434 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th9-3.pm5

=

16 8 328 3 0 298

19 8

..

.

×+×

= 323.5 K

i.e., t = 323.5 – 273 = 50.5°C. (Ans.)

(ii)Final pressure, p :

pV = mRT

or p =
mRT
V

19 8 0 297 10 323 5
1 437

..^3.

.

×××
= 13.24 × 10^5 N/m^2
i.e., p = 13.24 bar. (Ans.)
Example 9.13. Calculate the increase in entropy when 3 kg of O 2 at 50°C are mixed with
9 kg of N 2 at the same temperature. The initial pressure of each constituent is 11 bar and is the
same as that of the mixture.

Solution. xO 2 =

p

p

O (^2) =^332
332 928
/
//+
= 0.225
xN 2 =
p
p
N (^2) =^928
332 928
/
//+ = 0.774
Increase of entropy due to diffusion
∆S = – mROO 22 loge
p
p
O (^2) – mR
NN 22 loge^
p
p
N 2
= – 3 ×
8314
32
F.
HG
I
KJ loge 0.225 – 9 ×
8314
28
F.
HG
I
KJ × loge 0.774
= 1.1626 + 0.6846 = 1.8472 kJ/kg K.
Example 9.14. 2.5 kg of N 2 at 15 bar and 40°C is contained a rigid vessel. Adequate
quantity of O 2 is added to increase the pressure to 20 bar while the temperature remains con-
stant at 40°C.
Calculate the mass of O 2 added.
Solution. mN 2 = 2.5 kg, MN 2 = 28
Initial pressure, pN 2 = 15 bar
Final pressure, ptotal ()=+ppNO 22 = 20 bar
Temperature = 40°C
In this process, the number of nitrogen moles remains constant at
nN 2 =
m
M
N
N
2
2

25
28
.
= 0.0893
After the oxygen is added, pN 2 = 15 bar, since the volume remains unchanged
pO 2 = ptotal – pN 2 = 20 – 15 = 5 bar