice are in equilibrium. Triple point of water can also be defined as the temperature at which liquid water, gaseous water and solid water coexist in equilibrium Kelvin Scale contd. The difference in the values of triple point and ice point temperatures is due to: (1) pressure difference: ice point pressure is 760 mmHg while triple point pressure is 4.6 mmHg (2) Removal of dissolved air from distilled water used for triple point. The steam point on this scale is 373.15 K The scale is divided into 100 equal degrees between the ice and steam points. 0 K (or its equivalence, -273.150C) is referred to as absolute zero. It is the temperature at which the pressure of a gas is zero. At a lower temperature, the pressure will become negative.  Celsius Scale Characteristics of Celsius Scale Ice point is defined as 0oC Steam point is 100oC at 760mmHg The scale is divided into 100 equal degrees between the two points. The number of division on this scale is the same with that of kelvin scale. Relationship between kelvin (TK) scale and Celsius (TC)scale TK = TC +273.15 or TC = TK -273.15 (1) NOTE: The difference of two temperatures on Celsius scale is the same as the difference of their corresponding temperature on kelvin scale and vice-versa. This can be shown as follows: If two temperature T1C and T2C on Celsius scale have their corresponding kelvin scale temperature as T1K and T2K respectively, then, using eqn (1) above T1C = T1K - 273.15 (i) T2C = T2K - 273.15 (ii) Subtract equation (i) from (ii) T2C - T1C = T2K - T1K (iii)  Fahrenheit Scale Characteristics of fahrenheit Scale Ice point is defined as 32oF Steam point is 212oF at 760mmHg The scale is divided into 180 equal degrees between the two points. Relationship between fahrenheit (TF) scale and Celsius (TC) scale 9 𝑇𝐹 = 𝑇 + 32 (2) 5 𝐶 Making 𝑇𝐶 the subject of the relation, we have, 5 𝑇𝐶 = (𝑇𝐹 −32) (3) 9 9 NOTE: The difference of two temperatures on the Fahrenheit scale is times the 5 difference of their corresponding temperature on Celsius scale. This can be shown as follows: 9 𝑇1𝐹 = 𝑇1𝐶 + 32 (i) 5 9 𝑇2𝐹 = 𝑇2𝐶 + 32 (ii) 5 Subtracting (i) from (ii), we have, 9 𝑇2𝐹 − 𝑇1𝐹 = (𝑇2𝐶 −𝑇1𝐶 ) (4) 5 Conversion formula between two Temperature scales Examples of temperature conversion formulae are equations (1) and (2) 9 TC = TK - 273.15 and 𝑇𝐹 = 𝑇 + 32. 5 𝐶 These can be derived using the following simple method: Step 1  Draw two equal vertical lines to represent the two scales whose conversion Formula you wish to establish  Step 2  On one of the lines, label the lower point (TAL) and upper point (TAU) for the ice and steam points of one scales respectively.  On the second line, label the lower point (TBL) and upper point (TBU) for the ice and steam points of the second scales respectively as shown in the figure below Since the scales are of the same length, the reference temperature TA on the scale A and its corresponding TB on the scale B should be at the same distance from TAL and TBL so also from TAU and TBU respectively. Therefore we can write 𝑇𝐴 −𝑇𝐴𝐿 𝑇𝐵 −𝑇𝐵𝐿 = (5) 𝑇𝐴𝑈 −𝑇𝐴𝐿 𝑇𝐵𝑈 −𝑇𝐵𝐿 Examples Let us consider kelvin and Celsius temperature scale For Celsius, the ice (lower) point and the steam (upper) point are respectively 00C and 1000C. On kelvin scale, the two points are 273.15k and 373.15K respectively. 𝑇𝐴 −𝑇𝐴𝐿 𝑇𝐵 −𝑇𝐵𝐿 Therefore we can say = correspond to 𝑇𝐴𝑈 −𝑇𝐴𝐿 𝑇𝐵𝑈 −𝑇𝐵𝐿 𝑇𝐶 −0 𝑇𝐾 −273.15 = 100−0 373.15−273.15 𝑇𝐶 𝑇𝐾 − 273.15 = 100 100 100𝑇𝐶 = 100(𝑇𝐾 − 273.15) Therefore, 𝑇𝐶 = 𝑇𝐾 − 273.15 Which is the formula to convert Celsius temperature to its corresponding kelvin scale temp. DIY 1: NOTE: In this derivation we have considered the reference temperatures 𝑇𝐶 and 𝑇𝐾 with the ice points of the two scales. (a) Use the reference temperatures and the steam points of the two scales to derive the conversion equation. (b) Derive the conversion formula to convert Celsius temperature to Fahrenheit scale temperature and vice-versa. (i) use the reference and the lower (ice) points (ii) also, use the reference temperatures and the upper (steam) points.  Worked Example on Temperature Scales Find the difference of two temperatures 400C and 900C in Fahrenheit. There are two approaches to this: Approach 1 Use conversion formula in equation (2) to determine the Fahrenheit equivalent of each of the two temperatures and then find their difference 𝑇1𝐹 =(9/5)*40 + 32 = 1040F 𝑇2𝐹 =(9/5)*90 + 32 =1940F Δ𝑇𝐹 = 194 – 104 = 900F Approach 2 Find the difference of the two temperatures in Celsius and then use equation (4) to find their difference in Fahrenheit Δ𝑇𝐶 = 90 – 40 = 500C Δ𝑇𝐹 = (9/5)*Δ𝑇𝐶 = (9/5)*50 = 900F As expected, the two approaches gave the same result. DIY 2: (1) Find the equivalent of the following on a Celsius scale (a) 253 K (b) 100°F. (2) A pan of water is heated from 50°F to 160°F. (a) What is the change in its temperature on the Kelvin scale (b) What is the change in its temperature on the Celsius scale? use the two approaches as in the previous example. (3) If the difference between two temperatures given in Celsius is 45°C, find the difference between the two temperatures in Fahrenheit (4) If the difference between two temperature in fahrenheit is 45°F, find the difference between the two temperatures in Celsius  Fixed points If different thermometers are used to measure the temperature of an object, perfect agreement (yielding exactly the same temperature value) are only at the fixed points (because they are the points which are used for their calibrations) The fixed points are: (a) Triple point of water (is the single combination of temperature and pressure at which liquid water, gaseous water, and ice (solid water) coexist in equilibrium). On kelvin scale the temperature is 273.16 K (0.010C) (b) Ice point (mixture of water and ice in thermal equilibrium at atmospheric pressure). On the Celsius temperature scale, the mixture is defined to have a temperature of 00C and 273.15K on Kelvin scale (b) Steam point (mixture of water and steam in thermal equilibrium at atmospheric pressure). On the Celsius temperature scale, the mixture is defined to have a temperature of 1000C and 373.15 K on Kelvin scale.  Thermometer Calibration Calibration is the process of standardizing a temperature monitoring instrument to ensure that it will measure within a specific temperature range in which it is designed to operate. It is a way of relating the thermometric quantity (the quantity which changes as temperature changes) with temperature. Thermometer can be calibrated by placing it in thermal contact with any of the natural systems (i.e. the systems that defined the fixed points). Problems with Thermometers calibration. Extreme accuracy is not achievable. E.g. reading given by alcohol thermometer calibrated at the ice and steam points of water might agree with those given by a mercury thermometer only at the calibration points (for instance if one gives a reading of 40°C, the other may give a slightly different value). Reason is because the two liquids have different thermal expansion properties. The discrepancies between thermometers are especially large when the temperatures to be measured are far from the calibration points.  Problems with Thermometers calibration Contd Limited range of temperatures over which they can be used. The limitation is usually associated with the physical properties of the substance in the thermometer. For example, a mercury thermometer cannot be used below -39°C which is the freezing point of mercury. Types of Thermometers There are different types of thermometer, Some of them are:  Liquid-in-glass thermometers Constant-Volume gas thermometer Electrical thermometers  Liquid-in-glass thermometers Examples are Mercury as the liquid (can measure between -35°C and 350°C). Alcohol as the liquid (can measure between -80°C and 70°C. Thermometric property of this type of thermometer is the length of the liquid (mercury and alcohol) that expands in the glass. Thermometric property is the physical property that changes with change in temperature  The Constant-Volume gas thermometer and absolute zero temperature At pressures of the order of 760 mmHg (i.e. 1atm), different gases give slightly different temperature because none of them obeys gas law perfectly. As the pressure is reduced, the gases approach close to ideal gas and their temperature scales agree well together. Therefore, the thermometer’s readings are independent of the Substance (i.e. the type of gas) used in it and therefore it is free of the problems mentioned in a previous slide. For all the gases the pressure is zero at -273.15°C. This is the basis for absolute temperature scale. -273.15°C is the zero point of this scale and it is often referred to as absolute zero. The Constant-Volume gas thermometer and absolute zero temperature It is calibrated using ice and steam points of water. They cannot be used to monitor rapidly changing temperatures They are usually bulky and cumbersome and required lots of expertise to operate It is used to calibrate electrical thermometers It can measure a wide range of temperature. With the use of different gases (Hydrogen, Helium and Nitrogen) it can measure between -270 °C to +1500 °C Electrical thermometers Examples are 1. Resistance thermometers 2. Thermocouples 3. Thermoelectric (consisting of two thermocouples that are series connected with a potentiometer and a constant-temperature bath). Properties of Electrical thermometers They are more accurate than others with exception of gas thermometers They are quicker in action and less cumbersome  Resistance thermometers It cannot measure a local or rapidly changing temperature It can be used to measure temperatures between -200° to 1000°C if the resistance wire is platinum. The resistance wire can also be nickel and copper Thermometric property is the resistance of a metal wire. Thermocouples There are different types of thermocouple They can be used to measure temperature between -270°C and approximately 2000°C The wire of thermocouples are either platinum or patinum-rhodium Thermometric property is electromotive force (emf)  Thermoelectric Its measuring element is the welded junction of two fine wires It can measure temperature almost at a point. Measuring element is very small It can follow rapidly changing temperature because it has very small heat capacity Thermometric property is electromotive force. Thermal expansion of solids and liquids If the temperature of a liquid increases, its volume increases. For solids, as temperature increases, its dimensions increase. This phenomenon is known as thermal expansion. It has many applications. For example, thermal joints are included in buildings, railroad tracks, concrete highways and bridges to compensate for dimensional changes that occur as temperature changes. Explanation It is a consequence of the change in the average separation between the constituents atoms in an object. Consider the figure below. Thermal expansion of solids and liquids CONTD. At absolute zero temperature the atoms in an object oscillate about their equilibrium position with an average spacing between atoms as 10-10m. As the temperature increases, the amplitude of the oscillation increases, as a result the average separation between atoms increases. Consequently the objects expands. If thermal expansion is sufficiently small relative to the initial dimensions of the object, then L is proportional to T Li Thermal expansion of solids and liquids where T, L and Li are change in temperature, change in length associated with T and the original length respectively. Hence  is the coefficient of linear expansion L  T Li  L f  Li  Li T f  Ti  Thermal expansion of solids and liquids Since all the linear dimensions (e.g. length, breadth and height) of an object change with temperature, it implies the surface area and volume change as well. Similar to the change in the length, V is proportional to T Vi V and  T Vi the change in volume is V  ViT  V f  Vi  Vi T f  Ti   is the coefficient of volume expansion Similarly, surface area of a solid change with increase in temperature, the change in area is A  A T i  is the coefficient of area expansion TO BE COntinueD….. in the next lecture PHY 103: Basic Principle of Physics II Heat and Thermodynamics Lecture 2 By: Dr. E. Oyeniyi Mr. A. O. Ayoola Thermal expansion of solids and liquids If the temperature of a liquid increases, its volume increases. For solids, as temperature increases, its dimensions increase. This phenomenon is known as thermal expansion. It has many applications. For example, thermal joints are included in buildings, railroad tracks, concrete highways and bridges to compensate for dimensional changes that occur as temperature changes. Explanation It is a consequence of the change in the average separation between the constituents atoms in an object. Consider the figure below. Thermal expansion of solids and liquids CONTD. At absolute zero temperature the atoms in an object oscillate about their equilibrium position with an average spacing between atoms as 10-10m. As the temperature increases, the amplitude of the oscillation increases, as a result the average separation between atoms increases. Consequently the objects expands. If thermal expansion is sufficiently small relative to the initial dimensions of the object, then L is proportional to T Li Thermal expansion of solids and liquids where T, L and Li are change in temperature, change in length associated with T and the original length respectively. Hence  is the coefficient of linear expansion L  T Li  L f  Li  Li T f  Ti  Thermal expansion of solids and liquids Since all the linear dimensions (e.g. length, breadth and height) of an object change with temperature, it implies the surface area and volume change as well. Similar to the change in the length, V is proportional to T Vi V and  T Vi the change in volume is V  ViT  V f  Vi  Vi T f  Ti   is the coefficient of volume expansion Similarly, surface area of a solid change with increase in temperature, the change in area is A  A T i  is the coefficient of area expansion Thermal expansion of solids and liquids Recall that, For a solid,  =3 and =2α(this assumes that  is the same in all directions.. To show this, consider the box below a temperature Ti. If the temperature changes to Ti + T, its volume will also change from Vi + V as a result of the change in all of its dimensions. Therefore Thermal expansion of solids and liquids Contd. Initial volume, Vi = Li*Wi* Di Final volume (after expansion) = Vf = (Li+L)(Wi+ W)(Di+ D) ….. (*) If the coefficient of linear expansion is the same in the three directions then L = Li T, W = Wi T, D = Di T Substituting these in (*), one obtains Vf = (Li+ Li T)(Wi+ Wi T)(Di+ Di T) Thermal expansion of solids and liquids Contd. Vf = LiWiDi(1+ T)(1+ T)(1+ T) Vf = LiWiDi(1+ T)3  Vf = LiWiDi[1+3 T+3( T)2+ ( T)3] V= Vf - Vi = LiWiDi[1+3 T+3( T)2+ ( T)3] – LiWiDi V=LiWiDi[3 T+3( T)2+( T)3] Thermal expansion of solids and liquids Contd. Divide both sides by Vi yields V/Vi = 3 T+3( T)2+( T)3 ……. (**) Since  is usually very small, then ( T)2 and ( T)3 will be small compared to T. Therefore they can be neglected. So that equation (**) therefore becomes V/Vi = 3 T  3 = V/(Vi* T)   = 3 Thermal expansion of solids and liquids Contd. In a similar manner, it can be shown that for a rectangular plate, the change in area is A = 2 AiT Where  = 2 Thermal expansion of solids and liquids Solved problems (1)A constant-volume gas thermometer is calibrated in dry ice (that is, carbon dioxide in the solid state, which has a temperature of -80.0°C) and in boiling ethyl alcohol (78.0°C). The two pressures are 0.900 atm and1.635 atm. (a) What Celsius value of absolute zero does the calibration yield? b) What is the pressure at (i) the freezing point of water and (ii) the boiling point of water? (2) A segment of steel railroad track has a length of 30.00m when the temperature is 0.00C. (a) what is its length when the temperature is 40.00C? (b) suppose the ends of the rail are rigidly clamped at 0.00C so that expansion is prevented. What is the thermal stress set up in the rail if its temperature is raised to 40.00C if the Young modulus of the steel is 20 x 1010Nm-2 and coefficient of linear expansion is 11 x 10-6(0C)-1 c) What is the length of the unclamped segment if the temperature drops to - 40.00C? (3)The active element of a certain laser is a glass rod 30.0 cm long by 1.50 cm in diameter. If the temperature of the rod increases by 65.0°C, what is the increase in (a) its length, (b) its diameter, and (c) its volume? (Assume that = 9.00x10-6 °C-1.) Solution to question (1) 𝑃𝐿 −𝑃𝑟 𝑇𝐿 −𝑇𝑟 (a) = 𝑃𝑈 −𝑃𝑟 𝑇𝑈 −𝑇𝑟 9.09𝑥104−0.0 −80.0−𝑇𝑎𝑏 = 1.65𝑥105−0.0 78.0−𝑇𝑎𝑏 9.09𝑥104(78.0 − 𝑇𝑎𝑏 )= 1.65𝑥105(−80.0 − 𝑇𝑎𝑏 ) 7090200 − 90900𝑇𝑎𝑏 =-13,200,000 -165000𝑇𝑎𝑏 −90900𝑇𝑎𝑏 + 165000𝑇𝑎𝑏 = −13,200,000- 7090200  74100𝑇𝑎𝑏 = −20290200 −20290200 𝑇𝑎𝑏 = 74100 𝑇𝑎𝑏 = −273.82°C b) (i) 𝑃0 − 𝑃𝐿 𝑇0 − 𝑇𝐿 = 𝑃𝑈 − 𝑃𝐿 𝑇𝑈 − 𝑇𝐿 𝑃0 − 90900 0 − (−80.0) = 165000 − 90900 78.0 − (−80.0) (𝑃0 − 90900)(78.0 + 80.0) = 74100𝑥80 74100𝑥80 𝑃0 − 90900 = 78.0 + 80.0 74100𝑥80 𝑃0 = + 90900 = 37519.98 + 90900 78.0 + 80.0 𝑃0 = 128419.98𝑃𝑎 = 1.27𝑎𝑡𝑚  𝑃𝑈 −𝑃𝐿 𝑇𝑈 −𝑇𝐿 b) (ii) = 𝑃100 −𝑃𝐿 𝑇100 −𝑇𝐿 165000 − 90900 78.0 − (−80.0) = = 𝑃100 − 90900 100.0 − (−80.0) 74100 158.0 = 𝑃100 − 90900 180.0 (𝑃100 −90900)158.0 = 74100𝑥180.0 74100𝑥180.0 𝑃100 = + 90900 158.0 𝑃100 = 84417.72 + 90900 𝑃100 = 175317.72𝑃𝑎 = 1.736𝑎𝑡𝑚 Solution to question (2) (a) Recall that, ∆𝐿 =⍺∆T 𝐿𝑖 so that ∆𝐿 = 𝐿𝑖 ⍺∆T = 30.0 x 11 x 10-6 x 40.0 ∆L = 0.013m 𝐿𝑓 = 30.0 + 0.013 = 30.013m (b) The thermal stress is the same as the tensile stress in the situation in which the ray expands freely. 𝐹 ∆𝐿 0.013 Tensile stress, 𝐴 = 𝑌 𝐿 = 20 𝑥 1010 ( 30.0 ) 𝑖 = 8.7 𝑥 107 𝑁/𝑚2 c) ∆𝐿 = 𝐿𝑖 ⍺∆T In this case there will be a decrease in length when the temperature decreases. We assume ⍺ is constant over the entire range of temperature. Therefore, if there is increase in the length by 0.013m when the temperature increases by 40.00C, there will be a decrease in the length by 0.013m if the temperature decreases by 40.00C. The new length at the colder temperature is 30.0 – 0.013 = 29.987m 4) A poorly designed electronic device has two bolts attached to different parts of the device that almost touch each other in its interior as shown in the figure. The steel and brass bolts are at different electric potentials, and if they touch, a short circuit will develop, damaging the device. The initial gap between the ends of the bolts is 5.0μm at 27.00C. At what temperature will the bolts touch? Take coefficient of linear expansion for steel and brass respectively to be 11 𝑥 10−6 /0C and 19 𝑥 10−6 /0C. Solution At a particular change in temperature, ∆T, The steel will increase in length by ∆𝐿𝑠 = 𝐿𝑖,𝑠 ⍺𝑠 ∆T so also the brass will increase by ∆𝐿𝑏 = 𝐿𝑖,𝑏 ⍺𝑏 ∆T at the same temperature change. Therefore, ∆𝐿𝑠 + ∆𝐿𝑏 = 5.0 𝑥10−6𝑚 𝐿𝑖,𝑠 ⍺𝑠 ∆T +𝐿𝑖,𝑏 ⍺𝑏 ∆T = 5.0 𝑥10−6 (𝐿𝑖,𝑠 ⍺𝑠 +𝐿𝑖,𝑏 ⍺𝑏 )∆T = 5.0 𝑥10−6 5.0 𝑥10−6 ∆𝑇 = 𝐿𝑖,𝑠 ⍺𝑠 +𝐿𝑖,𝑏 ⍺𝑏 5.0 𝑥10−6 ∆𝑇 = (0.010 x 11 x 10−6)+(0.030 x 19 x 10−6) ∆𝑇 = 7.40C The temperature at which the bolts will touch is now 270C+7.40C = 34.40C Macroscopic Description of an Ideal Gas The volume V, temperature T and pressure P of a gas are related by an equation called “equation of state”. The equation is very complicated, but simple for gases maintained at a very low pressure (or low density). Such low-density gas is referred to as Ideal gas Macroscopic Description of an Ideal Gas Suppose an ideal gas is confined in a cylindrical container whose volume can be varied by means of moving piston. Experiments have shown that, for such a system: (1)When the gas is kept at a constant temperature, its pressure is inversely proportional to its volume (Boyle’s law) i.e P  1/V (2)When the pressure of the gas is kept constant, its volume is directly proportional to its temperature(Charles law) i.e. V  T These two observations are summarised by the equation of state for an ideal gas. i.e PV = nRT (n = m/M) R is a constant (Gas constant) The equation is also known as ideal gas law. Macroscopic Description of an Ideal Gas Definition of an Ideal gas An ideal gas is one for which PV/nT is constant for all pressures. Given that n is also equal to N/NA , then PV = (N/NA)RT. R and NA are constants, R/NA is the Boltzmann’s constant, k. Therefore PV = N kT Heat and 1st Law of Thermodynamics The first law of thermodynamics is the law of conservation of energy. It describes systems in which the only energy change is that of internal energy, which is due to transfers of energy by heat or work. The first law makes no distinction between the results of heat and the results of work. According to the first law, a system’s internal energy can be changed either by an energy transfer by heat to or from the system or by work done on or by the system. Heat and 1st Law of Thermodynamics Internal energy This is all the energy of a system that is associated with its microscopic components—atoms and molecules—when viewed from a reference frame at rest with respect to the object. Internal energy includes translation, vibration and rotation of molecules, potential energy within molecules and potential energy between molecules Kinetic energy of the system due to its motion through space is not included in internal energy. For a monoatomic ideal gas, the only type of energy available for its microsopic component is associated with translational motion Heat and 1st Law of Thermodynamics More generally, in solids, liquids, and molecular gases, internal energy includes other forms of molecular energy. Heat This is defined as the transfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings. When you heat a substance, you are transferring energy into it by placing it in contact with surroundings that have a higher temperature. Both heat and work are ways of changing the energy of a system. Heat when the transfer is as a result of temp. difference, Example of energy transfer by work is when a gas is compressed by piston or when the piston is released and the gas expands. Heat and 1st Law of Thermodynamics Different units of energy include calorie and Joule. 1 Calorie = 4.186 J This equality is known as the mechanical equivalent of heat. Heat Capacity and Specific Heat The heat capacity C of a particular sample of a substance is defined as the amount of energy needed to raise the temperature of that sample by 1°C. e.g. if the energy needed to raise the temp. of a sample by T is Q, then Q C   Q  CT T Heat and 1st Law of Thermodynamics Dividing the above equation by the mass, m, of the substance produces a quantity called specific heat, c. C Q i.e c  Q  mcT m mT Specific heat capacity is a measure of how thermally insensitive a substance is to the addition of energy. The higher the c, the more energy must be added to a given mass of the material to cause a particular temperature change. Heat and 1st Law of Thermodynamics The above equation means if temperature increases, Q is +ve and heat flows into the system. If temperature decreases , Q is -ve and heat flows out of the system. c varies with temperature, however for small temp. interval the variation is small and can be neglected. For example, for change in temp. of water from 0oC to 100oC the variation is only about 1%. When the variation can not be neglected then Tf Q  m  cT Ti Heat and 1st Law of Thermodynamics Conservation of Energy Qcold = -Qhot i.e. mcccTc= -mhchTh If the initial temps. of the cold and hot substances are T1 and T2, and when in contact reached a final temp. Tf. Then mccc(Tf-T1)= -mhch(Tf-T2) Heat and 1st Law of Thermodynamics Latent Heat There are situations in which the transfer of energy to a substance does not result in a change in temp. This is always the case when phase change occurs. Two common phase changes are: Solid to liquid (melting) and liquid to gas (boiling). Heat and 1st Law of Thermodynamics The amount of energy transfer during the phase changes is also dependent on mass. The energy transfer per unit mass during a phase change is called latent heat, L. i.e. L = Q/m  Q = mL L depends on the nature of phase change, so there are: Latent heat of fusion for melting and Latent heat of vaporization for boiling Heat and 1st Law of Thermodynamics Example Heat and 1st Law of Thermodynamics Solution Part A: Part B Part C Part D Part E The total amount of energy that must added to change 1g of ice at - 30oC to steam at 120oC is the sum of the five energies = 3.11 x103 J TO BE CONTINUED………. PHY 103: Basic Principle of Physics II Heat and Thermodynamics Lecture 3 By: Dr. E. Oyeniyi Mr. A. O. Ayoola Heat and 1st Law of Thermodynamics Example. Calculate the amount of energy that must added to change 1g of ice initially at -30oC to steam at 120oC as illustrated in the figure below. Heat and 1st Law of Thermodynamics Solution Part A: Part B Part C Part D Part E The total amount of energy that must added to change 1g of ice at - 30oC to steam at 120oC is the sum of the five energies = 3.11 x103 J Worked Examples 1. A 0.05kg ingot of metal is heated to 200.00C and then dropped into a light insulated beaker containing 0.400kg of water initially at 20.00C. the final equilibrium temperature of the mixed system is 22.40C. Find the specific heat capacity of the metal assuming the energy absorbed by the container is negligible. Take the specific heat capacity of water to be 4186J(kg0C)-1 2. What mass of steam initially at 1300C is needed to warm 200.0g of water in a 100.0g glass container from 200C to 500C? Latent heat of fusion of ice = 3.33 x 105Jkg-1 Latent heat of vaporization of water = 2.26 x 106Jkg-1 Specific heat capacity of ice = 2.09 x 103Jkg-1K-1 Specific heat capacity of water = 4.19 x 103Jkg-1K-1 Specific heat capacity of steam = 2.01 x 103Jkg-1K-1 Specific heat capacity of the metal = 8.37 x 102Jkg-1K-1 solution (1) Recall that, Qcold = -Qhot Assumptions: we assume that the system is sealed and some of water that may vaporize when the ingot is dropped cannot escape; Since the container is light and insulated, we assume the mass and heat absorbedcapacity of the container is negligible. i.e. mcccTc = -mhchTh mwcwTw = -mmcmTm mwcw(Tf-Ti)= -mmcm(Tf-Ti) mccc(Tf −Tw) cm = −mm (Tf −Tm) 0.40 x 4186(22.4−20.0) cm = −0.05(22.4−200.0)  0.40 x 4186(22.4−20.0) 1667.2 x2.4 4018.56 cm = = = 0.05(200.0−22.4) 0.05 x 177.6 0.05 x 177.6 cm = 452.5Jkg−1K−1 = 4.53x102Jkg-1K-1 2) Recall that, Qcold = -Qhot In this case, the gaseous water i.e. the steam (hot body) undergoes three processes: A decrease in temperature from 1300C to 1000C; Condensation into liquid water at 1000C and lastly; A decrease in temperature of the water from 1000C to 500C Energy involved in the first stage, 𝑄1 = 𝑚𝑠 𝑐𝑠 T1 𝑄1 = 𝑚𝑠 𝑐𝑠 (100 − 130) 𝑄1 = 𝑚𝑠 𝑥2.01 x 103(-30) 𝑄1 = −6.03𝑥104𝑚𝑠 J Energy involved in the second stage, 𝑄2 = 𝑚𝑠 𝑐𝑠 T2 𝑄2 = −𝑚𝑠 𝑥2.26 x 106 𝑄2 = −2.26 x 106𝑚𝑠 J Energy involved in the last stage, 𝑄3 = 𝑚𝑠 𝑐𝑠 T3 𝑄3 = 𝑚𝑠 𝑐𝑠 (50 − 100) 𝑄3 = 𝑚𝑠 𝑥4.19𝑥103(−50) 𝑄3 = −2.095𝑥105𝑚𝑠 J Energy transfer three stages 𝑄ℎ𝑜𝑡 = 𝑄1 + 𝑄2 + 𝑄3 𝑄ℎ𝑜𝑡 = −6.03𝑥104𝑚𝑠 + −2.26 x 106𝑚𝑠 + −2.095𝑥105𝑚𝑠 𝑄ℎ𝑜𝑡 = −2.53𝑥106𝑚𝑠 𝐽 The water and the container undergo only one process, an increase in temperature from 20.00C to 50.00C The amount of energy transfer in this process is: 𝑄𝑐𝑜𝑙𝑑 = 𝑚𝑤 𝑐𝑤 T + 𝑚𝑐 𝑐𝑐 T 𝑄𝑐𝑜𝑙𝑑 =0.2x4.19 x 103(50.0−20.0)+0.1x8.37 x 102(50.0−20.0) 𝑄𝑐𝑜𝑙𝑑 =2.514x104+ 2.511 x 103=2.77 x 104J Recall that, Qcold = -Qhot 2.77 x 104= − −2.53𝑥106𝑚𝑠  2.77𝑥104 𝑚𝑠 = 2.53𝑥106 𝑚𝑠 = 1.09𝑥10−2 𝑘𝑔 = 10.9𝑔 WORKED EXAMPLE 1. In an x-ray tube, 1018 electrons per second arrive with a speed of 2 x106ms-1 at a metal target of mass 0.2kg and specific heat capacity 500Jkg-1K-1. If the mass of an electron is 9.0 x 10-31Kg and assuming 98% of the incident energy is converted into heat, find how long the target will take to rise in temperature by 500C assuming no heat loss. 2. Water flow at the rate of 0.150kgmin-1 through a tube and is heated by heater dissipating 25W. The inflow and outflow water temperature are 15.20C and 17.40C respectively. When the rate of flow is increased to 0.2318kgmin-1 and the rate of heating to 37.8W, the inflow and outflow temperature remain unchanged. Find (a) the specific heat capacity of water (b) the rate of loss of heat from the tube SOLUTION 1 1) The K.E of an object is 𝑚𝑉 2 , 2 1 K.E of an electron arriving at the target in a second = 𝑥9.1𝑥10−31 𝑥(2.0𝑥106 )2 ) 2 K.E 1 of all the electrons arriving at the target in a second = 𝑥9.1𝑥10−31 𝑥(2.0𝑥106 )2 )𝑥1018 = 1.82𝐽 2 In time t, the total K.E of all the electrons arriving at the target will be 𝐾. 𝐸 = 1.82𝑡𝐽 98 but remember that 98% of 1.82𝑡𝐽 is converted to heat= 𝑥1.82 = 1.8𝑡𝐽 100 The amount of energy transferred by the electrons = amount of energy absorbed (gained) by heat by the metal target i.e. 1.8𝑡 = 𝑚𝑐𝑇 1.8𝑡 = 0.2𝑥500𝑥50 0.2 𝑥 500 𝑥 50 5000 𝑡= = 1.8 1.8 𝑡 = 2777.7 = 2780𝑠 = 46.3𝑚𝑖𝑛  𝐸𝑛𝑒𝑟𝑔𝑦 2) recall that Power = 𝑡𝑖𝑚𝑒 (a) 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 ℎ𝑒𝑎𝑡𝑒𝑟 = 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑤𝑎𝑡𝑒𝑟 + 𝑅𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑙𝑜𝑠𝑠 𝑃 = 𝑚𝑐𝑇 + 𝐻𝐿 where 𝐻𝐿 is the rate of heat loss In the first case, 0.150 25.2 = × 𝑐 × 17.4 − 15.2 +𝐻𝐿 60 In the second case, 0.2318 37.8 = × 𝑐 × 17.4 − 15.2 + 𝐻𝐿 60 0.0025 × 𝑐 × 2.2 +𝐻𝐿 = 25.2 0.0025 × 𝑐 × 2.2 +𝐻𝐿 = 25.2 i 0.0039 × 𝑐 × 2.2 +𝐻𝐿 = 37.8 ii 0.0055𝑐+𝐻𝐿 = 25.2 iii 0.0085𝑐+𝐻𝐿 = 37.8 iv  Solve equations (iii) and (iv) simultaneously Subtracting equation (iii) from equation (iv), we have 0.0030𝑐 = 12.6 12.6 𝑐= = 4200𝐽 = 4.2 × 103 𝐽𝑘𝑔−1 𝐾 −1 0.0030 (b) Substitute the value of c into equation (iii) 0.0055 × 4200 + 𝐻𝐿 = 25.2 23.1 +𝐻𝐿 =25.2 𝐻𝐿 = 2.1𝑊 Work and Heat in Thermodynamics Processes In thermodynamics, the state of a system is described using variables like pressure, volume, temperature and internal energy. As a result of that, these quantities belong to a category called state variables. Macroscopic state of an isolated system can be specified only if the system is in thermal equilibrium internally. In the case of a gas in a container, internal thermal equilibrium requires that every part of the gas be at the same pressure and temperature. Assumption: we assume that the expansion or contraction of the gas is quasi-static. Quasi-static expansion or contraction is the one that is slow enough to allow the system to remain essentially in thermal equilibrium at all times. Work and Heat in Thermodynamics Processes Consider a gas contained in a cylinder fitted with a movable piston The volume of the gas at equilibrium is V, and it exerts a pressure P on the cylinder’s wall and on the piston. Work and Heat in Thermodynamics Processes If the cross sectional area of the piston is A, the force exerted on the piston is F=P*A The workdone by the gas on the piston when it moves up a distance x is w = F * x = P*A*x If the change is infinitesimally small, the work done when the gas expands becomes; dw= P*A*dx But A*dx is the same as change in volume of the gas, dV. Then dw = P * dV The total work done by the gas as its volume changes from Vi to Vf is Vf W   PdV Vi Work and Heat in Thermodynamics Processes Please Note the following In expansion, dV is positive, therefore the work done by the gas is +ve. We say in this case that the gas does work. In compression, dV is negative, therefore the work done by the gas is - ve. This can be interpreted as, work is done on the gas. If the volume remains unchanged, dV is zero and the work done is zero In thermodynamics, positive work represents a transfer of energy out of the system. PV-DIAGRAM If the pressure and volume are known at each step of the process, the state of the gas at each step can be plotted on a graphical representation called a PV diagram. The curve on a PV diagram is called the path between the initial and final states. The area under the curve represent the workdone. Work and Heat in Thermodynamics Processes NOTE: The work done by a system depends on the initial and the final states and on the path followed by the system from initial and the final states. Energy transfer by heat like the work done depends on the initial state, the final states and intermediate states of the system.  To explain this consider the figures below.  A A B A B B C C In the three cases above, the system moves from initial state i to final state f In the second case (b) but followed different paths. In the first case (a) W = 𝑊𝐴𝐵 + 𝑊𝐵𝐶 = 𝑃𝑖 𝑉𝑓 − 𝑉𝑖 + 0 W = 𝑊𝐴𝐵 + 𝑊𝐵𝐶 = 0 + 𝑃𝑓 𝑉𝑓 − 𝑉𝑖 = 𝑃𝑖 𝑉𝑓 − 𝑉𝑖 = 𝑃𝑓 𝑉𝑓 − 𝑉𝑖  For the last case (c), let’s take the path followed to be straight, then the equation relating P and V can be written just like y = mx + c, so we have P = mV + c Where m is the slope and c is the intercept on p axis. Vf Recall that, W   PdV Vi 𝑉𝑓 𝑊= 𝑉𝑖 𝑚𝑉 + 𝐶 𝑑𝑉 Integrating this, we have 𝑚𝑉 2 𝑉𝑓 𝑊=[ + 𝐶𝑉]𝑉 𝑚 2 𝑖 𝑊= 𝑉𝑓 2 − 𝑉𝑖 2 + 𝐶(𝑉𝑓 − 𝑉𝑖 ) 2  We see that in the three cases, the work done are not the same though the system has moved from the initial state, i to the final state, f. Therefore, the work done by/on a system depends on the initial and final states and on the path followed between the two states. 1st Law of Thermodynamics The conservation of energy law is stated as the change in the energy of a system is equal to the sum of all the energies transferred across the boundary of the system. The 1st law of thermodynamics is a special case of the law of conservation of energy that describes the processes in which only the energy change is the internal energy and the only energy transfers are by heat and work In other words, It is known that energy can be exchanged between a system and its surrounding through the following: i. By work done and ii. By Heat Both work done and heat result in change in the internal energy of the system. If the energy transfer by heat to a system when it changes from an initial state to a final state is Q and work W is done by the system. First Law of Thermodynamics Contd. If the quantity Q – W is measured for various paths connecting the initial and final equilibrium states, we find that it is the same for all paths connecting the two states. Then one can state the 1st law of thermodynamics in the following way: When any closed (control mass) system is alter adiabatically, the net work associated with the change of state is the same for all the possible processes between two given equilibrium states. This difference between heat and work i.e.the quantity, Q – W is known as the change in internal energy. i.e 𝐸𝑖𝑛𝑡 = Q – W 1 This means that although Q and W both depend on path, the quantity Q – W is independent of the path. First Law of Thermodynamics When the system undergoes an infinitesimal change in state, then small amount of energy dQ will be transferred and a small work dW will be done by the system. The first-law in this case is 𝑑𝐸𝑖𝑛𝑡 = Q – W 2 Equations (1) and (2) are known as First-law equations. The following must be taken note of (i) Q is positive when energy enters the system and negative when energy leaves the system (ii) and W is positive when the system does work on the surroundings and negative when work is done on the system. (iii) The first-law specify that internal energy is the only type of energy that change in the system. 1st Law of Thermodynamics Contd. Let us investigate some special cases in which the first law can be applied: a. An isolated system: This a type From 𝑑𝐸𝑖𝑛𝑡 = Q – W of system in which there is no Q=0, and W =0 interaction between the system and the surrounding. The Therefore, In this case the change in change in internal energy of this internal energy, kind of system is zero i.e. 𝑑𝐸𝑖𝑛𝑡 = 0 𝑑𝐸𝑖𝑛𝑡 =0. 𝐸𝑖𝑛𝑡_𝑓 = 𝐸𝑖𝑛𝑡_𝑖 The implication of this is that, the internal energy of an isolated system remains constant. 1st Law of Thermodynamics Contd. b. Cyclic process: This is a procces From the first law equation, that starts and ends at the same 𝑑𝐸𝑖𝑛𝑡 = Q – W state. The change in internal energy 𝑑𝐸𝑖𝑛𝑡 = 0 is again zero, so the energy added to the system equal the work, W Q=W done on the system. i.e. On a PV diagram, a cyclic process appears as a closed curve. 1st Law of Thermodynamics Contd. c. Process where work is zero: From the equation of the 1st law of thermodynamics, 𝐸𝑖𝑛𝑡 = Q – W. Let us investigate how the 1st law of thermodynamics can be applied to If the work done W=0, then the case when there is no work 𝐸𝑖𝑛𝑡 = Q done either by or on the system. The implication of this is that all the heat added to the system is used to increase the internal energy of the system. This means that the final internal energy of the system will be 𝐸𝑖𝑛𝑡_𝑓 = 𝐸𝑖𝑛𝑡_𝑖 + 𝑄  From the equation of the first law, d. Process where heat is zero 𝐸𝑖𝑛𝑡 = Q – W. If work is done on the system, W If Q = 0, then 𝐸𝑖𝑛𝑡 = – W. is -ve which implies increase in If W is +ve as in expansion, 𝐸𝑖𝑛𝑡 < internal energy i.e. E > 0 0 But if the system does work, W is If W is –ve as in compression, +ve which implies that there will 𝐸𝑖𝑛𝑡 > 0 be decrease in internal enrgy of For instance, If a gas is compressed the system i.e. E < 0 by a moving piston in an insulated cylinder, no energy is transferred by heat and the work done by the gas is negative; thus, the internal energy increases because kinetic energy is transferred from the moving piston to the gas molecules. APPLICATION OF FIRST LAW OF THERMODYNAMICS in considering the application of thermodynamics, we shall look at four (4) possible processes. 1. Adiabatic Process: This is a process where no energy enters or leaves the system by heat. i.e. Q = 0. So Adiabatic process can be achieved in one of the following ways: i. by thermally insulating the system from its surroundings. ii or by performing the process rapidly, so that there is a neglible time for energy to transfer by heat. For this process, the first-law becomes 𝐸𝑖𝑛𝑡 = – W.  NOTE the following: This result shows that if a gas is compressed adiabatically, such that the work done is negative, then 𝐸𝑖𝑛𝑡 is positive. Conversely, that if a gas expands adiabatically, such that the work done is positive, then 𝐸𝑖𝑛𝑡 is negative. This means that: i. adiabatic expansion of a gas leads to decreases in temperature ii. Adiabatic compression of gases leads to increase in temperature In adiabatic free expansion, both Q and W are zero. That is the initial and final internal energies of a gas are equal in an adiabatic free expansion ADIABATIC FREE EXPANSION The process is described by the figures on the left hand side, it is a unique expansion process. It is adiabatic because it takes place in a insulated container. Also, because the gas expands into a vacuum when the membrane is broken, it does not apply a force on any piston, so no work is done on or by the gas. i.e. W =0 from the first law equation, since Q=0 then 𝐸𝑖𝑛𝑡 = 0. we can now see that, 𝐸𝑖𝑛𝑡_𝑓 = 𝐸𝑖𝑛𝑡_𝑖 i.e. initial and final internal energies of a gas in adiabatic free expansion are the same. 2. Isobaric Process: this is a process that takes place at a constant pressure. In such a process the values of the heat and work are both usually nonzero. In an isobaric process, the work done is given by 𝑊 = 𝑃(𝑉𝑓 − 𝑉𝑖 ) 3. Isovolumetric Process: this is a process that takes place at a constant volume. This process is sometimes know as Isochoric process. Since the volume remains constant, W=0 and 𝐸𝑖𝑛𝑡 = 𝑄. This is shows that if energy is added by heat to a system kept at a constant volume, all the transferred energy remain in the system as an increase in internal energy of the system. 4. Isothermal process: this is a process that takes place at a constant temperature. NOTE: Since internal energy is a function of temperature, this means that change in internal energy of a system undergoing isothermal process will be zero. So from first law we will have that 𝑄 = 𝑊. Therefore any energy Q that enters the system by heat is transferred out of the system by work. For ideal gas, 𝑛𝑅𝑇 PV = nRT so that P = 𝑉 𝑉𝑓 𝑊= 𝑃𝑑𝑉 = 𝑉𝑖 𝑉𝑓 𝑉𝑓 𝑛𝑅𝑇 1 𝑊= 𝑑𝑉 = 𝑛𝑅𝑇 𝑑𝑉 𝑉𝑖 𝑉 𝑉𝑖 𝑉 𝑉𝑓 𝑊= 𝑛𝑅𝑇𝑙𝑛[𝑉]𝑉 𝑖 𝑊 = 𝑛𝑅𝑇𝑙𝑛(𝑉𝑓 − 𝑉𝑖 ) 𝑉𝑓 𝑊 = 𝑛𝑅𝑇𝑙𝑛( ) 𝑉𝑖  A A B A B B C C In fig. (a), process AB is an isovolumetric In fig. (c), process AB is an isothermal process while Process BC is an isobaric process process In fig. (b), process AB is an isobaric process while BC is a isovolumetric process TO BE CONTINUED………. PHY 103: Basic Principle of Physics II Heat and Thermodynamics Lecture 4 By: Dr. E. Oyeniyi Mr. A. O. Ayoola  QUESTIONS 1 1.0-mol sample of an ideal gas is kept at 0.00C during an expansion from 3.0L to 10.0L. (a) How much work is done on the gas during expansion? (b) How much energy transfer by heat occurs between the gas and its surroundings in this process? (c) if the gas returns to the original volume by a means of an isobaric process, how much work is done on the gas?  SOLUTION 1) This is an isothermal expansion, a) work done is given by 𝑉𝑓 𝑊 = 𝑛𝑅𝑇𝑙𝑛( ) 𝑉𝑖 10.0 𝑊 = 1.0 𝑥 8.31 𝑥 273𝑙𝑛 3.0 𝑊 = 2729.099 = 2.7 × 103 𝐽 b) To get how much energy transferred by heat, we use equation of the first law of thermodynamics i.e. ∆𝐸𝑖𝑛𝑡 = 𝑄 − 𝑊 but since the process is an isothermal and internal energy is a function of temperature, ∆𝐸𝑖𝑛𝑡 = 0 so the first law equation becomes, 𝑄 = 𝑊 = 2.7 × 103 𝐽  If the gas returns to its original volume through isobaric process, W = 𝑃∆𝑉 𝑊 = 𝑃(𝑉𝑓 −𝑉𝑖 ) 𝑛𝑅𝑇 1.0 𝑥8.31 𝑥 273 But P = = ( −3 ) 𝑉𝑖 10.0 𝑥10 5 P = 2.27 x10 𝑃𝑎 𝑊 = 2.27 𝑥105 (3 𝑥10−3 − 10.0 𝑥10−3 ) 𝑊 = 1589𝐽 = 1.6 𝑥103 𝐽 QUESTION 2 A 1.0kg bar of copper is heated at atmospheric pressure so that its temperature increases from 200C to 500C. (Take coefficient of linear expansion of copper as 1.7 x 10-5 0C-1, its specific heat capacity as 387Jkg-1 0C-1 and the density to be 8.92 x 103kgm-3). a) What is the work done on the copper bar by the surrounding atmosphere? b) How much energy is transferred to the copper bar by heat? c) What is the change in internal energy of the copper bar?  SOLUTION This process takes place at a constant pressure we consider it as an isobaric. Since this example involves solid, change in volume due to thermal expansion is very small. The work done, W = 𝑃∆𝑉, Recall that ∆𝑉 = (𝑉𝑓 −𝑉𝑖 ) = γ𝑉𝑖 ∆𝑇 ∆𝑉 = 3⍺𝑉𝑖 ∆𝑇 = 3 × 1.7 × 10−5 × 𝑉𝑖 50 − 20 𝑚 But, 𝑉𝑖 = ⍴ 1.0 ∴ ∆𝑉 = 3 × 1.7 × 10−5 × 3 30 8.92 × 10 ∆𝑉 = 1.7 × 10−7 𝑚3  W = 𝑃∆𝑉 = 1.013 × 105 × 1.7 × 10−7 𝑊 = 1.7 × 10−2 𝐽 b) Energy transferred by heat, 𝑄 = 𝑚𝑐∆𝑇 𝑄 = 1.0 × 387 50 − 20 = 11610 𝑄 = 1.2 × 104 𝐽 c) ∆𝐸𝑖𝑛𝑡 = 𝑄 − 𝑊 ∆𝐸𝑖𝑛𝑡 = 1.2 × 104 − 1.7 × 10−2 = 119999.98 ∆𝐸𝑖𝑛𝑡 = 1.2 × 104 𝐽 This shows that most of the energy transferred into the system by heat goes into increase the internal energy of the copper bar. Hence when the thermal expansion of a solid or liquid is analyzed, the small amount of work done is usually ignored.  QUESTION 2 A thermodynamic system undergoes a process in which its internal energy decreases by 500J. If at the same time 200J of work is done on the system, what is energy transferred to or from it by heat? SOLUTION ∆𝐸𝑖𝑛𝑡 = −500𝐽, 𝑊 = −200𝐽 Applying ∆𝐸𝑖𝑛𝑡 = 𝑄 − 𝑊 −500 = 𝑄 − −200 𝑄 = −500 − 200 𝑄 = −700𝐽 QUESTION 4 A gas is compressed from 9.00m3 to 2.00m3 at a constant pressure of 0.8atm. in the process 400J of energy leaves the gas by heat. (a) what is the work done on the gas (b) What is the change in internal energy? SOLUTION a) 𝑊 = 𝑃(𝑉𝑓 −𝑉𝑖 ) = 0.8𝑥1.01𝑥105 (2.00 − 9.00) = − 5.66𝑥105 𝐽 b) From ∆𝐸𝑖𝑛𝑡 = 𝑄 − 𝑊 ∆𝐸𝑖𝑛𝑡 = −400 − (−5.66𝑥105 ) ∆𝐸𝑖𝑛𝑡 = 5.65𝑥105 𝐽  Kinetic Theory of Gases Molecular model of an Ideal Gas The model shows that the pressure that a gas exerts on the walls of its container is a consequence of the collisions of the gas molecules with the walls. There are some assumptions made in the development of this model: (a) The number of molecules is large, and the average separation between molecules is great compared with their dimensions. This means that the volume of the molecules is negligible compared with the volume of the container.  Kinetic Theory of Gases (b) The molecules obey Newton’s laws of motion, but as a whole they move randomly. By “randomly” we mean that any molecule can move in any direction with equal probability. We also assume that the distribution of speeds does not change in time, despite the collisions between molecules. That is, at any given moment, a certain percentage of molecules move at high speeds, a certain percentage move at low speeds, and a certain percentage move at speeds intermediate between high and low. (c) The collision between molecules and between the wall of the container is elastic. Thus, in the collisions, both kinetic energy and momentum are constant.  Kinetic Theory of Gases (d) The forces between molecules are negligible except during collision. The forces between molecules are short- range, so the molecules interact with each other only during collisions. (e) The gas under consideration is a pure substance. That is, all of its molecules are identical. Kinetic Theory of Gases: Pressure of an Ideal Gas To obtain the pressure of an ideal gas, Consider N number of of the gas in a container of volume V Let the container be a cube of length L For a molecule moving with a velocity v towards the right-hand face of the box makes a collision. The velocity components of the molecule are vx, vy and vz. As the molecule collides with the wall elastically, its x-component of velocity is reversed, while the y and z components remain unaltered.  The change in momentum of the molecule is therefore Px  mv x  mv x   2mv x Py  0 Pz  0 From Impulse-momentum theorem 𝐹𝑤 ∆𝑡 = ∆𝑃𝑥 = −2𝑚𝑉𝑥 Where 𝐹𝑤 is the component of the average force that the wall exerts on the molecule during the collision and ∆𝑡 is the duration of the collision.  Before the molecule can make another collision with the same wall, the molecules needs to travel to and fro covering a total distance of 2L in the direction (x-direction). Hence, the time interval between the collision; 2𝐿 ∆𝑡 = 𝑉𝑥 So, the force 𝐹𝑤 can be expressed as −2𝑚𝑉𝑥 𝐹𝑤 = ∆𝑡 −2𝑚𝑉𝑥 𝑚𝑉𝑥2 𝐹𝑤 = 2𝐿 = 𝑉𝑥 𝐿 From Newton’s third law, the average force exerted by the molecule on the wall is equal in magnitude and opposite in direction to the average farce exerted on the molecule by the wall.  the force that the molecule exerted is written as; 𝑚𝑉𝑥2 𝐹𝑚 = 𝐿 F  F1  F2 ....  FN  L  m 2 vx1  vx22 ....vxN 2  Therefore the total force exerted by all the molecules on the wall is obtained by adding the forces exerted by individual molecules 2 𝑁 𝑚𝑉𝑥𝑖 𝑚 𝑁 2 𝐹𝑇 = 𝑖=1 𝐿 = 𝑉 𝑖=1 𝑥𝑖 𝐿  Average value of the square of velocity of all molecules can be expressed as 2 𝑁 𝑉𝑥𝑖 𝑉𝑥2 = 𝑖=1 𝑁 If we define the average value of the square of the velocity in the x direction for N molecules as 2 𝑁 𝑉𝑥𝑖 𝑉𝑥2 = 𝑖=1 𝑁 Therefore, 𝑚 𝐹= 𝑁 𝑉𝑥2 1 𝐿  The average value of V2 for all the molecules in the container (in all directions) is related to the average values of 𝑉𝑥2 , 𝑉𝑦2 𝑎𝑛𝑑 𝑉𝑧2 according to the expression 𝑉 2 = 𝑉𝑥2 + 𝑉𝑦2 + 𝑉𝑧2 But 𝑉𝑥2 = 𝑉𝑦2 = 𝑉𝑧2 (i.e. the average values of the speed in each direction are equal since the motion is random) 1 So, 𝑉2 = 3𝑉𝑥2 (i.e. 𝑉𝑥2 = 𝑉2 ) 3 The total force exerted by the molecules on the wall 1 𝑉2 is now 𝐹 = Nm 3 𝐿  recall that 𝐹 𝑝= 𝐴 𝐴 = 𝐿2 1 𝑁𝑚𝑉 2 𝑝= 3 𝐿3 1 𝑁 𝑝= 𝑚𝑉 2 3 𝑉 2 𝑁 1 𝑝=. 𝑚𝑉 2 2 3 𝑉 2 This means the pressure is proportional to the number of molecules per unit volume and to the average translational kinetic energy of the molecules.  IMPLICATIONS The equation means that the pressure inside a container can be increased by (1) increasing the number of molecules per unit volume in the container. This is what happens when we add air to the tyre of automobile. (2) increasing the average translational kinetic energy of the molecules in the container. This is why pressure inside the tyre increases as the tyre warms up while travelling and sometimes bursts if the tyre could not withstand the pressure any longer.  MOLECULAR INTERPRETATION OF TEMPERATURE From equation (2) above, we have that; 2 1 𝑝𝑉 = 𝑁 𝑚𝑉 2 3 2 By comparing equation (3) with the equation of state of an ideal gas: 𝑝𝑉 = 𝑁𝐾𝐵 𝑇 We can write; 2 1 𝑁𝐾𝐵 𝑇 = 𝑁 𝑚𝑉 2 3 2 Therefore, 2 1 𝑇= 𝑚𝑉 2 4 3𝐾𝐵 2  The result of this equation shows that the temperature is a direct measure of average translational kinetic energy. Equation (4) implies that temperature is proportional to average kinetic energy.  That is, temperature is a direct measure of average molecular kinetic energy. By rearranging equation (4) above, we get 1 3 𝑚𝑉 2 = 𝐾 𝑇 5 2 2 𝐵 3 This means average kinetic energy per molecule is 𝐾 𝑇 2 𝐵 Use the fact that 𝑉 2 = 3𝑉𝑥2 to obtain 1 1 1 1 this means that, 𝑚𝑉𝑥2 = 𝑚𝑉𝑦2 = 𝑚𝑉𝑧2 = 𝐾 𝑇 2 2 2 2 𝐵  The generalization of this result is known as equipartition of energy which states that each translational degree of freedom contributes an equal amount of energy 1 ( 𝐾𝐵 𝑇 ) to the system. 2 Degree of freedom means an independent way (means) by which a molecule (particle) can acquire energy. The total translational kinetic energy of N molecules of gas is simply N times the average energy per molecule i.e. 1 3 3 𝑘𝑡𝑜𝑡𝑎𝑙 𝑡𝑟𝑎𝑛 = 𝑁 𝑚𝑉 2 = 𝑁𝐾𝐵 𝑇 = 𝑛𝑅𝑇 2 2 2  3𝐾𝐵 𝑇 3𝑅𝑇 𝑉𝑟𝑚𝑠 = 𝑉2 = = 7 𝑚 𝑀  WORKED EXAMPLE A tank used for filling helium balloons has a volume of 0.3m3 and contains 2.0moles of helium gas at 200C. (a) What is the total translational K.E of the gas molecules? (b) What is the average K.E per molecule? SOLUTION n=N/NA, and T =200C = 20 + 273 =293K and R= 8.31Jmol-1K-1 1 3 applying 𝑘𝑡𝑜𝑡𝑎𝑙 𝑡𝑟𝑎𝑛𝑠 = 𝑁 𝑚𝑉 2 = 𝑛𝑅𝑇 2 2  3 𝐾. 𝐸𝑡𝑜𝑡_𝑇𝑟𝑎𝑛𝑠𝑙 = 𝑥 8.31 𝑥 293 2 = 7304.5 = 7.3 𝑥 103 𝐽 We can also apply: 3 𝐾. 𝐸𝑡𝑜𝑡_𝑇𝑟𝑎𝑛𝑠𝑙 = 𝑁𝐾𝐵 𝑇 2 𝐾𝐵 = 1.381 𝑥 10−23 𝐽/𝐾 and 𝑁𝐴 = 6.02 𝑥 1023 𝑚𝑜𝑙 −1 3 (b) average K.E per molecule = 𝐾𝐵 𝑇 2 3 𝑥 1.381 𝑥 10−23 𝑥 293 2 6.07 𝑥 10−21 𝐽  Questions 2) The temperature of the system decreases in the process of (A) Free expansion (B) Adiabatic expansion (C) Isothermal expansion (D) Isothermal compression Ans = B 3) In an Isothermal expansion, the pressure is determined only by what? Ans: Volume 4) 2500 J of heat are added to two moles of an ideal monatomic gas, initially at a temperature of 300 K, while the gas performs 5000 J of work. (a) What is the final temperature of the gas? (b) calculate the final internal energy of the system.  Solution (a) Using equation of first law of thermodynamics, ∆𝐸𝑖𝑛𝑡 = 𝑄 − 𝑊 ∆𝐸𝑖𝑛𝑡 = 2500 − 5000 ∆𝐸𝑖𝑛𝑡 = −2500 3 𝐸𝑖𝑛𝑡 = 𝑛𝑅𝑇 2 3 3 ∆𝐸𝑖𝑛𝑡 = 𝑛𝑅∆𝑇 = 𝑛𝑅(𝑇𝑓 − 𝑇𝑖 ) 2 2 3 −2500 = 𝑥2.0𝑥8.31(𝑇𝑓 − 300) 2 2500𝑥2 𝑇𝑓 = − + 300 3𝑥2.0𝑥8.31 𝑇𝑓 = 200𝐾  3 𝐸𝑖𝑛𝑡_𝑓= 𝑛𝑅𝑇𝑓 2 3 𝐸𝑖𝑛𝑡_𝑓 = 𝑥2𝑥8.31𝑥200 2 𝐸𝑖𝑛𝑡_𝑓 = 4986𝐽 = 5.0𝑥104 𝐽 OR Use, 𝐸𝑖𝑛𝑡_𝑓 = ∆𝐸𝑖𝑛𝑡 + 𝐸𝑖𝑛𝑡_𝑖 3 𝐸𝑖𝑛𝑡_𝑓 = −2500 + 𝑥2𝑥8.31𝑥300 = 5.0𝑥104 𝐽 2  DIY 1) A cylinder contains 8.0 g of Oxygen gas. How much work must done to compress the gas at constant temperature of 100oC un volume is halved? Ans = 1,074 J 2) A gas at a pressure of 1000 Pa and initially occuping a volume 8 m3 expands isobarically. If the work done by the gas in the expansion is 20 kJ, what is the final volume attained by the gas? Ans = 28 m3. 3) 4KJ of heat is given off by a gas when it is compressed from 0.08m3 to 0.05m3 under a pressure of 200KPa. What happens to its internal energy? TO BE CONTINUED………. PHY 103: Basic Principle of Physics II Heat and Thermodynamics Lecture 5 By: Dr. E. Oyeniyi Mr. A. O. Ayoola  KINETIC THEORY OF GASES Contd. Molar Specific Heat of an Ideal Gas We have seen that energy required to raise the temperature of n moles of gas from Ti to Tf depends on the path taken between the initial and final states  Consider the case of an ideal gas taken from one isotherm at T to another isotherm at T+T. In this case Q that takes the gas from T to T+T at constant volume is different from the one that takes it from T to T+T at constant pressure. But But Q = 𝑛𝐶𝑣 T, since n and T are constants. Then the only quantity that can differentiate the two paths is 𝐶𝑣 , the molar specific heat. Kinetic Theory of Gases At a constant volume, Q = 𝑛𝐶𝑣 T 𝐶𝑣 = molar specific heat capacity at constant volume. At constant pressure Q = 𝑛𝐶𝑝 T 𝐶𝑣 = molar specific heat capacity at constant pressure. We shall see later that Q at a constant pressure > Q at constant volume Relationship between CV and CP Consider a monoatomic gas.  If energy is added to the gas at constant volume, from 1st law of thermodynamics (𝐸𝑖𝑛𝑡 = Q – W) we know that all the energy is used to increase the internal energy. In this case all the added energy goes to increase translational kinetic energy of the molecules. 𝑉𝑓 Therefore, Since at constant volume, 𝑊 = 𝑉𝑖 𝑃𝑑𝑉 =0 𝐸𝑖𝑛𝑡 = 𝑄, Using Q = 𝑛𝐶𝑣 T 𝐸𝑖𝑛𝑡 = 𝑛𝐶𝑣 T This equation 𝐸𝑖𝑛𝑡 = 𝑛𝐶𝑣 T is valid for all ideal gas  𝐸𝑖𝑛𝑡 So that 𝐶𝑣 =. 𝑛T In the limit of infinitesimal change, d𝐸𝑖𝑛𝑡 1 d𝐸𝑖𝑛𝑡 𝐶𝑣 = = 𝑛dT 𝑛 dT 3 recall that 𝐸𝑖𝑛𝑡 = 𝑛𝑅𝑇 diffentiating 𝐸𝑖𝑛𝑡 with respect to T, we have 2 3 𝐶𝑣 = 𝑅 2  Now consider the path when the gas is taken from isotherm T to isotherm T+T, at a constant pressure. Energy transferred to the gas in this process is given by Q = 𝑛𝐶𝑝 T The change in internal energy for the process 𝑖→𝑓 ′ , however is equal to that for the process 𝑖→f because internal energy depends only on temperature for an ideal gas and T is the same for both processes. So, 𝐸𝑖𝑛𝑡 = 𝑛𝐶𝑣 T in the two cases. In this path, work PdV is done and this can be obtained using PV=nRT, differentiating this we have, 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑛𝑅𝑑𝑇 𝑉𝑑𝑃 = 0 since, the process is an isobaric, W = 𝑃𝑑𝑉 = 𝑛𝑅𝑑𝑇  From first law equation, 𝐸𝑖𝑛𝑡 = Q – W, we have 𝑛𝐶𝑣 T=𝑛𝐶𝑝 T − 𝑛𝑅𝑑𝑇 𝐶𝑣 dT=𝐶𝑝 dT − 𝑅𝑑𝑇 𝐶𝑝 − 𝐶𝑣 = 𝑅 This is valid for any ideal gas. It predicts that the molar specific heat of an ideal gas at constant pressure is greater than the molar specific heat at constant volume by an amount R. 3 𝐶𝑝 = 𝑅+ 𝑅 2 5 𝐶𝑝 = 𝑅 2 3 5 For a monoatomic gas where 𝐶𝑣 = 𝑅 and 𝐶𝑝 = 𝑅, 2 2 𝐶𝑝 = = 1.67 𝐶𝑣 Ideal Gas PV Diagram for an adiabatic process Adiabatic expansion process, from state a to state b  Kinetic Theory of Gases: Adiabatic Processes for an Ideal Gas contd. As noted before in an adiabatic process, no energy is transferred by heat between a system and its surrounding. In an adiabatic process, pressure and volume are related by the expression 𝑃𝑉  = 𝑐 * Where c is a constant Prove of equation (*) In adiabatic process, recall Q = 0, W = PdV, the change in internal energy will be the same as for isovolumetric process between the same temperatures, d𝐸𝑖𝑛𝑡 = 𝑛𝐶𝑣 dT (because change in internal energy depends only on temperature). Therefore the 1st law equation, (dEint = Q – W), Q=0 d𝐸𝑖𝑛𝑡 = 𝑛𝐶𝑣 dT= − PdV (**) Taking the total differential of PV=nRT , we have PdV+VdP=nRdT (***) nRT But P = , using this in equation (**), we have 𝑉 nRT 𝑛𝐶𝑣 dT = − 𝑑𝑉 (+) 𝑉 RT dV By making dT the subject of the relation (+), dT= − and 𝐶𝑣 𝑉 substituting in equation(***) we have, RT dV PdV+VdP= −nR which can be written as; 𝐶𝑣 𝑉 𝑅 nRT PdV+VdP= − 𝑑𝑉, 𝐶𝑣 𝑉 we now have 𝑅 PdV+VdP= − 𝑃𝑑𝑉 (++) 𝐶𝑣 Recall that R=𝐶𝑝 − 𝐶𝑣 , so equation (++) becomes 𝐶𝑝 − 𝐶𝑣 PdV+VdP= − 𝑃𝑑𝑉 𝐶𝑣 Divide both sides by PV we have 𝑑𝑉 𝑑𝑃 𝐶𝑝 − 𝐶𝑣 𝑑𝑉 + = −( ) 𝑉 𝑃 𝐶𝑣 𝑉 𝑑𝑉 𝑑𝑃 𝑑𝑉 + = 1− 𝑉 𝑃 𝑉 Opening the bracket and rearranging, we arrive at  𝑑𝑉 𝑑𝑃  + =0 𝑉 𝑃 By integrating we get  ln 𝑉 + ln 𝑃 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃𝑉  = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 1.0 i.e.   𝑃𝑖 𝑉𝑖 = 𝑃𝑓 𝑉𝑓 1.1 Which gives the relationship between pressure and volume in an adiabatic process.  Relationship between temperature and volume in adiabatic processes nRT Using PV=nRT so that P = in equation 1.0, we have 𝑉 nRT  𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉 This becomes, 𝑛𝑅𝑇𝑉 −1 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Since n and R are constants, we have 𝑇𝑉 −1 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 1.3 i.e. −1 −1 𝑇𝑖 𝑉𝑖 = 𝑇𝑓 𝑉𝑓 1.4 Which gives the relationship between temperature and volume in adiabatic processes.  Relationship between temperature and volume in adiabatic processes nRT From equation of state of an ideal gas, we have V = 𝑃 substitute this expression in equation (1.0) 𝑛𝑅𝑇  𝑃[ ] = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃 𝑃1− 𝑇  (𝑛𝑅) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Since n and R are constant, we have, 𝑃1− 𝑇  = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 1.5 i.e. 1−  1−  𝑃𝑖 𝑇𝑖 = 𝑃𝑓 𝑇𝑓 1.6 This is the relationship between pressure and temperature in an adiabatic process.  Work done in adiabatic processes Recall that 𝑃𝑉  = 𝑐 so that c P=  𝑉 𝑉𝑓 But 𝑊 = 𝑉𝑖 𝑃𝑑𝑉 𝑉𝑓 𝑉𝑓 𝑉𝑓 c 𝑊=  𝑑𝑉 = 𝑐𝑉 − 𝑑𝑉 = 𝑐 𝑉 − 𝑑𝑉 𝑉𝑖 𝑉 𝑉𝑖 𝑉𝑖 Integrating to have, 𝑉 1− 𝑉𝑓 𝑊 = 𝑐[ ] 1− 𝑉𝑖 c 1−  1− 𝑊= (𝑉𝑓 − 𝑉𝑖 ) 1− 𝑐 = 𝑃𝑉   𝑃𝑖 𝑉𝑖1−  1− 𝑊 = 1− (𝑉𝑓 − 𝑉𝑖 ) 1.7 Equation (1.7) is the formula to calculate work done of an adiabatic process. WORKED EXAMPLES 1. Consider the adiabatic process represented by the curve on the right side. (a.) Determine the pressure in stage 2 in the process (b) calculate the work done in each process (take =2).  SOLUTION P1 = 2 x 105N/m2 P2 = ? V1 = 0.05m3 V2 = 0.1m3 V3 = 0.05m3 𝑝1 𝑉1ϒ = 𝑝2 𝑉2ϒ 𝑝1 𝑉1ϒ 𝑝2 = ϒ 𝑉2 2 𝑥 105 𝑥 0.052 𝑝2 = 0.12 𝑝2 = 5 𝑥 104 𝑁/𝑚2 Stage 1 to 2 is adiabatic process (expansion) 1 𝑊12 = 𝑝1 𝑉1ϒ (𝑉21−ϒ − 𝑉11−ϒ ) 1−ϒ  1 𝑊12 = 𝑥2 𝑥 105 𝑥 0.052 (0.1(1−2) − 0.05(1−2) ) 1−2 1 𝑊12 = − 𝑥 500(10 − 20) 1 𝑊12 = −500(−10) 𝑊12 = 5000𝐽 STAGE 2 to 3 is an isobaric process 𝑊 = 𝑝2 𝑉3 − 𝑉2 𝑊23 = 5 𝑥 104 0.05 − 0.1 𝑊23 = −2500𝐽 STAGE 3 to 1 is an isochoric process 𝑊31 = 0𝐽 2. One mole of an ideal monatomic SOLUTION gas is at an initial temperature of a) For an isovolumetric process, 300 K. The gas undergoes an Q = 𝑛𝐶𝑣 T Q = 𝑛𝐶𝑣 (𝑇𝑓 − 𝑇𝑖 ) isovolumetric process, acquiring 3 500 J of energy by heat. It then n = 1, recall that 𝐶𝑣 = 𝑅, 𝑇𝑖 = 300𝐾 2 3 undergoes an isobaric process, 500 = 1𝑥 𝑥8.31(𝑇𝑓 − 300) 2 losing this same amount of energy by heat. Determine: 500𝑥2 𝑇𝑓 = + 300 1𝑥3𝑥8.31 (a) the new temperature of the 𝑇𝑓 = 40.1 + 300 = 340.1𝐾 gas and b) For isobaric process, Q = 𝑛𝐶𝑝 T (b) the work done on the gas. Q = 𝑛𝐶𝑝 (𝑇𝑓 − 𝑇𝑖 ) 5 Q = −500J, 𝐶𝑝 = 𝑅 2 𝑇𝑖 = 340.1𝐾, n = 1 5 −500 = 1𝑥 𝑥8.31(𝑇𝑓 − 340.1) 2 −500𝑥2 𝑇𝑓 = + 340.1 1𝑥5𝑥8.31 𝑇𝑓 = −24.1 + 340.1 = 316𝐾 Using the first law equation, 𝐸𝑖𝑛𝑡 = Q – W 3 𝐸𝑖𝑛𝑡 = 2 𝑛𝑅𝑇 and since Q absorbed by the system during the first stage (isovolumetric process) has been lost during the second process (isobaric process), consequently, Q = 0, we now have, W= −𝐸𝑖𝑛𝑡 3 W = − 𝑛𝑅𝑇 2 3 W= − 𝑛𝑅(𝑇𝑓 − 𝑇𝑖 ) 2 In this case, 𝑇𝑖 = 300𝐾, 𝑇𝑓 = 316𝐾 3 W = − × 1 × 8.31(316 − 300) 2 W = −199.4𝐽 Question Two moles of an ideal gas ( = 1.40) expands slowly and adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L. (a) What is the final pressure of the gas? (b) What are the initial and final temperatures? (c) Find Q, W, and Eint. TO BE CONTINUED……….AFTER THE TEST……….. ALL THE BEST  PHY 103 Heat and Thermodynamics Energy Transfer Mechanism-Conduction I A metal rod subjected to heat on one end soon become hot (increase in temperature) on the other. Such a transfer of heat energy is by conduction. I It occurs as molecules in the hotter region collide with and transfer energy to those in the adjacent cooler region. I The rate of energy transfer by conduction depends properties of the substance being heated Conduction The thermal energy conduction rate; Q Th − Tc P= = kA (1) ∆t ∆x Q -thermal energy (Joules), P- power in Watts, ∆t - time interval in secs, Th and Tc - temperatures of opposite faces of the slab (Th > Tc ) in 0 C , k - thermal conductivity of the slab (W0 C/m), A - cross-sectional area (m2 ) of the slab. P can be expressed in terms of thermal resistance, R: P = Th −T c ∆x R , where R = kA in KW −1 I Materials with high thermal conductivity values are good thermal conductors. kAg > kCu > kAl > kPb > kglass Conduction through a composite material The figure shows a slab made up of two different materials with different thicknesses and thermal conductivities. For the first material, k2 A(Th − T ) P2 = L2 and for the second, k1 A(T − Tc ) P1 = L1 Assuming a steady-state, thermal energy transfer rate through the two materials in the slab must be equal: k2 A(Th − T ) k1 A(T − Tc ) = L2 L1 Conduction through a composite material Make T the subject of the formula: k1 L2 Tc + k2 L1 Th T = k1 L2 + k2 L1 Substitute T into A(Th − Tc ) P2 = L1 L2 k1 + k2 Generally, if the slab consist of n materials, A(Th − Tc ) P= Pn Lj j=1 kj I < = kL called the