IDEAL AND REAL GASES 395

dharm

\M-therm\Th8-2.pm5

(iv)Evaluate the increase in specific internal energy, the increase in specific enthalpy, increase
in specific entropy and magnitude and sign of heat transfer.
Solution. Mass of N 2 , m = 1 kg mole i.e., 28 kg
Capacity of the vessel, V 1 = 3 m^3
Temperature, T 1 = 90 + 273 = 363 K
(i)Pressure (p 1 ) and specific volume (v 1 ) of the gas :
Using the relation
p 1 V 1 = mRT 1

p 1 × 3 = 28 ×

8314

28

F

HG

I

KJ × 363

Q R R

M

L ==

NM

O

QP

0 8314

28

∴ p 1 = 1005994 J/m^2 or 10.06 bar. (Ans.)

Specific volume, v 1 =

V

m

(^1) =^3
28 = 0.107 m
(^3) /kg. (Ans.)
(ii)cp = ?, cv =?
c
c
p
v
= 1.4 (given) ...(i)
But cp – cv = R =
8314
28 ...(ii)
Solving for cp and cv between (i) and (ii)
cp = 1.039 kJ/kg K ; cv = 0.742 kJ/kg K. (Ans.)
(iii)Final pressure of the gas after cooling to 20°C :
Initially After cooling
p 1 = 10.06 bar p 2 =?
V 1 = 3 m^3 V 2 = 3 m^3
T 1 = 363 K T 2 = 20 + 273 = 293 K
Now,
pV
T
11
1

pV
T
22
2
or
p
T
1
1

p
T
2
2
(as V 1 = V 2 )
∴ p 2 = pT
T
12
1
= 10 06 293
363

. × = 8.12 bar. (Ans.)

(iv)∆u, ∆h, ∆s, Q :

For a perfect gas,

Increase in specific internal energy

∆u = cv(T 2 – T 1 ) = 0.742(293 – 363) = – 51.94 kJ/kg. (Ans.)

Increase in specific enthalpy,

∆h = cp(T 2 – T 1 ) = 1.039(293 – 363) = – 72.73 kJ/kg. (Ans.)

Increase in specific entropy,

∆s = cv loge T

T

2

1

F

HG

I

KJ

+ R loge

v

v

2

1

F

HG

I

KJ