Solution manual for physical chemistry for the life sciences 1st edition by engel

Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ Chapter 1: Fundamental Concepts of Thermodynamics Questions on Concepts Q1.1) The location of the boundary between the system and the surroundings is a choice that must be made by the thermodynamicist Consider a beaker of boiling water in an airtight room Is the system open or closed if you place the boundary just outside the liquid water? Is the system open or closed if you place the boundary just inside the walls of the room? If the system boundaries are just outside of the liquid water, the system is open because water can escape from the top surface The system is closed if the boundary is just inside the walls, because the room is airtight Q1.2) Real walls are never totally adiabatic Order the following walls in increasing order with respect to their being diathermal: 1-cm-thick concrete, 1-cm-thick vacuum, 1-cm-thick copper, 1-cm-thick cork 1-cm-thick vacuum < 1-cm-thick cork < 1-cm-thick concrete < 1-cm-thick copper Q1.3) Why is the possibility of exchange of matter or energy appropriate to the variable of interest a necessary condition for equilibrium between two systems? Equilibrium is a dynamic process in which the rates of two opposing processes are equal However, if the rate in each direction is zero, no exchange is possible, and the system cannot reach equilibrium Q1.4) At sufficiently high temperatures, the van der Waals equation has the form P ≈ RT/(Vm – b) Note that the attractive part of the potential has no influence in this expression Justify this behavior using the potential energy diagram of Figure 1.7 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics E V(r) r In this case, the energy well depth is small compared to the total energy of the particle Therefore, the particle is unaffected by the attractive part of the potential Q1.5) Parameter a in the van der Waals equation is greater for H2O than for He What does this say about the form of the potential function in Figure 1.7 for the two gases? This means that the depth of the attractive potential is greater for H2O than for He Problems Problem numbers in RED indicate that the solution to the problem is given in the Student Solutions Manual P1.1) A sealed flask with a capacity of 1.00 dm3 contains 5.00 g of ethane The flask is so weak that it will burst if the pressure exceeds 1.00 × 106 Pa At what temperature will the pressure of the gas exceed the bursting pressure? With pV = nRT and n = m : M Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ T= ( ) ( Chapter 1/Fundemental Concepts of Thermodynamics ) ( ) p V M 1.00 × 10 Pa × 0.001 m × 30.08 × 10 -3 kg mol −1 = = 723.6 K mR (0.005 kg ) × 8.314472 J K −1 mol −1 ( ) P1.2) Consider a gas mixture in a 2.00-dm3 flask at 27.0°C For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the composition of the mixture in mole percent: a 1.00 g H2 and 1.00 g O2 b 1.00 g N2 and 1.00 g O2 c 1.00 g CH4 and 1.00 g NH3 To calculate the partial pressures we use the ideal gas law: a) p H2 = p O2 = n H2 R T V n O2 R T V = = m H2 R T M H2 V m O2 R T M O2 V = (0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg ) = 6.18 × 10 Pa (2.00 × 10 -3 ) ( m × 2.02 × 10 -3 kg mol −1 ) ( 0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg ) = (2.00 × 10 -3 ) ( -3 m × 32.0 × 10 kg mol −1 ) = 3.90 × 10 Pa p total = p H + p O2 = 6.57 × 10 Pa   0.001 kg   -3 −1  2.02 × 10 kg mol  mol H  = 100 × = 94.1 % mol % H = 100 × (mol H + mol O )     0.001 kg 0.001 kg       2.02 × 10 -3 kg mol −1  +  32.0 × 10 -3 kg mol −1         0.001 kg   -3 −1  mol O  32.0 × 10 kg mol  mol % O = 100 × = 5.9 % = 100 × (mol H + mol O )     0.001 kg 0.001 kg       2.02 × 10 -3 kg mol −1  +  32.0 × 10 -3 kg mol −1        b) Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics p N2 = p O2 = n N2 R T V n O2 R T V = = m N2 R T M N2 V m O2 R T M O2 V = (0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg ) = 4.45 × 10 Pa = (0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg ) (2.00 × 10 -3 (2.00 × 10 -3 ) ( m × 28.02 × 10 -3 kg mol −1 ) ( -3 m × 32.0 × 10 kg mol −1 ) ) = 3.90 × 10 Pa p total = p H + p O2 = 8.35 × 10 Pa   0.001 kg   −1  -3 28.02 × 10 kg mol  mol N  = 100 × = 53.3 % mol % N = 100 × (mol N + mol O )     0.001 kg 0.001 kg       28.02 × 10 -3 kg mol −1  +  32.0 × 10 -3 kg mol −1          0.001 kg   -3 −1  mol O  32.0 × 10 kg mol  mol % O = 100 × = 46.7 % = 100 × (mol N + mol O )     0.001 kg 0.001 kg       28.02 × 10 -3 kg mol −1  +  32.0 × 10 -3 kg mol −1       c) p NH = p CH = n NH R T V n CH R T V = = m NH R T M NH V m CH R T M CH V = (0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg ) = 7.32 × 10 Pa = (0.001 kg ) × (8.314472 J K −1 mol −1 )× (300.15 kg ) (2.00 × 10 (2.00 × 10 -3 -3 ) ( m × 17.03 × 10 -3 kg mol −1 m × 16.04 × 10 kg mol p total = p H + p O2 = 1.51 × 10 Pa Full file at https://TestbankDirect.eu/ ) ( -3 −1 ) ) = 7.77 × 10 Pa Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ mol % NH = 100 × Chapter 1/Fundemental Concepts of Thermodynamics mol NH (mol NH + mol CH )   0.001 kg   -3 −1  17.03 × 10 kg mol   = 100 × = 48.5 %     0.001 kg 0.001 kg       16.04 × 10 -3 kg mol −1  +  17.03 × 10 -3 kg mol −1        mol % CH = 100 × mol CH (mol NH + mol CH )   0.001 kg   -3 −1  17.03 × 10 kg mol   = 100 × = 51.5 %     0.001 kg 0.001 kg       16.04 × 10 -3 kg mol −1  +  17.03 × 10 -3 kg mol −1        P1.3) Approximately how many oxygen molecules arrive each second at the mitochondrion of an active person? The following data are available: oxygen consumption is about 40 mL of O2 per minute per kilogram of body weight, measured at T = 300 K and P = 1.0 atm An adult with a body weight of 64 kg has about × 1012 cells Each cell contains about 800 mitochondria With pV = nRT the number of moles per minute and per kg of body weight is: ( ) ( pV 101325 Pa ) × 4.0 × 10 -5 m n= = 1.6249 × 10 -3 mol −1 = −1 −1 R T (300 K ) × 8.314472 J K mol × (1 ) ( ) For a body weight of 64 kg and per second the number of moles is:   -3 n = 1.6249 × 10 -3 mol −1 × (64 kg ) ×   = 1.7332 × 10 mol  60 s  ( ) Converting to molecules: ( ) ( ) molecules O = 1.7332 × 10 -3 mol × 6.02214 × 10 23 molecules mol -1 = 1.04 × 10 21 molecules Wit h × 1012 cells in a 64 kg body, and 800 mitochondria in each cell: (1.04 × 10 ) = 1.25 × 10 = (1.0 × 10 )× (800) 21 molecules O 12 molecules P1.4) In a normal breath, about 0.5 L of air at 1.0 atm and 293 K is inhaled About 25.0% of the oxygen in air is absorbed by the lungs and passes into the bloodstream For a respiration rate of Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics 18 breaths per minute, how many moles of oxygen per minute are absorbed by the body? Assume the mole fraction of oxygen in air is 0.21 Compare this result with Example Problem 1.1 We use the ideal gas law to calculate the number of moles of air inhaled every minute: n air ( ) ( pV 101325 Pa ) × 5.0 × 10 -4 m = = = 0.0208 mol R T (2930 K ) × 8.314472 J K −1 mol −1 ( ) We then use the mole fraction to determine the number of moles of O2 for 18 breaths: ( n O = (0.0208 mol) × (0.21) × (0.25) × 18 −1 ) = 0.01966 mol −1 P1.5) Suppose that you measured the product PV of mol of a dilute gas and found that PV = 22.98 L atm at 0.00°C and 31.18 L atm at 100°C Assume that the ideal gas law is valid, with T = t(°C) + a, and that the value of R is not known Determine R and a from the measurements pro- vided The ideal gas law and solving for R: n= pV R (T + a ) R= pV n (T + a ) Using the information for the two different conditions and solving for a yields a: p1 V1 p V2 = n (T1 + a ) n (T2 + a )  p V   T1 - T2 1  p V2  a= =  p1 V1   − 1  p V2    (22.98 L atm )   C - 100  C (31.18 L atm )   = 280.2  C  (22.98 L atm )   − 1  (31.18 L atm )  ( )( ) Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ Chapter 1/Fundemental Concepts of Thermodynamics R is then: R= p1 V1 (22.98 L atm ) = n (T1 + a ) (1 mol) ×  C + 280.2  C (( ) ( )) = 0.0820 L atm mol -1  C -1 P1.6) Devise a temperature scale, abbreviated G, for which the magnitude of the ideal gas con- stant is 1.00 J G–1 mol–1 The G-temperature scale is given by: G = T(K ) × 8.314472 G K -1 So that on the Kelvin scale: ( ) R T = 8.314472 J mol -1 K -1 × (293 K ) = 2436.14 J mol -1 And on the G scale: ( ) ( ) R G T = 1.0 J mol -1 G -1 × (293 K ) × 8.314472 G K -1 = 2436.14 J mol -1 P1.7) A rigid vessel of volume 0.500 m3 containing H2 at 20.5°C and a pressure of 611 × 103 Pa is connected to a second rigid vessel of volume 0.750 m3 containing Ar at 31.2°C at a pressure of 433 × 103 Pa A valve separating the two vessels is opened and both are cooled to a temperature of 14.5°C What is the final pressure in the vessels? We need to first calculate the number of moles of H2 and Ar using the ideal gas law: n H2 = (611× 10 Pa )× (0.500 m ) pV = 125 mol = R T (273.15 K + 20.5 K ) × (8.314472 J K −1 mol −1 ) n Ar = (433 × 103 Pa )× (0.750 m ) pV = 128 mol = R T (273.15 K + 31.2 K ) × (8.314472 J K −1 mol −1 ) Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics Using the final number of moles, the final pressure at 14.5ºC is then: p= n total R T (128 mol + 125 mol) × (273.15 K + 14.5 K ) × (8.314472 J K −1 mol −1 ) = = 4.84 × 10 Pa 3 Vtotal (0.500 m + 0.750 m ) P1.8) In normal respiration, an adult exhales about 500 L of air per hour The exhaled air is saturated with water vapor at body temperature T = 310 K At this temperature water vapor in equilibrium with liquid water has a pressure of P = 0.062 atm Assume water vapor behaves ideally under these conditions What mass of water vapor is exhaled in an hour? Using the ideal gas law, and solving for m yields: pV =nRT = m= m RT M M p V (18.02 × 10 -3 kg mol −1 )× (0.062 atm ) × (101325 Pa atm -1 )× (0.5 m ) = = 0.0227 kg h -1 −1 −1 RT (300 K ) × (8.314472 J K mol )× (1 h ) P1.9) At T = 293 K and at 50.% relative humidity, the pressure of water vapor in equilibrium with liquid water is 0.0115 atm Using the information in Problem P1.8, determine what mass of water is inhaled per hour and the net loss of water through respiration per hour Using the ideal gas law, and solving for m, with the volume from P1.8 yields: pV =nRT = m= m RT M M p V (18.02 × 10 -3 kg mol −1 )× (0.0115 atm ) × (101325 Pa atm -1 )× (0.5 m ) = × 0.5 = 2.15 × 10 -3 kg h -1 −1 −1 RT (293 K ) × (8.314472 J K mol )× (1 h ) Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ Chapter 1/Fundemental Concepts of Thermodynamics With the results from P1.8, the net loss of water is then: net loss = 0.0227 kg h -1 - 2.15 × 10 -3 kg h -1 = 0.0205 kg h -1 P1.10) A compressed cylinder of gas contains 1.50 × 103 g of N2 gas at a pressure of 2.00 × 107 Pa and a temperature of 17.1°C What volume of gas has been released into the atmosphere if the final pressure in the cylinder is 1.80 × 105 Pa? Assume ideal behavior and that the gas temperature is unchanged Let ni and nf be the initial and final number of moles of N2 in the cylinder, respectively: ni R T n RT = f pi pf Solving for nf yields: nf = ni pf (1.50 kg ) (1.80 × 10 Pa ) = 0.482 mol = × pi (28.01× 10 -3 kg mol −1 ) (2.99 × 105 Pa ) The initial number of moles is: ni = (1.50 kg ) (28.01× 10 -3 kg mol −1 ) = 53.55 mol The volume of gas that is released is: V= (n f − n i )RT (53.55 mol - 0.482 mol) × (290.25 K ) × (8.314472 J K −1 mol −1 ) = = 1.26 m (101325 Pa ) p P1.11) As a result of photosynthesis, 1.0 kg of carbon is fixed per square meter of forest As- suming air is 0.046% CO2 by weight, what volume of air is required to provide 1.0 kg of fixed carbon? Assume T = 298 K and P = 1.00 atm Also assume that air is approximately 20.% oxy- Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics gen and 80.% nitrogen by weight We first calculate the number of moles necessary to provide one kg of CO2: n (CO ) = (1 kg ) m(CO ) = = 22.722 mol M(CO ) 44 × 10 -3 kg mol-1 ( ) With the composition of air, a certain volume of air containing one kg of CO2 also contains: n (O ) = n (N ) = m(O ) M(O ) m(N ) M (N ) (1 kg ) ×  0.2    0.00046  = 13586.96 mol = (32 × 10-3 kg mol-1 ) (1 kg ) ×  0.8   0.00046   = (28 × 10-3 kg mol-1 ) = 62111.8 mol The volume of air can then be obtained by considering the total number of moles: Vair = = n tot R T {n (CO ) + n (O ) + n (N )} R T = p p {(22.722 mol) + (13586.962 mol) + (62111.8 mol)}× (8.314472 J K -1 mol-1 )× (298 K ) (101325 Pa ) = 1851.6 m P1.12) A balloon filled with 10.50 L of Ar at 18.0°C and atm rises to a height in the atmos- phere where the pressure is 248 Torr and the temperature is –30.5°C What is the final volume of the balloon? We first calculate the number of moles of Ar at atm using the ideal gas law: n= -1 p V (1 atm ) × (101325 Pa atm ) × ( 0.0105 m ) = = 0.4395 mol RT ( 291.15 K ) × (8.314472 J K −1 mol −1 ) We can then determine the volume after the balloon raised using the number of moles: 10 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics b The percent relative humidity is defined as %RH = PH pressure of water in the sample and PH* 2O 2O P*H 2O where PH 2O is the partial = 0.197 atm is the equilibrium vapor pressure of water at 60.°C The gas is compressed at 60.°C until the relative humidity is 100% What volume does the mixture contain now? c What fraction of the water will be condensed if the total pressure of the mixture is isother- mally increased to 200 atm? a) mol % N = 100 × mol % O = 100 × mol % Ar = 100 × p N2 p total p O2 p total = 100 × = 100 × (0.78 × 0.88 atm ) = 68.6 % (1.00 atm ) (0.21× 0.88 atm ) = 18.5 % (1.00 atm ) p Ar (0.01 × 0.88 atm) = 0.9 % = 100 × (1.00 atm ) p total mol % H O = 100 × p H 2O p total = 100 × (0.12 atm) = 12.0 % (1.00 atm ) b) p H 2O V = n H 2O R T V p ′H O V ′ = p H 2O V , where the prime refers to 100% RH V′ = p H 2O V p ′H O = (0.12 atm )× (20.0 L ) (0.197 atm ) = 12.2 L If all the water remained in the gas phase, the partial pressure of water at a total pressure of 200 atm would be: 12 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ Chapter 1/Fundemental Concepts of Thermodynamics p H O = p total × mol fraction H O = 200 atm × 0.12 = 24.0 atm However, the partial pressure of water cannot be greater than 0.197 atm, and the excess will condense The fraction that condenses is given by: frcation condensed = (24.0 atm ) − (0.197 atm ) (24.0 atm ) = 0.992 P1.16) A mixture of 2.50 × 10–3 g of O2, 3.51 × 10–3 mol of N2, and 4.67 × 1020 molecules of CO is placed into a vessel of volume 3.50 L at 5.20°C a Calculate the total pressure in the vessel b Calculate the mole fractions and partial pressures of each gas a) The pressure in the vessel can be calculated by using the total number of moles: n (O ) = 4.67 × 10 20 molecules 2.5 × 10 -3 g −5 ( ) n CO = = 7.7547 × 10 − mol = 7.8125 × 10 mol 6.02214 × 10 23 molecules mol -1 32.0 g mol -1 n tot R T = V 7.8125 × 10 −5 mol + 7.7547 × 10 − mol + 3.51 × 10 −3 × 8.314472 J K −1 mol −1 × (278.35 K ) = 0.0035 m 2885.375 Pa = 0.00288 bar p= ( ( b) The mole fractions, x i = ) ) ( ) ni , and partial pressures, p i = n i × p , for the gases are: n tot x (O ) = 7.8125 × 10 −5 mol = 0.0179 4.3636 × 10 −3 mol p(O ) = 0.00288 bar × 0.0179 = 5.16 × 10 −4 bar x (N ) = 3.51 × 10 −3 mol = 0.8044 4.3636 × 10 −3 mol p(N ) = 0.00288 bar × 0.8044 = 2.32 × 10 −3 bar x (CO ) = 7.7547 × 10 −4 mol = 0.1777 4.3636 × 10 −3 mol p(CO ) = 0.00288 bar × 0.1777 = 5.12 × 10 −4 bar P1.17) Carbon monoxide (CO) competes with oxygen for binding sites on the transport protein hemoglobin CO can be poisonous if inhaled in large quantities A safe level of CO in air is 50 13 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics parts per million (ppm) When the CO level increases to 800 ppm, dizziness, nausea, and unconsciousness occur, followed by death Assuming the partial pressure of oxygen in air at sea level is 0.20 atm, what ratio of O2 to CO is fatal? Converting the partial pressure of O2 in the atmosphere to ppm using xi = x (O ) = pi : p 0.2 atm = 0.2 = 20% = 200000 ppm atm Therefore, the fatal O2/CO ratio is: x (O ) 200000 = = 250 x (CO ) 800 P1.18) A normal adult inhales 0.500 L of air at T = 293 K and 1.00 atm To explore the surface of the moon, an astronaut requires a 25.0-L breathing tank containing air at a pressure of 200 atm How many breaths can the astronaut take from this tank? We first need to calculate the number of moles inhaled at atm using the ideal gas law: n breath = ( ) (101325 Pa ) × 0.0005 m pV = = 0.0208 mol R T 8.314472 J K −1 mol −1 × (293 K ) ( ) Next we determine how many moles of air are in the tank: n tank ( 200 x 101325 Pa ) × ( 0.025 m3 ) pV = = = 207.96 mol R T ( 8.314472 J K −1 mol −1 ) × ( 293 K ) Therefore, the number of breaths the astronaut can take is: number of breaths = n tank (207.96 mol) = 9998 = n breath (0.0208 mol) P1.19) Liquid N2 has a density of 875.4 kg m–3 at its normal boiling point What volume does a balloon occupy at 18.5°C and a pressure of 1.00 atm if 2.00 × 10–3 L of liquid N2 is injected into 14 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ Chapter 1/Fundemental Concepts of Thermodynamics it? We can use the density of liquid N2 to calculate the mass of N2 that is injected into the balloon: m N = d LN2 VLN2 = ( 875.4 kg m -3 ) × ( × 10-6 m3 ) = 1.7508 × 10-3 kg We calculate the volume using the ideal gas law: V= nRT mRT = p Mp (1.7508 × 10 kg ) × ( 291.15 K ) × (8.314472 J K = ( 28 × 10 kg mol ) × (101325 Pa ) -3 −1 -3 −1 mol −1 ) = 1.50 × 10-3 m3 = 1.50 L P1.20) Yeast and other organisms can convert glucose (C6H12O6) to ethanol (CH3CH2OH) by a process called alcoholic fermentation The net reaction is C6H12O6(s) → 2C2H5OH(l) + 2CO2(g) Calculate the mass of glucose required to produce 1.0 L of CO2 measured at P = 1.00 atm and T = 300 K First, we calculate the number of moles of CO2 under the conditions given: n CO ( ) Vp × 10 -3 m × (101325 Pa ) = = R T (300 K ) × 8.314472 J K −1 mol −1 ( = 0.0406 mol ) From the equation of the reaction, we can see that the number of moles of glucose is half the number of moles of CO2, and the mass of glucose necessary is: m glucose = M n glucose = M n CO  0.0406 mol  = 180.18 g mol −1 ×   = 3.66 g 2   ( ) P1.21) A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 3.00 times the amount needed to completely oxidize the propane to CO2 and H2O at con15 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics stant temperature Calculate the mole fraction of each component in the resulting mixture after oxidation assuming that the H2O is present as a gas The reaction we have to consider is: C H (g ) + O (g ) ⎯ ⎯→ CO (g ) + H O (g ) If m moles of propane are present initially, there must be 15 m moles of O2 After the reaction is complete, there are m moles of CO2, m moles of H2O, and 10 m moles of O2 Therefore: x CO = 3m = 0.176 ; 17 m x H 2O = 4m = 0.235 , 17 m x O2 = 10 m = 0.588 17 m P1.22) Calculate the volume of all gases evolved by the complete oxidation of 0.25 g of the amino acid alanine (NH2CHCH3COOH) if the products are liquid water, nitrogen gas, and carbon dioxide gas and the total pressure is 1.00 atm and T = 310 K The reaction equation is: NH 2CHCH 3COOH (s ) + O (g ) ⎯ ⎯→ N (g ) + H 2O (l ) + CO (g ) We use the mass of alanine to calculate the number of moles of alanine that were oxidized: n alanine = (0.25 g ) = 2.8387 ×10-3 mol m alanine = M alanine 88.07 g mol−1 ( ) From the reaction equation we know that: n N2 = n alanine = 1.4194 × 10-3 mol And the volume of N2 consumed in the reaction is: 16 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ VN2 = ( Chapter 1/Fundemental Concepts of Thermodynamics ) ( ) n R T 1.4194 × 10-3 mol × (310 K )× 8.314472 J K −1 mol−1 = = 3.61× 10-5 m = 0.0361 L (101325 Pa ) p n CO2 = n alanine = 8.5161× 10-3 mol VCO2 = ( ) ( ) n R T 8.5161× 10-3 mol × (310 K )× 8.314472 J K −1 mol−1 = = 2.166 × 10-4 m = 0.2166 L (101325 Pa ) p Then the total volume of gases produced in the reaction is: Vtotal = VN2 + VCO2 = 0.2166 L + 0.0361 L = 0.253 L P1.23) A gas sample is known to be a mixture of ethane and butane A bulb having a 200.0-cm3 capacity is filled with the gas to a pressure of 100.0 × 103 Pa at 20.0°C If the weight of the gas in the bulb is 0.3846 g, what is the mole percent of butane in the mixture? With pV = nRT the total number of moles of moles of the mixture is: n tot = ( ) ( ) pV 100 × 10 -3 Pa × 0.0002 m = = 8.2055 × 10 -3 mol R T 8.314472 J K −1 mol −1 × (293.15 K ) ( ) The total number of moles can also be expressed as: n tot = n ethane + n butane = m ethane m butane + M ethane M butane With methane = mtot – mbutane: n tot = ((m tot − m butane ) × M butane + m butane M ethane ) M ethane × M butane Solving for mbutane and dividing by Mbutane yields after some rearrangement: n butane = (n tot × M ethane − m tot ) ((8.2055 × 10 −3 mol) × (30.08 g mol -1 ) − (0.3846 g )) = = 4.9101 × 10 −3 mol An -1 -1 (M ethane − M butane ) ((30.08 g mol ) − (58.14 g mol )) d finally: x butane = ( ( ) ) n butane 4.9101 × 10 −3 mol = = 0.599 = 59.9% n total 8.2055 × 10 −3 mol 17 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics P1.24) A glass bulb of volume 0.136 L contains 0.7031 g of gas at 759.0 Torr and 99.5°C What is the molar mass of the gas? With M = m and pV = nRT the molar mass of the gas is: n -3 −1 −1 m R T ( 0.7031 ×10 kg ) × (8.314472 J K mol ) × ( 372.65 K) M= = = 0.1583 ×10 -3 kg mol -1 = 158.3 g mol -1 pV (101191.68 Pa ) × ( 0.000136 m ) P1.25) The total pressure of a mixture of oxygen and hydrogen is 1.00 atm The mixture is ig- nited and the water is removed The remaining gas is pure hydrogen and exerts a pressure of 0.400 atm when measured at the same values of T and V as the original mixture What was the composition of the original mixture in mole percent? We start by constructing the following table: H2 (g) + O2 (g) initial moles n H2 n O2 at equilibrium n H2 − 2α n O2 − α → H2O (  ) 2α If O2 is consumed completely, n O − α = , or n O = α The number of moles of H2 remaining is 2 n H2 − 2α = n H2 − 2n O2 Let p1 be the initial total pressure and p2 be the total pressure after all O2 is consumed ( p1 = n H2 + n O2 ) RVT and p = (n  H2 + n O2 ) RVT Dividing the second equation by the first: n H2 n O2 p2 = −2  = x H2 − 2x O2 = − x O2 − 2x O2 = − 3x O2 p1 n H2 + n O2 n H2 + n O2 ( ) ( ) 18 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ Chapter 1/Fundemental Concepts of Thermodynamics  p   (0.400atm)   = 0.20 , and x H2 = 0.80 x O2 = 1 −  = 1 − (1.00 atm)   p1   P1.26) The photosynthetic formation of glucose in spinach leaves via the Calvin cycle involves 4− the fixation of carbon dioxide with ribulose 1-5 diphosphate C5 H P2 O11 (aq) to form 3- phosphoglycerate C3 H PO 3− (aq) : 4− (aq) + H 2O(l) + CO (g) C5 H P2 O11 + → 2C 3H PO 3− (aq) + 2H (aq) If 1.00 L of carbon dioxide at T = 273 K and P = 1.00 atm is fixed by this reaction, what mass of 3-phosphoglycerate is formed? First we need to calculate the number of moles of fixed CO2: nCO2 = (101191.68 Pa ) × ( 0.001 m ) pV = = 0.0446 mol RT (8.314472 J K−1 mol −1 ) × ( 273 K ) The number of moles of 3-phosphoglycerate formed is given by: n3− phos = nCO2 = 0.0892 mol And finally, the mass of 3-phosphoglycerate formed is: m = n M = ( 0.0928 atm ) × (182.97 g mol −1 ) = 16.32 g P1.27) Calculate the pressure exerted by Ar for a molar volume of 1.42 L mol–1 at 300 K using the van der Waals equation of state The van der Waals parameters a and b for Ar are 1.355 bar dm6 mol–2 and 0.0320 dm3 mol–1, respectively Is the attractive or repulsive portion of the potential dominant under these conditions? 19 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics To determine what portion of the potential is dominant we need to compare the pressure predicted by the van der Waals equation of state with that predicted by the ideal gas law The van der Waals equation of state yields: p vdW = ( ) ( ) RT a 8.314472 × 10-2 bar dm K −1 mol−1 × (300 K ) 1.355 bar dm mol−2 − = − = 17.3 bar (Vm − b ) Vm 1.42 dm mol−1 − 0.0321 dm mol−1 1.42 dm mol −1 (( ) ( )) ( ) The ideal gas law gives: p ideal = ( ) R T 8.314472 × 10-2 bar dm K −1 mol−1 × (300 K ) = = 17.6 bar Vm 1.42 dm mol−1 ( ) Because pvdW < pideal, the attractive part of the potential dominates P1.28) Calculate the pressure exerted by benzene for a molar volume of 1.42 L at 790 K using the Redlich–Kwong equation of state: P= a nRT n2 a 1 RT − = − Vm − b T Vm (Vm + b) V − nb T V (V + nb) The Redlich–Kwong parameters a and b for benzene are 452.0 bar dm6 mol–2 K1/2 and 0.08271 dm3 mol–1, respectively Is the attractive or repulsive portion of the potential dominant under these conditions? The exerted benzene pressure is calculated using p = (8.314 × 10 L bar K mol )× (790 K ) − (1.42 L mol − 0.08271 L mol ) (452.0 L bar K mol ) (1.42 L mol )× (1.42 L mol ( 790K ) -2 p= -1 −1 -1/2 RT a : − (Vm − b ) T Vm × (Vm + b ) −1 −1 −2 −1 −1 − 0.08271 L mol −1 ) = 41.6 bar To determine whether the attractive or repulsive portion of the potential is dominant we need to compare the pressure from above with that obtained by using the ideal gas law: 20 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ p ideal = ( Chapter 1/Fundemental Concepts of Thermodynamics ) ) ( n R T (1 mol) × 8.314472 J K −1 mol −1 × (790.0 K ) × (1 bar ) = 46.3 bar = V 1.42 × 10 -3 m × 10 Pa ( ) Since the pressure calculated with the ideal gas law is higher than the pressure calculated with the Redlich–Kwong equation of state, the attractive forces are dominant P1.29) When Julius Caesar expired, his last exhalation had a volume of 500 cm3 and contained 1.00 mol % argon Assume that T = 300 K and P = 1.00 atm at the location of his demise Assume further that T and P currently have the same values throughout the Earth’s atmosphere If all of his exhaled Ar atoms are now uniformly distributed throughout the atmosphere (which for our calculation is taken to have a thickness of 1.00 km), how many inhalations of 500 cm3 must we make to inhale one of the Ar atoms exhaled in Caesar’s last breath? Assume the radius of the Earth to be 6.37 × 106 m (Hint: Calculate the number of Ar atoms in the atmosphere in the simplified geometry of a plane of area equal to that of the Earth’s surface and a height equal to the thickness of the atmosphere See Problem P1.30 for the dependence of the barometric pressure on the height above the Earth’s surface.) The total number of Ar atoms in the atmosphere is: ∞ ~ ~ ~ N Ar =  N Ar A dz , where N Ar is the number Ar atoms per m3 at the surface of the earth N Ar is given by: ( ) ( ) N p 6.023 × 10 23 mol -1 × 0.0100 × × 10 Pa ~ N Ar = A Ar = = 2.41× 10 23 m -3 −1 −1 RT 8.314472 J K mol × (300 K ) ( ) The total number at Ar atoms in the atmosphere is: 21 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics N Ar ∞ ∞ 0 − ~ ~ =  N Ar A dz = N Ar  e (2.41×10 = 23 ) M Ar g z RT RT ~ A dz = N Ar A M Ar g ( ( ) ( ) ( ) m -3 × 4π × 6.37 × 10 m × 8.314472 J K −1 mol −1 × (300 K ) = 7.85 × 10 41 39.9 × 10 −3 kg × 9.81 m s -1 ) The fraction of these atoms that came from Caesar’s last breath, f, is given by: f= ( ) ( ) ~ N Ar V 2.41×10 23 m -3 × 0.500 ×10 −3 m = = 1.53 ×10 − 22 41 N Ar 7.85 ×10 ( ) The number of Ar atoms that we inhale with each breath is: ) ( ( ) ( ) pV 0.0100 × 1× 10 Pa × 0.500 × 10 −3 m 23 -1 N = NA = 1.21× 10 20 = 6.023 × 10 mol × −1 −1 RT 8.314472 J K mol × (300 K ) ( ) The number of these that came from Caesar’s last breath is f × N: f×N = 1.53 × 10-22 × 1.21 × 1020 = 1.85 × 10-2 The reciprocal of this result, or 54, is the number of breaths needed to inhale one Ar atom that Caesar exhaled in his last breath P1.30) The barometric pressure falls off with height above sea level in the Earth’s atmosphere − M i g RT as Pi = Pi0 e where Pi is the partial pressure at the height z, Pi is the partial pressure of component i at sea level, g is the acceleration of gravity, R is the gas constant, T is the absolute temperature, and Mi is the molecular mass of the gas Consider an atmosphere that has the composition x N = 0.600 and xCO = 0.400 and that T = 300 K Near sea level, the total pressure is 2 1.00 bar Calculate the mole fractions of the two components at a height of 50.0 km Why is the composition different from its value at sea level? 22 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/  N2 pN2 = p e − Chapter 1/Fundemental Concepts of Thermodynamics M N2 g z RT  ( 28.04 × 10−3 kg mol -1 ) × ( 9.81 m s-1 ) × ( 50 × 103 m )  = ( 0.600 ) × (1.01325 × 105 Pa ) × Exp  −  = 248 Pa (8.314472 J K −1 mol −1 ) × ( 300 K )   pCO2 = p  CO2 e − M CO2 g z RT  ( 44.04 × 10−3 kg mol -1 ) × ( 9.81 m s-1 ) × ( 50 × 103 m )  = ( 0.400 ) × (1.01325 × 105 Pa ) × Exp  −  = 7.02 Pa (8.314472 J K −1 mol −1 ) × ( 300 K )   x CO2 = (p pCO2 CO2 + p N2 ( 7.02 Pa ) = ) ( 7.02 Pa + 248 Pa ) = 0.028 x N = − x CO = 0.972 P1.31) Assume that air has a mean molar mass of 28.9 g mol–1 and that the atmosphere has a uniform temperature of 25.0°C Calculate the barometric pressure at Denver, for which z = 1600 m Use the information contained in Problem P1.30 − Mi g z   the barometric pressure at 1600 m is:  RT  Using Pi = Pi0 × Exp  ( ) ( )  − 28.9 × 10 −3 kg mol -1 × 9.80665 m s -2 × (1600 m )  Pi = atm × Exp   = 0.833 atm = 8.44 × 10 Pa −1 −1 ( ) 8.314472 J K mol 298.15 K ×   ( ) P1.32) A mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a drying tube Hydrogen reduces the CuO according to the reaction CuO + H2 → Cu + H2O, and oxygen reoxidizes the copper formed according to Cu + 1/2 O2 → CuO At 25°C and 750 Torr, 100.0 cm3 of the mixture yields 84.5 cm3 of dry oxygen measured at 25°C and 750 Torr after passage over CuO and the drying agent What was the original composition of the mixture? 23 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics We need to consider two equilibria: CuO (s) + H2 (g) → H2O (l) + Cu (s) at equilibrium α −β n H2 − α Cu (s) + ½ O2 (g) → CuO (s) n O2 − β α −β at equilibrium In the final state, only O2 is present Therefore, n O − α In an excess of O2, all the copper is oxi2 dized, and α − β = and n O = β That means that n O = n O − n H 2 2 Let V1 and V2 be the initial and final volumes: ( V1 = n H2 + n O2 ) RpT and V2 =  n O + n H  R T  2  p Dividing the second equation by the first yields: n O n H2 V2 1 =   − = x O2 − x H2 = − x H2 = − x H2   V1 n H2 + n O2 n H2 + n O2 2 ( ) ( ) ( ( ) = 0.103, and x )  2 V  2 84.5 cm3 x H2 = 1 −  = 1 −  V1   100 cm3  O2 = − x H2 = 0.897 P1.33) Aerobic cells metabolize glucose in the respiratory system This reaction proceeds ac- cording to the overall reaction 6O2(g) + C6H12O6(s)→6CO2(g) + 6H2O(l) Calculate the volume of oxygen required at STP to metabolize 0.010 kg of glucose (C6H12O6) STP refers to standard temperature and pressure, that is, T = 273 K and P = 1.00 atm Assume 24 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ Chapter 1/Fundemental Concepts of Thermodynamics oxygen behaves ideally at STP The number of moles of O2 according to the stoichiometry of the equation is: n (O ) = × n (glucose) = × (10 g ) (180.18 g mol ) −1 = 0.0555 mol Therefore, the volume of O2 required is: VO2 = n O2 R T p = (0.0555 mol) × (8.314472 J K −1 mol−1 ) × (273 K ) = 7.47 × 10-3 m = 7.47 L (101325 Pa ) P1.34) Consider the oxidation of the amino acid glycine (NH2CH2COOH) to produce water, carbon dioxide, and urea (NH2CONH2): NH2CH2COOH(s) + 3O2(g) →NH2CONH2(s) +3CO2(g) + 3H2O(l) Calculate the volume of carbon dioxide evolved at P = 1.00 atm and T = 310 K from the oxidation of 0.0100 g of glycine The number of moles of CO2 according to the stoichiometry of the equation is: n (CO ) = × n (glycine) = × (0.0100 g ) (75.05 g mol ) = 3.997 × 10 −1 −4 mol Therefore, the volume of CO2 evolved is: VCO2 = n CO2 R T p = (3.997 × 10 −4 ) ( ) mol × 8.314472 J K −1 mol −1 × (310 K ) = 1.0168 × 10 -5 m = 1.02 × 10 -2 L (101325 Pa ) P1.35) An initial step in the biosynthesis of glucose (C6H12O6) is the carboxylation of pyruvic acid (CH3COCOOH) to form oxaloacetic acid (HOOCCOCH2COOH): CH3COCOOH(s)+CO2(g)→HOOCCOCH2COOH(s) If you knew nothing else about the intervening reactions involved in glucose biosynthesis other than that no further carboxylations occur, what volume of CO2 is required to produce 0.50 g of glucose? Assume P = atm and T = 310 K The number of moles of CO according to the stoichiometry of the equations is: 25 Full file at https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics n (CO ) = n (glucose) = (0.50 g ) (180.18 g mol ) −1 = 2.775 × 10 −3 mol Therefore, the volume of CO2 required is: V= ( ) ( ) n R T 2.775 × 10 −3 mol × 8.314472 J K −1 mol −1 × (310 K ) = = 7.059 × 10 -5 m = 0.0706 L (101325 Pa ) p 26 Full file at https://TestbankDirect.eu/ ... https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ Chapter 1/Fundemental Concepts of Thermodynamics R is then: R=... https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file Chapter at https://TestbankDirect.eu/ 1/Fundemental Concepts of Thermodynamics Using the final... https://TestbankDirect.eu/ Solution Manual for Physical Chemistry for the Life Sciences 1st Edition by Engel Full file at https://TestbankDirect.eu/ Chapter 1/Fundemental Concepts of Thermodynamics With the results