TITLE.PM5

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.)