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# Old Exam 1 - solutions CHEM3330

Texas State

GPA 2.9

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This 10 page Bundle was uploaded by mythili venkateswaran on Friday September 4, 2015. The Bundle belongs to CHEM3330 at Texas State University taught by Dr. Easter in Fall 2014. Since its upload, it has received 58 views. For similar materials see Physical Chemintry 1 in Chemistry at Texas State University.

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Date Created: 09/04/15

SOLUTIONS Write your name and your Texas State ID number leginy BOTH at the top of this page AND the following page For full credit final answers must be expressed to the correct number of significant digits and with the correct units Circle your final answer to make it easy to identify Show all of your work on all problems including units in all steps and unit conversions Provide a brief explanation of each equation Incorrect answers may receive partial credit ifthe work you show is valid and complete but will receive no credit if work is missing invalid andor unexplained Exam Conditions Closed Book Closed Notes Calculator You may use any scientific calculator including a programmable andor graphing calculator Except for a pencil and eraser no other items are permitted on the exam Exam Policies A minimum of 10 points will be deducted from the exam score for each violation of exam rules whenever there is no apparent attempt to gain unfair advantage Any violation involving an attempt to gain an advantage will be prosecuted as a deliberate honor code violation Cell phones and all personal devices other than a calculator are strictly prohibited Before the exam begins turn your cell phone OFF not just on quotsilentquot or quotvibrate and store it with all other personal items out of sight A 10 point penalty will automatically be assessed if your phone rings during an m There are no excuses or exceptions All personal items that are not permitted on the exam must be stored on the floor You may not share anything with any other student by giving or receiving during the exam If you arrive unprepared ie by not bringing an appropriate calculator you must work the exam without the materials that you failed to bring You will not receive credit for any questions or problems that require those materials This is a 75minute exam Your exam must be turned in by the final cutoff time that is announced A minimum of 10 points will be deducted from the total score for exams that are not turned in by the cutoff time NOTE for situations when one part of a problem requires a value that was calculated in a previous part of the problem If you are able to calculate the value and you believe it to be correct use that value If you are not able to calculate the value or if you believe the value that you calculated is in error state a reasonable assumed value with appropriate units for the variable in question state your assumption clearly and then answer the question using your assumed value You may receive partial credit ifthe assumed value is physically reasonable in the context of the problem and ifyour subsequent result is consistent with the assumed value Chem 3330 Exam 1 Fall 2014 Page 1 SOLUTIONS Chemistry 3330 Fall 2014 Exam 1 Exam Score Summary Question Number Points Possible Points Earned 1 2 3 4 Total Points raw Exam Score percent Chem 3330 Exam 1 Fall 2014 Page 2 Selected Constants Conversion Factors Data and Equations Numbers have uncertainty only in the last digit shown The number of digits used in an ongoing calculation should be one more than the correct number of significant digits required for the final answer Most of these values were taken from the US Government NIST web site in 2010 2011 Physical Constant Symbol Numerical Value Units Universal Gas Constant R 831447 J K391 mol391 Universal Gas Constant R 820574 X 10392 L atm K391 mol391 Boltzmann s Constant k 138065 X 103923 J K391 Avogadro s Constant NA 60221418 X 1023 mol391 Speed of Light c 299792458 exact m s391 Atomic Mass Unit u 166053878 X 103927 kg Standard acceleration of gravity g 9806 65 exact m s392 Conversion Factors 1 cal 4184 J exact 1 Pa 1kg m391 3392 1 atm 101325 bar exact 1bar 100 kPa 14696 psi 1 atm 1J1kgm2s392 1 atm 101325 Pa exact 1 atm 760 torr exact 1 hr 60 min exact 1 in 254 cm exact 1 1b 45359237 g 1 min 60 s exact 1 mi 1609344 m 11b16oz Average Periodic Table Atomic Masses of Selected Elements Element Atomic Number Atomic Mass u H 1 100794 He 2 4002602 C 6 120107 N 7 1400674 0 8 15 9994 F 9 18998404 Ne 10 201797 C1 17 354527 Ar 18 39948 Br 35 79904 Kr 36 838 I 53 12690447 Xe 54 13129 Second and Third Virial Coefficients of Several Real Gases at 29815 K M mol L391 Gas B 10393 M391 C 10396 M392 B39 10394 atm391 C39 10398 atm392 H2 141 350 576 253 He 118 121 482 305 N2 45 1100 184 180 02 161 1200 658 157 Ar 158 1160 646 152 CO 86 1550 352 247 Volume Expansion Pressure Expansion Chem 3330 Exam 1 Fall 2014 Page 3 van der Waals Coefficients of Several Real Gases nRT n 2 RT a p V nb a V quotp Vm b an Gas aatm M392 b10392 M 1 Ar 1337 320 CH4 2273 431 C2H4 4552 582 C02 3610 429 H2 02420 265 He 00341 238 NH3 4169 371 N2 1352 387 02 1364 319 Xe 4137 516 Selected molar constantpressure heat capacities at 1 bar and 29815 K formula CIDm J K391 mol391 carbon dioxide C02g 3711 water liquid H20I 75291 ammonia NH3g 3506 nitrogen N2g 29125 hydrogen H2g 28824 carbon graphite Csgr 8527 oxygen 02g 29335 Coefficients for TemperatureDependent Heat Capacities of Selected Substances Cwquot U K391 m0 a bT cT392 Substance a b10393 K39l c 105 K2 NH3g 2975 251 155 N2g 2858 377 050 H2g 2728 326 050 Csgr 1686 477 854 029 2996 418 167 c02g 4422 879 862 Chem 3330 Exam 1 Fall 2014 Page 4 Major error failure to explain in short words and phrases what your equations are supposed to mean When you do not clearly explain what you are doing you disqualify yourself from most if not all partial credit Major error failure to explicitly show all unit conversions and to work through to make sure they work out algebraically Major error mindlessly pugging numbers into memorized equations without understanding the derivation of the equation what the equation means or how it is used You are REQUIRED to provide a brief explanation in words for every equation you write The explanation must be in line with the equation not off somewhere to the side You are REQUIRED to show units with every number in every step of every problem Waiting until the end means that you have not checked units and are only guessing Many students left out p in denominator of Virial expansion Many students did not understand the importance of using an expansion in p when p is given Chem 3330 Exam 1 Fall 2014 Page 5 Problem 1 You may assume that all gases behave ideally in this problem B OPEN StSTEM l Consider the cartoon of a manometer shown above The system consists of Krg A manometer reading is taken at 223 C in which the liquid used was sodium polytungstate p 3100 g cm393 The ambient pressure in the laboratory was 7223 Torr and the difference between liquid heights high minus low was 1152 cm Use this information to calculate the system pressure Express your final answer in units of atm Identical inform to Prob 13 pleft pambient 7223 Torr1 atm 7600 Torr 095039 atm 1 pliquid pgh 1 3100 g cm393980665 m s392 1152 cm 350215 g m cm392 s392 1 Converting units pnquid 350215 g m cm392 s392 10393 kgg100 cmm2 35022 kg m391 5391 Converting units pnquid 35022 kg m391 s3921 Pa kg m391 s3921 atm 101325 Pa pliquid 034564 atm 2 Mechanical equilibrium requires pleft pright pright pliquid p where p is the pressure of the system Rearrange p pleft pnquid 095039 atm 034564 atm 060475 atm 1 p 06047 atm or 06048 atm 1 Calculate the molar volume of the Krg in units of L mol39l Similar to Prob 23 Using the exterior pressure from Part A in place of the system pressure is a fatal error even if you illogically confused it with the system pressure in part A C For any ideal gas Vm RTp 1 T 223 27315 29545 K 1 Vm 820574 X 10392 L atm molquot1L K39129545 060475 atm 40089 L molquot1 1 Vm 4009 L mol391 1 Calculate the mass density of the Krg in units of g L391 Variation on Prob 28 Using any values related to the manometer liquid is fatal in this part of the problem because the calculation is supposed to focus on the system p MWVm 1 For Kr MW 838 g mol391 p 838 g mol391 40089 L mol39l 2090 g L391 1 p 209 g r1 1 Chem 3330 Exam 1 Fall 2014 Page 6 Problem 2 All molar volumes in this problem should be expressed in units of L mol39l A Calculate Vm for O2g treating it as a Virial gas when T 29815 K and p 149 atm Identical inform to Prob 34 Part B and to Prob 36 Virial Equation in p me RT1 B39p C39p2 therefore RT Vm 1 B39p C39pz Using series in Vm is fatal 9 Given in the table Gas B39 10394 atm391 C39 10398 atm392 02 658 157 Pressure Expansion B39p 658 x 10394 atm391149 atm 009804 1 C39p2 157 x 10398 atm392 149 atm2 003485 1 Some students did not know what p2 means 2 1 B39p C39pz 1 009804 003485 093682 1 Z 09368 V m RTp 820574 x 10392 L atm K391 mol391 29815 K 149 atm 01642 L mol391 1 Vm 01642 L mol391 093682 01535 L mol391 1 Vm 0154 L mol391 1 B Calculate the compression factor 2 for O2g under the conditions specified in Part A Identical inform to Prob 36 Credit can be earned for only one of the three methods shown If you attempt more than one method you score is based on the least correct if there is a difference Method 1 Z meRT 1 B39p C39pz Z 09368 see above Using series in Vm is fatal Method 2 Z VmV m both of which are calculated above 2 01535 L mol391 01642 L mol391 09367 Z 0937 Method 3 Z meRT 149 atm 01538 L mol391 820574 x 10392 L atm K391 mol391 29815 K 09367 Z 0937 2 C Calculate the pressure of 02g in atm treated as a van der Walls gas when Vm 205 L mol391 at 200 C Identical inform to Prob 39 G en RT a IV p Vm b V Gas aatm M392 b10392 M391 02 1364 319 T 2000 27315 29315 K 1 RT 29315 K 820574 x 10392 L atm K391 mol39l 240551 L atm mol391 1 Vm b 205 L mol391 00319 L mol39l 2015 L mol391 1 RTVmb 240551 L atm mol391 2018 L mol39l 1192 atm 1 aVm2 1364 atm M392 205 M12 03245 atm 1 Some students did not know what sz means p 1192 atm 03245 atm 1165 atm p 116 atm 11 Chem 3330 Exam 1 Fall 2014 Page 7 Problem 3 04348 g of diphenyl ether C12H100s MW 1702072 was combusted in an adiabatic bomb calorimeter The calorimeter constant was measured beforehand to be 1189 kJ K39l To initiate the combustion 00094 grams of ignition wire were combusted The heat released by combustion of the wire is known to be 586 k g39l The initial and final temperatures of the calorimeter water bath were 22903 C and 24221 C respectively Complete the following two calculations in a 2gical order not necessarily in the order listed Identify and derive as appropriate all equations used and state the conditions for which each equation is valid Express all answers in units of kl mol391 Note the use of an invalid equation in one part of the problem will disqualify any subsequent calculation that is logically based on the outcome of the invalid equation Calculate the molar enthalpy of combustion AHW and the molar internal energy of combustion Augquot for diphenyl ether consistent with the data provided Express your final answers in units of k mall to the number of significant digits warranted by the data provided Show all work Derive the equations with explanation needed from basic starting equations Clearly state any assumptions used 10 points Calculation of AUijdenticaI inform to Prob 64 and 76 To receive any credit for this problem AU must be calculated before AH Indicated steps of the derivations must be shown to receive credit It is a fatal error in this problem to try calculating AU as anything except a calorimeter problem q qsur ern AU constant volume AU nAUm internal energy is extensive 1 q Qwire ern Qwire nAUm 1 qwire mqwires 00094 g586 kJ g39l 00551 k 1 n mMW 04348 g1702072 g mol39l 00025545 mol 1 QSur CATsur 1 AT 24221 C 22903 C 1318 C 1318 K 1 39CATsur Qwire nAUm 1 nAUm 39CATsur 39 qwire AUm 39CATsur Clwireln 1 AUm 1189 kJ K391 1318 00551 kJ00025545 mol 1 AUm 15671 kJ 00551 kJ00025545 mol 15615 kJ00025545 mol Au 451131 kJ mol1 Aum 6113 kJ mol391 1 Chem 3330 Exam 1 Fall 2014 Page 8 10 points Calculation of AHijdenticaI inform to 76 To receive credit the following calculations must be based on a reasonable AUm value It is a fatal error in this problem to try calculating AH before calculating AU Derive relationship between AH and AU H U pV definition of enthalpy AH AU ApV consider macroscopic changes AH AU ApVgas assume volumes of gases are so large compared to volumes of liquids and solids that changes in pV can be approximated by the changes for the gases pVgas ngRT Assume all gases behave ideally AH AU AngRT assume volumes of gases are so large compared to AH AU RTAng Assume isothermal pull constants outside the A Derivation must be shown 3 Write the balanced combustion reaction equation for one mole of C12H100s C12H100S 029 C02g 5 H20 1 Calculate Ang 12 14 Ang 2 2 The remaining points are contingent on credible values for AUm and for Ang Calculate RTAng and convert to k mol39l The initial temperature must be used for T T 22903 273150 296053 K RTAng 8314471 K391 mol391 296053 K2 492305J mol1 kJ 1000 J RTAng 492305 kJ mol391 2 AH AU RTAng 61131 kJ mol391492305 kJ mol39l 61180 kJ mol391 AH 6118 kJ mol391 2 Chem 3330 Exam 1 Fall 2014 Page 9 Problem 4 2150 moles of an ideal diatomic gas Cum 52 R are initially confined in a volume of 1260 L at 31523 K The sample is then expanded against the ambient pressure of 09761 atm until the system achieves mechanical equilibrium with the surroundings In the process the temperature of the system gas decreases to 28372 K Complete the following four calculations in a 2gical order not necessarily in the order listed Identify and derive as appropriate all equations used and state the conditions for which each equation is valid Express all answers in units of k Note the use of an invalid equation in one part of the problem will disqualify any subsequent calculation that is logically based on the outcome of the invalid calculation Calculate w q AU and AH for the process Place a box around your final answers Identical in form to Prob 714 8 points Calculate work It is a fatal error to calculate work via the isothermal reversible equation Doing this automatically disqualifies all points for w and for q Expansion against constant external pressure not reversible not isothermal Derive equation for work w lpexth Pulling constant outside the integral then integrating W pext AV 2 Calulate Vf nRTfpc ideal gas Vf 2150 moles 820574 x 10392 L atm K391 mol391 28372 K 09761 atm Vf 51280 L 2 Calculate AV Vf Vi 51280 L 1260 L 3868 L 1 Calculate W 39pext AV 09761 atm3868 L 37755 L atm 1 convert to k 37755 L atm0101325 kJ Latm 38255 k 1 W 38255 k 1 6 points Calculate AU AU CV AT n Cvm AT 1 AT 28372 31523 3151 K 1 C n Cvm 2150 moles25 831447 J K391 mol391 44690 K391 2 AU 446901 K3913151 K1 kJ1000 J 14082 k 1 AU 1408 k 1 5 points Calculate AH AH Cp AT n CIDm AT 1 cpm CVm R 35 R 1 Cp n CIDm 2150 moles35 831447 J K391 mol391 62565 K391 1 AH 625651 K3913151 K1 kJ1000 J 197145 k 1 AH 1971 k 1 3 points Calculate q It is a fatal error to calculate work via the isothermal reversible equation Doing this automatically disqualifies all points for w and for q First Law AU q w therefore q AU w 1 Reasonable values for AU and w must be calculated above for credit here q 14082 kJ 38255 k1 24174 k 1 q 2417 k 1 Chem 3330 Exam 1 Fall 2014 Page 10

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