I want a detailed explanation of how he was able to solve a and b and what is the difference between them. The law is only without the numbers. I want the method of the law

Values of Tc and Pc for n-butane from App. B yield: Tr=425.1350=0.8233andPr=37.969.4573=0.2491 Parameter q is given by Eq. (3.51) with ,, and (TF) for the RK equation from Table 3.I: q=TrTr1/2=Tr1/2=0.086640.42748(0.8233)3/2=6.6048 Parameter is found from Eq. (3.50): =TFPr=0.8233(0.08664)(0.2491)=0.026214 (a) For the saturated vapor, write the RK form of Eq. (3.48) that results upon substitution of appropriate values for and from Table 3.1: Z=1+qZ(Z+)(Z) or Z=1+0.026214(6.6048)(0.026214)Z(Z+0.026214)(Z0.026214) 3.6. Cubic Equations of State Solution yields Z=0.8305, and Vv=PZRT=9.4573(0.8305)(83.14)(350)=2555cm3mol1 An experimental value is 2482cm3mol1, (b) For the saturated liquid, apply Eq. (3.49) in its RK form: Z=+Z(Z+)(q1+Z) 6r2 Z=0.026214+Z(Z+0.026214)(6.6048)(0.026214)(1.026214Z) Solution yiclds Z=0.04331, and Vf=PZRT=9.4573(0.04331)(83.14)(350)=133.3cm3mol1 An experimental value is 115.0cm3mol1