MATSCI 204 THERMODYNAMICS AND PHASE EQUILIBRIA Winter ...
MATSCI 204 THERMODYNAMICS AND PHASE EQUILIBRIA Winter 2013 Midterm Solutions 2/12/2013
Write your answers neatly and clearly. I will not give credit to statements that are not clearly justified (in other words, make sure your handwriting is better than mine!). Class average: 76/100 (=125/165) Standard deviation: 14 (=23)
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Problem 1: (10 points) Answer the following questions with True or False. If the statement is False, write a 1line explanation of why you think the statement is false or you won’t get credit. Write the explanation on this sheet and don’t forget to turn it in! T/F
For any observable process, ΔSUniverse>0 (2 points).
True T/F
When 2 phases are in equilibrium, their chemical potentials are equal only if the equilibrium occurs at constant (P,T) (2 points).
False: chemical equilibrium is true no matter what the control variables T/F
If a system goes through a cycle with an irreversible leg, its entropy increases (2 points)
False: entropy is a state function T/F
There cannot be a critical point between a solid and a liquid (2 points)
True
T/F
For an arbitrary process of an arbitrary system, dSsystem ! 0 (2 points)
!Q so if δQ 0 , which would violate the Second Law. Notice that this is different from the situation where a higher pressure than the equilibrium value is first established and then liquid water is introduced in the room.
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Problem 3: (20 points) Petr and Saahil are having a tea party. Saahil is in charge of picking the flavor while Petr is in charge of boiling the water. Vapor bubbles form but the water is just gently simmering. Hovering, Saahil instructs: "Just raise the temperature – that will make the water boil more." Petr says "there is no way I can raise the temperature!" 1. Is Saahil correct? Why or why not? (5 points) After a while Saahil turns up the gas and the water starts boiling furiously. 2. So, was Saahil correct? Explain why water boiled more vigorously (5 points) The quarter is finally over and Petr and Saahil decide to celebrate with a trip to Lake Tahoe. At the end of a hard day of skiing, they enjoy a cool glass of water at the bottom of the slopes. The outside temperature is exactly 0°C (you can neglect the effect of atmospheric pressure on the melting point of ice). As they look at their glass of water, which has ice cubes floating in it, Petr observes: “Don’t hold your water in your hands, it will make it warmer!” Saahil, who can’t stop thinking about thermo answers: “The temperature won’t change even if I hold it in my hands”. 3. Is Saahil correct? Why or why nor? (5 points) Saahil and Petr then go grab something to eat and leave their water with ice outside. On the way back, Petr says: “I bet the whole glass froze!”. Saahil answers: “No, I bet all the ice melted!”. Prof. Salleo overhears them and states: “I bet you are both wrong!”. 4. Who is right and why? How will they find the ice water when they come back? (5 points) Answer 1. If the water is boiling, there are 2 phases at constant pressure (1 atm) therefore as long as 2 phases are present, the temperature is fixed at 100°C. Saahil was wrong. 2. Saahil was still wrong, the water boils more vigorously because the heating rate went up by turning up the gas and as a result the rate of evaporation went up as well. Do not confuse heat with temperature! 3. Saahil is wright: as long as water and ice coexist in the glass, the temperature is 0°C. The heat flowing from Saahil’s hands will progressively melt the ice at constant temperature (0°C). 4. Prof. Salleo is right: the water is at 0°C, the ambient is at 0°C so there is no heat transfer and the iced water remains in the same state as it is at equilibrium.
4
Problem 4: (20 points) A system can undergo a transformation from state A to state B following two alternative reversible paths, shown below in the T-S plane. For both cases where the work W is positive or negative, discuss along what path is W larger? Make sure you unambiguously discuss magnitude and value.
Answer In a reversible transformation, dQ=TdS. Along path 1, the integral of TdS is larger than that along path 2 therefore the system absorbed more heat (the integral is positive and is equal to the heat absorbed by the system). ΔU does not depend on the path. If W>0, then W1<W2 If W|W2|
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Problem 5: (50 points) Lead(II) sulfide (PbS) is a small band gap (Eg = 0.37 eV) material that has been used in the past for infrared detectors - it's typically sensitive enough to be used as a photon detector in the near-IR up to 2500 nm. Given the toxicity of lead to humans, it's important to have a strong understanding of the equilibria of stochiometric PbS as a function of P and T. Use the vapor pressures for PbS given in the following table to: a. Estimate the enthalpy of sublimation (Δhsub) b. Estimate the enthalpy of vaporization (Δhvap) c. Estimate the enthalpy of fusion (Δhfus) Phase Solid Solid Liquid Liquid
T [°C] 1048 1108 1221 1281
(10 points) (10 points) (10 points) P [Torr] 40.0 100 400 760
d. Estimate the temperatures and pressure of the triple point
(20 points)
You can get the majority of the credit for this problem by solving for each quantity analytically. Answer Note that T is provided in Celsius and P in Torr. This question is very similar to a problem from 204-2013, PS 3. a. Estimation of Δhsub Use the provided vapor pressures of the solid phase and the Clausius-Clapeyron equation: !P $ (h ! 1 1 $ ln # f & = ' sub ## ' && R " Tf Ti % " Pi % Rearranging to solve for Δhsub yields: $ Pf ' #Rln& ) % Pi ( "hsub = $ 1 1' && # )) % Tf Ti (
Δhsub = 231.7 [kJ mol-1]
!
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b. Estimation of Δhvap Use the provided vapor pressures of the liquid phase and the Clausius-Clapeyron equation. This is similar to Part A, but this time solving for Δhvap.
"hvap
$ Pf ' #Rln& ) % Pi ( = $ 1 1' && # )) % Tf Ti (
Δhvap = 206.5 [kJ mol-1]
!
c. Calculation of Δhfus Based on our calculations in parts (a) and (b), and since enthalpy is a state function, we can calculate Δhfus accordingly.
"h fus = "hsub # "hvap "h fus = 231.7[kJ$ mol #1 ] # 206.5[kJ$ mol #1 ] "h fus = 25.2[kJ$ mol #1 ]
!
d. Calculating the triple point pressure and temperature At the triple point, the solid and liquid have the same vapor pressures and temperatures since we have three phases in equilibrium. Thus, using the Clausius-Clapeyron relations for S-V and L-V equilibria, we have a system of two equations and two unknowns (PTP, TTP). Considering the S-V equilibrium line, we start with Clausius-Clapeyron:
!P $ (h ! 1 1$ ln # TPS & = ' sub # ' S& R " TTP T i % " Pi % (h ! 1 (h ! 1 1$ 1$ ' sub # ' S & + ln ( Pi S ) = ' vap # ' L & + ln ( Pi L ) R " TTP T i % R " TTP T i % and thus,
"hsub # 1 1& ! S ( + ln ( Pi S ) % R $ TTP T i ' S S where (Pi ,Ti ) are initial conditions from the S-V equilibrium line which can be read from the provided data table. ln ( PTP ) = !
Similarly, for the L-V equilibrium line, we start with Clausius-Clapeyron:
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!P $ (h ! 1 1$ ln # TPL & = ' vap # ' L& R " TTP T i % " Pi % and thus,
"hvap # 1 1& ! L ( + ln ( Pi L ) % R $ TTP T i ' L L where (Pi ,Ti ) are initial conditions from the L-V equilibrium line which can be read from the provided data table. ln ( PTP ) = !
Now, equate the two expressions derived for ln(PTP) and solve for TTP : "h # 1 "hsub # 1 1& 1& ! S ( + ln ( Pi S ) = ! vap % ! L ( + ln ( Pi L ) % R $ TTP T i ' R $ TTP T i ' &# "h & 1 # 1 "hsub L S ((% vap = %% ! + R ln P ! R ln P (i) ( i )(' TTP $ "hvap ! "hsub '$ Ti L Ti S
!
!1
)# &# "h &, 1 "hsub L S ((% vap TTP = +%% ! + R ln P ! R ln P (i) ( i )('.. L Ti S +*$ "hvap ! "hsub '$ Ti -
TTP = 1398 [K ] Now, plug the solution for TTP back in and solve for PTP. This can be done using the Clausius-Clapeyron relationship for both the S-V or L-V to get the answer: L
PTP = Pi e
!
"hvap # 1 1 & % ! ( R %$ TTP T iL('
S
= Pi e
!
"hsub # 1 1 & % ! ( R %$ TTP T iS('
PTP = 128 [Torr]
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Problem 6 (50 points) We have discussed in class how the entropy maximum principle describes the direction of spontaneous change. We have also noticed how most students are more familiar with an “energy minimum principle” than with an “entropy maximum principle”. In your notes, I describe how the “energy minimum principle” is still a form of the Second Law, from which the entropy maximum principle descends. 1) What special conditions must apply for the “energy minimum principle” to apply? (2.5 points) 2) In those conditions, if the “energy minimum principle” is violated, is the Second Law violated as well? (2.5 points) You will now solve a problem that should convince you that the entropy maximum principle is the most fundamental way to determine equilibrium. In order to make sure you are convinced, I have enlisted Newton and Boltzmann…
Newton sets-up a pendulum, located in an unstable equilibrium situation (see Figure): he calls this the initial state A. The pendulum is located in a room at 300K, which you can consider a heat bath. Newton is a brilliant instrument maker and is able to make pendulum that exhibits no friction at all, not at the pivot point and not with the atmosphere. 3) What is the stable equilibrium point B of the pendulum in terms of θ? (5 points) 4) What is ΔSpendulum of the process that takes the pendulum from A to B? (5 points) 5) What equilibrium condition applies to the pendulum? (5 points) Now Prof. Salleo, who is not such a good engineer, modifies the pendulum. After the modification, the pivot point generates friction (you can neglect the friction with the atmosphere). Friction is typically proportional to the swinging angular speed. Newton gives a little push to the pendulum to perturb it away from A. During the swinging process the amount of heat generated at the pivot point until the pendulum comes back to rest at its equilibrium position is Q.
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6) Did the equilibrium state of the pendulum change from the frictionless case? What do you conclude about its condition of equilibrium when friction is considered compared to the frictionless case? (5 points) 7) What is ΔSpendulum in this case? (5 points) 8) How would you calculate ΔSUniverse after the pendulum is at rest? (5 points) Newton and Boltzmann start discussing about energy conservation. Newton states that everything he understands about mechanics leads him to believe that if he provided the same amount of heat Q to the pivot point, the pendulum could absorb the heat and start swinging as energy would be conserved. 9) Does the process imagined by Newton violate the First Law? If so, how? If not, why do you think Newton is wrong? (5 points) Boltzmann explains to Newton that a different principle must apply in order to predict what processes occur spontaneously. 10) Can you state the principle Botzmann refers to? (2 points) 11) If you are siding with Boltzmann, show that if the pendulum started swinging once Newton provided heat Q, this would violate the Second Law. (5 points) 12) Bonus Question: is it always impossible to convert heat into work? Explain. (3 points) Answer 1. Constant S and V 2. Yes, the energy minimum principle is derived from the Second Law 3. θ=180° 4. The pendulum does not exchange any heat because there is no friction. Whether the process is spontaneous or not, this is always true so ΔSpendulum=0 5. S and V are constant: energy minimum principle 6. The state of equilibrium is the same (θ=180°) and so the condition of equilibrium is the same for the pendulum. 7. Same initial and final state as for the frictionless case: ΔSpendulum=0 (S is a state function) 8. The surroundings are a thermal reservoir ΔSsurroundings=ΔSUniverse=Q/T>0 9. The process imagined by Newton conserves energy so it does not violate the First Law. Newton is wrong because the First Law does not allow him to predict possible processes, he needs to apply the Second Law for that. 10. It’s the Second Law or the entropy maximum principle 11. If the pendulum started swinging, it would still have ΔSpendulum=0 but ΔSsurroundings=ΔSUniverse=-Q/T
Write your answers neatly and clearly. I will not give credit to statements that are not clearly justified (in other words, make sure your handwriting is better than mine!). Class average: 76/100 (=125/165) Standard deviation: 14 (=23)
1
Problem 1: (10 points) Answer the following questions with True or False. If the statement is False, write a 1line explanation of why you think the statement is false or you won’t get credit. Write the explanation on this sheet and don’t forget to turn it in! T/F
For any observable process, ΔSUniverse>0 (2 points).
True T/F
When 2 phases are in equilibrium, their chemical potentials are equal only if the equilibrium occurs at constant (P,T) (2 points).
False: chemical equilibrium is true no matter what the control variables T/F
If a system goes through a cycle with an irreversible leg, its entropy increases (2 points)
False: entropy is a state function T/F
There cannot be a critical point between a solid and a liquid (2 points)
True
T/F
For an arbitrary process of an arbitrary system, dSsystem ! 0 (2 points)
!Q so if δQ 0 , which would violate the Second Law. Notice that this is different from the situation where a higher pressure than the equilibrium value is first established and then liquid water is introduced in the room.
3
Problem 3: (20 points) Petr and Saahil are having a tea party. Saahil is in charge of picking the flavor while Petr is in charge of boiling the water. Vapor bubbles form but the water is just gently simmering. Hovering, Saahil instructs: "Just raise the temperature – that will make the water boil more." Petr says "there is no way I can raise the temperature!" 1. Is Saahil correct? Why or why not? (5 points) After a while Saahil turns up the gas and the water starts boiling furiously. 2. So, was Saahil correct? Explain why water boiled more vigorously (5 points) The quarter is finally over and Petr and Saahil decide to celebrate with a trip to Lake Tahoe. At the end of a hard day of skiing, they enjoy a cool glass of water at the bottom of the slopes. The outside temperature is exactly 0°C (you can neglect the effect of atmospheric pressure on the melting point of ice). As they look at their glass of water, which has ice cubes floating in it, Petr observes: “Don’t hold your water in your hands, it will make it warmer!” Saahil, who can’t stop thinking about thermo answers: “The temperature won’t change even if I hold it in my hands”. 3. Is Saahil correct? Why or why nor? (5 points) Saahil and Petr then go grab something to eat and leave their water with ice outside. On the way back, Petr says: “I bet the whole glass froze!”. Saahil answers: “No, I bet all the ice melted!”. Prof. Salleo overhears them and states: “I bet you are both wrong!”. 4. Who is right and why? How will they find the ice water when they come back? (5 points) Answer 1. If the water is boiling, there are 2 phases at constant pressure (1 atm) therefore as long as 2 phases are present, the temperature is fixed at 100°C. Saahil was wrong. 2. Saahil was still wrong, the water boils more vigorously because the heating rate went up by turning up the gas and as a result the rate of evaporation went up as well. Do not confuse heat with temperature! 3. Saahil is wright: as long as water and ice coexist in the glass, the temperature is 0°C. The heat flowing from Saahil’s hands will progressively melt the ice at constant temperature (0°C). 4. Prof. Salleo is right: the water is at 0°C, the ambient is at 0°C so there is no heat transfer and the iced water remains in the same state as it is at equilibrium.
4
Problem 4: (20 points) A system can undergo a transformation from state A to state B following two alternative reversible paths, shown below in the T-S plane. For both cases where the work W is positive or negative, discuss along what path is W larger? Make sure you unambiguously discuss magnitude and value.
Answer In a reversible transformation, dQ=TdS. Along path 1, the integral of TdS is larger than that along path 2 therefore the system absorbed more heat (the integral is positive and is equal to the heat absorbed by the system). ΔU does not depend on the path. If W>0, then W1<W2 If W|W2|
5
Problem 5: (50 points) Lead(II) sulfide (PbS) is a small band gap (Eg = 0.37 eV) material that has been used in the past for infrared detectors - it's typically sensitive enough to be used as a photon detector in the near-IR up to 2500 nm. Given the toxicity of lead to humans, it's important to have a strong understanding of the equilibria of stochiometric PbS as a function of P and T. Use the vapor pressures for PbS given in the following table to: a. Estimate the enthalpy of sublimation (Δhsub) b. Estimate the enthalpy of vaporization (Δhvap) c. Estimate the enthalpy of fusion (Δhfus) Phase Solid Solid Liquid Liquid
T [°C] 1048 1108 1221 1281
(10 points) (10 points) (10 points) P [Torr] 40.0 100 400 760
d. Estimate the temperatures and pressure of the triple point
(20 points)
You can get the majority of the credit for this problem by solving for each quantity analytically. Answer Note that T is provided in Celsius and P in Torr. This question is very similar to a problem from 204-2013, PS 3. a. Estimation of Δhsub Use the provided vapor pressures of the solid phase and the Clausius-Clapeyron equation: !P $ (h ! 1 1 $ ln # f & = ' sub ## ' && R " Tf Ti % " Pi % Rearranging to solve for Δhsub yields: $ Pf ' #Rln& ) % Pi ( "hsub = $ 1 1' && # )) % Tf Ti (
Δhsub = 231.7 [kJ mol-1]
!
6
b. Estimation of Δhvap Use the provided vapor pressures of the liquid phase and the Clausius-Clapeyron equation. This is similar to Part A, but this time solving for Δhvap.
"hvap
$ Pf ' #Rln& ) % Pi ( = $ 1 1' && # )) % Tf Ti (
Δhvap = 206.5 [kJ mol-1]
!
c. Calculation of Δhfus Based on our calculations in parts (a) and (b), and since enthalpy is a state function, we can calculate Δhfus accordingly.
"h fus = "hsub # "hvap "h fus = 231.7[kJ$ mol #1 ] # 206.5[kJ$ mol #1 ] "h fus = 25.2[kJ$ mol #1 ]
!
d. Calculating the triple point pressure and temperature At the triple point, the solid and liquid have the same vapor pressures and temperatures since we have three phases in equilibrium. Thus, using the Clausius-Clapeyron relations for S-V and L-V equilibria, we have a system of two equations and two unknowns (PTP, TTP). Considering the S-V equilibrium line, we start with Clausius-Clapeyron:
!P $ (h ! 1 1$ ln # TPS & = ' sub # ' S& R " TTP T i % " Pi % (h ! 1 (h ! 1 1$ 1$ ' sub # ' S & + ln ( Pi S ) = ' vap # ' L & + ln ( Pi L ) R " TTP T i % R " TTP T i % and thus,
"hsub # 1 1& ! S ( + ln ( Pi S ) % R $ TTP T i ' S S where (Pi ,Ti ) are initial conditions from the S-V equilibrium line which can be read from the provided data table. ln ( PTP ) = !
Similarly, for the L-V equilibrium line, we start with Clausius-Clapeyron:
7
!P $ (h ! 1 1$ ln # TPL & = ' vap # ' L& R " TTP T i % " Pi % and thus,
"hvap # 1 1& ! L ( + ln ( Pi L ) % R $ TTP T i ' L L where (Pi ,Ti ) are initial conditions from the L-V equilibrium line which can be read from the provided data table. ln ( PTP ) = !
Now, equate the two expressions derived for ln(PTP) and solve for TTP : "h # 1 "hsub # 1 1& 1& ! S ( + ln ( Pi S ) = ! vap % ! L ( + ln ( Pi L ) % R $ TTP T i ' R $ TTP T i ' &# "h & 1 # 1 "hsub L S ((% vap = %% ! + R ln P ! R ln P (i) ( i )(' TTP $ "hvap ! "hsub '$ Ti L Ti S
!
!1
)# &# "h &, 1 "hsub L S ((% vap TTP = +%% ! + R ln P ! R ln P (i) ( i )('.. L Ti S +*$ "hvap ! "hsub '$ Ti -
TTP = 1398 [K ] Now, plug the solution for TTP back in and solve for PTP. This can be done using the Clausius-Clapeyron relationship for both the S-V or L-V to get the answer: L
PTP = Pi e
!
"hvap # 1 1 & % ! ( R %$ TTP T iL('
S
= Pi e
!
"hsub # 1 1 & % ! ( R %$ TTP T iS('
PTP = 128 [Torr]
8
Problem 6 (50 points) We have discussed in class how the entropy maximum principle describes the direction of spontaneous change. We have also noticed how most students are more familiar with an “energy minimum principle” than with an “entropy maximum principle”. In your notes, I describe how the “energy minimum principle” is still a form of the Second Law, from which the entropy maximum principle descends. 1) What special conditions must apply for the “energy minimum principle” to apply? (2.5 points) 2) In those conditions, if the “energy minimum principle” is violated, is the Second Law violated as well? (2.5 points) You will now solve a problem that should convince you that the entropy maximum principle is the most fundamental way to determine equilibrium. In order to make sure you are convinced, I have enlisted Newton and Boltzmann…
Newton sets-up a pendulum, located in an unstable equilibrium situation (see Figure): he calls this the initial state A. The pendulum is located in a room at 300K, which you can consider a heat bath. Newton is a brilliant instrument maker and is able to make pendulum that exhibits no friction at all, not at the pivot point and not with the atmosphere. 3) What is the stable equilibrium point B of the pendulum in terms of θ? (5 points) 4) What is ΔSpendulum of the process that takes the pendulum from A to B? (5 points) 5) What equilibrium condition applies to the pendulum? (5 points) Now Prof. Salleo, who is not such a good engineer, modifies the pendulum. After the modification, the pivot point generates friction (you can neglect the friction with the atmosphere). Friction is typically proportional to the swinging angular speed. Newton gives a little push to the pendulum to perturb it away from A. During the swinging process the amount of heat generated at the pivot point until the pendulum comes back to rest at its equilibrium position is Q.
9
6) Did the equilibrium state of the pendulum change from the frictionless case? What do you conclude about its condition of equilibrium when friction is considered compared to the frictionless case? (5 points) 7) What is ΔSpendulum in this case? (5 points) 8) How would you calculate ΔSUniverse after the pendulum is at rest? (5 points) Newton and Boltzmann start discussing about energy conservation. Newton states that everything he understands about mechanics leads him to believe that if he provided the same amount of heat Q to the pivot point, the pendulum could absorb the heat and start swinging as energy would be conserved. 9) Does the process imagined by Newton violate the First Law? If so, how? If not, why do you think Newton is wrong? (5 points) Boltzmann explains to Newton that a different principle must apply in order to predict what processes occur spontaneously. 10) Can you state the principle Botzmann refers to? (2 points) 11) If you are siding with Boltzmann, show that if the pendulum started swinging once Newton provided heat Q, this would violate the Second Law. (5 points) 12) Bonus Question: is it always impossible to convert heat into work? Explain. (3 points) Answer 1. Constant S and V 2. Yes, the energy minimum principle is derived from the Second Law 3. θ=180° 4. The pendulum does not exchange any heat because there is no friction. Whether the process is spontaneous or not, this is always true so ΔSpendulum=0 5. S and V are constant: energy minimum principle 6. The state of equilibrium is the same (θ=180°) and so the condition of equilibrium is the same for the pendulum. 7. Same initial and final state as for the frictionless case: ΔSpendulum=0 (S is a state function) 8. The surroundings are a thermal reservoir ΔSsurroundings=ΔSUniverse=Q/T>0 9. The process imagined by Newton conserves energy so it does not violate the First Law. Newton is wrong because the First Law does not allow him to predict possible processes, he needs to apply the Second Law for that. 10. It’s the Second Law or the entropy maximum principle 11. If the pendulum started swinging, it would still have ΔSpendulum=0 but ΔSsurroundings=ΔSUniverse=-Q/T
