Give all answers in MKS units: energy in Joules, pressure in Pascals, volume in m 3, etc. Only work the number of problems required. Chose wisely.

Transcription

1 Chemistry 45/456 0 July, 007 Midterm Examination Professor G. Drobny Universal gas constant=r=8.3j/mole-k=0.08l-atm/mole-k Joule=J= Nt-m=kg-m /s 0J= L-atm. Pa=J/m 3 =N/m. atm=.0x0 5 Pa=.0 bar L=0-3 m 3. Watt= W= Js Midterm Average =0.70 Give all answers in MKS units: energy in Joules, pressure in Pascals, volume in m 3, etc. Only work the number of problems required. Chose wisely. Part : (8 points total) Define 3 out of the following 6 terms as they are used in the fields of thermodynamics and/or mechanics. Keep your definitions brief and to the point. Use of equations, when appropriate, is encouraged, but you need not perform calculations..) Compare and contrast the reversible and irreversible adiabatic expansions of an ideal gas. In particular, explain how the entropy changes in each case differ and explain the difference. Answer: An adiabatic process occurs in the absence of heat transfer: q=0. For a qrev reversible adiabatic process qrev = 0 S = = 0. For an irreversible adiabatic process qrev qrev > qirrev = 0 S = > 0.) Define the following properties in terms of partial derivatives of state functions or state variables: the heat capacity at constant pressure, the isothermal compressibility, the thermal expansivity, and the chemical potential. Note in each case under what conditions the partial derivative is calculated. Answer: H Heat capacity at constant pressure CP = P V hermal Expansivity: β = V P V Isothermal Compressibility: κ = V P G A U H Chemical Potential: µ i = = = = ni ni ni ni Pn,, j i Vn,, j i SVn,, j i SPn,, j i

2 Note: For a multi-component system, any of the four relationships above for the chemical potential will be sufficient..3) Describe the two types of perpetual motion machines: i.e.perpetual motion machines of the first and second kind. Give an example of each and explain why each cannot exist according to the laws of thermodynamics. Answer: A perpetual Motion Machine of the first kind violates the first Law of hermodynamics. It does not conserve energy, e.g. produces work out of nothing. A perpetual motion machine of the second kind violates some aspect of the second law of thermodynamics. For example, a heat engine might conserve energy, but it might h l transform heat into work with an efficiency that exceeds the second law limit ε = h.4) State and explain the third law of thermodynamics. State a practical consequence of this law. Answer: In any thermodynamic process involving only pure phases at equilibrium the entropy change S approaches zero at the absolute zero of temperature, i.e. =0K. Because of this limiting behavior the absolute entropy (at constant pressure can be CP ( ) measured, i.e. S( ) = d 0.5) Explain the difference between the Helmholtz and Gibbs energies. In particular relate these two quantities to reversible work. Answer: he Helmholtz Energy is da = Sd PdV + dwother = Sd + dwrev. he Helmholtz energy therefore measures the isothermal reversible work. For the Gibbs energy: dg = Sd + VdP + dwother so the Gibbs eerngy measures the reversible nonexpansion work at constant and P..6) Explain the Second Law of hermodynamics as it applies to the efficiency of heat engines. Answer: Assuming the Universe can be treated as an isolated system, as a result of any real process the entropy of the universe must increase. Also, any process whose sole outcome is the complete conversion of heat to work is impossible. he efficiency of any h l heat engine is therefore limited by ε = where the temperatures are related by l >. h l Part (0 points) Answer two out of the four questions below. Answers should be no longer than 00 words. In some cases a brief calculation might be useful and would save a lot of words..) Before the days of electricity, flatirons for ironing clothes were literally made out of solid iron. hey were heated on wood-burning stoves, and had to be quite heavy, despite the fact that anyone could show that it was the heat not the weight of the flatiron that took wrinkles out of clothes. Explain why flatirons had to be so heavy.

3 Answer: he heat transferred at constant pressure is MC where M is the mass, C is the heat capacity and is the temperature difference between the cloth and the iron. o maximize heat you have to either maximize or M. Making too high is not an option because you would burn the cloth so the mass had to be made large..) Air is about 80% nitrogen and 0% oxygen. Suppose a sample of air is placed in a chamber which is separated from a second, evacuated chamber by a wall. his entire system is mechanically and thermally isolated from the surroundings. A small pinhole is opened in the wall between the two chambers and the gases effuse through the pinhole into the second chamber. Because nitrogen effuses faster than oxygen, the second chamber fills with nitrogen faster than oxygen, and the two gases thus unmix irreversibly, even though the system is isolated from the surroundings. Does the entropy of the isolated system decrease, thus violating the second law? Explain. Answer: Associated with unmixing that results from effusion, the gases expand into a larger volume thus accomplishing a large net entropy increase.3) he equipartition principle predicts that the heat capacity C P of an ideal monatomic gas is 5R/ and the heat capacity of an ideal diatomic gas is 9R/. he experimental heat capacity of helium gas is C P =0. JK - mol - and the heat capacity of N gas is 9. J K- mol -. Explain any discrepancies between the experimental heat capacities and the predictions of the equipartition theorem. Answer: Assuming the gases behave ideally, the heat capacities at constant pressure follow the rule: CP = CV + R. his means that for a ideal monatomic 3R 5R - - CP = CV + R = + R = = 0.78JK mol, reflecting the three degrees of translational freedom possessed by a monatomic. For a diatomic molecule according to the 7R 9R - - equipartition principle CP = CV + R = + R = = 37.40JK mol, reflecting three translational, two rotational, and two vibrational degrees of freedom. While the experimental value for C P for He agrees with the predicted value, the experimental C P 7R - - value for N is much closer to = 9.09JK mol. he two degrees of vibrational freedom are removed because the vibrational energy levels are not thermally populated to any great extent at ambient temperatures. his is a quantum mechanical effect that is not accounted for in the equipartition theorm..4) Water is a polar liquid at ambient temperature with a molecular weight of 0.08 kg mol -. Liquid water has a heat capacity of 75.9 J K - mol - and water ice has a heat capacity of about 35.3 JK - mol -. Argon is a noble gas at ambient temperature and has an atomic weight of 0.08 kg mol -. Liquid argon has a heat capacity of 4.8 JK-mol-, and solid argon has a heat capacity of about 30.0 J K - mol -. Why is the difference between

4 the heat capacities of solid and liquid water so much greater that the difference between the heat capacities of solid and liquid argon? Answer: he large difference in the heat capacities of liquid and solid water are believed by most to arise from the fact that water is a polar molecule and as a consequence is a highly structured liquid. In the liquid state water molecules are strongly hydrogen bonded, and when heat is absorbed by liquid water, this energy can break hydrogen bonds. Another line of thought has it that in liquid water the addition o f heat can also distort hydrogen bonds, changing their length and altering the mutual orientation of water molecules. None of these effects can occur in argon, which forms a non-polar liquid. herefore, while the heat capacity of liquid argon is higher than that of solid argon, the heat capacity difference is not so great as occurs between liquid and solid water. Part 3: (30 points) Perform two out of the four calculations given below. 3.) Suppose five moles of an ideal monatomic gas at an initial volume of.l and temperature of 98K expand adiabatically and reversibly until the volume has reached 48.8L. Calculate the temperature change, U, and the work done. 5R 5 3 Solution: For a reversible adiabatic change PV γ P = cons tan t., where γ = = =. C R V 3 Because the gas is ideal (PV=nR) it follows V γ = cons tan t. So γ /3 γ γ V.L V = V = = ( 98K) = 9.4K V 48.8L = = 78.6K U = w = nc = 5moles R 78.6K =,3J V 3 ( )( )( ) C 3.) Consider the heat engine in the diagram below which produces net work by exploiting the temperature gradient that exists between the top of a lake and the lake bottom. he sun sustains this temperature difference by warming the top layer of lake water. Assume the lake has a surface area of.00x0 5 m. Assume the water at the surface of

5 the lake has a temperature of 98K and the water at the bottom of the lake has a temperature of 80K. Assume the average solar flux at the lake surface is 500 Watts m -. It is claimed that this lake engine can generate work at an average rate of.00x0 7 Watts. Does this claim violate the First or Second Laws of hermodynamics? Support your answer with a calculation. 0 Watts 7 Watts m 5 m = Solution: Claimed efficiency= ( ) 500 / 0 heoretical efficiency= = Impossible because even though the engine conserves energy and does not violate the First Law, the efficiency promised exceeds the efficiency limit of heat engines set by the Second Law ) For copper metal at P=atm and =98K the thermal expansivity 6 6 β = K and the isothermal compressibility κ = bar. Calculate the difference CP CV for copper metal at =98K. he density of copper is 890 kg m -3. Solution: V M C β β P C = V κ = ρκ where M is the atomic weight and ρ is the density. - - From the periodic table posted in the wall of the class M 63.6g mol kg mol hen

6 C = - 6 M β ( kg mol )( 98K)( K ) CV = = ρκ ( kgm )( bar ) - 6 ( kg mol )( 98K)( K ) ( kgm )( bar )(.00 0 barpa ) P K mol = = m N m JK mol 3.4) Ammonia is a common metabolic by-product. It is also very toxic so terrestrial animals convert ammonia to urea: NH CONH. Consider the synthesis of urea from its constituent elements: C gr + N g + O g + H g NH CONH s ( ) ( ) ( ) ( ) ( ) he table below shows entropy values at for =98K and heat capacity values for each reaction component. Substance 0 ( S JK mol 0 ) CP ( JK mol ) C(gr) N (g) O (g) H (g) Urea(s) Calculate the entropy change for this reaction at =30K. Assume all heat capacities are constant between 98K and 30K. S( ) = S( ) + CP ln = ( ) JK mol 30 + ( ) ln JK mol 98 = 456.3JK mol JK mol = 457.0JK mol Part 4: (3 points). Perform one out of the two multi-step calculations. 4.) Consider the formation of glucose from carbon dioxide and water, i.e. the reaction of the photosynthetic process: 6CO g + 6H O l C H O s + 6O g. ( ) ( ) ( ) ( ) 6 6 hermodynamic Information

7 =98K CO (g) H O C 6 H O 6 O H 0 f kj/mole S 0 f J/mole-K C 0 p J/mole-K Assume all heat capacities are constant between =98K and =330K. a) Calculate the entropy and enthalpy changes for this chemical system at =98K Solution: H = 6 H O g + H C H O s 6 H CO g 6 H H O l ( ( )) ( ( )) ( ) ( ) ( ) ( ) ( ) ( ( )) 6 6( ) 6 6 ( ) ( ) ( ) ( ) ( ( )) f f 6 6 f f = kJ kJ kJ = kJ ( ) ( ) ( ( )) S = S O g + S C H O s S CO S H O l = J / K +.3 J / K J / K = 59 J / K b) Calculate the entropy change for the surroundings at =98K. Based on your answer is the production of glucose spontaneous at =98? Explain. 0 0 H 80.65kJ Ssurr = = = 9.40kJK 98K herefore Suniv = Ssurr + Srxn = 9400JK mol 59JK mol = 9659JK mol he entropy change of the universe resulting from the production of glucose and oxygen from carbon dioxide and water is negative. Assuming the universe is an isolated system, this reaction will not occur. It is not spontaneous. c) Repeat the entropy and enthalpy calculations for =330K. Does photosynthesis become spontaneous at =330K? Explain. d) H ( ) = H ( 98K) + CP ( 98K) S ( ) = S ( 98K) + CP ln 98 K C = 6C O g + C C H O s 6C CO g 6C H O l P P( ( )) P( ( )) P ( ) P ( J K) J K ( J K) ( J K) ( ) ( ()) 6 6 = / / / / = 76.6 J / K J / K.78 J / K 45.8 J / K = 80.6 J / K hen

8 ( ) ( 98 ) ( 98 ) ( 80.6 / )( ) H = H K + CP = 79.03kJ K = J + J K K K K S ( ) = S ( 98K) + CP ln = 59 J / K + ( 80.6 J / K) ln 98K 98K = J / K 0 0 H 79kJ Ssurr ( ) = = = 8.46kJK 330K. S = S + S = 8460JK mol 88JK mol = 8748JK mol univ surr rxn Sa me result as the lower temperature. he entropy change of the universe is still negative. he process is not spontaneous. 4.) 8 grams of water ice at 30 K are placed in thermal contact with a large heating block maintained at a temperature of 350 K. Calculate the entropy change S, the energy change U, and the enthalpy change H of the water, when 8 grams of water ice at 30 K are converted to liquid water at 350K. Also calculate the entropy change of the heating block as a result of this process and the entropy change of the universe as a result of this process. For ice Cp=37.7 J/K. For liquid water Cp=75 J/K. For water Hfusion=6.0 kj/mole. Assume the density of ice is 0.9 gm/ml and approximately constant between 30K and 73K. he density of liquid water is gm/ml at 73K and approximately constant up to 350K. Assume only P-V work occurs. he external pressure is atm. Solution: Designate a three-step path from water ice at =30K to liquid water at =350K: ice( 30K) water( 350K) B ice( 73K) water( 73K) A ice Step : H = ncp = ( mole)( 37. 7J / K mole)( 73K 30K) = 6J ice F final 73 S = ncp H G I ln ( mole)( 377. J / K mole) ln 646. J / K 30 initial KJ = F H G I K J = U = qp Pext V qp = H Step : H = ( mole) H fusion = ( mole)( 600J / mole) = 600J H 600J S = = =. 0J / K fusion 73K U = qp Pext V qp = H = 600 J; V = Vliquid Vice V = 0. 08L ( 0. 08L/ 0. 9) =. 006L U = 600 J + (.006 L)( atm)(0 J / L atm) = 600.J liquid Step 3: H3 = nc = ( mole)( 75J / K mole)( 350K 73K) = 5775J P

9 liquid final 350 S3 ncp ln ( mole)( 75J / K mole) ln 8. 63J / K 73 = F H G initial As before U3 H3 = 5775J I F = KJ H G I K J = otals: H water = H+ H + H3 = 3, 406J S water = S+ S + S3 = J / K water water E H = 3, 406J water H J S for Surroundings: Ssurroundings = = 3, 406 = J / K 350K surroundings S for Universe: S = S + S = J / K J / K = 8. 79J / K universe surroundings water