TITLE.PM5

424 ENGINEERING THERMODYNAMICS

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\M-therm\Th9-1.pm5

i.e., Gas constant for the mixture = 0.2608 kJ/kg K. (Ans.)

(iv) To find specific volume of the mixture, v using the relation :

pv = RT

v =

RT

p =

0 2608 10 293

110

3

5

. ××
×
= 0.7641 m^3 /kg

i.e., Specific volume of the mixture = 0.7641 m^3 /kg. (Ans.)
Example 9.5. A mixture of hydrogen (H 2 ) and oxygen (O 2 ) is to be made so that the ratio of H 2
to O 2 is 2 : 1 by volume. If the pressure and temperature are 1 bar and 25°C respectively, calculate :
(i)The mass of O 2 required ; (ii)The volume of the container.
Solution. Pressure, p = 1 bar
Temperature, T = 25 + 273 = 298 K
Ratio of H 2 to O 2 = 2 : 1 by volume.
(i)The mass of O 2 required :
Let the mass of O 2 per kg of H 2 = x kg

Now, ni =

m

M

i

i

∴ nH 2 =

1

2

= 0.5

and nO 2 = x
32
From equation,
V
V

i = n

n

i, we have

V

V

H

O

2

2

=

n

n

H

O

2

2

and

V

V

H

O

2

2

= 2 (given)

∴^05

32

.

x/

= 2 ∴ x = 32 0 5

2

×. = 8

i.e., Mass of O 2 per kg of H 2 = 8 kg. (Ans.)
(ii)The volume of the container, V :
The total number of moles in the vessel per kg of H 2 is

n = nnHO 22 + = 0.5 +

x

32

= 0.5 +^8

32

= 0.75

Now using the relation,

pV = nR 0 T

∴ V =

nR T

p

(^0) = 0 75 8 314 10
110
3
5
.(.××)
×
× 298 = 18.58 m^3
i.e., The volume of the container = 18.58 m^3. (Ans.)
Example 9.6. A gaseous mixture of composition by volume, 78% H 2 and 22% CO is con-
tained in a vessel. It is desired that the mixture should be made in proportion 52% H 2 and 48% CO
by removing some of the mixture and adding some CO. Calculate per mole of mixture the mass of
mixture to be removed, and mass of CO to be added.
Assume that the pressure and temperature in the vessel remain constant during the
procedure.