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Chem340 Physical Chemistry for Biochemists Exam 2 Mar 16, 2011 Your Name _ I affirm that I have never given nor received aid on this examination. I understand that cheating in the exam will result in a grade F for the class. Signature . Sec 1 Multiple Choice [ /42] Sec 2 Calculation/MC [ /32] Sec 3 Long Question [ /45] Sec 4 Derivation [ /12] Total [ /100] Total points are 131 points. If you score more than 100, you will receive 100 points.

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Multiple Choice Questions (Some questions may be similar to those one in quiz, but not the same) Each 3 points [ /42]

Q1. One mole of an ideal gas with CP = 5R/2 undergoes the transformations from an initial state described by Ti = 300. K and Pi = 1.50 bar. The gas undergoes a reversible isothermal compression at 300 K until P becomes 1.0 bar. Choose a formula that best represents S.

(a) S = (5R/2)ln(1.5/1) (b) S = (5R/2)ln(1/1.5) (c) S = Rln(1/1.5) (d) S = Rln(1.5/1)

Q2. You have containers of pure H2 and He at 298 K and 1 atm pressure. Calculate Gmixing relative to the unmixed gases for a mixture of 1.5 mol of H2 and 0.5 mol of He. Choose the most appropriate formula for Gmixing

(a) RT{3/4ln(3/4) + 1/4ln(1/4)} (b) 2RT{3/4ln(3/4) + 1/4ln(1/4)} (c) 0

(d) RT{1/4ln(3/4) + 3/4ln(1/4)} (e) 2RT{1/4ln(3/4) + 3/4ln(1/4)} Q3. One mole of N2 at 600 K and one mole of N2 at 300K were thermally contacted in an adiabatic container of a fixed volume. At the end, the system reached the equilibrium at 450K. Choose two correct answers below.

(a) The entropy change is Cv{ln(450/600) + ln(450/300)}

(b) The entropy change is -Cv{ln(450/600) + ln(450/300)} (c) The entropy change is 0.

(d) The change in the internal energy is positive

(e) The change in the internal energy is negative (f) The change in the internal energy is zero. Q4. Consider the formation of glucose from carbon dioxide and water, that is, the reaction of the following photosynthetic process: CO2(g) + H2O(l) (1/6)C6H12O6(s) + O2(g). Choose a formula that best represents the reaction entropy for this chemical system at T = 348 K. The following table of information will be useful in working this problem:

(a) 50x (-213.8 -70 + (1/6) x 209.2 + 205.2) + (-6x 37.1 -75.3 + (1/6) x 219.2 + 29.4) Jmol-1K-1 (b) (-6x 213.8 -6x 70 + 209.2 + 6x 205.2) + 50x (-6x 37.1 -6x 75.3 + 219.2 + 6x 29.4) Jmol-1K-1 (c) (-213.8 -70 + (1/6)x 209.2 + 205.2)ln(348/298) + (-37.1 -75.3 + (1/6)x 219.2 + 29.4) Jmol-1K-1 (d) (-6x 213.8 -6x 70 + 209.2 + 6x 205.2) + (-6x 37.1 -6x 75.3 + 219.2 + 6x 29.4)ln(348/298) Jmol-1K-1

(e) None of them

Q5. Which formula best represents the reaction enthalpy at 298 K for the reaction in Q4 in kJmol-1 units? (a) (-6x 393.5 - 6x 285.8 - 1273.1) (b) (393.5 + 285.8 - (1/6)x1273.1)

(c) (-6x 393.5 - 6x 285.8 + 1273.1) (d) (-393.5 - 285.8 + (1/6)x1273.1

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Q6. Which relationship is correct for an irreversible process? Choose all correct answer(s).

(a) T

Dq leirreversib0 (b) 0dS (c) T

DqdS leirreversib

(d) T

DqdS leirreversib (e) 0dS (f)

T

DqdS leirreversib

Q7. We calculated G for two expansion processes of 10.00 mol of an ideal gas at 300 K from an initial pressure of 1.0 bar at 1.0 L to a final volume of 10.0 L. In process X, we obtained GX for the isothermal reversible expansion. In process Y, we obtained GY for the irreversible isothermal expansion at the external pressure of Pext = 0.05 bar? CP, m = (5/2)R. What is GY/GX?

(a) 7/5 (b) 5/3 (c) 3/2 (d) 1 (e) 2/3 (f) 3/5 (g) 5/7 Q8. Choose the most appropriate reaction for which a standard formation enthalpy for NH3 can be obtained as a reaction enthalpy.

a. NH3(g) N(g) + 3H(g) b. 1/2N2(g) + 3/2H2(g) NH3(g) c. 1/3NH3(g) 1/3N(g) + H(g) d. N2(g) + 3H2(g) 2NH3(g)

Q9. For the following reaction, ΔUreaction = ΔHreaction + X. What is X?

TiCl4(l) + 2H2O(l) TiO2(s) + 4HCl(g)

(a) 3RT (b) 4RT (c) -3RT (d) -4RT

Q11. In the case of this figure, melting point Tm and boiling point Tb are changed as follows when P is increased from P0 to P0 + P. Choose one correct answer. a) Tm and Tb are both decreased. b) Tm is increased while Tb is decreased. c) Tm is decreased while Tb is increased. d) None of all

Q12 Choose ALL correct equations. (a) U = TdS - PdV (b) H = TdS + VdP

(c) G = -SdT - PdV (d) A = -SdT + VdP

(e) (G/P)T = V (f) (A/T)V = -S Q12. A choice of (e), (f) is also correct since “d” was missing from dU & dH in (a) & (b) by our mistake.

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Q12. Choose ALL correct statements on the right diagram.

(a) Vm(sl) > 0 (b) Vm(sl) < 0 (c) Sm(sl) > 0

(d) Sm(sl) < 0

(e) Line b denotes a phase transition from liquid to solid through the melting point.

(f) Sublimation (s g) does not take place for this system when the temperature is above that for triple point.

(g) This system undergoes a phase transition as g l s when the pressure is increased at constant T which is a little below the triple point temperature. Q13 The figure right shows a DSC scan of a solution of a T4 lysozyme mutant. Which of the following of [A-D] represents (i) melting temperature Tm and (ii) intrinsic excess heat capacity? Choose the most suitable combination. (i, ii) =

(a) (A, B) (b) (D, A) (c) (D, B) (d) (D, C) (e) (A, C) (f) (A, B)

Q14. Figure in the right shows Gibbs energy (Gmixture) for a mixture of mol of N2O4 and 2(1-) mol of NO2 in a reaction of 2NO2(g) ↔ N2O4(g) (*). How much is the molar Gibbs energy of pure N2O4. Choose the closest.

(a) -81 kJmol-1 (b) -79.5 kJmol-1 (c) -77 kJmol-1

(d) -38 kJmol-1 (e) -4 kJmol-1 (f) 0 kJmol-1

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2. Calculation/Multiple Choice Question each 4 point [ /32 points]

2.1 One mole of N2 at 600K and 6.00 bar undergoes a transformation to the state described by 300K and 3.00 bar. Calculate S if Cp,m /(Jmol-1K-1) = a + b(T/K) + c(T/K)2. Choose the most appropriate formula that represents S.

(a) Rln2 + a(Tf –Ti)/K + b{(Tf/K)2 – (Ti/K)2}/2 + c{(Tf/K)3 – (Ti/K)3}/3

(b) - Rln2 + a(Tf –Ti)/K + b{(Tf/K)2 – (Ti/K)2}/2 + c{(Tf/K)3 – (Ti/K)3}/3

(c) Rln2 + a ln(Tf/Ti) + b{(Tf/K) – (Ti/K)} + c{(Tf/K)2 – (Ti/K)2}/2

(d) -Rln2 + a ln(Tf/Ti) + b{(Tf/K) – (Ti/K)} + c{(Tf/K)2 – (Ti/K)2}/2

(e) Rln2 + Cp,m[a ln(Tf/Ti) + b{(Tf/K) – (Ti/K)} + c{(Tf/K)2 – (Ti/K)2}/2]

(f) -Rln2 + Cp,m[a ln(Tf/Ti) + b{(Tf/K) – (Ti/K)} + c{(Tf/K)2 – (Ti/K)2}/2] (g) None of them

2.2. 2.3 Consider the reversible Carnot cycle (a b c d a) with 1 mol of an ideal gas with CV = 3/2R as the working substance. The initial isothermal reversible expansion occurs at the hot reservoir temperature of Thot = 600 K (Ta, Tb) from an initial volume of 5.0 L (Va) to a volume of 10.0 L (Vb). The system then undergoes an adiabatic reversible expansion until the temperature falls to Tcold = 300 K (Tc, Td). The system then undergoes an isothermal reversible compression until V= Vd. Finally, the system is subject to a subsequent adiabatic reversible compression until the initial state described by Ta = 600 K and Va = 5.0 L is reached. (from P5.2, 5.3)

2.2 How much is the sum of the work (wbc + wda) in the adiabatic process? Choose the closest.

(a) -11.0 kJ (b) -5.5 kJ (c) 0 kJ (d) 5.5 kJ (e) 11 kJ

2.3 How much is ΔHda? Choose the closest.

(a) -9.5 kJ (b) -3.6 kJ (c) 0 kJ (d) 3.6 kJ (e) 9.5 kJ

2.4 Calculate A and G for the isothermal irreversible compression of 1.00 mol of an ideal gas at 298 K from an initial volume of 35.0 L to a final volume of 12.0 L. Choose the closest value for each of A and G from (a-e) and (f-j), respectively. (2 points if only one is correct)

A (a) 5.5 kJ (b) 2.5 kJ (c) 0kJ (d) -2.5 kJ (e) -5.5 kJ

G (f) 5.5 kJ (g) 2.5 kJ (h) 0kJ (i) -2.5 kJ (j) -5.5 kJ

2.5 From the following data at 25°C, calculate the standard enthalpy of formation of Fe2O3(s) and choose the closest value.

Hreaction

kJ mol1 Fe2O3(s) + 3C(graphite) 2Fe(s) + 3CO(g) 492.6

FeO(s) + C(graphite) Fe(s) + CO(g) 155.8

C(graphite) + O2(g) CO2(g) –393.51

CO(g) + 1/2 O2(g) CO2(g) –282.98

(a) -900 kJmol-1 (b) -620 kJmol-1 (c) -400kJmol-1 (d) -150 kJmol-1 (e) 0 kJmol-1 (f) 150 kJmol-1

(f) 400 kJmol-1(g) 620 kJmol-1 (h) 920 kJmol-1

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2.6. Calculate O2

mixture 298.15 K, 1 bar for oxygen in air diluted by helium, assuming that the molar

fraction of O2 in air is 0.100 and O2 behave as an ideal gas. The molar entropy of O2, Sm0 is 205.2

Jmol-1K-1. Choose the closest. (a) -66.9 kJmol-1 (b) -64.9 kJmol-1 (c) -62.9 kJmol-1 (d) 62.9 kJmol-1 (e) 64.9 kJmol-1 (f) 66.9 kJmol-

2.7. 1 mole of steam (H2O(g)) at 500K was gradually cooled to 300K, while forming water, H2O(l) at the boiling point Tb. Assume that Hvap denotes a molar vaporization enthalpy of water, and heat capacities for steam and water, Cp(steam) and Cp(water) are constant over the temperature. Choose the most appropriate expression that corresponds to S during the process.

(a) bpbpbvap TKwaterCTKsteamCTH 300)(500)(/ (b) bpbpbvap TKwaterCKTsteamCTH 300)(500)(/ (c) bpbpbvap TKwaterCTKsteamCTH /300ln)(/500ln)(/ (d) bpbpbvap TKwaterCKTsteamCTH /300ln)(500/ln)(/

2.8. One mole of a van der Waals gas at T (300K) is expanded isothermally and reversibly from an initial volume of Vi to a final volume of Vf . For the van der Waals gas, U / V

T a / Vm

2. Assuming that b is 0,

choose the right formula that represents S. Use van der Waals gas equation P = RT/Vm – a/Vm2, where b=0.

(a) T/

ifi

f

VV a

V

VlnR

112 (b)

T/

ifi

f

VV a

V

VlnR

11 (c)

i

f

V

VlnR (d)

T/

ifi

f

VV a

V

VlnR

11

(e) T/

ifi

f

VV a

V

VlnR

112

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Calculation Question. Choose 3 questions among the following 4 questions. If you solve 4 without choosing, we will grade for 3 questions with the lowest scores. You will gain considerable points by indicating the CORRECT formula. [ /45]

Write answers in the specified place. You should show calculations for full credit and partial credit. Please include units in calculations. 3.1 [Circle if you choose this] (Original Question P4.18) [ /15]

P4.18) Given the data in the following information, calculate the single bond enthalpies and energies for C–F as follows: (a) List all the reactions needed for calculations with their formation enthalpies. (6 points) (b) By combining the reactions in (a) using Hess’s law, obtain the reaction from which you can elucidate the single bond enthalpy for C-F as a reaction enthalpy, and calculate the single bond enthalpy. (6 points) (c) Calculate the single bond energy for C-F (3 points). Assume that the standard temperature is 298 K (rather than 298.15K). Show equations for a major credit.

Substance F2(g) F(g) CF4(g) NF3(g) C(g) N(g)

H f

kJ m ol1 0 79 –925 –125 717 472

(a) g CF g F 2graphiteC 42 -1

reaction mol kJ 259ΔH

g F g 1/2F2 -1reaction molkJ 9ΔH 7

gC graphiteC -1reaction mol kJ 716ΔH

2 point each for showing a reaction with its formation enthalpy (total 6 points)

(b, c) g F 2graphite C g CF 24 -1reaction mol kJ 925ΔH

g F 4 g F 2 2 -1reaction mol kJ 974ΔH

g C graphiteC -1reaction mol kJ 717ΔH

Additional 2 points (1 each) for balancing as shown above for CF4(g) & F(g) to produce the reaction below

g F 4g C g CF4 -1reaction mol kJ 1959ΔH (2 points)

The average C-F bond enthalpy is then:

1--1

bondavrg mol kJ 490

4

mol kJ 1959H (equation 1 point; calculation 1 point)

(c) And the average C-F bond free energy:

1-

-1-1-1reactionreaction

mol kJ 1949

K 298.15mol K J 8.3144mol kJ 959.01T Rn ΔHΔU

(eq. 1 point;

calculation 1 point)

1--1

bondavrg mol kJ 487

4

mol kJ 0.1949U (1 point for equation)

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3.2 [Circle if you choose this] (Original Question P5.13) [ /15] One mole of an ideal gas with CP,m = 5/2 R undergoes the transformations described in the following list from an initial state described by Ti = 300. K and Pi = 4.00 bar. The gas undergoes an adiabatic expansion against a constant external pressure of Pext until the final pressure Pf ==1.00 bar. (a) Calculate q, w, and S when Pext = 0.00 (6 points). (b) Obtain the temperature after the expansion (Tf) and S when Pext = 1.00 bar. (6 points) Show a formula first and then calculate the value. (c) Calculate S when the gas undergoes an adiabatic reversible expansion (in place of a constant external pressure Pext) until the final pressure Pf ==1.00 bar. What is the final temperature Tf? (3 points)

(a) wU

i

i

f

fexternalifmV, p

T

p

T pR nTT C n

Since Pexternal=0, w = 0. (1 point) From this, U= 0 using the equation above. q = U –w = 0. (1 point) (Alternatively, q = 0 since this is adiabatic process. Thus, U= 0 Tf =Ti ) Because U= 0, Tf = Ti. S = -nRln(Pf/Pi)+nCPln(Tf/Ti) (1 point) = -nRln(1.00/4.00) ( 1 point; for the second term is zero) = nRln3. = (8.315 Jmol-1K-1)ln(4.00). = 11.57 JK-1 mol-1 (1 point) (b)

wU

i

i

f

fexternalifmV, p

T

p

T pR nTT C n (1 points)

ii

xtf

extmV, T

p

p R nT )

pnRC (n

mv

e

f

nCp , (1 points)

if T 2

3

4

1Rn T nR)(3/2nR

Rn Tf = 7Ti/10 = 210K. (2 points)

S = -nRln(Pf/Pi) + nCP,m ln(Tf/Ti) (eq. 1 points) = -nRln(1.00/4.00) + 5nR/2 ln(7/10) = 4.11 JK-1mol-1 (1 point) (c) Since this is a reversible adiabatic expansion = S = 0 (1 point)

1

f

i

i

f

p

p

T

T

1

f

i

i

f

p

p

T

T(1 point)

574.000.4bar 1.000

bar 4.00

T

T5

2

3

53

51

i

f

Tf = Ti * 0.574 = 172 K (1 point)

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3.3 [Circle if you choose this] Original Question P6.8) [ /15]

(a) Calculate Greaction

for the reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l) at T =T0 = 298 K

from the combustion enthalpy of methane (-890.3 kJ mol-1) and the entropies of the reactants and products. S0 (CO2, g) = 213.8 Jmol-1, S0 (H2O, l) = 70.0 Jmol-1, S0 (CH4, g) = 186.3 Jmol-1

,

S0 (O2, g) = 205.2 Jmol-1. (6 point)

(b) Give an equation to calculate Areaction

from Greaction you obtained in (a) at T=T0. Then,

calculate the values. (3 point)

(c) Define Kx for this reaction and give an expression that yields Kx of the reaction at 298K at a

constant total pressure P. Just show an equation (as KX = …) using Greaction obtained in (b), P,

T0, and P0. (4 points)

(d) Give an equation to obtain Kp at T =1000K from Kp(298K) assuming ΔH0 is constant. (2 points)

(a) All reactants and products are treated as ideal gases

2 2 4 2

1 1 1 1 1 1 1 1

1 1

3

CO , 2 H O, CH , 2 O ,

213.8 J mol K + 2 70.0 J mol K 186.3 J mol K 2 205.2 J mol K

242.9 J mol K

890.3 10 J m

combustion combustion combustion

combustion

combustion

G H T S

S S g S l S g S g

G

1 1 1 3 1ol 298.15 K 242.7 J mol K 817.9 10 J mol

(1 point for equation of G0combustion, 2 point for equation for S0

combustion; 2 point for the correct answer)

(b) Greaction

= Areaction

+ (PV) Areaction

= Greaction - (PV) = Greaction

- RTn (2 point)

= Greaction

- RT(-2) = -817.9 kJmol-1 - 8.315 JK-1mol-1 x 298 K x (-2) = -812.9 kJmol-1 (1 point)

(c) KX = (xCO2)/{(xCH4 (xO2)

2} (1 point)

Kp(298K, 1 bar) = exp(-Greaction0/RT0) (1 point)

Kp = Kx(P/P0)v Kx = exp(-Greaction

0/RT0) (P/P0)-v (1 point),

where v = 2. (1 point)

(d)

12

0

12

11)}(ln{)}(ln{

TTR

HTKTK pp (2 points)

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3.4 [Circle if you choose this] [ /15]

2 mole of NO2(g) (2 mole is the initial number of mole) produced N2O4(g) from the following reaction: 2NO2(g) ↔ N2O4(g) (*). A resultant mixture of NO2(g) and N2O4(g) is in its equilibrium between (2 -2 ξ) mole of NO2 and ξ mole of N2O4(g).

(a) Assume that the reaction takes place at a constant pressure and temperature at 1 bar and 298 K. Express formulae of Gpure and Gmixture using ξ and molar Gibbs energy for NO2, Gm

0(NO2, g) and N2O4, Gm

0(N2O4, g). (6 points).

(b) Express ΔGreaction for (*) using ξ, Gm0(NO2, g), and Gm

0(N2O4, g). You can divide ΔGreaction into two terms as ΔGreaction

0 + ΔGmixing (5 points)

(c) How do you find ξ in the equilibrium state (under constant P and T)? Report two methods. You do not have to calculate the value, but show the principle. (4 points)

(a) ),(G),(G )2-(2 G 42

om2

ompure gONgNO (3 points)

)2

ln22

22ln

2

22()-(2G

)lnln(GΔGGG

),(G),(G )2-(2 G

pure

puremixingpuremixture

42om2

ompure

424222

RT

xxxxnRT

gONgNO

ONONNONO

(4 points; Only ΔGmix is correct 3 points)

(b)

)lnlnln2(),(G),(G -2

ΔGGG

42242om2

om

mixing0

reactionreaction

ONNO xxRTgONgNO

(5 points; 3 Points if reactionG 0 is written) (c) Find ξ that minimizes mixtureG . Find ξ that satisfies 0Greaction (2 points each)

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4. Derivation Questions Q1-6 2 points each [ /12]

Fill a formula or equation in [Q1-Q4] and answer Q5. This section has some hint for other questions.

4.1 Now, we would like to derive ΔHreaction(308 K) for the reaction B + C → 2X at the standard temperature using CA (A

= B, C, X) and ΔHreaction0(298 K) for this reaction. For this purpose, we use dHA = CpAdT – Cp, AμJ-T dP + (∂H/∂n)T, P

dnA (A = B, C, or X) assuming that μJ-T and Cp,A (A = B, C, or X) are constant over the pressure range. (Also, assume that the J-T coefficients are the same for B, C, X). First, we consider a process 1: B(308 K) + C(308K) →B(298K) + C(298K). Since dP = 0 and dnA = 0, dHA = CpA dT. Thus, ΔH for this process (ΔH1) is given by [Q1 ]. Next, we consider a process 2: B(298K) + C(298K) →2X(298K).

ΔH for this reaction (ΔH2) is ΔHreaction0(298K). Lastly, we consider a process 3: 2X(298K) →2X(308K).

As shown for a process 1, ΔH for this process (ΔH3) can be obtained by using Cp,X. ΔHreaction(308K), which

can be obtained by ΔH1+ΔH2+ΔH3, is expressed as [Q2 ] using μJ-T, Cp,A (A = B, C, X) and ΔHreaction0(1.0 bar).

Q1: [ (Cp, B + Cp, C )(298K – 308K) ] (check the sign is correct.)

Q2: [ (2Cp, X - Cp, B - Cp, C )(308K – 298K) + ΔHreaction0(298K) ]

4.2 We prove that U

T

TA

V

)/1(

/ as follows. Fill appropriate formula in [Q3] and [Q4]. First,

2222

/]3[1

)(

/TQ

T

STA

T

A

T

S

T

A

T

A

TT

TA

VV

Then,

U

T

TAQ

T

T

T

TA

T

TA

VVVV

/

]4[)/1(

/

)/1(

/.

Q3: [ ΔU ] Q4: [ -T2 ]

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4.3 We examine the phase transition between solid, liquid, and gas states.

From

d(T, P) = -SmdT + VmdP. (1)

For a constant P,

d(T, P) = -SmdT . (2)

Assuming Sm is constant,

(T, P) = (0, P) -Sm T. (3)

The figure in the right shows chemical potential for a different temperature.

3 lines represent solid, liquid, gas. From (3), the slopes of these lines are given by molar entropies of solid (Sm, solid), liquid (Sm, liquid), and gas (Sm, gas). Among Sm, solid, Sm, liquid, Sm, gas, Sm is the greatest for [Q5 ] from the figure.

At the boiling point Tb,

liquid(Tb, P) = gas(Tb, P). (4)

Now, we define the change in the chemical from liquid to solid as

lg(T, P) = gas(T, P) - liquid(T, P). (5)

From (4) and (5)

lg(Tb, P) = 0. (6)

Thus, L and G are in the equilibrium at T = Tb. When the temperature is increased (ΔT >0),

lg(Tb+ ΔT, P) = [Q6 ] ΔT, (7)

where we used (Tb+ ΔT, P) = (Tb, P) - SmΔT for liquid and solid and eq. (4). Thus, lg(Tb+ ΔT, P) < 0. The gas state is more stable than the superheated liquid state.

[Q5: Sm, gas (or gas) ]

[Q6: - (Sm, gas - Sm, liquid) ]

μsolid(T, P)

μliquid(T, P)

μgas(T, P)

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dln(x)/dx = 1/x dCos(x)/dx = -Sin(x) dSin(x)/dx = Cos(x) de-x/dx = -e-x Constants

L = dm3 = 10-3 m3

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Calculation sheet (You can detach this sheet, if you like)