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# exam 1 solutions CHEM3330

Texas State

GPA 2.9

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This 5 page Study Guide was uploaded by mythili venkateswaran on Thursday October 1, 2015. The Study Guide belongs to CHEM3330 at Texas State University taught by Dr. Easter in Fall 2014. Since its upload, it has received 31 views. For similar materials see Physical Chemintry 1 in Chemistry at Texas State University.

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Date Created: 10/01/15

Problem 1 5 points The density of a gas was measured to be 147 g L391 at 594 K and 546 atm Calculate the apparent molecular weight of the gas SHOW AND EXPLAIN ALL WORK Equation Brief Explanation pV nRT 1 Ideal gas law n mMW 2 Moles expressed in terms of mass and mol wt pV mMWRT 3 Substitute n from 1 into 2 MW mRTpV 4 Rearrange 3 to isolate MW p mV 5 Definition of density massvolume MW pRTp 6 Substitute r for mV 5 into 4 Derivation with complete and valid explanations required 2 pts Correct final equation with or without derivation 1 pts Substitute values into the final equation 6 MW pRTp MW 147 g L391 820574 x 10392 L atm K391 mol391 594 K 546 atm 1312 g mor1 Final Answer MW 131 g mol391 2 pts Possible deductions from final answer 12 point for each digit in sig fig error 1 point for missing or invalid units Chem 3330 Exam 1 Solutions Fall 2015 Page 1 Problem 2 7 points A Treating carbon monoxide as a Virial gas calculate the molar volume of CO at 29815 K when p 223 atm B Use the result to calculate the compression factor Z under the same conditions C Are the net interactions between CO molecules attractive or repulsive under the stated conditions SHOW AND EXPLAIN ALL WORK Second and Third Virial Coefficients of Several Real Gases at 29815 K M mol L391 Gas B39 10394 atm391 C39 10398 atm392 CO 352 247 Because we are asked to calculate the molar volume dependent variable we must use the Virial expansion in pressure The coefficients from the table are duplicated here me RT 1 B p C pz Virial Equation in p to the third term Vm RTp1 Bquot C pz Divide both sides by p to isolate Vm Correct Equation MUST be a pexpansion cannot be a Vm expansion 2 points Substitute Values into the Equation Vm RTp1 B p C pz vm 820574 x 10392 L atm K391m01391 29815 K 223 atm1 352 x 10394 atm391223 atm 247 x 10398 atm392223 atm2 Vm RTp 01097 L moI391 0110 L mol391 2 points No points if this was calculated as Vm confusing actual Vm with ideal Vm Z meRT 1 B39p Up2 100000 007850 01228 10443 1044 1 point No points if the pexpansion equation and corresponding coefficients were not used Vm 2 V0 10443 01097 L mol391 01146 L moI391 0115 L mol391 1 point No points if the pexpansion equation and corresponding coefficients were not used Because 2 gt 1 net forces are repulsive 1 point You can earn credit if you calculated a value for Z and your answer is consistent with that value Possible deductions from final answer 12 point for each digit in sig fig error 1 point for missing or invalid units Chem 3330 Exam 1 Solutions Fall 2015 Page 2 Problem 3 10 points An adiabatic bomb calorimeter was calibrated and the calorimeter constant C was measured to be C 1118 kJ C391 05889 g of an organic compound C14H1202s MW 2122439 was combusted The amount of ignition wire combusted for which the energy of combustion is known to be 1400 calg was 00098 g The final and initial temperatures of the water bath were 2738 C and 2532 C respectively Calculate the experimental value of AHm for the compound SHOW AND EXPLAIN ALL WORK The heat absorbed by the water bath is given by CAT In the problem qsur 1118 kJ C391 206 C The heat absorbed by the water bath is equal to the system s q Because the reaction is at constant volume q for the reaction alone is AUcombustion The heat generated by combustion of the wire must be added Therefore q AU qwire qsur Eqn 1 qwire is the mass of the wire multiplied its energy of combustion per gram The units must be converted from cal to k qwire mwireAUcwire 00098 g1400 cag4184 Jca1 kJ1000 J Rearranging Eqn 1 AU qwire qsur 2303 kJ 00574 k1 2297 kJ 3 pts The number of moles combusted is n mMW 05889 g 2122439 g mol39l 00027745 mol The molar value of AU is AUm AUn 2297 kJ 00027745 mol 2 pts A balanced combustion equation is required to earn any of the remaining points that follow 1 Pts The change is moles of gas in the balanced equation is Ang 14 16 1 pts Conversion to AH is required assuming constant T and ideal conditions AH AU RTAng Eqn 2 Important Ang 2 ONLY if AU is molar because it is based on one mole of the reactant f AU is extensive and not molar the value you must use for Ang is 2 multiplied by the number ofmoles The initial temperature of the water bath must be used for T T 2532 27315 K RTAng 8314471 K391 mol391 29847 K2 mol 496324J x 1 kJ1000 J Finally using Eqn 2 AH 8279 k 496324 k 8284 k Final 3 pts Credit requires that the work be shown Deduct 12 for each digit of sig fig error 1 pt max deduct 1 point for invalid or missing units Chem 3330 Exam 1 Solutions Fall 2015 Page 3 Problem 4 10 points 0952 moles of benzene MW 7811 undergo the liquid vapor phase transition at 801 C and exactly 1 atm For benzene AHvap 308 k mol39l and the density of liquid benzene is 0815 g cm393 under the stated equilibrium phase change conditions Calculate AH q w and AU for the phase change assuming that the benzene vapor behaves ideally SHOW AND EXPLAIN ALL WORK The vaporization reaction is C5H5l gt C5H5g Calculation of AH 0952 moles 308 k mol391 2932 k 2 pts Reminder the final answer for a requested quantity must be rounded to the correct number of significant figures Subscripts may not be shown You must demonstrate that you know how to round correctly However if the value is use in a subsequent calculation all of the extra nonsignificant digits should be used for the calculation Calculation of q The process is at constant pressure and the system is in mechanical equilibrium with its surroundings Therefore Calculation of w 1pt Starting equation dw pext dV Integrate both sides using the fact that the pressure is constant AV Vgas 39 Vliquid T 801 27315 35325 K 0952 mol820574 x 10392 L atm K391 mol39135325 K1000 atm 0952 mol 7811 g mol391 0815 g cm3931000 cm3L AV 275954 L 009124 L w 1000 atm27504 L 27504 L atm w 27504 L atm101325 JLatm1 kJ1000 J 2787 kJ 5 pts Because information was provided to enable you to calculate Vliquid the volume of the liquid should not be neglected in the calculation The equation w RTAngis an approximation that neglects the volume of the liquid It is not correct given the information provided in the problem and can earn no more than 2 of the 5 points Calculation of AU First Law Au 293 k m k 265 k 2 pt Deduct 12 for each digit of sig fig error 1 pt max deduct 1 point for invalid or missing units Chem 3330 Exam 1 Solutions Fall 2015 Page 4 Problem 5 8 points 215 moles of N2 are initially at a volume of 175 L and a temperature of 3002 K The sample is then expanded reversibly and isothermally until the final pressure is 0924 atm Calculate w q AU and AH for the process assuming that N2 behaves ideally SHOW AND EXPLAIN ALL WORK Isothermal means that AT 0 Calculation of AU Because AU 0 The correct equation must be given and AT 0 must be explicitly stated to receive credit 1 pt Calculation of AH Because AH 0 The correct equation must be given and AT 0 must be explicitly stated to receive credit 1 pt Calculation of w th 1 Eqn 2 Must be explicitly stated in wordsin your derivation Eqn 3 Must be explicitly stated in wordsin your derivation Combining Eqns 23 and substituting into 1 gives dw nRT dVV Integrate both sides ldw l nRT dVV with the other constants n and R ldw nRT ldVV Must be explicitly stated in wordsin your derivation Integrations gives w nRT nV2V1 Eqn 4 Because n and T are constant Boyle39s Law applies p1V1 p2V2 Constant n and T must be explicitly stated in words Boyle39s Law rearranges to Eqn 5 Substitute Eqn 5 into Eqn 4 Final equation only if derivation is completely detailed in words and accurate 3 pts p1 is calculated from the ideal gas law p1 nRTVl I 215 moles 820574 x 10392 L atm K391 mol391 3002 K 175 L 1 pt Using Eqn 6 w 215 moles831447J K391 mol391 3002 K n30254 atm 0924 atm 48723 tpt Alternate path use w nRT nV2V1 and calculate V2 from n T and p2 Calculation of q First Law Because q w 1 pt Deduct 12 for each digit of sig fig error 1 pt max deduct 1 point for invalid or missing units Chem 3330 Exam 1 Solutions Fall 2015 Page 5

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