Exam (elaborations) TEST BANK FOR Thermodynamics Problem Solving in Physical Chemistry 1st Edition By Kathleen E. Murphy

Exam (elaborations) TEST BANK FOR Thermodynamics Problem Solving in Physical Chemistry 1st Edition By Kathleen E. Murphy (Study Guide and Map [Full Worked Solution] (2020, CRC Press) Thermodynamics Problem Solving in Physical Chemistry Study Guide and Map Kathleen E. Murphy Thermodynamics Problem Solving in Physical Chemistry – Full Solutions 1 Full Solutions: PART 1: Gases and Gas Laws 1.1 A) Calculate the values the equations have in common first: CH 4 = 25.0g 1.00molCH 4 16.0g é ë ê ê ù û ú ú = 1.163mol V m ,CH 4 = 2.0L 1.563mol = 1.28 L mol Given RT = 0.08206 L-atm mol-K (303K) = 24.86 L-atm mol , then calculate the pressure from each equation of state: (a) P ideal = RT V m = 24.86 L-atm mol 1.28 L mol = 19.4atm (b) P VdW ,CH 4 = RT V m - b - a V m 2 = 24.86 L-atm mol (1.28 - 0.04278) L mol - 2.283 L2-atm mol2 (1.28)2 L2 mol2 = (20.10 -1.394)atm = 18.7atm (c) P virial = RT V m 1 + B V m é ë ê ê ù û ú ú = 24.86 L-atm mol 1.28 L mol 1 + -43.9 cm3 mol ( ) 1.0L 1000cm3 1.28 L mol é ë ê ê ê ê ù û ú ú ú ú = 19.40(0.9657)atm = 18.7atm B) (a) V m ,CH 4 = 0.20L 1.563mol = 0.128 L mol P ideal = RT V m = 24.86 L-atm mol 0.128 L mol = 194atm (b) P VdW ,CH 4 = RT V m - b - a V m 2 = 24.86 L-atm mol (0.128 - 0.04278) L mol - 2.283 L2-atm mol2 (0.128)2 L2 mol2 = (291.7 -139.3)atm = 152atm (c) P virial = RT V m 1 + B V m é ë ê ê ù û ú ú = 24.86 L-atm mol 0.128 L mol 1 + -43.9 cm3 mol ( ) 1.0L 1000cm3 0.128 L mol é ë ê ê ê ê ù û ú ú ú ú = 194.0(0.657)atm = 127.5atm 1.2 A) P virial = RT V m 1 + B V m é ë ê ê ù û ú ú Multiply both sides by V m RT sothat : P virial ´ V m RT = V m RT ´ RT V m 1 + B V m é ë ê ê ù û ú ú Þ Z = 1 + B V m é ë ê ê ù û ú ú B) Z(in 2.0L) = 1 + -0.043.9 L mol ( ) 1.28 L mol é ë ê ê ù û ú ú = 0.966 and Z(in 200mL) = 1 + -0.043.9 L mol ( ) 0.128 L mol é ë ê ê ù û ú ú = 0.657 1.3 A) (a) PV = nRT = mass MW gas é ë ê ê ù û ú ú RT Þ d = mass V = P(MW gas ) RT (b) d gas = MW gas P RT æ è ç ö ø ÷ = MW gas , g mol 1 V m , mol L æ è ç ö ø ÷ = d gas , g L B) (a) Z = V m,obs V m,ideal = MW gas d obs é ë ê ê ù û ú ú ´ d ideal MW gas é ë ê ê ù û ú ú = d ideal d obs (b) d obs = d ideal Z so that d obs d ideal when Z 1.0 (c) d obs = d ideal Z so that d obs d ideal when Z 1.0 1.4 A) d gas = MW gas P RT æ è ç ö ø ÷ = 16.0g mol 130atm 0.08206 L-atm mol-K (323K) æ è ç ö ø ÷ = 78.5g / L B) d obs = d ideal Z = 78.5g / L 0.8808 = 89.1g / L Thermodynamics Problem Solving in Physical Chemistry – Full Solutions 2 1.5. A) P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 Þ P 1 P 2 = T 1 T 2 Þ P 2 = P 1 T 1 T 2 æ è ç ö ø ÷ = 100atm 500K 300K æ è ç ö ø ÷ = 167atm B) Need Vm in L/mol so need to convert mass to moles, and volume to liters. n N 2 = 92.4X103 g 1.00molN 2 28.0g é ë ê ê ù û ú ú = 3300mol V N 2 = 0.500m3 1000L 1.0m3 æ è ç ö ø ÷ = 500L V m ,N 2 = 500L 3300mol = 0.1515 L mol P VdW ,N 2 = RT V m - b - a V m 2 = 0.08206 L-atm mol-K (500K) (0.1515 - 0.0357) L mol - 1.352 L2-atm mol2 (0.1515)2 L2 mol2 = (354.3 - 58.9)atm = 295atm 1.6 A) Z(I) = PV m RT = 10.0atm(2.606 L mol ) 0.08206 L-atm mol-K (340K) = 0.934 Z(II) = PV m RT = 10.0atm(0.9082 L mol ) 0.08206 L-atm mol-K (340K) = 0.814 B) Both values less than 1.0 indicating attractive forces between NH3 molecules dominating, causing lower than ideal molar volumes. Since NH3 molecules hydrogen bond with each other, this behavior is not unexpected, since a strong attractive force. Increasing pressure causes an increase in attractive forces since number of collisions increases, making it more likely the molecules will aggregate or group C) T Boyle = a Rb = 4.169 L2-atm mol2 0.08206 L-atm mol-K (0.0371 L mol ) = 1368K 1.7 A) T Boyle = a Rb = a L2 - atm mol2 ( ) L - atm mol -K ( L mol ) so R must be 0.08206 L-atm mol-K in equation B) T B,CH 4 = 2.283 L2-atm mol2 0.08206 L-atm mol-K (0.0428 L mol ) = 650K T B,N 2 = 1.408 L2-atm mol2 0.08206 L-atm mol-K (0.03913 L mol ) = 438.5K T B,H 2 = 0.2476 L2-atm mol2 0.08206 L-atm mol-K (0.0266 L mol ) = 113K T B,Ar = 1.355 L2-atm mol2 0.08206 L-atm mol-K (0.0320 L mol ) = 516K So only TB for H2 comes close to tabled value, while in all others the calculation overestimated true value by about 100 K. 1.8 MW gas = mass(RT) PV = d gas RT P æ è ç ö ø ÷ = 1.881 g L 0.08206 L-atm mol-K (298K) 1.00 atm æ è ç ö ø ÷ = 46.0 g mol 1.9 MW gas = 3.71 g L 0.08206 L-atm mol-K (773K) 699torr 1.0 atm 760torr æ è ç ö ø ÷ æ è ç ç ç çç ö ø ÷ ÷ ÷ ÷÷ = 3.71(63.43) 0.9197 g mol = 256 g mol Then per molecule = MW gas AW = 256 g mol 32 g mol = 8.0 B) Molecule = S 8 1.10 A) To identify the diatomic gas, need to determine the molecular weight of the gas from the data. Given it is an ideal gas: MW gas = 3.864 g L 0.08314 L-bar mol-K (298K) 1.35bar æ è ç ö ø ÷ = 70.9 g mol So 2(AW) X = 70.9, and AW gas X = 35.45, so gas is Cl2(g). B) To define mass % will need mass of Ar(g) in 3.864 g. Know that: n mix = n Ar + n Xe = PV mix RT = mass Ar 1mol Ar 39.94g æ è ç ö ø ÷ + mass Xe 1mol Ar 131.3g æ è ç ö ø ÷ but then need second equation relating moles of Ar and Xe in mixture to solve for mass of Ar(g). Thermodynamics Problem Solving in Physical Chemistry – Full Solutions 3 Also true that: n mix = n Ar + n He = PV mix RT = 1.35 bar(1.0L) 0.08314 L-bar mol-K (298K) æ è ç ö ø ÷ = 0.0545mol so that: n mix = 0.0545mol = x g Ar 1mol Ar 39.94g æ è ç ö ø ÷ + (3.864 - (x g Ar)) 1mol Ar 131.3g æ è ç ö ø ÷ = 0.0254x + 0.02943 - 0.00752x 0.0545= (0.01742x + 0.02943) Þ x = 0.02507 0.01742 = 1.439g = mass Ar, then%Ar = 1.439g 3.864g ´100 = 37.2% 1.11 A) Reduced variables must have NO units since:Preduced = P P c ;Vreduced = V V c ;Treduced = T T c B) Both the terms in the reduced form of the Van der Waals equation must not have any units. C) V reduced = V V c = 15.0L /mol 0.0752L /mol = 199 T reduced = T T c = 300K 151.5K = 1.985 P reduced = 8T r 3V r -1 - 3 V r 2 = 8(1.985) 3(199) -1 - 3 (199)2 = 0.0266 - 7.57X10-5 = 0.0266 D) Since Tr ≈ 2.0 and Pr ≈ 0.03, the value of Z should be close to 1.0 and the gas is acting as an ideal gas. E) P ideal = RT V m = 0.08206 L-atm mol-K (300K) 15.0 L mol = 1.64atm 1.12 A) By definition: V c = 3b P c = a 27b2 T c = 8a 27Rb so that: V c = 3b = 3(0.0226 L mol) = 0.0678 L mol P c = a 27b2 = 0.751L2-atm mol2 27(0.0226)2 L2 mol2 = 54.5atm T c = 8a 27Rb = 8 0.751 L2-atm mol2 æ è ö ø 27(0.08206 L-atm mol-K )(0.0226) L mol = 120K B) Z c = P c V c RT c = 54.5 atm 0.0678 L mol ( ) 0.08206 L - atm mol - K ( )(120 K ) = 0.375 1.13 A) Need P VdW = RT V m - b - a V m 2 where b in L/mol so easiest to convert m3 → L and Pa→ kPa first. Since 1m3 = 1000L and 1 kPa = 1000 Pa then: V m = 5.00X10-4 m3 mol 1000L 1.0m3 æ è ç ö ø ÷ = 0.500 L mol and a = 0.500 m6 - Pa mol2 ´ 1000L 1.0m3 æ è ç ö ø ÷ 2 ´ 1 kPa 1000Pa = 500 L2 - kPa mol2 so that: 3000kPa = 0.08314 L-kPa mol-K (298K) 0.50 L mol - b æ è ç ö ø ÷ - 500 L2-kPa mol2 0.0025 L2 mol2 æ è ç çç ö ø ÷ ÷÷ Þ 5000kPa = 22.70 L-kPa mol (0.50 L mol - b) Þ (0.50 L mol - b) = 22.70 L-kPa mol 5000kPa = 0.454 L mol Þ (0.50 L mol - 0.454 L mol ) = b Þ b = 0.0460 L mol B) Z = PV m RT = 3000kPa(0.500 L mol ) 8.314 kPa-L mol-K (273K) = 0.661 1.14 A) V m ,Xe = 1.0L 131g Xe 1mol 131.3 æ è ç ö ø ÷ = 1.002 L mol P ideal = RT V m = 0.08206 L-atm mol-K (298K) 1.002 L mol = 24.4atm • So answer is NO, it is not an ideal gas since Pideal not close to 20 atm. Thermodynamics Problem Solving in Physical Chemistry – Full Solutions 4 B) Van der Waals constants for Xe(g): a = 4.25 L2 - atm mol2 b = 0.05105 L mol P VdW ,Xe(g) = RT V m - b - a V m 2 = 0.08206 L-atm mol-K (298K) (1.002 - 0.05105) L mol - 4.25 L2-atm mol2 (1.002)2 L2 mol2 = (25.72 - 4.24)atm = 21.5atm • Result is closer, but still not equal to 20 atm, so NOT a Van der Waals gas either. C) (a) T = PV m R = 20atm 1.002 L mol ( ) 0.08206 L-atm mol-K = 243.7K = -29.3°C (b) T = P + a V m 2 é ë ê ê ù û ú ú ´ V m - b R é ë ê ê ù û ú ú = 24.25atm 0.9490 L mol 0.08206 L-atm mol-K æ è ç ö ø ÷ = 280.4K = 7.3°C 1.15 A) Given: P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 Þ P 1 = P 2 V 2 V 1 = 3780torr 1atm 760torr æ è ç ö ø ÷ 4.65L 6.85L æ è ç ö ø ÷ = 3.37atm B) Cannot determine P1 as a Van der Waals gas since know only P2, V1 and V2 values. • Can’t divide equations to cancel like terms, but could possibly subtract equations to collect like terms: P 1 = RT V1 - b - a V1 2 and P 2 = RT V2 - b - a V2 2 If we assume bV1 or V2, then can say by subtracting P’s, Leads to: P 1 - P 2 = nRT V 1 - nRT V 2 é ë ê ê ù û ú ú + n2a V 2 2 - n2a V 1 2 é ë ê ê ù û ú ú Þ P 1 = P 2 + nRT V 2 -V 1 V 1 V 2 é ë ê ê ù û ú ú + an2 V 1 2 -V 2 2 V 1 2V 2 2 é ë ê ê ù û ú ú • Cannot cancel n and T, so must know T and number of moles to solve, and • Must also know chemical identity of gas since need value of a, the Van der Waals constant • Since these values not given, cannot solve for P1 if a Van der Waals gas. 1.16 A) First add 2nd term to both sides: P VdW = RT V m - b - a V m 2 ÞP VdW + a V m 2 = RT V m - b Subtract righthand term to produce zero on the righthand side: P VdW + a V m 2 - RT V m - b = 0 Multiply all terms by (Vm-b):P(V m - b) + a(V m - b) V m 2 - RT(V m - b) V m - b = 0 ÞPV m - Pb + a V m - ab V m 2 - RT = 0 And then multiply equation by Vm 2 and divide by P: PV m V m 2 P é ë ê ê ù û ú ú - Pb V m 2 P é ë ê ê ù û ú ú + a V m V m 2 P é ë ê ê ù û ú ú - ab V m 2 V m 2 P é ë ê ê ù û ú ú - RT V m 2 P é ë ê ê ù û ú ú = 0 Þ V m 3 - b + RT P æ è ç ö ø ÷ V m 2 + a P V m - ab P = 0 • Combining like terms leads to a polynomial in Vm as the final result (in blue). B) (a) V m,O 2 = RT P = (0.08314 L-bar mol-K)(298K) 200bar = 0.124 L mol (b) Given the form: Ax3 + Bx2 + Cx + D = 0 then A = 1.0, B = - b + RT P æ è ç ö ø ÷ , C = a P , D = - ab P For roots of equation need a, b values for O2(g) a = 1.378 L2-atm-mol-2, b = 0.03183 L-mol-1 Then: A = 1.0 L mol , B = - 0.03183 L mol + 0.08314 L-bar mol-K (298K) 200bar æ è ç ö ø ÷ = -(0.03183 + 0.124) = -0.1557 L mol C = a P = 1.378 L2-atm mol2 200bar ´ 1.0atm 1.013bar é ë ê ù û ú = 6.96X10-3 L2 mol2 D = - ab P = 1.378 L2-atm mol2 0.03183 L mol ( ) 200bar ´ 1.0atm 1.013bar é ë ê ù û ú = 2.215X10-4 L3 mol3 The only possible root that solves the equation is: V m,VdW = 0.110 L mol Thermodynamics Problem Solving in Physical Chemistry – Full Solutions 5 C) Since value for Van der Waals gas is very different than the ideal molar volume, expect that O2 is acting as real gas in the tank, not an ideal gas. 1.17 A) First assume b V then substitute for n as mass/MW in modified Van der Waals equation P VdW = mass MW gas é ë ê ê ù û ú ú RT V - mass MW gas é ë ê ê ù û ú ú 2 a V2 Þ mass V é ë ê ù û ú RT MW gas - mass V é ë ê ù û ú 2 a MW gas 2 Þ P VdW = RT MW gas é ë ê ê ù û ú ú d gas - a MW gas 2 é ë ê ê ù û ú ú d gas 2 The equation has the form of a quadratic equation: Ax2 + Bx + C = 0where x = -B ± B2 - 4AC 2A • BUT the terms A, B and C would have units of pressure, so we must divide all terms in equation by P BEFORE rearranging into quadratic so x = density in units of g/L and no units exist on either side of the equation, which produces: 1.0 = RT P(MW gas ) é ë ê ê ù û ú ú d gas - a P(MW gas 2 ) é ë ê ê ù û ú ú d gas 2 Þ a P(MW gas 2 ) é ë ê ê ù û ú ú d gas 2 - RT PMW gas é ë ê ê ù û ú ú d gas +1.0 = 0 Then: Ax2 + Bx + C = 0 and d = -B ± B2 - 4AC 2A where: A VdW = a P(MW gas )2 B VdW = - RT P(MW gas ) C VdW = 1.0 B) For the Virial equation use same approach as for Van der Waals equation to develop a unit-less quadratic, with d in g/L. The coefficients differ in that the second virial coefficient, B, replaces “a” in C, in the virial equation polynomial, the sign of the second term changes, and the 1.0 is subtracted. P virial = nRT V 1 + nB V é ë ê ù û ú = nRT V + n2BRT V2 = mass V é ë ê ù û ú RT MW gas + (mass)2 V2 é ë êê ù û úú BRT MW gas ( )2 Þ P virial = d gas RT MW gas é ë ê ê ù û ú ú + d gas 2 BRT MW gas ( )2 é ë ê ê ê ù û ú ú ú Þ1.0 = d gas RT P(MW gas ) é ë ê ê ù û ú ú + d gas 2 BRT P(MW gas )2 é ë ê ê ù û ú ú Þ d gas 2 BRT P(MW gas )2 é ë ê ê ù û ú ú - d gas RT P(MW gas ) é ë ê ê ù û ú ú -1.0 = 0 The coefficients of the quadratic equation are: A virial = BRT P(MW gas )2 é ë ê ê ù û ú ú B virial = RT P(MW gas ) é ë ê ê ù û ú ú C virial = -1.0 C) Ideal gas density: d Cl 2 (g) (ideal) = MW gas P RT æ è ç ö ø ÷ = 71.0g mol 1.0atm 0.08206 L-atm mol-K (323K) æ è ç ö ø ÷ = 2.88g / L For the Van der Waals gas density: A VdW = a P(MW gas )2 = 6.579 L2-atm mol2 1.0atm(71.0 g mol )2 = 1.305X10-3 L2 g2 B VdW = - RT P(MW gas ) = - 1 d ideal = -0.347 L g C VdW = 1.0 d VdW = -B ± B2 - 4AC 2A = -(-0.347) ± (0.347)2 - 4(1.0)(1.305X10-3) 2(1.305X10-3) = 0.347 ± 0.3394 2.61X10-3 = 263 g L or 2.91 g L For the Virial gas density: A Virial = BRT P(MW gas )2 = -0.314 L mol (0.08206 L-atm mol-K (300K)) 1.0atm(71.0 g mol )2 = -1.53X10-3 L2 g2 B VdW = RT P(MW gas ) = - 1 d ideal = 0.347 L g C VdW = -1.0 d Virial = -B ± B2 - 4AC 2A = -(0.347) ± (0.347)2 - 4(-1.0)(-1.53X10-3) 2(-1.53X10-3) = -0.347 ± 0.338 -3.06X10-3 = 2.94 g L or 224 g L • The density values are very close which proves the derivations are valid equations. 1.18 A) As ideal gases: Only 2nd root makes sense Only 1st root makes sense. Thermodynamics Problem Solving in Physical Chemistry – Full Solutions 6 Cl 2 = 250g 1.00mol 71.0g é ë ê ù û ú = 3.52molCl 2 P Cl 2 = nRT V = 3.52mol(0.08206 L-atm mol-K )(300K) 10.0L = 8.75atm C 2 H 6 = 150g 1.00mol 30.0g é ë ê ù û ú = 5.00molC 2 H 6 P C 2 H 6 = nRT V = 5.00mol(0.08206 L-atm mol-K )(300K) 10.0L = 12.43atm P total = P Cl 2 + P C 2 H 6 = 8.75 +12.43 = 21.2atm B) As Van der Waals gases (real gases) need a, b and Vm: V m ,Cl 2 = 10.0L 3.52mol = 2.84 L mol a Cl 2 = 6.579 L2 - atm mol2 b Cl 2 = 0.05632 L mol P VdW ,Cl 2 (g) = RT V m - b - a V m 2 = 0.08206 L-atm mol-K (303K) (2.84 - 0.05622) L mol - 6.579 L2-atm mol2 (2.84)2 L2 mol2 = (8.93 - 0.816)atm = 8.11atm V m ,C 2 H 6 = 10.0L 5.00mol = 2.00 L mol a C 2 H 6 = 5.562 L2 - atm mol2 b C 2 H 6 = 0.0638 L mol P VdW ,C 2 H 6 (g) = RT V m - b - a V m 2 = 0.08206 L-atm mol-K (303K) (2.00 - 0.0638) L mol - 5.562 L2-atm mol2 (2.00)2 L2 mol2 = (12.84 -1.39)atm = 11.45atm P total,VdW = P Cl 2 + P C 2 H 6 = 8.11+11.45 = 19.6atm 1.19 Since the volume is constant, the partial pressure ratio will equal the mole ratio in the mixture. n CO n total = P CO P total and n O 2 n total = P O 2 P total so that n O 2 n CO = P O 2 P CO • Also need to convert the ppm units to g/L for partial pressure of CO, assuming ideal gas: 50 ppm = 50gCO 1.0X106 gsoln = 50gCO 1000L = 0.050 gCO L 800 ppm = 0.800 gCO L 3200 ppm = 3.200 gCO L P CO @50 ppm = d gas RT MW gas é ë ê ê ù û ú ú = 0.050 g L 0.08206 L-atm mol-K (298K) 28.0 g mol é ë ê ê ù û ú ú = 0.0437atm n O 2 n CO = 0.200atm 0.0437atm = 4.58@50 ppm P CO @800 ppm = 0.800 g L 0.08206 L-atm mol-K (298K) 28.0 g mol é ë ê ê ù û ú ú = 0.699atm n O 2 n CO = P O 2 P CO = 0.200atm 0.699atm = 0.286@800 ppm P CO @3200 ppm = 3.200 g L 0.08206 L-atm mol-K (298K) 28.0 g mol é ë ê ê ù û ú ú = 2.81atm n O 2 n CO = P O 2 P CO = 0.200atm 2.81atm = 0.0712@3200 ppm • So to be safe we need over 4.6 molecules of O2 for every 1 molecule of CO in the air we breathe. 1.20 n H 2 O = 0.062 atm(500L) 0.08206 L-atm mol-K (310K) æ è ç ö ø ÷ = 1.22mol 18.0g 1mol æ è ç ö ø ÷ = 21.9gH 2 O 1.21 Reaction: 2H2(g) + O2(g) → 2 H2O(l) Know: P total - P excess = P reacted = (1.0-0.40)atm= 0.60 atm= P H 2 reacted + P O 2 reacted and P O 2 P H 2 = 1 2 Þ P O 2 = 1 2 P H 2 then: 0.60 atm = P H 2 reacted + P O 2 reacted = P H 2 reacted + 0.5P H 2 reacted = 1.5P H 2 reacted ÞP H 2 reacted = 0.60 1.5 = 0.40atm P H 2 initial = P H 2 excess + P H 2 reacted = 0.40- 0.40 = 0.80atm so that P O 2 initial = 1.0 - 0.80 = 0.20atm • The molar ratio will equal the molecule ratio Thermodynamics Problem Solving in Physical Chemistry – Full Solutions 7 Mol %O 2 = c O 2 ´100 = P O 2 P total ´100 = 20.0% Mol %H 2 = 80.0