solution by iteration starting from z= 1 yields z = 0.8305

Can you please show me how to get z= 0.8305

Solution by iteration starting from Z=1 yields Z=0.8305, and V=PZRT=9.4573(0.8305)(83.14)(350)=2555cm3mol1 An experimental value is 2482cm3mol1. (b) For the saturated liquid, we apply Eq. (3.49) in its RK form: Z=+Z(Z+)(q1+Z) or Z=0.026214+Z(Z+0.026214)(6.6048)(0.026214)(1.026214Z) Solution by iteration yields Z=0.04331, and Vl=PZRT=9.4573(0.04331)(83.14)(350)=133.3cm3mol1 An experimental value is 115.0cm3mol1. umes of (a) saturated-vapor and (b) saturn Z=0.026214+Z(Z+0.02 Solution by iteration yields Z=0.04331 Vl=PZRT=9.457.(0.04331)(83. An experimental value is 115.0cm3mo Solution 3.9 Values of Tc and Pc for n-butane from App. B yield: For comparison, values of Vv and Vlc Tr=425.1350=0.8233andPr=37.969.4573=0.2491 Parameter q is given by Eq. (3.51) with , , and (Tr) for the RK equation from Table 3.1: q=TrTr1/2=Tr3/2=0.086640.42748(0.8233)3/2=6.6048 Corresponding States; Acentric Parameter is found from Eq. (3.50): The dimensionless thermodynamic coorc corresponding-states correlations: =TrPr=0.8233(0.08664)(0.2491)=0.026214 All fluids, when compared at th pressure, have approximately deviate from ideal-gas behavic (a) For the saturated vapor, we write the RK form of Eq. (3.48) that results upon Two-parameter corresponding-states co substitution of appropriate values for and from Table 3.1: parameters Tc and Pc. Although these c (argon, krypton, and xenon), systemat Appreciable improvement results from or Z=1+qZ(Z+)(Z) eter (in addition to Tc and Pc ), charact parameter is the acentric factor, , intr Z=1+0.026214(6.6048)(0.026214)Z(Z+0.026214)(Z0.026214)