TITLE.PM5

IDEAL AND REAL GASES 397

dharm

\M-therm\Th8-2.pm5

But γ =

c

c

p

v

= 1.4 (given) ...(i)

and cp – cv = R (= 0.287 kJ/kg K for air) ...(ii)

Solving for cv between (i) and (ii),

cv = 0.717 kJ/kg K

∴∆s = 0.717 loge 290 8

373

F.

HG

I

KJ

+ 0.287 loge 04637

01338

.

.

F

HG

I

KJ

= – 0.1785 + 0.3567 = 0.1782 kJ/kg K

i.e., Increase in entropy, ∆s = 0.1782 kJ/kg K. (Ans.)

(ii)Work done and heat transfer :

The work done in a polytropic process is given by,

W =

pv p v

n

11 2 2

1

−

− =

RT T

n

() 12

1

−

−

=

0 287 373 290 8

12 1

.( .)

(. )

−

− = 117.96 kJ/kg

i.e., Work done = 117.96 kJ/kg. (Ans.)
Heat transfer, Q = ∆u + W
where ∆u = cv(T 2 – T 1 )
= 0.717 (290.8 – 373) = – 58.94 kJ/kg
∴ Q = – 58.94 + 117.96 = 59.02 kJ/kg
Hence heat transfer = 59.02 kJ/kg. (Ans.)
(b) (i) Though the process is assumed now to be irreversible and adiabatic, the end states are
given to be the same as in (a). Therefore, all the properties at the end of the process are the
same as in (a). (Ans.)
(ii) As the process is adiabatic, Q (heat transfer) = 0. (Ans.)
∆u = ∆u in (a)
Applying first law for this process
Q = ∆u + W
0 = ∆u + W
or W = – ∆u
= – (– 58.94) = 58.94
∴ Work done = 58.94 kJ/kg. (Ans.)
Example 8.6. Two spheres each 2.5 m in diameter are connected to each other by a pipe
with a valve as shown in Fig. 8.11. One sphere contains 16 kg of air and other 8 kg of air when the
valve is closed. The temperature of air in both sphere is 25°C. The valve is opened and the whole
system is allowed to come to equilibrium conditions. Assuming there is no loss or gain of energy,
determine the pressure in the spheres when the system attains equilibrium.
Neglect the volume of the pipe.

Solution. Volume of each sphere =

4

3 πR

(^3) =^4
3 π ×
25
2

.^3
F
HG

I

KJ = 8.18 m

3