IDEAL AND REAL GASES 403
dharm
\M-therm\Th8-2.pm5
(i)Using perfect gas equation :
Characteristic gas constant, R =
R
M
(^0) =^8314
44 = 188.95 Nm/kg K (for CO^2 )
Using perfect gas equation
pV = mRT
∴ p =
mRT
V =
10
3
××188 95 300.
= 188950 N/m^2 or 1.889 bar. (Ans.)
(ii)Using Van der Waals’ equation :
p a
v
F +
HG
I
(^2) KJ
(^) ()vb− = R 0 T
p =
RT
vb
0
−
- a
v^2
From Table 8.1
For CO 2 : a = 362850 Nm^4 /(kg-mol)^2
b = 0.0423 m^3 /(kg-mol)
v = Molar specific volume =
344
10
×
= 13.2 m^3 /kg-mol
Now substituting the values in the above equation, we get
p = 8314 300
13 2 0 0423
×
..−
- 362850
(.)13 2^2
= 189562 – 2082.5 = 187479.5 N/m^2 or 1.875 bar. (Ans.)
(iii)Using Beattie Bridgeman equation :
p = RT e
v
0
2
() 1
()
− ()vBA
v
+− 2
where p = pressure, A = A 0 F 1 −
HG
I
KJ
a
v
, B = B 0 F 1 −
HG
I
KJ
b
v
and e =
c
vT^3
From Table 8.2 A 0 = 507.2836, a = 0.07132
B 0 = 0.10476, b = 0.07235
C = 66 × 10^4
∴ A = 507.2836 1
0 07132
13 2
FHG − IKJ
.
.
= 504.5
B = 0.10476^1
0 07235
13 2
F −
HG
I
KJ
.
. = 0.1042
C =
66 10
13 2 300
4
3
×
.( )×
= 0.001852
Now substituting the various values in the above equation, we get
p =
8314 300 1 0 001852
13 2^2
×−(. )
(.)
(13.2 + 0.1042) –
504 5
13 2^2
.
(.)
= 190093 – 2.89 ~_ 1.9 × 10^5 N/m^2 = 1.9 bar. (Ans.)