Chemistry 420/523 Chemical Thermodynamics (Spring ) Examination 1

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Stanley Atkinson
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1 Chemistry 420/523 Chemical hermodynamics (Sring ) Examination 1 1 Boyle temerature is defined as the temerature at which the comression factor Z m /(R ) of a gas is exactly equal to 1 For a gas obeying the equation of state P m R " µ # 2 c c show that the Boyle temerature deends only on the critical temerature From the equation of state, we have Z P m R " µ # 2 c c he right hand side is equal to unity when 0, 0, or when 14( c / ) 2 Since the first two conditions are unhysical, we get B 2 c herefore, the Boyle temerature deends only on the critical temerature 2 Using the relationshi H U +, and obtaining transformation relationshis by considering H as a function of any two of the variables (,, ), show that µ µ µ µ H U H + From H U +, differentiating with resect to temerature holding the volume constant (to get the first term in the exression) we get µ µ µ H U + (1) Considering enthaly as a function of temerature and ressure, H(,), we write µ µ H H dh d + d (2) Differentiating Eq (2) with resect to at constant, weget µ µ µ µ H H H + (3) From Eq (3), µ µ µ µ H H H Substituting for the first term on the right hand side from Eq (1), we get µ H µ U + c 1 c µ H µ

2 3 A flask initially containing benzene (C 6 H 6 ) at its freezing temerature of 2786 K is brought into contact with an ice-water bath until 1 mole of benzene is frozen For benzene, fus H 9, 832 Jmol 1 (a) Find the entroy change for the benzene (b) Find the entroy change for the ice-water bath (c) Find the entroy change for the Universe Is this rocess sontaneous? (d) If DfusH for ice is 6,025 J mol-1, how many grams of ice would melt or freeze (state which) as a result of the rocess taking lace in the flask? (a) Entroy change for the freezing rocess for the benzene: S m fush 9832 J 2786 K J K 1 (b) Entroy change for the ice-water bath: S surr q surr 9832 J K JK 1 (c) Entroy change of the Universe: S univ JK 1 Since this is ositive, the rocess is sontaneous (d) It takes 6, 025 J to melt one mole (or g) of ice herefore, the mass of ice melted as a result of 9, 832 J being exelled into the ice-water bath will be w 9832 J gmol Jmol g 4 (Chem 420) One mole of ideal gas initially at 298 K is allowed to exand adiabatically and reversibly from 100 bar to 050 bar Find the final temerature, and U, H, q, w, and S for this rocess Since this is an ideal gas, m R Also, for an adiabatic rocess, 1 γ 1 2 γ 2, where γ C,m /C,m 5/3 herefore, 2 µ 1 2 hen, the final temerature is herefore, 3/5 µ R1 1 µ 10 bar 050 bar R 5/3 2 1 (R 1 / 1 ) 5/3 Ã 3/ LbarK 1 mol 1! K 10 U C,m ( 2 1 ) 3 2R ( ) 898 J H C,m ( 2 1 ) 5 2R ( ) 1497 J q 0by definition, for an adiabatic rocess w U 898 J S 0by definition, for an adiabatic rocess L

3 4 (Chem 523) An ice cube at 0 Cweighing100gisdroedinto1kgofwaterat20 C [ fus H for ice is 6,025 J mol 1, C,m for liquid water is 753 J K 1 mol 1 ] (a) Does all of it melt? If not, how much remains? What is the final temerature? (b) What is the entroy change of the ice + water system during this rocess? (a) Let us assume that all of the ice melts and solve for the final temerature that would result If this comes out to be below 0 C, we will have to conclude that all of the ice does not melt he basic rincile of energy conservation is sufficient to find the final temerature: heat gained heat lost, or q ice q water he heat is gained by 100 g of ice which first melts and then heats u to the final temerature: q ice 100 g gmol 1 " fus H + Z f C,m d At the same time, the water sulies this heat and cools down from 20 C (29315 K) as a result: 1000 g q water gmol 1 C,m d Using the energy conservation relationshi, we have 100 g h i gmol Jmol JK 1 mol 1 ( f 27315) K herefore, all the ice does melt (b) Entroy changes: S ice Z f 1000 g h i gmol JK 1 mol 1 ( f 29315) K K " 100 g 6025 Jmol gmol JK 1 mol 1 ln # µ # µ JK g S water gmol JK 1 mol 1 ln JK 1 S ice+water JK 1 5 A gas obeys the equation of state m R + B, whereb is a constant he C,m for this gas is a constant, equal to (3/2)R (a) Starting with the first law, derive an exression for the entroy change ds m,andthus,for S m during a reversible exansion/comression (b) If B 0015 L mol 1, what is the entroy change of the system if the gas is allowed to exand in an isobaric rocess from 100 L at 298 K to 1000 L? (a) From the first law, we get du dq + dw By definition, du C,m d Since the rocess is reversible, dq rev ds he work of exansion of a gas is dw d herefore, C,m d ds d d ds m C,m + d m d C,m + R d m m B 3

4 Integrating both sides from initial to final states, we have µ µ 2 2 B S m C,m ln + R ln 1 B 1 (b) From the equation of state, the final temerature is given by 2 ( 2 B) Sinceremains constant R during the rocess, we may substitute for using the initial conditions: R 1 /( 1 B) herefore, 2 R 1 1 B 2 B R 2 B 1 1 B K herefore, S m 15R ln µ µ R ln JK 1 mol 1 6 A reversible Joule cycle consists of four stes: isobaric exansion, adiabatic exansion, isobaric comression and, finally, adiabatic comression Sketch the ath taken by the system in such a cycle schematically using the following coordinates: (a) (, ) (b) (U, ) (c) (,S) (d) Sketch the ath taken by the system in a Carnot cycle using (,S) for coordinates See next age for the lots 4

5 Plots for Question 6: (a) (b) (c) U S (d) S

δq T = nr ln(v B/V A )

δq T = nr ln(v B/V A ) hysical Chemistry 007 Homework assignment, solutions roblem 1: An ideal gas undergoes the following reversible, cyclic rocess It first exands isothermally from state A to state B It is then comressed adiabatically