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Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 3. WORK, INTERNAL ENERGY, HEAT & THE FIRST LAW OF THERMODYNAMICS. Work dW PdV 1st LAW Clausius (1850) Heat dQ TdS dU dQ dW Internal Energy U 3 3 PV NK BT 2 2 3.1 Work. One of the most important ideas in thermodynamics is the concept of work, of a system’s capability to perform mechanical work in a system transition between equilibrium states. Continuing with use of the fluid/gas system as the paradigm to stand for general systems the concept of work can be explored with the aid, once again, of the piston now acting as a 38 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan moving wall with which the contained gas particles may exchange energy, something that has already been looked at. Consider the piston system sketched below. P, V A -F F= PA x dx x The force acting on the piston due to the pressure P of the gas is F = PA And this acts to the right and in mechanical equilibrium (the piston not moving) an equal but opposite external force –F must act to the left to maintain the position of the piston. If the position of the piston changes under the external force exerted by the environment from x to x - dx, thus changing the volume of the gas from Vi xA to Vf x dx A with dV Vf Vi Adx The infinitesimal work done on the gas by the external force is dW Fdx PAdx PdV The infinitesimal work done on a fluid during an infinitesimal change of volume, dV is then; đW =-PdV Note that because in this case (compression) dV is negative the value of –PdV is positive. 39 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan This is an example of a very important sign convention; work done on a system is positive To compress the gas work was done ON it by the surroundings. Reversible processes It is important to note that in defining the work in this way it is essential to ensure that the pressure remains well defined at all times and this requires that the process must be performed quasi statically, that is, it must proceed such that at any instant the system is in an equilibrium state. This further means that at any instant the change could be reversed and the system returned to its original state (along with the surroundings in as much as they are affected by the system). Such a quasi-static process would need to be performed infinitely slowly and is more usually called a reversible process. To be reversible there cannot be any loss or dissipation in the system and therefore the piston must be frictionless. In actuality a reversible process can only be approximately achieved but the concept is extremely important in enabling the theoretical description of the thermodynamic process. In the equation, đW = -PdV , where an infinitesimal amount of work, đW has been performed (in this case on the system) the bar through the đ indicates the fact that the amount of work for a given volume change depends on the path chosen ie how the change dV is made. The change from an initial equilibrium state with volume Vi to a final equilibrium state with volume Vf can be made in an infinity of ways and any one of them can be performed quasi-statically and reversibly but requiring different amounts of work to be performed. NB The basic difference between đW and dV is simply stated as; dV represents a change of state from one equilibrium state to another and its value is path independent. whereas đW represents a process and its value is path dependent dV is said to be a perfect differential whereas đW is an imperfect differential. 40 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan It is generally the case that an infinitesimal change in a function of state (U, T, V, P…) is represented by a perfect differential while an imperfect differential represents how that change was carried out (a process or flow). NB W is NOT a function of state! If a large change is made in the volume it is necessary to find the net effect through integration and care has to be taken in evaluating any integral in accounting for how the large changes were made, ie. the path followed. The volume change in going from A to B is easily evaluated V Vf Vi Vf dV Vi The work performed has to be more carefully evaluated because in the integral Vf Pd V W Vi In the integrand the value of P will also change and its value at each value of V (or equilibrium state) will depend on the path taken. Furthermore there is no “initial work” Wi or “final work” Wf the concept is meaningless as it is a process that is being considered. There will be an amount of work done during the process represented by the integral but caution is required here as depending on the path chosen to get from i to f the value of P inside the integral will change in different ways and needs to be included as part of the integral. A P A P PB=PA/2 C1 C3 C2 C3 O B V VA 41 VB=2VA Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan The PV diagram shown above, includes the start and end point of a process taking an ideal gas from equilibrium point A (PA , VA) to the equilibrium point B (PA/2, 2VA). There are three processes (pathways) indicated and each is examined in turn to find the work done in going from A to B. Example 1. C1 shows the direct route from A to B (on the diagram anyway). To find the work the first requirement is to know how P can be written in terms of V in order to evaluate the integral of PdV. In general the equation of the straight line joining (x1 , y1 ) to (x2, y2) may be written as y = y1 + [(y2 - y1) / (x2 - x1)]·(x - x1) Using this the equation of the line C1 is PA P A P P 3 2 P PA (V VA ) PA A (V VA ) PA A V 2V V 2VA 2 2VA A A Therefore the integral may be written VB W VB 3 PdV PA 2 VA VA VB P dV A 2VA P 3 VdV PA VB VA A 2 2VA VA V 2 V 2 B A 2 Knowing VB = 2VA this can be tidied up to obtain 3 3 W PAVA nRT 4 4 Example 2 C2 shows the isotherm. The equation of state for the ideal gas may be used to find P as a function of V. As the temperature is constant along C2, The integral is then 42 P nRT V Thermal & Kinetic Physics: Lecture Notes VB W © Kevin Donovan VB PdV nRT VA dV V V nRT lnV VB nRT ln B V A V A VA W nRT ln 2 Example 3 The path C3 is in fact an isochore (V = VA = constant) followed by an isobar (P = PA/2 = constant). This can be broken into two integrals and VA W VA VB P PdV A 2 dV VA The first integral is zero as V = 0 and the second is easily evaluated P P V nRT W A VB VA A A 2 2 2 In summary C1 W 1.5nRT C2 W nRT ln 2 0.69nRT C3 W 0.5nRT Going from A to B necessitated the system doing work on the surroundings (all processes have negative values for the work). This was of necessity the case as the gas expanded! The work expended by the system was different for each path. Being the least for C3 and the greatest for C1. 43 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan VB Another view of the work can be seen from consideration of the integral PdV and the VA PV diagram. Isochore P Isotherm Isobar V VB The integral PdV is seen to be the negative of the area under the line of the path VA between the path and the P = 0 volume axis and looking at the above pressure volume indicator graph it is easy to see the results obtained earlier for C1 and C3 by geometric area evaluation. This is not so easily done for path C 2 the isotherm. To get the sign correct it is necessary to take note of the direction in which the path is traversed. If going from left to right the sign convention, positive for work done on a system, the area is taken as negative. Proceeding from right to left or high to low volumes the area is positive as the gas is compressed and work is therefore done on the system. Frequently in this module cyclic processes will be of interest and importance and these may be represented on the PV diagram and the work carried out by or done on the system can be calculated over a cycle where the system at an equilibrium state A is caused to pass through a series of points equilibrium points before returning to A. 44 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan Example 4. Isotherm P 3 Isochore Isobar 2 2P1 P1 1 V1/2 V V1 An ideal gas is compressed isothermally from V1 to V1/2 on the path 1 2. It is then expanded isobarically at 2P1 back to its original volume on path 2 3 before finally being returned isochorically to its original equilibrium state on the path 3 1 What is the work done by/on the gas?. We need to evaluate and add three integrals, one for each path. Path 1 2 is an integral evaluated in an earlier example V2 W1 2 V1 V2 PdV nRT dV V V nRT lnV V2 nRT ln 1 nRT ln 2 1 V V2 V1 Path 2 3 is isobaric and P is constant (again this was done earlier) V W2 3 P2 V1 V2 P2 1 P1V1 nRT 2 And path 3 1 is isochoric, V = 0 and therefore no work is done 45 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan W3 1 0 The total work done in the closed cycle is therefore Wtot W12 W23 W3 1 nRT ln 2 nRT 0 Wtot nRT (0.69 1) 0.31nRT Note the signs of W in each of the arms of the cycle recalling that positive means work is done on the gas. The net work is negative meaning that overall in the cycle work is carried out by the gas. If the cycle is now reversed; W13 0 V W3 2 P2 V2 V1 P2 1 P1V1 nRT 2 V1 W2 1 V1 PdV nRT V2 dV V V nRT lnV V1 nRT ln 1 nRT ln 2 2 V V2 V2 This time the net work is WTot W13 W3 2 W21 0 nRT 0.69nRT WTot nRT 1 0.69 0.31nRT and is positive, ie. the system has work done on it by the surroundings. In fact, when the work extracted in one direction is equal to the work done in the other direction. The cycle is said to be reversible. 46 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 3.2 Internal Energy. Internal energy changes during the cycle may now be dealt with having looked at the work. Previously it was found for an ideal monatomic gas (no rotation or vibration) that the internal energy was given by the expression U 3 PV 2 This may be used to find the changes in U as the cycle is traversed, first in the clockwise sense from 1 2 3 U1 2 U 2 U1 3 P2V2 P1V1 3 nRT2 nRT1 0 2 2 Because 1 and 2 lie on an isotherm and T1 = T2 For the next stage U 2 3 U 3 U 2 3 P2V1 P2V2 3 2P1V1 P1V1 3 2nRT1 nRT1 3 nRT1 2 2 2 2 And for the final stage U 3 1 U1 U 3 3 P1V1 P2V1 3 P1V1 2P1V1 3 nRT1 2 2 2 The net change in internal energy around the cycle is then U1 2 U 2 3 U 3 1 0 3 3 nRT1 nRT1 0 2 2 In performing this analysis two important things have been learned about the changes in internal energy after undergoing processes of a certain type; i) Around a closed loop there is no change in internal energy. This is because there is no net temperature change and therefore the molecules cannot be going slower or faster otherwise continuous looping would lead to infinite or zero velocities. More succinctly, there is no change in U because it is a STATE VARIABLE and its 47 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan value is representative of the particular equilibrium state. Unlike W which is not a state variable and is not therefore associated with any particular equilibrium state but rather with the process of going from one state to another. ii) Also note that equilibrium states on an isotherm all have the same internal energy and U12 0 along the isotherm. It is obviously true that U is zero if the loop is traversed in the opposite sense (try it) for the same reason, U is a state variable. In fact it is possible to be completely general, whatever the closed loop from A back to A there will be no change in U because it is a state variable. P CU 2 1 CL V V1 The above arbitrary closed cycle shows the general situation V2 U dU 0 C For W we can find the work done after one cycle by finding the area of the upper and lower curve and subtracting the lower curve, work done by the system, (going from left to right with volume increasing) from the upper curve, work done on the system, (going from 48 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan right to left with volume decreasing) to obtain the net work done on the system. It is the area enclosed by the circuit. If the circuit is traversed in the other sense then the upper and lower curves would have their signs altered and work done on the system would be minus the area enclosed. W = +Area Enclosed (counter clockwise) W = -Area Enclosed (clockwise) Around a closed cycle; i) The work done depends on direction traversed and on the particular cycle chosen. It is path dependent and dW is an imperfect differential. ii) The internal energy is unaltered independent of the closed cycle. U is a state variable and dU a perfect differential. 49 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 3.3 Heat. In the eighteenth century the popular view of heat was that it had the nature of a fluid and would flow from hot bodies to cold bodies. This fluid, caloric, could neither be created or destroyed and every body contained a certain amount of caloric depending on its temperature. This view allowed explanations of many phenomena and was almost universal although a few notable scientists including Boyle and Hooke held a contrary position that heat was an expression of some form of motion, the so called kinetic theory of heat. Indeed the phrase heat flow persists to this day and a Calorie is an alternative unit of energy much favoured still by chemists. Benjamin Thompson aka Count Rumford (1753 – 1814), became the most well known opponent to the caloric viewpoint due to his experiments on frictional heat developed during the boring of cannons in the Munich arsenal which was published in 1798. In this work he used a blunt borer and with the heat produced was able to boil large quantities of water implying, from the caloric viewpoint, a limitless supply of the fluid. These experiments along with other later experiments finished off the idea of heat as an actual physical fluid. It was the careful quantitative work of Joule that finally put paid to the idea of caloric and enabled the kinetic theory of heat to be placed on its pedestal. James Prescott Joule (1818 – 1889) Demonstrated the relationship between heat and mechanical work. In so doing the foundations of the law of conservation of energy were finally established almost 150 years after Leibnitz first proposed it in his theory of vis-viva. Adiabatic Work In 1843 Joule demonstrated in a classic set of carefully carried through experiments that if he enclosed a system within adiabatic walls (insulating walls allowing no heat transfer into or out of the system) then by doing a known amount of work on the system, that starting from a particular equilibrium state the final end state depended only on the work done and not on how the system was taken from the initial to the final state. eg. stirring with paddles or heating with a known current and resistance 50 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan i i He showed that the temperature rise T U = W the adiabatic work done on the system. In fact this is the empirical definition of the internal energy WAdiabatic Ufinal Uinitial 3.4 Heat and the First Law. The case of a thermally/adiabatically isolated system is a special one and the observations noted previously have important implications when followed through. When the system is not adiabatically isolated then it is the case that U W This is of great consequence since at first sight it implies non-conservation of energy! By now the conservation of energy had taken its place in the pantheon of physical principles as an inviolable requirement and this position could only be reversed at great cost to the existing understanding of the universe. It was now an item of faith that the amount of energy in the universe was a constant and unchanging, only the form in which it was 51 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan present altered, it’s actual quantity was God given, from the start of the universe and could not be added to or subtracted from by any mechanism. In this respect it had been a late comer to the group of conserved quantities which included momentum and angular momentum (Newton), again fixed quantities which were present at the creation and continued unaltered, and mass (Lavoisier). To maintain this inviolability of energy conservation in the face of the evidence of careful thermodynamic measurements where heat flows were allowed to occur it was recognised that in thermodynamic systems the concept of heat flow, Q was essential and that Q was just another form of energy. Heat flow like work however was a form of energy in transit. No body could be said to hold a quantity of work or heat, the two only existed as transient forms while energy changes from one form to another. Thus the First Law of Thermodynamics was born. The change in internal energy after some process is carried through is given by U Q W W is the work done on the system in that process and Q is the heat transferred to the system by that process. A simple and early statement of the first law is that the change in the internal energy of a closed system is equal to the amount of heat absorbed by the system, minus the amount of work done by the system on its surroundings. Given the sign convention that we are using this statement corresponds to the preceding formulation of the first law in the form of an equation. The first comprehensive statement of this most important of laws was made by Clausius in 1850 some seven years after Joule’s experiments. Note that Q like W is not a state variable and should be formally written with the strikethrough across the or d. dQ and dW are both path dependent and imperfect differentials unlike dV, dP , dT and dU. The sign convention must be upheld when using the first law. i) Earlier the convention that W was positive was established when the system has work done on it and negative when it is the system that does work. 52 Thermal & Kinetic Physics: Lecture Notes ii) © Kevin Donovan Similarly the convention for heat flow is that Q is positive when it flows into the system and negative when it flows from the system. The first law may also be written in infinitesimal form as dU dQ dW In the infinitesimal form the first law holds for all processes reversible and irreversible. When the process is reversible the expression(s) for dW may be used to cast the first law in the form Fluid dU dQ PdV (Reversible) The heat flow must be path dependent as the work is path dependent and the first law must hold for any path. B Q dQ A Returning to the previous example to look at heat flow around a closed cycle using the first law. 53 Thermal & Kinetic Physics: Lecture Notes Isobar 2 P Kevin Donovan 3 Isochore 2P1 © Isotherm P1 1 V1/2 V V1 V Q1 2 U1 2 W1 2 0 nRT1 ln 1 nRT1 ln 2 V2 This is negative on going from 1 2 and means that heat flows out from the system. Q2 3 U 2 3 W2 3 3 5 5 P1V1 ( P1V1) P1V1 nRT1 2 2 2 This is positive and on going from 2 3 heat flows into the gas. 3 3 Q3 1 U 3 1 W3 1 P1V1 0 nRT1 2 2 This is negative and therefore on going from 3 1 heat flows out of the gas. The net heat flow around the cycle is Qnet Q12 Q2 3 Q3 1 nRT ln 2 54 5 3 nRT1 nRT1 nRT1(1 ln 2) 2 2 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan This is positive and therefore there is net heat flow into the gas if this closed cycle is traversed. Previously the net work was found to be; Wnet nRT (ln 2 1) 0 The net work was done by the gas and is equal to the net flow of heat into the gas. They balance one another leaving U = 0 as required after traversing a closed cycle and in order to satisfy the first law of thermodynamics. A Test for Perfect Differentials The differences between perfect and imperfect differentials have been mentioned frequently with respect to state variables such as internal energy which are examples of the former and work and heat, which are associated with particular processes therefore depending on the path taken from initial to final state and are examples of the latter. An important test for a perfect differential may now be established as follows; Suppose z = z(x, y) is a function of state which depends on two independent variables, x and y. Then z z dz dx dy x y y x And hence dz a( x, y )dx b( x, y )dy For dz to be a perfect differential requires by definition that 2z 2z yx xy implying a( x, y ) b( x, y ) y x 55 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan Given a differential in the form; dz a( x, y )dx b( x, y )dy Such that a b y x Then the differential is perfect and can be integrated to give z independent of the path, and z is a function of state. If a b y x Then the differential is imperfect and the integral of dz depends on the path. z is not a function of state. Example. U An ideal monatomic gas, i) 3 PV 2 3 U U dU dP dV VdP PdV 2 P V V P 3 a V 2 b 3 P 2 a 3 b 3 V 2 P 2 U is a perfect differential. ii) dW PdV First law dQ dU dW 3 VdP PdV PdV 3 VdP 5 PdV 2 2 2 56 Thermal & Kinetic Physics: Lecture Notes 3 a V 2 b © Kevin Donovan 5 P 2 a 3 b 5 V 2 P 2 dQ is not a perfect differential Applications of the First Law may now be examined. i) Heat Capacities. There are many simple applications of the first law for example it may be used to define heat capacity. If an amount of heat Q is introduced to a finite system there will be an increase in temperature, T and the heat capacity is defined as QRe v dQRe v dT T 0 T C limit (Reversible change) It is defined when the change in temperature is a reversible one, that is to say at each intermediate temperature the new state is an equilibrium state obeying the state equation. In this case the input or output of heat will raise or lower the temperature by the same amount. The specific/molar heat capacity is the heat capacity per unit mass/mole, c 1 dQRe v C m dT m cm 1 dQRe v C n dT n The value of C depends on the type of process in which the heat was transferred. a) Constant Volume Heat capacity QRe v T 0 T V const CV limit 57 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan From the first law U Q PdV Q (V = const, dV = 0) This allows for a measure of the specific heat capacity for a gas at constant volume that involves only state functions; U 3 U nR 2 T 0 T V const T V CV limit b) Constant Pressure Heat Capacity Q T 0 T P const CP limit From the first law U Q PdV Q (PV ) so for the heat Q U (PV ) (U PV ) A new state function, Enthalpy, H is defined H U PV The infinitesimal change in H may be found as follows; dH = dU + PdV + VdP And for isobaric processes where dP = 0 the heat change is given by; dQ dU PdV dH Therefore a useful definition of the specific heat capacity for a gas at constant pressure that involves only state functions is 58 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan Q H H CP limit Re v limit T T 0 T P const T P U H U PV 3 3 PV nRT 2 2 3 5 5 PV PV PV nRT 2 2 2 3 U CV nR T V 2 5 H CP nR T P 2 And the relationship between CP and CV is then CP CV nR It should be noted that all of the above only applies to a monatomic gas, (no potential, rotational or vibrational contributions to U) where U 3 3 PV nRT . 2 2 Taking a closer look at ENTHALPY the newly defined state function H U PV 3 5 5 PV PV PV nRT 2 2 2 dH dU PdV VdP In situations where the pressure is constant, and these situations arise frequently in liquids or solutions which are often at atmospheric pressure; dH dU PdV dQ 59 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan And the natural variables of the state function, H, in an isobaric situation are U (or T ) and V where by natural variables it is meant those variables which when held constant lead to no further change occurring in that state function ie. for the state function enthalpy if U and V are constant, dU = dV = 0 and the enthalpy is constant. A further example is for the internal energy of an ideal gas where U 3 3 PV and dU PdV VdP . 2 2 Clearly it is V and P that are the natural variables of U as when dV = dP =0 then dU = 0. If there is heat flow during a chemical reaction it is equal to the enthalpy of that reaction and the enthalpy is widely used in physical chemistry and is tabulated for common reactions. In the definition of heat capacity the need for a reversible heat flow process was stressed. This is similar to reversible work previously encountered as it means the heat flow is induced by infinitesimal temperature differences rather than by finite difference in temperature, ie. it occurs quasi statically. An example of irreversible heat flow is the dropping of a hot metal (metals conduct heat well and therefore the temperature of the metal will be uniform from inside to out and remain so) into a bowl of water at room temperature. The metal will rapidly transfer heat to the water in an irreversible process passing through non-equilibrium intermediate states. There is no way to take the metal back to its original (hot) state by infinitesimal changes to the environment. An example of reversible heat flow is to place a finite system eg. the bowl of water in contact with a heat reservoir at a slightly different temperature (higher or lower). This could raise/lower the temperature of the water by dT before iterating with a reservoir at a slightly higher/lower temperature than the previous one. In this way the water could be taken from one equilibrium state to another at a higher/lower temperature in incremental steps, reversibly. At this point the HEAT RESERVOIR which is a very important and useful concept in thermodynamics, may be defined. Recall that, as stated earlier, it is wrong to think of an object as containing heat, where heat like work is the result of a process and is correctly termed heat flow. The heat reservoir is a body of such large heat capacity that a flow of heat Q to or from the system of interest, whilst changing the temperature of that system 60 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan will have no effect in changing the temperature of the reservoir. Using a series of heat reservoirs at different temperatures heat may be added to or removed from a system reversibly at any specified temperature. ii) Adiabatic Process for an Ideal Gas. Beginning with the analysis of an ideal gas undergoing a reversible adiabatic change (expansion/compression) and starting with the first law, dU dQ dW dQ PdV In an adiabatic process dQ = 0 by definition. whence dU PdV For an ideal monatomic gas it has already been established that U And therefore dU 3 PdV VdP PdV 2 Re-arranging 5 3 PdV VdP 0 2 2 and 5 dV 3 dP 2 V 2 P 5 dV dP 3 V P By integration 5 lnV ln P const 3 Rewritten using n ln x ln( x n ) 5 ln P lnV 3 const Combining the logs 61 3 PV 2 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 5 PV 3 PV PV V 1 nRTV 1 const We have then demonstrated that at all points on the adiabatic PV curve (where in this case T does vary) are linked through the relationship; 5 PV 3 PV const and TV 1 const that remain true. NB the constants in the two equations are not the same! This problem could have been approached in a different fashion; At constant volume U U(T ) Meaning that U dU dT CV dT PdV T V CV dT CV (1st Law with dQ = 0) nRT dV V dT dV nR T V Integrating and using the relationship CP CV nR CV lnT nR lnV const (CV CP ) lnV const C lnT P 1 lnV const CV 62 Thermal & Kinetic Physics: Lecture Notes CP TV CV 1 © Kevin Donovan TV 1 const With the important exponent given by CP CV From a previous result; 3 5 U H CV nR and CP nR T V 2 T P 2 The exponent CP 5 CV 3 5 is thus placed on a more physical basis. It should be noted that the 3 equation of state U 3 PV is only true for monatomic gases and therefore the exponent, 2 , will be different for non-monatomic gases. 3.5 Diatomic Gases. All of the foregoing was related to monatomic gases where the relations U 3 3 PV nRT were found by considering the translational momentum exchange of a 2 2 molecule at a container wall and how this contributed to the pressure. Turning now to the case of diatomic gases in order to be able to compare and contrast their behaviour with that of the monatomic gas. Previously it was shown that for the monatomic gas, where there is only translational energy to be considered; U N Note that 1 mv x2 2 1 1 2 1 1 mv N mv x2 mv y2 mv z2 2 2 2 2 3 PV 2 represents the energy available in one of the 3 degrees of freedom available to a spherically symmetric atom which comprises a monatomic gas. 63 Thermal & Kinetic Physics: Lecture Notes 1 1 1 U N mv x2 mv y2 mv z2 2 2 2 n N NA and © Kevin Donovan 3 3 PV nRT 2 2 R kB NA 1 1 1 U N mv x2 mv y2 mv z2 2 2 2 3 Nk BT 2 We note here without formal proof the equipartition theorem, that it is the case that for each degree of freedom available to the atom there is 1 k BT average thermal energy. Ie. 2 1 1 mv x2 k BT 2 2 1 1 With s degrees of freedom U s NkBT s nRT 2 2 With this reminder it is now possible to ask what happens if more degrees of freedom are allowed or in other words if more independent or orthogonal ways in which a constituent part of a system may hold energy are included. Examples of this are diatomic gases in two special circumstances; i) A gas composed of rigid diatomic molecules where by rigid it is meant that there is a constant internuclear separation. z y x 64 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan The molecule shown as a dumbbell is now able to rotate about the x, y and z axes with consequent rotational kinetic energies; 1 I z 2 2 1 I x 2 2 1 I y 2 2 respectively where I is the moment of inertia. With the molecule aligned along the z axis and the atoms treated as point like objects the moment of inertia Iz << Ix = Iy and there are effectively two new degrees of freedom added to each molecule. For each of these 1 1 I x 2 k BT 2 2 and 1 1 I y 2 k BT 2 2 The internal energy of a collection of N of these rigid diatomic molecules is U N ETrans E Rot N 1 1 1 1 1 mv x2 mv y2 mv z2 I x 2 I y 2 2 2 2 2 2 The number of degrees of freedom is now s = 5 and the equations obtained previously for a monatomic gas where s = 3 are now modified; U s 1 1 1 5 NkBT s nRT s PV PV 2 2 2 2 CV is now given by; 1 5 U CV s nR nR 2 2 T V And for H = U + PV H U PV 5 7 7 PV PV PV nRT 2 2 2 So for CP s2 7 H CP nR nR 2 2 T P and 65 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan CP s 2 7 CV s 5 For an adiabatic process T and V are always related by CP TV ii) CV 1 2 TV 5 TV 1 const A gas composed of vibrating diatomic molecules where the condition of rigidity were to be relaxed and the diatomic molecule was allowed to vibrate. z y x The vibration occurs along the axis joining the two atoms and this motion has potential energy and kinetic energy associated with it, adding a further two degrees of freedom and a further k BT per molecule to the internal energy of the gas which is now U 7 7 7 Nk BT nRT PV 2 2 2 The number of degrees of freedom is now s = 7 And the changes to CV , H and CP that follow this change in s and in internal energy are now; 1 7 U CV s nR nR 2 2 T V and 66 Thermal & Kinetic Physics: Lecture Notes H U PV © Kevin Donovan 7 s2 9 PV PV PV nRT 2 2 2 with ` s2 9 H CP nR nR 2 2 T P and CP s 2 9 CV s 7 and finally for an adiabatic process on a gas of non rigid diatomic molecules CP TV CV 1 2 TV 7 TV 1 const 67 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 3.5 Heat capacity of classical ideal gases Equation of state PV = nRT Monatomic Gas 3 nRT 2 U CV H 5 nRT 2 U 3 nR 2 H 5 nR 2 U 7 nRT 2 CV 7 nRT 2 H CP CP 5 CV 3 Vibrating Diatomic Gas 5 nR 2 CV 5 nRT 2 CP Rigid Diatomic Gas 7 nR 2 9 nRT 2 CP CP 7 CV 5 7 nR 2 9 nR 2 CP 9 CV 7 Internal energy due to Internal energy due to Internal energy due to 3 translational DoF . 3 translational DoF 3 translational DoF s=3 plus 2 rotational DoF. plus 2 Rotational DoF s=5 plus 2 Vibrational DoF. s=7 For more complex gases there may be even more degrees of freedom and the measurement of CP CV may provide valuable insights into the make up of the gas molecules. It should be remarked upon here that classically all is fine with the foregoing and there are nice relationships between specific heat capacities and the numbers of degrees of freedom. However if the behaviour of the heat capacity of an actual diatomic gas is examined it would be found that the heat capacity would depend on temperature, a fact that is not reflected in the preceding classical model. To explain this quantum mechanics must be appealed to where it is known that the energies of rotation and vibration are in fact quantised; For vibrational states; 1 EVib n 2 68 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan Where is the frequency of oscillation of the simple harmonic oscillator With the spring constant (related to bond strength) and m1m2 is the reduced m1 m2 mass and n the vibrational quantum number or occupancy of the vibrational level. The quantised vibration behaves as a boson and therefore it is not restricted by Pauli’s exclusion principle and any state (vibrational frequency) may have multiple occupancy, the quantum equivalent of an increased classical amplitude of oscillation. For rotational states; E Rot J (J 1) 2 2I where J is the rotational quantum number and I the moment of inertia. For both of these quantised energies, if the unit of quantisation is larger than 1 k BT , the 2 process is effectively unavailable to the molecule and the degree of freedom therefore unavailable also. This means that as temperature is lowered the extra processes are “frozen” out and the internal energy is dropped in an almost stepwise fashion causing the specific heats to drop “suddenly”. A non rigid diatomic molecule will first lose the two degrees of freedom due to vibration as temperature is lowered before losing the two rotational degrees of freedom. There is always the zero point vibrational energy 1 , but 2 this does not contribute to the heat capacity as it doesn’t contribute to a rise in internal energy as heat is introduced into the system. ie. it is always there irrespective. 69 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 3.6 Adiabatic Processes. Adiabatic processes, as seen, are described by the equation PV const P const P nRT V V cf Isothermal processes PV nRT const For adiabatic changes the rate of change of pressure with volume or slope of the adiabatic line on a P V diagram is found; dP const PV P dV V V 1 V 1 cf Isothermal changes where; dP nRT PV P dV V V2 V2 Adiabatic Change P Isothermal Change V It is to be noted that the magnitude of the local slope in a PV diagram is always greater for adiabatic change than for isothermal change, P P V Adiabatic V 70 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan Looking now at actual gases to see how well the foregoing discussion describes them; Gas CV CP Oxygen, O2 0.658 0.981 1.49 Nitrogen, N2 0.743 1.04 1.40 Hydrogen, H2 10.2 14.3 1.402 Carbon Monoxide, CO 0.744 1.04 1.398 Water, H2O 1.4 1.86 1.33 Carbon Dioxide, CO2 0.657 0.846 1.287 Sulphur DioxideSO2 0.47 0.60 1.276 Ammonia, NH3 1.66 2.15 1.295 Benzene, C6H6 0.67 0.775 1.157 Table of specific heat capacities for several monatomic, diatomic, triatomic and more complex gases. The values are taken at high temperature where all the degrees of freedom are allowed to play a role, ie. kBT > Equantum 71 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 70 CO2 H2O NO2 60 CP(kJ/m/K) 50 O2 H2 CO 40 30 N O H 20 10 0 1000 2000 3000 4000 5000 T (K) Graph showing variation of CP with temperature for a variety of monatomic, diatomic and triatomic gases. 72 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 3.7 The Van Der Waals Gas. Having looked at other types of “ideal” gas, (rigid diatomic and non rigid diatomic) and their behavior’s, attention may now be turned to real gases. The most well known of these is the van der Waals gas, where the modified equation of state discovered empirically by Dutch physicist Johannes van der Waals (1837 – 1923) was found to describe well the P-V behavior of real gases with the ideal equation of state, PV = nRT , being replaced by 2 P n a V nb nRT V 2 In this equation of state designed to describe real gases with potential interactions the pressure,, P, has a modification; PP n 2a V2 Atoms/molecules in the bulk of the gas have no net force imposed upon them by the presence of the intermolecular force as they are pulled equally from all sides but the atoms/molecules adjacent to the walls will have a net force pulling on them into the bulk of N 1 a the gas/fluid and therefore e reduction in P. The term 2 V NA V n 2a 2 represents a decrease on the pressure by an amount proportional to the square of the number density of particles in the system. If it is added to the actual pressure P the adjusted pressure is what would have been exerted by an ideal gas and allows the ideal gas equation of state to be applied. 73 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan In the second term, the volume V is reduced by an amount b per mole in order to account for the finite size of the atoms themselves and their collisions with each other. The atoms/molecules have a size assigned to them described by the van der Waals radius rvdW . Recalling that an atom is largely empty space this radius is determined by the extent of the interatomic potential. If the centres of two molecules approach within a distance 2rvdW they are deemed to have collided and each molecule has an “excluded” volume assigned to it Vex 3 1 4 4 3 2rvdW 4 rvdW . 2 3 3 It should be noted that the excluded volume per molecule is 4x the volume of the molecule. The total excluded volume is 4 3 VTot ex 4N rvdW 3 and the actual volume of the container is effectively reduced by this amount and V V VTot ex V nb in the van der Waals equation allowing the ideal gas equation thus modified to be used. b may be found by experimental determinastion of the P-V diagram and then rvdW 4 3 N VTot ex 4N rvdW nb b 3 NA 1 3 1 3 rvdW b 16 N A This has been the most successful modification made to the ideal gas equation of state and is still widely used. The internal energy of the van der Waals gas. When T = 0 and kinetic energy is by definition zero, in an ideal gas the pressure would go to zero as well as it is the direct result of the exchange of molecular momenta with the walls in which the gas is contained. However, the vdW equation of state reduces to 74 Thermal & Kinetic Physics: Lecture Notes P © n 2a Kevin Donovan (T = 0) V2 Where it is seen that the pressure is finite and negative at zero temperature and is proportional to the square of the density of molecules, and this is physically the result of the net attraction of surface molecules towards those in the bulk. Beginning with an extreme dilution whose energy is taken as the reference energy level and compressing this gas the work required to achieve the compression may be simply calculated, V V W PdV n 2a dV / V /2 integrating, V n 2a W n a V V/ 2 1 ie. after doing this work at T = 0 the gas has some energy which is taken to be the stored molecular potential energy at a volume V, not available to ideal gases! This means that the total internal energy of a vdW gas has both, kinetic and potential energy contributions 3 n 2a UvdW nRT 2 V This is for monatomic VdW (real) gases with three translational degrees of freedom. NB, from the above expression for UVdW and the equation for the heat capacity at constant volume U CV (vdW ) vdW T 75 3 nR V 2 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan is unchanged from that of a real gas. This is of course as it should be as the increase in heat flow should only affect the kinetic part of the internal energy and not the potential energy part. It is to be noted in passing that the internal energy is now a function of 2 state variables unlike previously; UvdW U(T ,V ) UIdeal U(T ) The internal energy of the vdW gas may be recast using the equation of state to obtain it in terms of pressure and volume; U vdW 2 3n n 2 a V n a P b 2 V V 2 n NB the factor n has been taken outside of the second bracket on the RHS. Now expanding the brackets[; UvdW 3n PV na n 2 ab n 2a Pb 2 n V V 2 V Simplify 3 1 n 2 a 3 n 3 ab UvdW P V nb 2 2 V 2 V2 and UvdW U(P,V ) 76 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan The Enthalpy of a van der Waals gas The enthalpy is defined in the usual way, HVdW UVdW PV The terms in the equation of state may be multiplied out and the equation rewritten as PV nRT n 2a n 2a (nb) nbP V2 V The enthalpy may then be written as; HVdW 5 n 2a n 3 ab nRT 2 nbP 2 V V2 The adiabatic law for van der Waals gases. With the equation of state of the Van der Waals gas ; 2 P n a V nb nRT V 2 it is sensible to ask if an equation describing an adiabatic process may be found, 5 equivalent to that which has already found for an ideal gas, PV 3 PV const (where 5/3 indicates that it is a monatomic gas that is being considered). Previously the internal energy of the van der Waals gas, UvdW 3 n 2a was nRT 2 V obtained. To find the adiabatic law the first law is used as follows; dU dQ dW 0 PdV Finding dU from the equation for UVdW , and P from a rearranged equation of state with pressure as the subject 77 Thermal & Kinetic Physics: Lecture Notes P © Kevin Donovan nRT n 2a V nb V 2 and using these in the first law 3 n 2a nRT n 2a dU nRdT dV PdV dV dV 2 V nb V2 V2 3 nRT nRdT dV 2 V nb 3 dT dV 2 T V nb By integrating the above 3 lnT fi ln(V nb)fi 2 Re-arranging T ln f Ti 3 2 V nb ln f Vi nb Therefore re arranging 3 ln Tf 2 Vf nb 3 Ti 2 Vi nb const And finally 3 T 2 V nb const is the adiabatic rule for a van der Waals gas. 78 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 3.8 Free expansion and Joule Kelvin effect. i) Joule Free Expansion. Consider now another irreversible process in order to see its consequences for ideal and real gases. This is the Joule free expansion illustrated below, where a gas is allowed to expand without constraint into a vacuum. V V 2V Vacuum Gas Gas The free expansion is represented in the above diagrams where initially a gas is contained in one half of a container, with adiabatic and rigid external walls, by a separating internal wall. These constraints (adiabatic and rigid respectively) mean that; W = 0 and that Q = 0. If the internal wall is suddenly removed/broken the gas is free to expand into the other half of the container. Applying the first law to this process is straight forward and U Q W 0 This means that for an ideal gas, where U ~ T , there is no change in temperature and for an isothermal process as V doubles the pressure halves. It is an irreversible process clearly as it can’t run backwards. This is approximately what was found for air by Joule but in fact there is a very slight cooling. The molecules on the right hand side are on average further apart and therefore their potential attraction drops meaning that their potential energy goes up (recall that the attractive potential is a negative quantity). If the potential increases then the kinetic must decrease in order that the total, U = 0. This is the origin of the slight cooling. It is an effect that can be expected for real gases where U = U(T, V) but not for an ideal gas where U = U(T). However that the effect is small shows that the ideal approximation is a good one. 79 Thermal & Kinetic Physics: Lecture Notes ii) © Kevin Donovan Joule Kelvin Effect. The Joule Kelvin effect is a way to effect an expansion accompanied by cooling that is more effective than the free expansion and used in the liquefaction of gases, also known as the throttling effect. Pi +dP Pi , Ti , Vi Pf Porous Plug Pf , Tf , Vf Pf - dP Pi Beginning with the situation depicted in the top figure where a porous plug divides a chamber with adiabatic walls into two with the porous plug allowing high pressure on the left hand side and lower pressure on the right hand side (a nozzle or constriction between the two may also be used). There is a piston either side, on the left hand side it is withdrawn and there is gas in the left hand side at initial pressure and temperature, Pi and Ti The pressure is slightly greater on the outside of the piston forcing it slowly in and pushing the gas through the porous plug into the right hand chamber where the pressure, Pf , is lower. The right hand piston slowly withdraws (the pressure is slightly lower on the outside of this piston to allow this) to accommodate the arriving gas at the final pressure and temperature, Pf and Tf . Because Pf is lower than Pi the volume Vf must be greater than Vi , an expansion has occurred. This is an isobaric process where the pressure remains unchanged on either side of the plug and the work done on the gas in forcing it through the plug is; 80 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 0 WLHS Pi dV PiVi Vi The work done by the gas expanding into the right hand side is Vf WRHS Pf dV Pf Vf 0 We can apply the first law recalling that the walls are adiabatic and therefore Q = 0 U Uf U i Q W PiVi Pf Vf Or re-expressed as U f Pf Vf H i U i Pi Vi H f and therefore this is an isenthalpic process with H 0 . The pressure and volume have changed but the enthalpy has remained unchanged. In general there will be a temperature change in the process and either heating or cooling can occur. Pf T dT dP JT dP P H T Pi T dP P H Pf JT dP Pi or T dT V Vf dV J dV H T Vi T V dV H Vf dV J Vi Where JT is called the Joule Kelvin or Joule Thomson coefficient (Lord Kelvin and Thomson are of course the same person!) and J is the Joule coefficient. An everyday example of this effect is the cooling of a propellant gas as it is released from a high pressure container into a room at ambient (low) pressure as with a deodorant spray. To proceed further the enthalpy of real gases must used for the Van der Waals gas as found earlier; HVdW 5 n 2a n 3 ab nRT 2 nbP 2 V V2 81 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan T The partial differential is required to get the Joule Thomson coefficient JT ’ P H Since T and P are the state variables of interest the enthalpy is required in terms of these two variables. The above expression is in terms of T, V and P and it is necessary to substitute for V and to simplify the expression as follows; 1. The van der Waals correction terms involving a or b are small first order corrections and therefore the final term on the RHS involving the product ab is of second order in smallness and may be dropped, HVdW 2. It is necessary to replace 5 n 2a nRT 2 nbP 2 V a in favour of P and T removing V. We do this by V approximating to an ideal gas for the third term on the RHS; a aP aP V PV nRT Thus HVdW 5 2naP nRT nbP 2 RT With reference to the throttling process it has been demonstrated that while P changes the H enthalpy remains constant and therefore 0 . It is then possible to obtain the Joule P T Thomson coefficient by differentiating both sides of the equation wrt P. P H 5 2na 2naP T H T 0 nR nb 2 RT P H P H RT 2 P H T Now solving for JT P H 2a b T RT JT P H 5 R 2aP 2 RT 2 82 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan All of the terms in the denominator on the RHS have positive values and it is therefore positive, however the numerator may take either sign dependant on the temperature and so then may JT . For T 2a coefficient is positive, ie. for low enough T the coefficient JT > 0 and a drop Rb in pressure produces a drop in temperature whereas if T 2a a drop in pressure will Rb cause an increase in temperature. The different behaviors arise because of the two effects of molecular interactions in a real gas. This can be seen by considering the molecular interaction via eg. the Lennard-Jones potential. 12 6 r rmin min U LJ r 2 r r U(r) rmin r The proximity of neighboring molecules as r (or V) becomes smaller from the right) causes a lowering of potential energy and as their proximity increases still further collisions (“touching” of molecules) causes a rise in potential energy. i) At low temperatures where there are low collision rates between gas molecules the effect of increasing the intermolecular distance (as described for the Joule free expansion) increases the molecule-molecule potential interaction as represented by the rising part of the potential interaction curve between the minimum and infinity thus increasing the potential energy part of the internal energy. For the internal energy to 83 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan remain constant the kinetic part of the internal energy must decrease and this is the origin of the decrease in temperature. ii) Collisions between molecules, as represented by the rising part of the potential interaction curve between the minimum and the origin results in a temporary conversion of kinetic to potential energy. At high temperatures where the molecular velocity and therefore collision rate is high this may be a large factor and an increase in volume resulting in a reduction in the collision rate will consequently result in a reduction of mean potential energy and therefore an increase in kinetic energy with resultant rise in temperature. TInv 2a is known as the inversion temperature and divides the two regimes. Rb This process is used in a continuous flow process (rather than the single shot process schematically described earlier) called the Linde liquefaction process. For good measure the Joule coefficient may be found in similar fashion; Starting with the van der Waals enthalpy; 5 n 2a n 3 ab HVdW nRT nRT 2 nbP 2 V V2 T The partial differential V is required to get the Joule coefficient J ’ H Since T and V are now the state variables of interest the enthalpy is required in terms of these two variables. The above expression is in terms of T, V and P and it is necessary to eliminate P and simplify as follows; 1. As previously the final term on the RHS involving the product ab is of second order in smallness and may be dropped, HVdW 5 n 2a nRT 2 nbP 2 V 84 Thermal & Kinetic Physics: Lecture Notes © Kevin Donovan 2. It is necessary to replace P in favour of V and T. We do this again, by approximating to an ideal gas for the third term on the RHS; P nRT V Thus HVdW 5 na 2 n 2 bRT nRT 2 2 V V T It is now possible to obtain the Joule coefficient V by differentiating both sides of the H equation wrt V. 5 T n 2a n 2 bR T n 2 bRT H 0 nR 2 2 V H V V H V H V2 V2 T Now solving for J V H n 2 bR n 2 2a bRT T 5 nR V V H 2 V2 n 2 2a bRT T J 5 nb V H V 2 nR 2 V All of the terms in the denominator on the RHS have positive values and it is therefore positive, however the numerator may take either sign dependant on the temperature and so then may J . For bRT 2a or T 2a the coefficient is positive and an increase in V bR (expansion) will result in an increase in T. The inversion temperature TInv 2a is as Rb previously found for JT The expansion of a gas causing a drop in temperature is frequently used in refrigeration processes and requires a negative Joule coefficient and therefore operation of the system above the inversion temperature of the particular gas used. 85