Question: 0 . 2 m 3 1 1 ( 4 4 g / mol Tc = 3 6 9 . 9 K , Pc =

"0.2 m311(44 g/mol Tc=369.9 K , Pc =42.5)25 C QM15
P RT a Cig 65.4 J/mol-K VbV2 V
(a)(20) P1 Z1
(b)(5)
U (c)(20)U=0 in (b) vdW T2 P2
: : R=8.314 J/molK,1bar=100000 Pa=0.987 atm.,1 m
3
=1000 L 1.0.2
3
.11(44/)
c
=369.9 K,P
c
=42.5)25
C. P=
Vb
RT
-
2
a
ig
=65.4 J/molK(a)(20) P
1
Z
1
.(b)(5) U (c)(20)U=0 in (b) T
2
2
vdW :

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