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)

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