Chemistry 45 July 3, 014 Enter answers in a Blue Book Examination Midterm Useful Constants: 1 Newton=1 N= 1 kg m s 1 Joule=1J=1 N m=1 kg m /s 1 Pascal=1Pa=1N m 1atm=10135 Pa 1 bar=10 5 Pa 1L=0.001m 3 Universal Gas Constant R=8.31 J K 1 mol 1 =0.081L atm K 1 mol 1 Avagadro s Number N A =6.04x10 3 mol 1 All answers must be in SI units (i.e. units of meters, seconds, kilograms, Joules, Pascals, Kelvin degrees etc.) Helpful math relationships: 1) dax dx n anx n = 1 ; a is a constant, n is a positive or negative number ) n+ 1 n x x dx = dx + c where c is a constant and n 1 n + 1 dx 3) ln x c x = + where c is a constant x ln xy = ln x+ ln y; ln = ln x ln y; xln y = ln y x y 4) ( )
Part 1 (18 points) Answer HREE out of the following SIX questions. Limit definitions to less than 00 words. Use equations where helpful or required, but detailed calculations are not necessary. Question 1.1. In Chemistry 45 and Chemistry 456 the First Law of hermodynamics is given as U = q+ w. But in thermal physics and engineering thermodynamics courses the First Law has the form U = q w. Explain the reasons for these two versions of the First Law. he different versions of the first law arise from differences in the sign conventions for work In the convention used in chemistry work done by the system is negative w<0 and work done on the system is positive w>0. In this convention U = q+ w. In engineering/physics work done by the system is positive w>0 and work done on the system is negative w<0. he conventions for heat transfer are identical. For heat transfer out of the system q<0, and transfer into the system q>0. his results in a sign change for work in the first law expression U = q w Question 1. he internal energy U is an exact differential and has the form for a closed system: du = d + dv. Explain the physical meanings of and V V V V. What are the values of these terms for an ideal monatomic gas? Explain. is the heat capacity, the degree to which internal energy changes per a unit V change in temperature. A high heat capacity indicates a high capacity to absorb heat. For an ideal monatomic gas =3R/. is the internal pressure, the change of internal V V energy per unit change in volume. he internal pressure reflects the magnitude of intermolecular interactions. For an ideal gas V =0. Question 1.3 Which of the following types of work are quantified by the Gibbs energy at constant pressure and temperature? A) Pressure volume work; B) electrical work (i.e. moving charge through a electrical potential gradient); C) osmotic work (i.e. moving solute mass through a concentration gradient) ; D) elastic work (i.e. stretching an elastic fiber). Explain your answer.
he definition of the Gibbs function: dg=dh d(s)=ds PdV+dwother+PdV+VdP ds Sd=dwother Sd+VdP. If d=dp=0 then dg=dwother, herefore PV work is not reflected by the Gibbs function while electrical, osmotic and elastic work are included in the Gibbs function. Question 1.4 Explain how the entropy limits the efficiency of a heat engine. You may use the Carnot engine as an example in your explanation. You may use equations to assist in your explanation, but detailed calculations are not necessary. A Carnot Cycle consists of 1) an isothermal expansion at H ; ) an adiabatic expansion; 3) an isothermal compression at C ; 4) an adiabatic compression. herefore the entropy change over one cycle q H / H +q C / C =0. For the entropy to sum to zero over a cycle q C =q H C / H must be nonzero. Question 1.5: For an adiabatic reversible expansion the entropy change is zero. But for an adiabatic irreversible change the entropy is not zero. Explain. For any adiabatic change q=0. If the adiabatic change is reversible then q rev =0 and so S=q rev /=0. For an irreversible change q rev >q irrev =0 and in this case S=q rev />0. Question 1.6 As the temperature of an atomic crystal is increased, what value does the heat capacity approach? Explain. he heat capacity approaches 3R. here are two degrees of freedom per dimension so the heat capacity per dimension is xr/=r. Foe vibration in three dimensions the heat capacity is 3R. Part : (0 points) Answer WO of the FOUR questions. Answers longer than 00 words are acceptable but numerical calculations are not required. Physical reasoning based on thermodynamic principles is recommended. Equations may be used to illustrate your points. Problem.1 Explain the following observation. Burns caused by contact between skin and steam at 373K are far more severe than burns caused by contact between the skin and liquid water at 373K. Use the appropriate state function in your explanation. Assume heat is transferred at constant pressure. A burn from hot water results from the transfer of heat from the water to the skin. A burn from steam results first from heat transfer from the steam to the skin, then the steam
condenses on the skin transferring more heat. Further heat transfer results from the liquid water after it condenses on the skin. Problem. Suppose a liquid is converted reversibly to its vapor at its normal boiling point. State whether the following thermodynamic quantities are positive, negative, or zero: q, w, V,, H, S, G. q>0, w<0, V>0, =0, H>0, S>0, G=0. Problem.3) Differential scanning calorimetry (DSC) measures directly the enthalpy of denaturation of a protein. Call the enthalpy of denaturation measured by DSC H DSC. A second way to obtain the enthalpy of denaturation supposes at any temperature the equilibrium constant fd fd is K = f = N 1 f where f D is the fraction of denatured protein and f N is the fraction of D structured protein. he van t Hoff equilibrium equation states that if the equilibrium constant is measured at two different temperatures, i.e. by measuring f D at two different temperatures, the enthalpy can be calculated from: K H 1 1 ln = vh K1 R 1 where H vh is called the van t Hoff enthalpy of denaturation. he enthalpy of denaturation of the same protein is measured repeatedly by these two methods and it is found consistently that H vh << H DSC. Assuming absence of systematic experimental errors, explain why this might be so. DSC is a direct measure of the heat absorbed by all protein forms in solution within a given temperature range. he van t Hoff equation given above assumes only two protein forms are in solution and only two forms of the protein are assumed to absorb heat. herefore the van t Hoff equation may under-estimate the amount of heat absorbed. Problem.4) he solar flux is the rate at which energy from the sun passes through a unit area on the surface of the earth per unit time. Suppose the solar flux warms the water at the surface of a lake to =300K. Because water at the bottom of the lake is at =80K, a temperature gradient is produced. Heat transferred from the warmer surface water to the cooler water at the lake bottom can used to run a heat engine which produces energy at a rate of 5x10 6 Watts (i.e. 5 MW). Assume the solar flux at the lake surface is about 500 Watts per squared meter (i.e. 500Wm - ) and the lake has an area of 10000m. Can the engine function as described? Does the engine obey the first law of thermodynamics? Does it obey the second law of thermodynamics? Explain. he total average energy flux into the lake is (500 W/m ) (10000m )=5x10 6 m. herefore the engine only outputs 0% of the input energy so it obeys the First Law. On the opther hand the efficiency of the engine should be: (300-80)/300=0.067 or 6.7%. So the heat engine violates the Second Law.
Part 3: (30 points) Answer WO of the FOUR questions. Problem 3.1: Determine the Gibbs energy change for converting graphite to diamond at P=15,000 atm and =98K. he following table provides some useful thermodynamic data for graphite and diamond at =98K and P=1.00 atm. Assume molar volumes are constant over pressure ranges specified. Based on your answer, can diamonds be produced in this way? G ( 1 f kjmol 3 1 ) Vm ( cm mol ) C ( s, graphite ) 0 5.33 C ( s, diamond ).900 3.4 A diagram is helpful in solving this problem but is not required. G (, ) C( dia, s) C gr s P= 15000 atm, = 98K P P G (, ) f C( dia, s) C gr s P= 1 atm, = 98K hen f ( ) ( ) f ( ( ) ( )) G = V gr P+ G V dia P G = G + V dia V gr P ( )( )( ) = 900Jmol + 3.4 10 m 5.33 10 m 15000atm 10135Pa atm 6 3 6 3 = 900Jmol 903Jmol = 3Jmol G<0 so this approach works.
Problem 3.. Between =73K and =373K mercury Hg(l) has a heat capacity, in units of 1 JK mol 3, that is given by: CP ( Hg, ) = 30.093 4.944 10. Calculate H and S if the K temperature of one mole of mercury is raised from =73K to =373K. 373 373 3 H CP ( ) d 30.093 4.944 10 d K 73 73 = = 3 = 30.093JK mol 373K 73K 4944 10 = 3009.3J 159.7J = 849.6J ( ) ( JK mol ) ( ) ( ) C = = = 373 373 P 30.093 4.944 10 3 S d d 73 73 373 73 373 30.093 ln 4.944 10 3 JK 1 mol 1 373K 73K 73 = 9.39JK 0.4944JK = 8.899JK ( )( ) Problem 3.3 For water at =300K and P=1 atm, the coefficient of thermal expansion 4 0 β = 3.04 10 K and the isothermal compressibility κ = 4.46 10 mn. Calculate V V for water at =300K and P=1 atm. Will this value of V for water at =383K and P=1atm? Explain. be greater than or less than 4 P β 3.04 10 K = P = P = ( 300K) 0135Nm 0 V κ 4.46 10 m N V =.045 10 Nm 1.01 10 Nm.045 10 Nm 8 5 8 Water boils at =373K so at =383K the water is in vapor form. Intermolecular interactions are smaller in the vapor versus the liquid so would be smaller at =383K than at =300K V
Problem 3.4 Suppose 1 mole of an ideal monatomic gas at an initial temperature of 1 =98K and pressure P 1 =1.01x10 6 Pa, expands adiabatically and reversibly until the pressure has dropped to P =10135Pa. Calculate the final volume, final temperature, U, and H. ( 8.31JK )( 98K ) nr PV = nr V = = =.45 10 m Pa 1 1 1 1 1 6 P1 1.01 10 3 3 1/ γ 6 3/5 γ γ P 1 3 3 1.01 10 3 3 1 1 = = 1 = (.45 10 ) 9.73 10 5 = P 1.0135 10 PV PV V V m m 3 3 ( 10135Pa)( 9.80 10 m ) PV = = = 119K nr 8.31JK 3R U = ncv = ( 119K 98K) = ( 1.5)( 8.31JK )( 179K) = 31J 5R H = ncp = ( 119K 98K ) = (.5)( 8.31JK )( 179K ) = 3719J Part 4 (3 points) Perform ONE out of the WO multi-step calculations Problem 4.1. In certain bacteria, the amino acid glycine H NCH COOH is synthesized by a reaction between ammonia NH 3, methane CH 4, and oxygen: 5 ( ) + ( ) + ( ) ( ) + 3 ( ) NH g CH g O g H NCH COOH s H O 3 4 hermodynamic data for this reaction are given in the table below for =98K and P=1.00 atm.. H ( 1 f kjmol ) S ( JK 1 mol 1 ) CP ( JK mol ) NH 3 (g) 46.19 19.51 35.66 CH 4 (g) 74.85 186.19 35.73
O (g) 0 05.03 9.36 H NCH COOH (s) 537.3 103.51 99.0 H O (l) 85.84 69.94 75.30 a) Using the data in the table, calculate the enthalpy change H reac and S reac reaction at =98K and P=1.00 atm. for the 5 H reac 3 H f ( HO, ) H f ( glycine, s) H f ( NH3, g ) H f ( CH4, g ) H f ( O, g ) 5 ( 3)( 85.84kJ ) 537.3kJ 46.19kJ ( )( 74.85kJ ) ( )( 0) 1199kJ = + = + + = ( )( ) 5 ( ) ( ) ( ) ( ) ( ) S = 3 S H O, + S glycine, s S NH, g S CH, g S O, g reac 3 4 = 3 69.94JK + 103.51JK 9.51JK = 764JK ( )( ) ( 5 186.19JK )( 05.03JK ) b) Suppose this type of bacterium exists near ocean floor steam vents where the temperature is =360K. Calculate the enthalpy change and entropy change for this reaction at =360K. Assume all heat capacities are constant between =98K and =360K. 5 ( ) ( ) ( ) ( ) ( ) C = 3 C H O, + C glycine, s C NH, g C CH, g C O, g P P P P 3 P 4 P 1 ( 3)( 75.30JK ) 99.0JK 35.66JK ( )( 35.73JK ) ( 5 )( 9.36JK ) = + = 144.6JK ( ) ( ) ( )( ) H 360K = H 98K + C = 1199kJ + 144.6JK 360K 98K =190kJ reac reac reac P ( 360 ) S K = S reac 360 98K + CPln = 764JK + 144.6JK 0.189 = 737JK 98 ( ) ( )( ) 1 1 1 c) Calculate the entropy change of the surrounding and the entropy change for the universe for this reaction at =360K ( 360) 1190000J H Ssurr = = = 3306JK 360K 360K
S = S + S = 3306JK 737JK = 569JK universe reac surr d) Based on your answer in part d, will the biosynthesis of glycine in these bacteria proceed spontaneously in the direction written above at =360K? Explain. S>0 for universe so reaction will proceed as written Problem 4. : he figure at the right is a four step cycle plotted as a V diagram for a Stirling Heat Engine. he Stirling Engine runs reversibly in a clockwise direction (i.e. from point 1 to, then to 3, etc.) and operates between a high temperature =400K, and a low temperature 1 =300, and between a maximum volume V =0.050m 3 and a minimum volume V 1 =0.010m 3. Assume an ideal monatomic gas is the working fluid. a) For each of the four steps of the Stirling Engine cycle calculate the heat q and the work w. 3R w1 = 0 q1 = U1 = ncv = ( 400K 300K) = ( 1.5)( 8.31JK )( 100K) = 147 J; V 0.05 U3 = 0 q3 = w3 = nr ln = ( 8.31JK )( 400K) ln = 5350J V1 0.01 3R w34 = 0 q34 = U34 = ncv = ( 300K 400K) = ( 1.5)( 8.31JK )( 100K) =47J V 1 0.01 U41 = 0 q41 = w41 = nr1 ln = ( 8.31JK )( 300K) ln = 401J V 0.05 b) Using your results from part a, calculate the total heat absorbed by the Stirling engine in one cycle, the total heat released by the Stirling engine in one cycle, and the net work performed by the engine in one cycle.
q = q + q = 147J + 5350J = 6597J in 1 3 q = q + q =47J 401J = 559J out 34 41 w = w + w + w + w = 0 5350J + 0 + 401J =338J net 1 3 34 41 c) Using your results from part b, calculate the efficiency of the Stirling heat engine. Compare this number to the efficiency of a Carnot Engine operating between H =400K and L =300K. ε ε Stirling Carnot ( 1338J ) w net = = = 0.0 qin 6597J H L 400K 300K = = = 0.5 400K H d) Suppose we run the Stirling engine backwards so that it functions as a refrigerator. Calculate the effectiveness of this refrigerator in removing heat from the low temperature reservoir. Compare this to the effectiveness of a Carnot refrigerator run between the same two temperatures. ε ε eff, Stirling eff, Carnot 559J = = 3.93 1338J C 300K = = = 3.00 100K H L