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Solutions Exercises Lecture 4 – AES1030 1. Two blocks consisting of different kinds of metal, A and B, have different temperatures, TA en TB. The mass of the blocks is also different, namely mA en mB. The specific heat capacity of the two blocks are given by C*P,A en C*P,B. The blocks are brought in thermal contact so that heat is exchanged at a constant pressure. There is no heat exchange with the surroundings. a. Derive the expression allowing the computation of the end temperature of the two blocks in terms of TA, TB, mA, mB, C*P,A and C*P,B. Use the first law of thermodynamics for the derivation of the equation. b. In a next step derive the equation to compute the total change of the entropy of the system, ∆Ssystem, and give the expression in terms of TA, TB, mA, mB, C*P,A and C*P,B. c. What is the change in entropy of the surroundings for the above described process? Support your answer. Solution: First, the system is defined and all given data are collected. The system is isolated (no heat exchange between the system and the surroundings)and consists of A and B with a total mass of (mA + mB). The complete system is divided into two subsystems, namely A and B. Initially the two subsystems have different temperatures, namely TA and TB. At the end of the process the two systems have the same temperature; the colder systems has heated up while the initial warmer system has cooled down. The end temperature or mixing temperature lays between the temperatures TA and TB. Ad a) Wanted: end temperature = mixing temperature For the computation of the end temperature it is assumed that the complete process is adiabatic and isobaric. Note that the process is only adiabatic for the complete system (subsystem A and B) but not for the subsystems A and B. For an isobaric process commonly first the change of the enthalpy is computed and not the change of the internal energy because the equations describing the change of enthalpy simplify for isobaric processes and the relationship between the enthalpy and the internal energy is well defined…. We start with the first law dU = dq + dw. The first law can be combined with the definition of the enthalpy H = U + p·V and with the total differential dH = dU + p·dV + V·dp. The two subsystems A and B are incompressible so that dV = 0 and thus dw = - pex dV = p · dV = 0 (note: pex=p for a reversible process). From this it follows: dq = dH. An equation for the computation the change of the enthalpy can be derived starting with H = H(P,T) giving ∂H ∂H dH = dp + dT ∂T p ∂p T , ∂H = dp + C p dT ∂p T And for an isobaric process dH = CP ·dT = m CP* ·dT. As already mentioned above, the complete system consisting of the subsystem A and the subsystem B, undergoes an adiabatic and isobaric process. However, there is heat exchange between the subsystem A and the subsystem B. The enthalpy is a state function, that means that the enthalpy of the complete system can be computed from the enthalpy of the subsystems forming the complete system. This gives: dHtot = dqtot = 0 = dHA + dHB = (CP dT)A+ (CP dT)B. From this equation an expression allowing the computation of the mixing temperature can be derived: T T TA TB ∆H = ∫ dH = ∫ C p , A ⋅ dT + ∫ C p , B ⋅ dT 0 = mA ⋅ CP* , A · (T − TA ) + mB ·CP* , B · (T − TB ) mA ·CP* , A · (T − TA ) = − mB ·CP* , B · (T − TB ) ( ) T · mA ·CP* , A + mB ·CP* , B = mB ·CP* , B ·TB + mA ·CP* , A ·TA T= mB ·CP* , B ·TB + mA ·CP* , A ·TA ( m ·C A * P, A + mB ·CP* , B ) Ad b) Next the equations for the computation of the entropy change need tob e derived. Generally the entropy change can be computed by: dS = dqrev / T. The complete system and het complete process is adiabatic: ∆Ssystem = 0. However, the entropy of the subsystems A and B changes. Just as the enthalpy the entropy is a state function. Thus the change of the entropy of the complete system can be computed from the entropy changes of the subsystems: ∆Ssystem = ∆SA + ∆SB. In general the computation of the entropy change is based on a reversible process, even if the real process is not reversible. In such a case a reversible process needs to be defined which proceeds from the same initial to the same end state as the irreversible process. For the given exercise it is not defined whether the process is reversible or irreversible. However, the process is isobaric so that the derivation of the equations based on the enthalpy is advised. Thus the relation between the internal energy and the enthalpy needs to be given: dU = dq + dw and H = U + p·V and dH = dU + p·dV + V·dp = dq + dw + p·dV + V·dp. For a reversible process further it is given that: pex = p so that dw = - pex dV = -p ·dV. With this one gets for a reversible isobaric process: dqrev = dH. With this we get the equation describing the entropy change for an arbitray system undergoing an isobaric and reversible process: T T dq C ⋅ dT dS = rev = P ⇒ ∆S = CP ⋅ ln 2 = m ⋅ CP* ⋅ ln 2 T T T1 T1 With the above derived equation the entropy change of the subsystems A and B can be computed, leading to the entropy change of the complete system ∆Ssystem ∆S system = ∫ dS = T ∫ C p, A TA with T = T T dT + ∫ TB C p,B T dT T T = mA ⋅ CP* , A ⋅ ln + mB ⋅ CP* , B ⋅ ln TA TB * * mB · CP , B · TB + mA· CP , A· TA (m ·C A * P, A + mB · CP* , B ) Ad c) Next it is asked to derive the equation to describe the entropy change of the surroundings. The given process is adiabatic. This means there is no heat exchange between the complete system and the surroundings (note; the only heat exchanged observed here is between the two subsystems but never with any part of the system with the surroundings). Thus, there is no ‘disturbance’ of the surroundings due to heat exchanged with the system, so that we can conclude that ∆Ssurroundings = 0 J/K, and for the complete change of entropy during the process: ∆Stotal = ∆Ssystem + ∆Ssurroundings=∆Ssystem. 2. 55.5 mol ice of -100C is brought into the state of steam at 1100C at a constant pressure of 1 bar. The following data are given: J J • H20(S): ∆ H •fus = 6010 T fus = 273,15 K CP = 37, 66 mol K ⋅ mol J J • • H20(L): ∆ H vap = 4400 Tvap = 373,15 K CP = 75, 32 mol K ⋅ mol J H20(V): CP = 30,54 + 1, 03 ⋅10−2 ⋅ T K ⋅ mol Please compute the change of entropy for the complete process. ( Solution: ) The process is schematically drawn in the p,T diagram below (note: this p,T diagram is for water as you can recognize on the slight negative slope of the melting curve): p S L 1 atm V T [ oC] -10 110 The system consists solely of water. The water undergoes an aggregation change so that the complete process needs to be split into five sub-processes: Step 1: Isobaric heating of ice from -100C = 263 K to 00C = 273 K. Step 2: Isobaric and isothermal transition from ice to liquid water at 273K and 1 atm. Step 3: Isobaric heating of liquid water from 00C = 273 K to 1000C = 373 K. Step 4: Isobaric and isothermal transition of liquid water into steam at 373K and 1atm. Step 5: Isobaric heating of steam from 1000C = 373 K to 1100C = 383 K. The determination of the change of entropy for the steps 1, 3 and 5 are analogous and for the steps 2 and 4 also: In general the equations to compute the entropy change are derived from the dq following equation: dS = rev . Note: even though the real process is not reversible, T for the computation of the entropy change a reversible process needs to be assumed from the same initial to the same end state as the real process. Step 1, 3, 5 are all isobaric (dp = 0) and reversible (pex = p) processes. For an isobaric process the enthalpy is used to derive the equations (see also exercise above): dU rev = dqrev + dwrev → dw = − pex ⋅ dV reversible ⇒ pex = p ⇒ dw = − p ⋅ dV H ≡ U + p ⋅ V ⇒ dH = dU + p ⋅ dV + V ⋅ dp ⇒ dU = dH − p ⋅ dV − V ⋅ dp ⇒ dU rev = dqrev + dwrev = dqrev − p ⋅ dV = dH − p ⋅ dV − V ⋅ dp dp = 0 ⇒ dqrev = dH ∂H ∂H H = H ( p, T ) ⇒ dH = ⋅ dp + ⋅ dT ∂T p ∂p T dp = 0 ⇒ dH = C p ⋅ dT = dqrev This is the general description of the exchanged heat during the process. In order to compute the entropy change for this process step, one needs to incorporate the respective heat capacity for the respective process step: Step 1: T = 273 K T = 273 K 2 2 dq dT dT C S ⋅ dT n ⋅ CPS ⋅ dT dS = rev = P = ⇒ ∆S stap1 = ∫ dS = ∫ n ⋅ CPS ⋅ = n ⋅ CPS ⋅ ∫ T T T T T T1 = 263 K T1 = 263 K T ⇒ ∆S stap1 = n ⋅ CPS ⋅ ln 2 T1 ⇒ ∆S stap1 = 55.5mol ⋅ 37.66 J J 273K ⋅ ln = 77.99 mol ⋅ K K 263K Step 3: T = 373 K T = 373 K 2 2 dq C L ⋅ dT n ⋅ CPL ⋅ dT dT dT dS = rev = P = ⇒ ∆S stap 3 = ∫ dS = ∫ n ⋅ CPL ⋅ = n ⋅ CPL ⋅ ∫ T T T T T T1 = 273 K T1 = 273 K T ⇒ ∆S stap 3 = n ⋅ CPL ⋅ ln 2 T1 J J 373K ⋅ ln = 1304.69 mol ⋅ K K 273K Step 5: For step 5 we have to account for the temperature dependence of the heat capacity of the steam. This needs to be done before the integration. ⇒ ∆S stap 3 = 55.5mol ⋅ 75.32 T = 383 K dS = T =383 K 2 2 dqrev CPV ⋅ dT n ⋅ CPV ⋅ dT dT dT = = ⇒ ∆S stap 5 = ∫ dS = ∫ n ⋅ CPV ⋅ = n ⋅ ∫ ( A + B ⋅T ) T T T T T T1 = 373 K T1 = 373 K T ⇒ ∆S stap 3 = n ⋅ A ⋅ ln 2 + B ⋅ (T2 − T1 ) T1 J 383K −2 ⇒ ∆S stap 3 = 55.5mol ⋅ 30.54 ⋅ ln + 1.03 ⋅10 ⋅ ( 383 − 373) K = 50.56 K 373K Step 2 and step 4 describe the actual phase transition which are isothermal and isobaric. Still, the exchanged heat can be described by: dqrev = dH. (Note: assuming that the process is isobaric and isothermal and therefore dH=0 is wrong for the description of a phase transition. The enthalpy of evaporation or fusion are also called ‘latent heat’ for this reason…there is indeed heat required if liquid is transferred into vapor). For the description of the entropy change of the phase transitions we get: dq dH dH 1 ∆H dS = rev = ⇒ ∆S = ∫ dS = ∫ isothermal ⇒ ∆S = ⋅ ∫ dH = T T T T T Step 2: ∆S stap 2 = ⇒ ∆S stap 2 ∆H •fus • T fus = n⋅ ∆ H •fus • T fus J mol = 1221.81 J = 55.5mol ⋅ 273K K 6010 Step 4: ∆S stap 4 = • ∆H vap • Tvap = n⋅ • ∆ H vap • Tvap J mol = 654.69 J ⇒ ∆S stap 4 = 55.5mol ⋅ 373K K The entropy change of the system (55.5. mol water) for the complete process is computed by summing up the entropy changes of all the sub-processes: ∆S system = ∆S step1 + ∆S step 2 + ∆S step 3 + ∆S step 4 + ∆S step 5 4400 J J J J J + 1221.81 + 1304.69 + 654.69 + 50.56 K K K K K J = 3309.74 K = 77.99 ∆S systeem 3. 3 kg pure liquid water of 400 K is mixed adiabatically at a constant pressure, which is higher than 1 bar, with 4kg pure liquid water of 300 K. Compute the total change in entropy, thus of the surroundings and the water, for the given mixing process. Assume for the computations that the specific heat capacity of water is constant in the temperature region of interest: CP* = 4180 J K-1 kg-1. Solution: First, the system is defined. The complete system consists of 7 kg water. Initially, 3 kg of the water has a temperature of 400K and 4 kg a temperature of 300 K. After mixing the complete system, thus 7kg water, has a mixing temerature T, between 300K and 400 K. The computation of the entropy is done via: dS = dqrev / T. The process of the complete system is adiabatic. However, the system is divided into two subsystems of different initial temperature. There is heat exchange between the two sub-systems. For the computation of the entropy and enthalpy change of the system is split up into two systems – this is possible as the entropy and the entropy are a state functions. Sub-system 1: 3kg water of an initial temperature of 400K Sub-system 2: 4kg water with an initial temperature of 300K The entropy change of the water is computed by: ∆SH2O = ∆SH2O,1 + ∆SH2O,2 and the enthalpy change by ∆HH2O = ∆HH2O,1 + ∆HH2O,2. The process is isobaric so that dq = dH (see for the derivation the earlier exercises). The equation for the computation of the enthalpy change is derived starting with H = H(P,T) and ∂H ∂H dH = dp + dT ∂T p ∂p T , ∂H = dp + C p dT ∂p T So that for an isobaric process dH = CP dT = m CP* dT. The complete process is adiabatic, this means that the complete system (consisting of the sub-system 1 and 2) does not exchange heat with the surrounding. However, sub-system 1 and 2 exchange heat. With this we can write: dHtot = dqtot = 0 = dHsys1 + dHsys2 = (CP dT)sys1+ (CP dT)sys2. With this the mixing temperature can be computed: ∆H = ∫ dH = T ∫ T C p , H 2O ,1 ⋅ dT + T1 = 400 K ∫ C p , H 2O ,2 ⋅ dT T2 =300 K 0 = mH 2O ,1 ⋅ CP* , H 2O ⋅ (T − T1 ) + mH 2O ,2 ⋅ CP* , H 2O ⋅ (T − T2 ) mH 2O ,1 ⋅ CP* , H 2O ⋅ (T − T1 ) = −mH 2O ,2 ⋅ CP* , H 2O ⋅ (T − T2 ) T ⋅ CP* , H 2O ⋅ ( mH 2O ,1 + mH 2O ,2 ) = CP* , H 2O ⋅ (T1 ⋅ mH 2O ,1 + T2 ⋅ mH 2O ,2 ) T= T1 ⋅ mH 2O ,1 + T2 ⋅ mH 2O ,2 mH 2O ,1 + mH 2O ,2 = 400 K ⋅ 3kg + 300 K ⋅ 4kg = 342.86 K ( 3 + 4 ) kg Knowing the mixing temperature allows the computation of ∆SH2O with ∆SH2O = ∆SH2O,1 + ∆SH2O,2 . In general for both sub-systems we set T1= initial temperature; T2= final temperature = mixing temperature: T T dq C ⋅ dT dS = rev = P ⇒ ∆S = CP ⋅ ln 2 = m ⋅ CP* ⋅ ln 2 T T T1 T1 An explicitly T T C p , H 2O ,1 C p , H 2O ,2 ∆S H 2O = ∫ dS = ∫ ⋅ dT + ∫ ⋅ dT T T T1 = 400 K T2 =300 K T T = mH 2O ,1 ⋅ CP* , H 2O ⋅ ln + mH 2O ,2 ⋅ CP* , H 2O ⋅ ln T1 T2 J 342.86 K 342.86 K = 4180 ⋅ 3kg ⋅ ln + 4kg ⋅ ln K ⋅ kg 400 K 300 K J K For the determination of ∆Ssurroundings it needs to be analysed if there is heat exchange with the system. As the process is adiabatic, there is no heat exchange = 299.84 between the system and the surroundings so that ∆Ssurroundings = 0 J/K and thus: ∆Stotal = ∆SH2O + ∆Ssurroundings = 299.84 JK-1 + 0 = 299.84 JK-1. From this we can see that the entropy increases during the process. Thus the process is spontaneous which is according to expectations. 4. One mol of ideal gas is irreversibly and adiabatically compressed so that the pressure increases from 2 bar to 10 bar. De initial temperature of the gas is 200C. Further it is known that the irreversible process requires 40% more work than the respective reversible process. Please compute the change of the entropy of the irreversible process. Further it is given that 5 J CP = ⋅ R R = 8,314 2 mol ⋅ K Solution: Before we start with solving the problem I would like to draw your attention to the fact that for the given irreversible process the final state is not completely described. For the given process it is given that the pressure increases irreversibly and adiabatically from 2 to 10 bar. Further, it is given that for this irreversible process work needs to be done on the system. The energy required for this work is 1.4 times the energy required for the reversible process from the same initial to the same end state. With the help of this information the temperature at the end state can be computed. If the temperature at the end state is known, the entropy change of the process can be computed. The system is isolated– no exchange of material or work with the surroundings – and consists of 1 mol ideal gas. The process is irreversible (pex ≠ psystem) and adiabatic (dqirrev = 0) and: wirrev = 1.4 · wrev. For the computation of the entropy change of the irreversible process, the entropy change of the system and of the surroundings need to be known. First the computation of the entopry change of the system. Generally, we know that: dq dS system = rev . Thus for the computation of the enropy change of the system we T need to assume a reversible process from the same initial to the same end state. Remember, that this reversible process does not necessarily need to be adiabatic! But before the entropy change of the system can be computed, we need to determine the end temperature of the real (irreversible) process: dU irrev = dU = dqirrev + dwirrev adiabatic ⇒ dqirrev = 0 ⇒ dU = dwirrev ∂U ∂U U = U (V , T ) ⇒ dU = ⋅ dV + ⋅ dT ∂V T ∂T V ∂U ∂U ideal gas ⇒ = 0 ⇒ dU = ⋅ dT = CV ⋅ dT ∂V T ∂T V dU = dwirrev ⇒ CV ⋅ dT = dwirrev Further we know wirrev = 1.4 wrev and also dwirrev = 1.4 dwrev . For the reversible process the internal pressure is equal to the external pressure: dwrev = − p ⋅ dV And for an ideal gas we have C C CP = CV + n ⋅ R ⇒ P = V + R ⇒ CP = CV + R n n ⇒ CV = CP − R And explicitly: 5 3 CV = CP − R = ⋅ R − R = ⋅ R = const 2 2 With this we can compute the final temperature: dU = dwirrev = 1.4 ⋅ dwrev ⇒ n ⋅ CV ⋅ dT = 1.4 ⋅ ( − p ⋅ dV ) n ⋅ R ⋅T V n ⋅ R ⋅T dT dV ⇒ n ⋅ CV ⋅ dT = −1.4 ⋅ ⋅ dV ⇒ n ⋅ CV ⋅ = −1.4 ⋅ n ⋅ R ⋅ V T V p, T , V var y ⇒ p = T2 n, CV , R are const ⇒ n ⋅ CV ⋅ ∫ T1 V 2 dT dV = −1.4 ⋅ n ⋅ R ⋅ ∫ T V V1 T V ⇒ n ⋅ CV ⋅ ln 2 = −1.4 ⋅ n ⋅ R ⋅ ln 2 T1 V1 n ⋅ R ⋅ T2 p T n ⋅ R ⋅T 2 V= ⇒ n ⋅ CV ⋅ ln 2 = −1.4 ⋅ n ⋅ R ⋅ ln n ⋅ R ⋅ T1 T p 1 p 1 T ⇒ n ⋅ CV ⋅ ln 2 = −1.4 ⋅ n ⋅ R ⋅ ln T1 T2 p2 T1 p1 T ⋅ p = −1.4 ⋅ n ⋅ R ⋅ ln 2 1 T1 ⋅ p2 T T p ⇒ n ⋅ CV ⋅ ln 2 = −1.4 ⋅ n ⋅ R ⋅ ln 2 + ln 1 T1 p2 T1 p T ⇒ n ⋅ CV + 1.4 ⋅ n ⋅ R ⋅ ln 2 = −1.4 ⋅ n ⋅ R ⋅ ln 1 T1 p2 ( ) −1.4⋅n⋅ R T p p n⋅CV +1.4⋅n⋅R −1.4 ⋅ n ⋅ R ⇒ ln 2 = ⋅ ln 1 = ln 1 T1 n ⋅ CV + 1.4 ⋅ n ⋅ R p2 p2 −1.4⋅n⋅ R −1.4⋅n⋅ R −1.4⋅n⋅ R −1.4 T p n⋅CV +1.4⋅n⋅R p n⋅CV +1.4⋅n⋅R p n⋅1.5⋅R +1.4⋅n⋅R p 1.5+1.4 ⇒ 2 = 1 ⇒ T2 = T1 ⋅ 1 ⇒ T2 = T1 ⋅ 1 = T1 ⋅ 1 T1 p2 p2 p2 p2 −1.4 2bar 2.9 ⇒ T2 = 293K ⋅ = 637.24 K 10bar Now the final state is completely described. With this the heat exchanged between the system and the surroundings can be computed if the process is reversible using: wirrev = 1.4 wrev and dU= dwirrev (see above): dU = dqrev + dwrev ⇒ dqrev = dU − dwrev ⇒ dqrev = dU − dwirrev dU 1 = dU − = dU ⋅ 1 − 1.4 1.4 1, 4 0.4 ⋅ n ⋅ CV ⋅ dT 1.4 Incorporating into the equation for the computation of the entropy change of the system gives: ⇒ dqrev = T dS system 2 dq 0.4 dT 0.4 dT = rev = ⋅ n ⋅ CV ⋅ ⇒ ∆S system = ⋅ n ⋅ CV ⋅ ∫ T 1.4 T 1.4 T T1 T 0.4 0.4 3 J 637.24 K ⋅ n ⋅ CV ⋅ ln 2 ⇒ ∆S system = ⋅1mol ⋅ ⋅ 8.314 ln 1.4 1.4 2 mol ⋅ K 293K T1 J ⇒ ∆S system = 2.768 K For the computation of the entropy change of the surroundings we use: dqsurroundings −dqsystem dS surroundings = = with dqsystem the heat which is really exchanged Tsurroundings Tsurroundings between the system and the surroundings, thus dqirrev. The real irreversible process is adiabatic, dqirrev = 0 so that: ⇒ ∆S system = dS surroundings = −dqsystem Tsurroundings = −dqirrev J = 0 ⇒ ∆S surroundings = 0 Tsurroundings K ∆S process = ∆S system + ∆S surroundings = 1.846 J J J + 0 = 1.846 K K K From this exercise it can be seen that for an irreversible and spontaneous process the entropy change needs tob e positive (and definitely not equal to zero!). Further, the entropy only increases in the system because we deal here with an isolated system. Note: Based on the fact that the process irreversible and adiabatic is can be concluded that the reversible process from the same initial to the same end state is not adiabatic. If we would have assumed that also the reversible process is adiabatic, then the determined entropy change would have been zero. However, only for a reversible process there is no change in entropy and not for an irreversible process!! 5. In the p,T-diagram given below the vapour pressure curve of a pure component is given a. Give the names of the beginning and the end of the vapour pressure curve and the names of the regions above and below the curve. b. Draw in another graph the schematic curve of the (molar) Gibbs energy as function of the temperature for the process from point 1 to 2. Clealry state which parameter is kept constant. Support your answer c. Draw in another graph the schematic curve of the (molar) Gibbs energy as function of the pressure for the process from 3 to 4. Which parameter is kept constant? Support your answer. p 4 pi•= pvap 2 1 3 Tvap T Solution: Ad a: The curve is called the vapour pressure curve. The end point of the vapour pressure curve is called critical point. The critical point is characterized by the fact that the vapour and the liquid phase are identical. This means there is no difference in the density or enthalpy between the two phases. For temperatures higher than the critical temperature and pressures higher than the critical pressure, the component exists as so-called FLUID. The starting point of the vapour pressure curve (at the left) is called the triple point. Here, the three phases solid, vapour and liquid coexist. Ad b & c: The absolute (molar) Gibbs energy cannot be computed. However, it is known how the Gibbs energy changes with changes of temperature and pressure. This expression can be derived from the definition of the Gibbs energy: G ≡ H − T ⋅ S ⇒ dG = dH − d (T ⋅ S ) = dH − T ⋅ dS − S ⋅ dT H ≡ U + p ⋅ V ⇒ dH = dU + d ( p ⋅V ) = dU + p ⋅ dV + V ⋅ dp dU = T ⋅ dS − p ⋅ dV ⇒ dH = T ⋅ dS − p ⋅ dV + p ⋅ dV + V ⋅ dp ⇒ dH = T ⋅ dS + V ⋅ dp ⇒ dG = T ⋅ dS + V ⋅ dp − T ⋅ dS − S ⋅ dT ⇒ dG = V ⋅ dp − S ⋅ dT Analogous for the molar Gibbs energy: G G = ⇒ dG = V ⋅ dp − S ⋅ dT n The total differential of the Gibbs energy as given above forms the base for the description of phase transitions but also as base for the derivation of the equations to describe the change of the Gibbs energy during a process. Note, that the phase with the smallest Gibbs energy is the stable phase. Ad b: The process from 1 to 2 is an isobaric process (dp = 0) for which the temperature in the system is lowered. From the above derived equation, we see that the change of the (molar) Gibbs energy is described by: dp = 0 ⇒ dG = − S ⋅ dT From this it can be concluded that for an isobaric process for which the temperature in the system decreases (dT <0), the (molar) Gibbs energy increases (dG > 0). Note that the (molar) entropy is always larger than zero. Thus for an isobaric process, the (molar) entropy S describes the slope of the curve. For the construction of the curve …the (molar) entropy in the gas phase is larger than the (molar) entropy in the liquid phase. Thus, the absolute slope of the Gibbs energy changing with temperature is larger in the gas phase than in the liquid phase. From this it can be concluded that the (molar) Gibbs energy is smaller for the gas phase than for the liquid phase at high temperatures (remember that the real slope is negative as it is –S). Further, it is known that at the equilibrium line (vapour pressure curve) the (molar) Gibbs energy of the gas phase and of the liquid phase are equal. Note: the phase with the smallest Gibbs energy is the stable phase. With all these data we can schematically draw the curve of the (molar) Gibbs energy: p L pi•= pvap 2 1 V G L T Tvap V P=const T2 Tvap T1 T Ad c: The process from 3 to 4 is an isothermal process (dT = 0) while the pressure is increased. In this case, the derived equation for the description of the change of the (molar) Gibbs energy simplifies to: dT = 0 ⇒ dG = V ⋅ dp Thus, the (molar) Gibbs energy increases with increasing pressure as the (molar) volume is always positive. Now the (molar) volume describes the slope of the Gibbs energy changing with the pressure. In general, the (molar) volume of the vapour phase is larger than the (molar) volume of the liquid phase so that we know that the slope of the Gibbs energy describing the liquid phase is smaller than of the Gibbs energy describing the vapour phase. With this and the fact that the (molar) Gibbs energy of the vapour and the liquid phase are equal, the run of the Gibbs energy as function of the pressure can be deduced (remember that for a constant temperature at low pressures the vapour phase exists while at high pressures the liquid phase): p L 4 pi•= pvap 3 V T Tvap G V L T=const P3 6. pvap P4 p The molar heat of fusion and the molar entropy of fusion of water at 273.15 K and 1 atm are: ∆ H fus = 6009.3 J J en ∆ S fus = 22.0 . mol K ⋅ mol Show by using these data that a. ∆G fus = 0 J at 273.15 K and 1 atm mol b. Further, show that for T < 273.15 K ice does not spontaneously melt and that for T > 273 K ice spontaneously melts. Solution: a) Here actually the equilibrium of a solid and a liquid phase is described. Thus, a point on the melting curve in the p,T-diagram is described. The phase transition from completely solid to completely liquid is isothermal and isobaric.For the derivation of the above given expression we start with the definition of the Gibbs energy: G ≡ H − T ⋅ S ⇒ dG = d H − T ⋅ d S − S ⋅ dT INTERMEZZO: In the above given equation, the notation for molar state functions is used ( G = G ) n because only molar properties are given. The general expressions,e.g., given on the slides of the presentations, are also valid for molar properties. Molar properties are solely functions of p and T or V and T…. END INTERMEZZO For the isothermal phase transition the expression for the change of the molar Gibbs energy simplifies to: dG = d H − T ⋅ d S ⇒ ∫ dG = ∫ d H − T ⋅ ∫ d S ⇒ ∆G fus = ∆ H fus − T ⋅ ∆ S fus Based on this equation we can now show that ∆G fus = 0 J at 273.15 K and 1 atm. For this purpose, mol the given values of the molar enthalpy and entropy of fusion are incorporated into the expression: ∆G fus = ∆ H fus − T ⋅ ∆ S fus = 6009.3 J J J − 273.15 K ⋅ 22.0 =0 mol K ⋅ mol mol b) For the solution of exerecise b we recall that a process occurs spontaneously if the system moves to its minimum Gibbs energy. Thus if a process occurs spontaneously then the change in the Gibbs energy needs to be negative. In other words, we need to show that for T < 273.15 K ∆G fus > 0 J J and that for T > 273.15 K ∆G fus < 0 . mol mol For this we assume that the enthalpy and entropy of fusion are constant and equal to the given values. This is reasonable for states which are not too far from the state for which the values are given, thus for temperature slightly warmer or colder than 273 K and for pressures slightly higher or smaller than 1 atm. First we calculate the change of the Gibbs energy at a temperature of T = 270 K: ∆G fus = ∆ H fus − T ⋅ ∆ S fus = 6009.3 J J J − 270 K ⋅ 22.0 = 69.3 >0 mol K ⋅ mol mol The value is larger than 0, meaning that ice at 270 K and 1 atm does not melt spontaneously (just as our experience is). The change of the Gibbs energy at a temperature slightly higher than 273 K, namely 273.5 K, and 1 atm is equal to: ∆G fus = ∆ H fus − T ⋅ ∆ S fus = 6009.3 J J J − 273.5 K ⋅ 22.0 = −7.7 < 0 Dus mol K ⋅ mol mol we Thus here the Gibbs energy decreases, meaning that the process is spontaneous (just as it is known from observations). 7. 3 mol water is heated isobarically at 0.1 MPa from to 2500C. The molar heat of evaporation of water at 1000C and 0.1 MPa is 40,292 kJ mol-1. The molar heat capacity of liquid water is constant and given by 75.312 J mol-1 K-1 , the molar heat capacity of steam can be computed with the following equation: J CP , H 2O (V ) = 30.363 + 9.611 ⋅10−3 ⋅ T + 11.839 ⋅10−7 ⋅ T 2 . mol ⋅ K a. Draw schematically the phase diagram of water (in terms of p and T) and indicate the gas phase, the liquid phase and the solid phase and all phase equilibrium curves. Further, draw the above described process into the phase diagram and explain how the process is described thermodynamically. b. In a next step, calculate the change of entropy of water for the complete process, ∆SH2O. State all the assumptions made for the computations. Further, it is given that: R = 8.314 J mol-1 K-1, 00C = 273.15 K, 1 bar =105 N m-2 Solution: Ad a) p Solid (S) Liquid (L) Step 1 Step 3 Step 2 Gas (V) T=25 C T=250 C T A and b) The complete process needs to be divided in sub-processes for the thermodynamic description. Step 1 describes the isobaric heating of the (liquid) water from 25 C to 100C, up to the vapour pressure curve which described the phase transition from liquid to gas at a given pressure (equilibrium curve). Step 2 is the isothermal and isobaric phase transtion from completely liquid to completely gas. Step 3 describes the isobaric heating of the gas (steam) from 100 C to 250 C. The thermodynamic description of the process step 1 and 3 are in principle the same: The change of the entropy of the water is described by dS = dqrev . For the explicit computation we need T to define an expression with which we can calculate the value with the given data. The process is reversible and isobaric. The system consists of water. We start with the first law of thermodynamics: dU = dq + dw. Because the process is isobaric, we continue with the enthalpy: ∂H ∂H dH = ⋅ dp + ⋅ dT ∂T p ∂p T ∂H dp = 0 ⇒ dH = ⋅ dT = CP ⋅ dT ∂T p Which is related to the internal energy by H ≡ U + pV and further: dH = dU + d ( p ⋅ V ) = dU + V ⋅ dp + p ⋅ dV ⇒ dH = dqrev + V ⋅ dp dU = dqrev + dwrev = dqrev − p ⋅ dV dp = 0 ⇒ dH = dqrev The change of enthalpy for a reversible and isobaric process can be computed by: giving: dH = CP dT 2 dq dH CP ⋅ dT C ⋅ dT 2 n ⋅ CP ⋅ dT = ⇒ ∆S = ∫ dS = ∫ P =∫ dS = rev = T T T T T T1 T1 T T For step I: ∆S step1 = T2 ∫ T1 n ⋅ CP , H 2O ( L ) ⋅ dT T ∆S step1 = 3mol ⋅ 75.312 = n ⋅ CP , H 2O ( L ) ⋅ T2 =100 C ∫ T1 = 25 C T dT = n ⋅ CP , H 2O ( L ) ⋅ ln 2 T T1 J J 373.15 K ⋅ ln = 50.696 mol ⋅ K K 298.15K For step 3(note here one has to account for the fact that the heat capacity for a gas (steam) is not constant with changing temperature): CP , H 2O (V ) = 30.363 + 9.611⋅10−3 ⋅ T + 11.839 ⋅10−7 ⋅ T 2 T2 ∆S step 3 = ∫ T1 n ⋅ CP , H 2O (V ) ⋅ dT T 30.363 + 9.611 ⋅10−3 ⋅ T + 11.839 ⋅10−7 ⋅ T 2 ⋅ dT T T1 T2 = n⋅ ∫ T2 T2 T2 30.363 9.611⋅10−3 ⋅ T 11.839 ⋅10−7 ⋅ T 2 ∆S step 3 = n ⋅ ∫ ⋅ dT + ∫ ⋅ dT + ∫ ⋅ dT T T T T T1 T1 1 T2 T2 T2 1 −3 −7 ∆S step 3 = n ⋅ 30.363 ⋅ ∫ ⋅ dT + 9.611 ⋅10 ⋅ ∫ dT + 11.839 ⋅10 ⋅ ∫ T ⋅ dT T T1 T1 T1 T 11.839 ⋅10−7 ∆S step 3 = n ⋅ 30.363 ⋅ ln 2 + 9.611 ⋅10−3 ⋅ (T2 − T1 ) + ⋅ T22 − T12 T 2 1 ( ) For step 3 we know that T1= 100C = 373.15 K and T2=250C = 523.15K: 11.839 ⋅10−7 523.15K −3 ∆S step 3 = 3mol ⋅ 30.363 ⋅ ln + ⋅ ⋅ − + ⋅ 523.152 − 373.152 9.611 10 523.15 373.15 ( ) 2 373.15K J ∆S step 3 = 35.34 K ( Step 2 is the isobaric and isotheraml transformation of liquid water into gaseous water (steam). This means we need to describe the vapour-liquid equilibrium at 100C and 0.1 MPa. On the equilibrium line the Gibbs energy of the coexisting phases are equal, thus there is no change of the Gibbs energy during the phase transition ∆Gvap= 0J. The Gibbs energy is defined by: G≡ H – TS and the change of the Gibbs energy: ) dG = dH − d (T ⋅ S ) = dH − T ⋅ dS − S ⋅ dT dT = 0 ⇒ dG = dH − T ⋅ dS ⇒ ∫ dG = ∫ dH − T ⋅ ∫ dS ⇒ ∆Gvap = ∆H vap − Tvap ⋅ ∆Svap = 0 ⇒ ∆Svap = ∆H vap Tvap With this the change in entropy for step 2 can be computed: ∆Svap = ∆H vap Tvap = n ⋅ ∆ H vap Tvap ∆S stap 2 = ∆Svap = 323.93 = 3mol ⋅ 40292 J mol 373.15 K J K The change of the entropy of the water for the complete process is thus: ∆S water = ∆S stap1 + ∆S stap 2 + ∆S stap 2 = 50.696 ∆S water = 409.966 J K J J J + 323.93 + 35.34 K K K