TITLE.PM5

(Ann) #1

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.)
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