TITLE.PM5

GASES AND VAPOUR MIXTURES 433

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\M-therm\Th9-2.pm5

16 × 10^5 × VA = 0.6 × 8.314 × 10^3 × (55 + 273)

∴ VA = 1.022 m^3

The mass of gas in vessel A

mA = nAMA = 0.6 × 28 = 16.8 kg

Characteristic gas constant R of nitrogen

R =

8 314

28

.
= 0.297 kJ/kg K
Vessel B :
pBVB = mBRTB
6.4 × 10^5 × VB = 3.0 × 0.297 × 10^3 × (25 + 273)
∴ VB = 0.415 m^3
Total volume of A and B
V = VA + VB = 1.022 + 0.415 = 1.437 m^3
Total mass of gas
m = mA + mB = 16.8 + 3 = 19.8 kg
Final temperature after mixing
T = 30 + 273 = 303 K
(a) (i) Final equilibrium pressure, p :
pV = mRT
or p × 1.437 = 19.8 × 0.297 × 303 × 10^3 = 12.4 × 10^5 N/m^2
i.e., p = 12.4 bar. (Ans.)

Also cv =

R

γ− 1 =

0297

14 1

.

. − = 0.743 kJ/kg K.
(ii)Amount of heat transferred, Q :
As there is no work transfer, the amount of heat transfer,
Q = change of internal energy
= U 2 – U 1
Measuring the internal energy above the datum of absolute zero (at T = 0 K, u = 0 kJ/kg).
Internal energy U 1 (before mixing)
= mAcvTA + mBcvTB
= 16.8 × 0.743 × 328 + 3.0 × 0.743 × 298 = 4758.5 kJ
Final internal energy U 2 (after mixing)
= mcvT = 19.8 × 0.743 × 303 = 4457.5 kJ
∴ Q = U 2 – U 1 = 4457.5 – 4758.5 = – 301 kJ. (Ans.)
(b) If the vessel were insulated :
(i)Final temperature, t 2 :
If the vessel were insulated
Q = U 2 – U 1 = 0
i.e., U 1 = U 2
mAcvTA + mBcvTB = mcvT

∴ T =

mcT mcT

mc

Av A Bv B

v

+

=

mT mT

m

AA+ BB

Fig. 9.9

Valve

Vessel A Vessel B

p = 16 bar

t = 55 C

0.6 kg mole

º

N 2

p = 6.4 bar

t = 25 C

3.0 kg mole

º

N 2

×