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PRT 140 PHYSICAL CHEMISTRY PROGRAMME INDUSTRIAL CHEMICAL PROCESS SEM 1 2013/2014 THE FIRST LAW OF THERMODYNAMIC BY PN ROZAINI ABDULLAH SCHOOL OF BIOPROSES ENGINEERING ©RBA FTK RY 20 2013 LEARNING OUTCOME: In this chapter you may be able to: ©RBA FTK RY 20 2013 IMPORTANT CONCEPT IN THERMODYNAMIC HEAT ENERGY WORK ©RBA FTK RY 20 2013 ENERGY WORK Expansion Contraction HEAT ©RBA FTK RY 20 2013 INTERNTIONAL SYSTEM OF UNITS (SI) ©RBA FTK RY 20 2013 GAS CONSTANT ©RBA FTK RY 20 2013 WORK Work (w) is defined as the force (F) that produces the movement of an object through a distance (d): Work = force × distance w=Fxd (2.1) Work also has units of J, kJ, cal, kcal etc. 1 J = 1 Nm = 1 kg m2/s2 The two most important types of chemical work are: – the electrical work done by moving charged particles. – the expansion work done as a result of a volume change in a system, particularly from an expanding or contracting gas. This is also known as pressure-volume work, or PV work. PV work occurs when the force is the result of a volume change against an external pressure. ©RBA FTK RY 20 2013 The example of PV work – in the cylinder of an automobile engine The combustion of the gasoline causes gases within the cylinder to expand, pushing the piston outward and ultimately moving the wheel of the car. The relationship between a volume change (∆V) and work (w): W= -P ∆V (2.2) Where P is external pressure The units of PV work are L·atm; 1 L·atm = 101.3 J ©RBA FTK RY 20 2013 ΔV Sign Gas expands positive negative sign (work energy is leaving the system) Gas contracts negative positive sign (work energy is entering the system). No change in volume zero no work done (This occurs in reactions in which there is no change in the number of moles of gas) Please remember sign!!!! ©RBA FTK RY 20 2013 EXPANSION WORK The most common way work is done on a thermodynamic system is by a change in the system’s volume. Expansion Volume increase Volume decrease Contraction ©RBA FTK RY 20 2013 REVERSIBLE P-V WORK Therefore, the work done when the system expands by dV against a pressure pex is : closed system, reversible process dw > 0 In a contraction, the work done on the system is positive Dw<0 In an expansion, the work done the system is negative. (2.3) ©RBA FTK RY 20 2013 REVERSIBLE P-V WORK To obtain the total work done when the volume changes from Vi to Vf , we integrate this expression between the initial and final volume: Closed syst., rev. proc. (2.4) Equation 2.3 & 2.4 Only can be apply to reversible expansion & contraction. These equation is apply to mechanically reversible volume changes. The expansion at constant pressure. ©RBA FTK RY 20 2013 EXPANSION AGAINST CONSTANT PRESSURE Expansion against constant pressure – now suppose that the external pressure is constant throughout expansion. Therefore, if we write the change in volume as ∆V=Vf -Vi w = - Pex ∆V The work done by a gas when it expands against a constant external pressure Pex is equal to the shaded area in this example of an indicator diagram. (2.5) Constant pressure with no phase change ©RBA FTK RY 20 2013 ISOTHERMAL REVERSIBLE EXPANSION Consider the isothermal, reversible expansion of a perfect gas. The expansion is made isothermal by keeping the system in thermal contact with its surroundings (which may be a constanttemperature bath). pV = nRT , p = nRT/V , with V volume at the stage of the expansion The temperature, T is constant in an isothermal expansion so (together with n and R) it may be taken outside the integral. It follows that the work of reversible isothermal expansion of a perfect gas from Vi to Vf at a temperature T is (2.6) ©RBA FTK RY 20 2013 (2.7) • When the volume is greater than the initial volume, as in expansion, the logarithm in above equation is positive and hence w<0. •In this case the system has done work on the surroundings and there is a corresponding reduction in its internal energy. • From the equation also show that more work is done for a given change of volume when temperature is increased: at higher temperature the greater pressure of the confined gas needs a higher opposing pressure to ensure reversibility and the work done is correspondingly greater. ©RBA FTK RY 20 2013 Calculate the work done when 50 g of iron reacts with hydrochloric acid to produce FeCl (aq) and hydrogen in : (a) A closed vessel of fixed volume (b) An open beaker at 25oC. 1) We need to judge the magnitude of the volume and then decide how the process occur. 2) If there is no change in volume, there is no expansion work however the process takes place. 3) A general features of a process in which a condensed phase changes into a gas is that the volume of the former may usually be neglected relative to that of the gas it forms. ©RBA FTK RY 20 2013 (a)The volume cannot change, so no expansion work is done and w = 0. (b) W = -pex ∆V ≈ -pex x (nRT/pex) = - nRT Fe (s) + 2 HCl (aq) FeCl2 (aq) + H2 (g) W ≈ - (50 g / 55.85 g mol-1 ) x 8.3145 J K-1 mol-1 x 298 K ≈ -2.2 kJ ©RBA FTK RY 20 2013 Heat Heat Transfer ©RBA FTK RY 20 2013 TYPES OF HEAT TRANSFER CONDUCTION CONVECTION Most appearance in solid conduction Most appearance in liquid and gas RADIATION Only form that can transmit heat over a vacuum ©RBA FTK RY 20 2013 HEAT A + B A B Heat is a exchange of thermal energy between a system and its surroundings caused by temperature difference. Heat (q) System ∆T q α ∆T The constant of proportionality between q and ∆T is called the heat capacity. ©RBA FTK RY 20 2013 HEAT CAPACITY Heat capacity of the system at that temperature Heat capacity is an extensive property The internal energy of a system increase when its temperature is raised. Heat capacity at constant pressure can be define as: (2.8) Internal Energy Vs. Temperature Heat capacity at constant volume can be define as: Eqs. 2.8 & 2.9 apply only to reversible process (2.9) ©RBA FTK RY 20 2013 HEAT CAPACITY AT CONSTANT VOLUME (2.10) HEAT CAPACITY AT CONSTANT PRESSURE (2.11) For closed system in equilibrium, P-V work only ©RBA FTK RY 20 2013 MOLAR HEAT CAPACITY (2.12) Pure substance, constant pressure (2.13) Pure substance, constant volume ©RBA FTK RY 20 2013 ©RBA FTK RY 20 2013 SPECIFIC HEAT • The specific heat (c, or specific heat capacity, Cs) of an object, is the quantity of heat required to change the temperature of 1 gram of a substance by 1 °C (or K): - Specific heat has units of J / g °C, and is an intensive property, which is independent of the sample size. http://oceanservice.noaa.gov/education/pd/oceans_weather_climate/media/specifi c_heat.swf ©RBA FTK RY 20 2013 ENERGY Energy is define as the ability to do work. Work is done when a force is exerted through a distance. Force through distance; work is done. Energy is measured in Joules (J) or Calories (cal). 1 J = 1 kg m2 s-2 Energy may be converted from one to another, but it is neither created nor destroyed (conversion of energy). In generally, system tend to move from situations of high potential energy (less stable) to situations having lower energy (more stable). ©RBA FTK RY 20 2013 UNIT OF ENERGY • A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1°C. 1 cal = 4.184 J • The nutritional unit Calorie (Cal) is actually a kilocalorie (kcal): 1 Cal = 1000 cal = 1 kCal = 4184 J ©RBA FTK RY 20 2013 INTERNAL ENERGY •The internal energy, U of a system is the sum of the kinetic and potential energies of all the particles that compose the system or the total energy of a system. •Internal energy is the state function, which means that its value depends only the state of the system, not the how the system arrive at the state. •Some examples include energy (and many other thermodynamic terms), pressure, volume, altitude, distance, etc. • An energy change in a system can occur by many different combinations of heat (q) and work (w), but no matter what the combination, ΔU is always the same — the amount of the energy change does not depend on how the change takes place. ∆U = q + w ©RBA FTK RY 20 2013 A system can exchange energy with its surroundings through heat and work: Heat (q) System Work (w) Surroundings According to the first law thermodynamic, the change in the internal energy of the system (∆U) must be the sum of the heat transferred (q) and the work done (w): ∆U = q +w (2.14) Sign of conventions for q, w, and ∆U q (heat) w (work) ∆U (change in internal energy) (+) system gain thermal energy (-) System loses thermal energy (+) work done on the system (-) Work done by the system (+) energy flows into the system (-) Energy flows out of the system ©RBA FTK RY 20 2013 • For an isolated system, with no energy flowing in or out of the system, the internal energy is a constant. – First Law of Thermodynamics (restated): The total internal energy of an isolated system is constant. • It is impossible to completely isolate a reaction from its surroundings, but it is possible to measure the change in the internal energy of the system, ΔU, as energy flows into the system from the surroundings or flows from the system into the surroundings. ∆U = Uf - Ui (2.15) ©RBA FTK RY 20 2013 ENTHALPY •Most reactions are not done in sealed containers: they are carried out in open vessels at constant pressure, with the volume capable of changing freely, especially if the reactants or products of the reaction involve gases. • In these cases, ΔV ≠ 0, and the energy change may be due to both heat transfer and P-V work. • In order to eliminate the contribution from P-V work, a quantity called enthalpy, H is defined as internal energy (U) plus the product of pressure and volume: H = U + PV (2.16) P= pressure, V=volume Because p,V,U are all state function, the enthalpy is a state function too. • The change in enthalpy (ΔH) is: ∆H = qP (2.17) ©RBA FTK RY 20 2013 THE MEASUREMENT OF ENTHALPY CHANGE 1) Calorimeter Monitoring the temperature change that accompanies a physical or chemical change occurring at constant pressure. Example: thermally insulated vessel open to the atmosphere: the heat released in the reaction is monitor by measuring the change in temperature of the content. ©RBA FTK RY 20 2013 THE MEASUREMENT OF ENTHALPY CHANGE 2) Bomb Calorimeter Measuring the internal energy change. ©RBA FTK RY 20 2013 THE MEASUREMENT OF ENTHALPY CHANGE 3) Differential Scanning Calorimeter The most sophisticated way to measure enthalpy changes. Reference pan Sample pan pan Sample Reference material Persistence thermometer Heaters ©RBA FTK RY 20 2013 Example 2: When 2.0 mol CO2 is heated at the constant pressure of 1.25 atm, it’s temperature increase from 250 K to 277K. Given the molar heat capacity of CO2 at constant pressure is 37.11Jmol-1K-1, calculate q, ∆H and ∆U. ©RBA FTK RY 20 2013 Solution: 1) ∆ H = qp = (2.0 mol ) x (37.11 Jmol-1K-1) x (277-250) K = 2.0 x 103 J 2) ∆ U = 2.0 x 103 J – (2.0 mol x 8.3145 J mol-1 K-1) x (277 – 250) K = 1.983 X 103 J ©RBA FTK RY 20 2013 ADIABATIC CHANGES For the reversible adiabatic expansion of a perfect gas, pressure and volume are related by an expansion that depend on the ratio of heat capacities. Step 1: The volume changes and the temperature is constant at its initial value. The system expands at the constant temperature; there is no change in internal energy if the system consist of a perfect gas Step 2: The temperature of the system is reduced at constant volume. (2.18) ©RBA FTK RY 20 2013 ADIABATIC CHANGES q=0 ∆U = q + w = 0 +w ∆U=wad Therefore, by equating the two expansion we have obtained for ∆U we obtained Wad = Cv ∆T (2.19) ©RBA FTK RY 20 2013 Work done during adiabatic expansion of a perfect gas is proportional to the temperature difference between the initial & final state. So a change in internal energy arising from temperature alone is also expected to be proportional to ∆T. The initial and final temperature of a perfect gas that undergoes reversible adiabatic expansion can be calculated : ViTi = VfTf The pressure of a perfect gas that undergoes reversible adiabatic expansion from a volume Vi to a volume Vf is related to its initial pressure by; pfVf= piVi ©RBA FTK RY 20 2013 STANDARD ENTHALPY CHANGES ENTHALPY OF PHYSICAL CHANGE ENTHALPHY OF CHEMICAL CHANGE HESS LAW ©RBA FTK RY 20 2013 ENTHALPY OF CHEMICAL CHANGE CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) ∆rHɵ = -890kJmol-1 rH nH (products ) mH (reactants ) o o f o f ©RBA FTK RY 20 2013 USING ΔHfΘ TO CALCULATE ENTHALPY OF REACTION Calculate the enthalpy of reaction for the following reaction: 2Al(s) + Cr2O3(s) Al2O3(s) + 2 Cr(s) given : ΔHfθ (Cr2O3(s)) = -1669 kJ/mol ΔHfθ (Al2O3(s)) = -1128 kJ/mol ©RBA FTK RY 20 2013 USING ΔHfΘ TO CALCULATE ΔHθrxn 2Al(s) + Cr2O3(s) ΔHfθ = 0 ΔHfθ = -1669 Al2O3(s) + 2 Cr(s) ΔHfθ = -1128 ΔHfθ =0 ΔHrxnθ = Σ of ΔHfθ of products - Σ of ΔHfθ of reactants = [2(0) + (-1128)] – [(-1669) + 2(0)] = 541 kJ/mol ©RBA FTK RY 20 2013 HESS’S LAW The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which reaction may be divided. ©RBA FTK RY 20 2013 EXAMPLE 1: Hydrogen iodide can be prepared from hydrogen and iodine using two separate routes. Route I: Eqn 1: H2(g) + I2(s) → 2HI(g) Route II: eqn 2: I2(s) → I2(g) eqn 3: H2(g) + I2(g) → 2HI(g) ∆H = +52.1 kJ ∆H= +61.3 kJ ∆H= -9.2 kJ ©RBA FTK RY 20 2013 Eqn 1 can be obtained when eqn 2 is added to eqn 3; eqn 1 = eqn 2 + eqn 3 eqn 2: eqn 3: I2(s) → I2(g) H2(g) + I2(g) → 2HI(g) H2(g) + Hence ∆H1 ∆H= +61.3 kJ ∆H= -9.2 kJ I2(s) → 2HI(g) = ∆H2 + ∆H3 = 61.3 – 9.2 = +52.1 kJ The total enthalpy change for the route I is the same as that for route II. ©RBA FTK RY 20 2013 EXAMPLE 2: ΔH for formation of SO3 cannot be obtained directly but the enthalpy of these reactions are known: S(s) O 2 (g ) SO 2 (g ); H -297 kJ o 2SO 3 (g ) 2SO 2 (g ) O 2 (g ); H 198 kJ o The above data can be used to obtain the enthalpy change for the formation of SO3 according to the following reaction? 2S(s ) 3O 2 (g ) 2SO 3 (g ); H ? o ©RBA FTK RY 20 2013 The third equation can be obtained by multiplying the first equation by 2 and added to the reverse of the second equation, they will sum together to become the third. 2S(s) 2O 2 (g ) 2SO 2 (g ); H (-297 kJ) (2) o 2SO 2 (g ) O 2 (g ) 2SO 3 (g ); H o (198 kJ) (-1) 2S(s ) 3O 2 (g ) 2SO 3 (g ); H -792 kJ o ©RBA FTK RY 20 2013 EXERCISE 1: Determine the heat of reaction; Fe2O3(s) + FeO(s) Fe3O4(s) Using the information below: i) 2Fe(s) + O2(g) 2FeO(s) ∆Ho = -554.0 kJ ii) 4Fe(s) + 3O2(g) 2Fe2O3(s) ∆Ho = -1648.8 kJ iii) 3Fe(s) + 2O2(g) Fe3O4(s) ∆Ho = -1118.4 kJ (Answer : -22.0 kJ) ©RBA FTK RY 20 2013 ENTHALPY DIAGRAM ILLUSTRATING HESS’S LAW ©RBA FTK RY 20 2013 THE JOULE & JOULE – THOMSON EXPERIMENT What is Joule Thomson Experiment? The objective of this experiment is to quantitatively measure the nonideality of gases using the Joule-Thomson coefficient and relating it to the coefficients of equations for non-ideality and the Lennard-Jones potential. For an ideal gas, the internal energy is only a function of the absolute temperature so in an isothermal process ΔE = 0. The same is true for the enthalpy for such a process: ΔH = 0. Thus: H H 0 V T P T These are non-zero for a non-ideal gas. ©RBA FTK RY 20 2013 For the perfect gas, the internal energy is independent of the volume (at constant temperature). If repulsion is dominant, the internal energy decrease as the gas expand. If attraction is dominant in a real gas, the internal energy increase with volume because the molecules become farther apart on average. ©RBA FTK RY 20 2013 The internal pressure, πT is the slope of U with respect to V with the temperature T held constant. ©RBA FTK RY 20 2013 He thought that he can measure the πT by observing the change in temperature of a gas when it is allowed to expand in a vacuum. A schematic diagram of the apparatus used by Joule in an attempt to measure the change in internal energy when a gas expand isothermally. The heat absorbed by the gas is proportional to the change in temperature of the bath. Water bath ©RBA FTK RY 20 2013 The thermodynamic implication as follow: 1.No work was done in the expansion into a vacuum, so w=0. 2.No energy entered of left the system (the gas) as heat because the temperature of the bath did not change, so q = 0. Consequently, within the accuracy of the experiment, ∆U=0. Joule concluded that: U does not change when a gas expand isothermally and therefore that =0. The heat capacity of the apparatus was so large that the temperature change that gases do in fact cause was too small to measure. Nevertheless, from his experiment Joule had extracted an essential limiting property of a gas, a property of a perfect gas, without a small deviations characteristic of a real gas. ©RBA FTK RY 20 2013 OBSERVATION OF THE JOULE-THOMPSON EFFECT The analysis of the Joule-Thompson coefficient is central to the technological problems associate with the liquefaction of gases. We need to be able to interpret its physical and measured it. ©RBA FTK RY 20 2013 The pistons represent the upstream and downstream gases, which maintain constant pressure either side of throttle. The transition from the top to bottom diagram which represents the passage of a given amount of gas through the throttle, occurs without change of enthalpy. ©RBA FTK RY 20 2013 The gas on the left is compressed isothermally by the upstream gas acting as piston. The work done on the gas is: The gas expands isothermally on the right barrier against the pressure pf provided the downstream gas acting like piston to driven out. The volume changes from 0 to Vf, so the work done on the gas at this stage: The total work done on this stage: ©RBA FTK RY 20 2013 It follows the change of internal energy of the gas as it moves adiabatically from one side of the barrier to the other is Reorganization of this expression gives: or Therefore, the expansion occurs without change of enthalpy ©RBA FTK RY 20 2013 The partial derivative T P H is known as the Joule-Thomson coefficient, JT. ©RBA FTK RY 20 2013 +ve sign → dT is –ve when dp is –ve (in which case the gas cool on expansion) A gas typically has two inversion temperatures, one at high temperature and the other at low. ©RBA FTK RY 20 2013 Renungan: “Termodinamik merupakan satu subjek yang aneh. Kali pertama anda mempelajarinya, anda tidak faham langsung. Kali kedua anda mempelajarinya, anda fikir anda memahaminya kecuali satu dua perkara. Kali ketiga anda mempelajarinya, anda tahu bahawa anda tidak memahaminya, akan tetapi anda sudah terbiasa dengan subjek itu maka subjek itu tidak akan menjadi masalah lagi kepada anda” Arnorld Sommerfeld (1868-1951) ©RBA FTK RY 20 2013