In the van der Waals equation, for example, the two constants for the mixture
are determined from(13–12a, b)where expressions for aiand biare given in Chapter 3.amaaki 1yia^1 i>^2 b2
¬and¬bmaki 1yibiChapter 13 | 687EXAMPLE 13–2 P-v-TBehavior of Nonideal Gas MixturesA rigid tank contains 2 kmol of N 2 and 6 kmol of CO 2 gases at 300 K and
15 MPa (Fig. 13–10). Estimate the volume of the tank on the basis of
(a) the ideal-gas equation of state, (b) Kay’s rule, (c) compressibility factors
and Amagat’s law, and (d) compressibility factors and Dalton’s law.Solution The composition of a mixture in a rigid tank is given. The volume
of the tank is to be determined using four different approaches.
Assumptions Stated in each section.
Analysis (a) When the mixture is assumed to behave as an ideal gas, the
volume of the mixture is easily determined from the ideal-gas relation for the
mixture:since(b) To use Kay’s rule, we need to determine the pseudocritical temperature
and pseudocritical pressure of the mixture by using the critical-point proper-
ties of N 2 and CO 2 from Table A–1. However, first we need to determine the
mole fraction of each component:Then,Thus,VmZm Nm Ru Tm
PmZmVideal 1 0.49 21 1.330 m^32 0.652 m^3TR Tm
Tcr,¿m300 K
259.7 K1.16PRPm
P¿cr,m15 MPa
6.39 MPa2.35∂Zm0.49¬¬ 1 Fig. A–15b 2
1 0.25 21 3.39 MPa 2 1 0.75 21 7.39 MPa 2 6.39 MPaP¿cr,m ayi Pcr,iyN 2 Pcr,N 2 yCO 2 Pcr,CO 2 1 0.25 21 126.2 K 2 1 0.75 21 304.2 K 2 259.7 KTcr,¿m ayi Tcr,iyN 2 Tcr,N 2 yCO 2 Tcr,CO 2yN 2 NN 2
Nm2 kmol
8 kmol0.25¬and¬yCO 2
NCO 2
Nm6 kmol
8 kmol0.75NmNN 2 NCO 2 2 6 8 kmolVmNmRuTm
Pm1 8 kmol 21 8.314 kPa#m^3 >kmol#K 21 300 K 2
15,000 kPa1.330 m^32 kmol N 2
6 kmol CO 2
300 K
15 MPa
Vm =?FIGURE 13–10
Schematic for Example 13–2.cen84959_ch13.qxd 4/6/05 9:35 AM Page 687