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110a1h6k - Problem Set 1 ~ Soluh‘ons Mfg MW P7.1 A sample...

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Unformatted text preview: Problem Set 1%; ~ Soluh‘ons Mfg MW P7.1) A sample containing 35.0 g of Ar is enclosed in a container of volume 0.165 L at 390 K. Calculate P using the ideal gas, van der Waals, and Redlich-Kwong equations of state. Based on your results, does the attractive or repulsive contribution to the interaction potential dominate under these conditions? V», = 0.165 LX39. 5 g 35.0g mol _£_ 8.314x10-2LxbarK"mor1 x390 K = 0.1883 Ltnol‘l p _ _M =172b Mg“ V", 0.1883 Lmol“ ar P _ RT 61 _8.314><10‘2L barK‘Imol‘lx390K 1.355L2barmol‘2 _______~*__________ “W Vm —b V; 0.1883 Lmol‘1—0.0320Lmol'l (0_1833Lm01-1)2 =169 bar szfifli 1 _ 8.314X10‘2LbarK‘1mol‘1 x390 K _ 0.1883 Lmor‘ —0.02219 Lmol“ l _ 16.86 L2 bar morzKE 1 J390 K 0.1883 Lmol" (0.1883Lmor1 + 0.02219 Lmol") PM = 174 bar Because PvdW < Pidealgas, the attractive part of the interaction appears to dominate using the van der Waals equation of state. However, because PM > PM”; gm the repulsive part of the interaction appears to dominate using the Redlich-Kwong equation of state. This is a case where a more accurate equation of state is needed to answer the question. at 450 K and 85.0 bar. TX" 450 K T’“ =——=———=1.55; T”! =1.55TH= =1.5 . = . R T5: 289.74K C 5x32 98 K 512 K _ P’“ _ 85.0 bar PXe __ _ _—_ R P? 58.40 bar = 1.46; PHI =1.46Pg’2 =1.46x12.93 bar = 18.8 bar P7.-10) 1 mol of Ar initially at 298 K undergoes an adiabatic expansion against a pressure Pa,em,,1= O fiom a volume of 20.0 L to a volume of 65.0 L. Calculate the final temperature usmg the ideal gas and van‘der Waals equations of state. w = q = O. AU= O for an ideal gas and AT= 0 because Uis a function ofTonly. Usmg the results of Example Problem 3.5 for a van der Waals gas, AUTm=a 1 - 1 ‘ V V,” 105Pa 10*m3 x[ 1 1 j >< —_ bar dm6 65 .0X10‘3m3mol‘I 20.0X10‘3m3mol‘1 = 1.355 dmfibar 13301"2 x = -4.69 J AU”, _ 4.69 Jmol‘I CM _125 JK'lmol" If = 297.6 K AT = = —0.376 K P7.l4) A van der Waals gas has a value of z = 1.00084 at 298 K and 1 bar and the Boyle temperature of the gas 15.125 K. Because the density is low, you can calculate Vm from the ideal gas law. Use this mformation and the result of Problem P7.12 to estimate a and b. Vm RT Rb z—lz—b—[l—Zi] V," T z—l RT 0.00084 8.314X10'2dm3barm01_1K-1X298K b: ——-=————X _3, P 1_125K lbar T 298K = 0.0359 dm3 m01"1 = 3.59 x10‘5m3 mol" a = RbT, = 8.3 14Jmol'1K" x359 x10‘sm3 mol-1 x125 K = 3.73 ><10‘2m6 Pa mol-2 P7.15) The experimental critical constants of H20 are T c = 647.14 K, PC = 220.64 bar, and V; = 55.95 X 10‘3 L. Use the values of PC and T c to calculate VG. Assume that H20 behaves as (a) an ideal gas, (b) a van der Waals gas, and (c) a Redlich—Kwong gas at the critical point. For parts (b) and (c), use the formulas for the critical compression factor. Compare your answers with the experimental value. Assuming an ideal gas, —2 -1 -1 RTc _8.314><10 Lbarmol K x647.l4K =0.2438L V. _—__.__ pa 220.64 bar For avan der Waals gas, R 2.=RV‘ :3; V; :3 Tc =§xo.2438 L=91.4><10‘3L R1; 8 8 PC 8 . For a Redlich-Kwong gas, PV RT . _3 c ‘ *O.333; Vc =0.333—-i=0.333><0.2438 L=81.2XlO L 26 = RT P, C P7.17) For the Bertholet equation, V = R—T + b — a "' P RT2 ’ Boyle temperature in terms of a, b, and R. find an expression for the 8P RZT3 3P Therefore b— :1 3 =0 T3 a T = —— K Rb P 7.2 0) At What temperature does the sloPe of the 2 versus P curve as P -—> 0 have its maximum value for a van der Waals gas? What is the value of the maximum slope? 51? :RT RT a 32 1 £ a) 1 '1 [ 2a) ._ _ =_ b——— + 3=— 2 b-— 9T 313 T M RT2 RT RT RT RT Setting this derivative equal to zero gives [82 ] J—Lb ——€—) for a van der Waals gas T,P—)0 2a 20 _ :0 T =—' b RTM w Rb 1 a b 5 b2 . ' b_ =— b— —" =— The maximum slope 15 RTMI RT“) Zal: 420]] 4a P7.23) For a gas at a given temperature, the compressibility is described by the 2 empirical equation 2 =1. — 9.00 X 10—3}{:°+4.00 x 104(3) , Where P° = 1 bar. P" Calculate the activity coefficient for P = 100, 200, 300, 400, and 500 bar. For Which of these values is the activity coefficient greater than one? ' P _ P 1—9.00><10‘3P’+4.00x10‘5 P’ 2 —1 0 P 0 P, In 7=—9.00 x10‘3P+2.00 x 10—51DZ 7/: 0.497, 0.368, 0.406, 0.670, and 1.65 at 100, 200, 300, 400, and 500 bar, respectively. dPl ...
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