TITLE.PM5

402 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th8-2.pm5

(ii)Efficiency of the cycle, ηcycle :

ηcycle =

Heat received Heat rejected

Heat received

−

=

181 954 10319

181 954

..

.

−

= 0.433 or 43.3%

i.e., Efficiency of the cycle = 43.3%. (Ans.)

REAL GASES

Example 8.9. One kg of CO 2 has a volume of 1 m^3 at 100°C. Compute the pressure by

(i)Van der Waals’ equation

(ii)Perfect gas equation.

Solution. (i) Using Van der Waals’ equation :

Molar specific volume, v = 1 × 44 = 44 m^3 /kg-mol (Q M for CO 2 = 44)

Temperature, T = 100 + 273 = 373 K

The values of a and b for CO 2 (from Table 8.1)

a = 362850 Nm^4 /(kg-mol)^2

and b = 0.0423 m^3 /kg-mol

R 0 = 8314 Nm/kg-mol K

Van der Waals’ equation is written as

p a

v

HFG +^2 IKJ^ ()vb− = R^0 T

or p =

RT

vb

a

v

0

− −^2

F

HG

I

KJ

Substituting the values in the above equation, we get

∴ p = 8314 373

44 0 0423

×

−.

-^362850
442
= 70548 – 187 = 70361 N/m^2 or 0.7036 bar. (Ans.)
(ii)Using perfect gas equation :
pv = R 0 T

∴ p =

RT

v

(^0) = 8314 373
44
×
= 70480 N/m^2 or 0.7048 bar. (Ans.)
Example 8.10. A container of 3 m^3 capacity contains 10 kg of CO 2 at 27°C. Estimate the
pressure exerted by CO 2 by using :
(i)Perfect gas equation
(ii)Van der Waals’ equation
(iii)Beattie Bridgeman equation.
Solution. Capacity of the container, V = 3 m^3
Mass of CO 2 , m = 10 kg
Temperature of CO 2 , T = 27 + 273 = 300 K
Pressure exerted by CO 2 , p :