S A 0.6. Units of J/mol K S U /N

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Bryce Mosley
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1 Solutions to Problem Set 5 Exercise 2. Consider heating a body A, of constant heat capacity J/ C and initially at temperature K, to a nal temperature of 2K. The heating takes place by sequentially placing it in thermal contact with N dierent large thermal reservoirs (so large that there temperature does not change during thermal contact). For example, N= The body A is placed in contact with a single reservoir at 2K. N=2 The body A rst attains thermal equilibrium with a one reservoir at T =5K subsequently, it attains thermal equilibrium with the second at T =2K. N = M The body A rst attains thermal equilibrium with a one reservoir at T = ( + =M)K. Next, it attains equilibrium with a reservoir at T =(+2=M)K. :::. Next, it attains equilibrium with a reservoir at T = ( + j=m)k. :::. Finally, A attains equilibrium with a reservoir at T = ( + M=M)K. On the same graph, plot the change in () entropy of the body A and (2) the change in entropy of the universe as a function of =N. Solution: In general, at constant pressure with constant C P where A stands for the body A, u for the universe and r for the reservoir, T is the initial temperature and is the nal temperature, the change in entropy of the body Aisgiven by: 4S A = Z Tf T C P T dt = C P ln( T ) and the change in entropy of the universe is given by: 4S u = 4S A + 4S r = 4S A + 4H r T r Using First Law: 4H r = ;4H A = Z Tf T C P dt = C P ( ; T ) So, the nal expression for the change in entropy of the universe is: 4S u = 4S A ; Cp ; T We calculate rst the change in entropy of the body A varying the number of reservoirs between the K and 2 K interval:

2 For N =,4S A = C P ln( 2 K K )=C P ln(2) ; For N =2,4S A = C P ln( 5 K 2 K )+ln( ) = C K 5 K P ln( 52)=C 5 P ln(2) ; For N =4,4S A = C P ln( 25 K 5 K 75 K )+ln( )+ln( K 25 K 5 K )+ln(2 K ) = C 75 K P ln( )= C P ln(2) For N = M, 4S A = C P ln(2) This means that the change in entropy of system A is path independent, i.e. it is a state function whose value is given by the initial and nal conditions. For the change in entropy of the universe we have: For N =: 4S u = 4S A ; C P 2; 2 For N =2: 4S u = 4S A ; C 5; P + 2;5 5 2 For N =4: 4S u = 4S A ; C 25; P + 5; ;5 + 2; For N = M: 4S u = 4S A ; C P P N j (+ j j; );(+ P N N ) = 4S (+ j A ; C N P N ) j j+n In the limit when N is large, the sum in the above expression converges to the value of C P ln(2). Therefore, in the limit of a quasistatic process, the entropy change in the universe is zero. The change in the Gibbs free energy of the system is also zero (local condition for equilibrium). This indicates that a quasistatic process follows a path of equilibrium processes: S A Units of J/mol K S U /N Exercise 2.2 Consider the phase transformation of pure carbon graphite to pure carbon diamond at atmospheric pressure and at temperatures between 298 and 2K. Using the data: 2

3 graphite Standard molar enthalpy H(T =298 P =atm)= graphite Molar entropy S(T =298 P = atm) =5:694 J mole K graphite Molar heat capacity C P (T =298 P =atm)=7:2+4:27t ; J 8:79;5 T 2 mole K diamond Standard molar enthalpy H(T =298 P =atm) = 9 mole J diamond Molar entropy S(T =298 P = atm) =2:44 J mole K diamond Molar heat capacity C P (T =298 P =atm)=9:2 + 3:2T ; J 6:9;5 T 2 mole K. Draw an accurate plot of the molar change in molar enthalpy for graphite transforming to diamond at atm pressure for temperatures between 298 and 2K. 2. Draw an accurate plot of the molar change in Gibb's free energy for graphite transforming to diamond at atm pressure for temperatures between 298 and 2K. 3. Draw an accurate plot of each of the molar free energies for diamond and for graphite as a function of the entropy of the system at atm pressure for temperatures between 298 and 2K. Solution:. We rst must nd an expression for the enthalpies of diamond and graphite.using C we have:h = T i C P dt + H(T i ) H graphite =7:2 +2:35T 2 f + :879e ; 4 ; 94722:4 H diamond =9:2 +6:6T 2 f + :69e ; 4 ; :6 4H = ;8:8 +4:465T 2 f ; :26e ; 4 ; 39222:2 7 x T, K J/mol Figure : 4H vs. T 3

4 2. To calculate 4G we need rst to calculate 4S. By using S = R T i C P T dt + S T i : S graphite =4:27 + :4395e ; 4 T 2 f S diamond =3:2 + :395e ; 4 T 2 f 4S =8:93 ; :3e ; 4 T 2 f +7:2ln( ) ; 364: :2 ln( ) ; 3983:7493 ; 8:8 ln( ) ; 268:36485 The Gibbs free energies can be calculated using G = H ; TS: G graphite = 38:9568 ; 2:35T 2 f + :4395e ; 4 ; 94722:4 ; 7:2 ln( ) G diamond = 3992: ; 6:6T 2 f + :395e ; 4 ; :6 ; 9:2 ln( ) 4G = 26:28485 ; 4:465T 2 f ; :3e ; 4 ; 39222:2 + 8:8 ln( ) 4

5 .5 x Equilibrium Phase Transition J/mol.5 2 J/mol T, K T,K Figure 2: 4G vs. T It is evident from g 2 that there occurs a phase transition. This transition is located at the point where 4G =. From the algebraic expression for 4G we nd that T T ransition ' 323:755K. 3. The parametric plots of G(T )vs.s(t ): Exercise 2.3 A quantity of super-cooled liquid Mn at 8K is adiabatically solidied into its equilibrium solid phase at constant pressure. From the data below, calculate what fraction of the Mn will have solidied and the nal temperature of the system. Transition (phase! phase 2) H ( J mole )! 2 993! ! 8 49! liquid T trans eq (K) 5

6 G graphite (T) J/mol Graphite x S graphite (T) J/mol K G graphite (T) J/mol Diamond x S graphite (T) J/mol K Figure 3: G(T ) vs. S(T )forboth graphite and diamond phase of Mn C p ( J mole K ) temperature range (K) 2:6+5:9T :9+2:8T : : liquid 46: 57-T boil Solution: The rst thing to do is use H = R T i C P dt + H(T i ) to calculate the enthalpies for all the phases. Note that H(T i ) for all phases other than would be H previous phase (T T ransition )+ 4H T ransition : H = ;72428:6+2:6 T +7:95 T 2 for T =298:::993 K H = : :9 T +:4 T 2 for T = 993 :::373 K H = : :8 or T =373:::49 K H = 8366: :3 or T = 49 :::57 K H liquid = : : or T = 57 :::T K We are told that liquid is cooled to 8 K without allowing it to solidify, This means that 6

7 we are not allowing for any nucleation to happen. By using a linear t, an expression for the enthalpy of the liquid as a function of temperature can be obtained: H liq =8: e6+4:6 T It is possible to extrapolate this expression to 8 K, since this enthalpy is calculated using material properties. We are told that the liquid solidies adiabatically at 8 K, therefore, by using First Law: dh = dq + V dp = This means that once the liquid has reached 8 K, theenthalpy of the system remains constant. Many people used this expression to establish that since 4H =, there would be no phase transformation whatsoever, which is wrong, since we are allowing the liquid to solidify to the equilibrium phase: What 4H = means is that there will be no heat released to the environment, and therefore all the heat released by the solidication (;4H melt ) would remain in the system. Therefore, once the liquid solidies into, it needs to raise its temperature and further transform into other phases (if necessary) in order to keep the initial enthalpy unchanged. This might be better understood by examining the next gure: >From g 4 it can be seen that the liquid transforms entirely into the phase, and the nal temperature reached by the system is T 37 K. The nal temperature can be precisely calculated by equating the enthalpy of the metastable liquid at 8 K to the enthalpy of the phase: H liquid 8K =8:426295e6 = : :9 T +:4 T 2 By solving this quadratic equation, the nal temperature of the system is calculated to be T =37:2 K Note: The clever way to solve this problem was by graphically realizing where the nal state of the system was supposed to be and then estimate what the nal temperature of the system was. 7

8 9 x α β γ δ liquid 8.45 x α β γ δ liquid liq e xp H, J/mol 5 4 H, J/mol T,K T,K Figure 4: Change in enthalpy of the system as a function of Temperature. The arrow points to the nal state of the system. 8

3.012 PS Issued: Fall 2004 Due: pm

3.012 PS Issued: Fall 2004 Due: pm 3.012 PS 2 3.012 Issued: 09.15.04 Fall 2004 Due: 09.22.04 5pm Graded problems: 1. In discussing coordination numbers and deriving the permitted range of radius ratio, R A / R B, allowed for each ( where